https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Coolmath2017&feedformat=atom AoPS Wiki - User contributions [en] 2021-04-16T20:19:51Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_24&diff=144554 2004 AMC 10B Problems/Problem 24 2021-02-02T02:56:25Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt; we have &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;AC=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is on the circumscribed circle of the triangle so that &lt;math&gt;AD&lt;/math&gt; bisects angle &lt;math&gt;BAC&lt;/math&gt;. What is the value of &lt;math&gt;\frac{AD}{CD}&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Set &lt;math&gt;\overline{BD}&lt;/math&gt;'s length as &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\overline{CD}&lt;/math&gt;'s length must also be &lt;math&gt;x&lt;/math&gt; since &lt;math&gt;\angle BAD&lt;/math&gt; and &lt;math&gt;\angle DAC&lt;/math&gt; intercept arcs of equal length (because &lt;math&gt;\angle BAD=\angle DAC&lt;/math&gt;). Using [[Ptolemy's Theorem]], &lt;math&gt;7x+8x=9(AD)&lt;/math&gt;. The ratio is &lt;math&gt;\frac{5}{3}\implies\boxed{\text{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> import graph;<br /> import geometry;<br /> import markers;<br /> <br /> unitsize(0.5 cm);<br /> <br /> pair A, B, C, D, E, I;<br /> <br /> A = (11/3,8*sqrt(5)/3);<br /> B = (0,0);<br /> C = (9,0);<br /> I = incenter(A,B,C);<br /> D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));<br /> E = extension(A,D,B,C);<br /> <br /> draw(A--B--C--cycle);<br /> draw(circumcircle(A,B,C));<br /> draw(D--A);<br /> draw(D--B);<br /> draw(D--C);<br /> <br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, SE);<br /> label(&quot;$D$&quot;, D, S);<br /> label(&quot;$E$&quot;, E, NE);<br /> <br /> markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));<br /> markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;E = \overline{BC}\cap \overline{AD}&lt;/math&gt;. Observe that &lt;math&gt;\angle ABC \cong \angle ADC&lt;/math&gt; because they subtend the same arc, &lt;math&gt;\overarc{AC}&lt;/math&gt;. Furthermore, &lt;math&gt;\angle BAE \cong \angle EAC&lt;/math&gt; because &lt;math&gt;\overline{AE}&lt;/math&gt; is an angle bisector, so &lt;math&gt;\triangle ABE \sim \triangle ADC&lt;/math&gt; by &lt;math&gt;\text{AA}&lt;/math&gt; similarity. Then &lt;math&gt;\dfrac{AD}{AB} = \dfrac{CD}{BE}&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{CE}&lt;/math&gt;, so &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{9-BE}&lt;/math&gt;. This in turn gives &lt;math&gt;BE = \frac{21}{5}&lt;/math&gt;. Plugging this into the similarity proportion gives: &lt;math&gt;\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_24&diff=144553 2004 AMC 10B Problems/Problem 24 2021-02-02T02:55:46Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt; we have &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;AC=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is on the circumscribed circle of the triangle so that &lt;math&gt;AD&lt;/math&gt; bisects angle &lt;math&gt;BAC&lt;/math&gt;. What is the value of &lt;math&gt;\frac{AD}{CD}&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Set &lt;math&gt;\overline{BD}&lt;/math&gt;'s length as &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\overline{CD}&lt;/math&gt;'s length must also be &lt;math&gt;x&lt;/math&gt; since &lt;math&gt;\angle BAD&lt;/math&gt; and &lt;math&gt;\angle DAC&lt;/math&gt; intercept arcs of equal length (because &lt;math&gt;\angle BAD=\angle DAC&lt;/math&gt;). Using [[Ptolemy's Theorem]], &lt;math&gt;7x+8x=9(AD)&lt;/math&gt;. The ratio is &lt;math&gt;\frac{5}{3}\implies\boxed{\text{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> import graph;<br /> import geometry;<br /> import markers;<br /> <br /> unitsize(0.5 cm);<br /> <br /> pair A, B, C, D, E, I;<br /> <br /> A = (11/3,8*sqrt(5)/3);<br /> B = (0,0);<br /> C = (9,0);<br /> I = incenter(A,B,C);<br /> D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));<br /> E = extension(A,D,B,C);<br /> <br /> draw(A--B--C--cycle);<br /> draw(circumcircle(A,B,C));<br /> draw(D--A);<br /> draw(D--B);<br /> draw(D--C);<br /> <br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, SE);<br /> label(&quot;$D$&quot;, D, S);<br /> label(&quot;$E$&quot;, E, NE);<br /> <br /> markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));<br /> markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;E = \overline{BC}\cap \overline{AD}&lt;/math&gt;. Observe that &lt;math&gt;\angle ABC \cong \angle ADC&lt;/math&gt; because they subtend the same arc, &lt;math&gt;\overarc{AC}&lt;/math&gt;. Furthermore, &lt;math&gt;\angle BAE \cong \angle EAC&lt;/math&gt; because &lt;math&gt;\overline{AE}&lt;/math&gt; is an angle bisector, so &lt;math&gt;\triangle ABE \sim \triangle ADC&lt;/math&gt; by &lt;math&gt;\text{AA}&lt;/math&gt; similarity. Then &lt;math&gt;\dfrac{AD}{AB} = \dfrac{CD}{BE}&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{CE}&lt;/math&gt;, so &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{9-BE}&lt;/math&gt;. This in turn gives &lt;math&gt;BE = \frac{21}{5}&lt;/math&gt;. Plugging this into the similarity proportion gives: &lt;math&gt;\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 3 ==<br /> By the Angle Bisector Theorem, we have &lt;math&gt;DC = \frac{24}{5}&lt;/math&gt; and &lt;math&gt;BD = \frac{21}{5}.&lt;/math&gt;<br /> Let &lt;math&gt;AD =x.&lt;/math&gt; By Stewart's Theorem, we have &lt;math&gt;7^2 \cdot \frac{24}{5} +8^2 \cdot \frac{21}{5} = 9(x^2 + \frac{24}{5} \cdot \frac{21}{5}),&lt;/math&gt; which gives x=\frac{<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_24&diff=144549 2004 AMC 10B Problems/Problem 24 2021-02-02T02:45:10Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt; we have &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;AC=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is on the circumscribed circle of the triangle so that &lt;math&gt;AD&lt;/math&gt; bisects angle &lt;math&gt;BAC&lt;/math&gt;. What is the value of &lt;math&gt;\frac{AD}{CD}&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Set &lt;math&gt;\overline{BD}&lt;/math&gt;'s length as &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\overline{CD}&lt;/math&gt;'s length must also be &lt;math&gt;x&lt;/math&gt; since &lt;math&gt;\angle BAD&lt;/math&gt; and &lt;math&gt;\angle DAC&lt;/math&gt; intercept arcs of equal length (because &lt;math&gt;\angle BAD=\angle DAC&lt;/math&gt;). Using [[Ptolemy's Theorem]], &lt;math&gt;7x+8x=9(AD)&lt;/math&gt;. The ratio is &lt;math&gt;\frac{5}{3}\implies\boxed{\text{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> import graph;<br /> import geometry;<br /> import markers;<br /> <br /> unitsize(0.5 cm);<br /> <br /> pair A, B, C, D, E, I;<br /> <br /> A = (11/3,8*sqrt(5)/3);<br /> B = (0,0);<br /> C = (9,0);<br /> I = incenter(A,B,C);<br /> D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));<br /> E = extension(A,D,B,C);<br /> <br /> draw(A--B--C--cycle);<br /> draw(circumcircle(A,B,C));<br /> draw(D--A);<br /> draw(D--B);<br /> draw(D--C);<br /> <br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, SE);<br /> label(&quot;$D$&quot;, D, S);<br /> label(&quot;$E$&quot;, E, NE);<br /> <br /> markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));<br /> markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;E = \overline{BC}\cap \overline{AD}&lt;/math&gt;. Observe that &lt;math&gt;\angle ABC \cong \angle ADC&lt;/math&gt; because they subtend the same arc, &lt;math&gt;\overarc{AC}&lt;/math&gt;. Furthermore, &lt;math&gt;\angle BAE \cong \angle EAC&lt;/math&gt; because &lt;math&gt;\overline{AE}&lt;/math&gt; is an angle bisector, so &lt;math&gt;\triangle ABE \sim \triangle ADC&lt;/math&gt; by &lt;math&gt;\text{AA}&lt;/math&gt; similarity. Then &lt;math&gt;\dfrac{AD}{AB} = \dfrac{CD}{BE}&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{CE}&lt;/math&gt;, so &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{9-BE}&lt;/math&gt;. This in turn gives &lt;math&gt;BE = \frac{21}{5}&lt;/math&gt;. Plugging this into the similarity proportion gives: &lt;math&gt;\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> By the Angle Bisector Theorem, we have &lt;math&gt;DC = \frac{24}{5}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2004_AMC_10B_Problems/Problem_24&diff=144546 2004 AMC 10B Problems/Problem 24 2021-02-02T02:43:34Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> In triangle &lt;math&gt;ABC&lt;/math&gt; we have &lt;math&gt;AB=7&lt;/math&gt;, &lt;math&gt;AC=8&lt;/math&gt;, &lt;math&gt;BC=9&lt;/math&gt;. Point &lt;math&gt;D&lt;/math&gt; is on the circumscribed circle of the triangle so that &lt;math&gt;AD&lt;/math&gt; bisects angle &lt;math&gt;BAC&lt;/math&gt;. What is the value of &lt;math&gt;\frac{AD}{CD}&lt;/math&gt;? <br /> <br /> &lt;math&gt;\text{(A) } \dfrac{9}{8} \qquad \text{(B) } \dfrac{5}{3} \qquad \text{(C) } 2 \qquad \text{(D) } \dfrac{17}{7} \qquad \text{(E) } \dfrac{5}{2}&lt;/math&gt;<br /> <br /> == Solution 1==<br /> <br /> Set &lt;math&gt;\overline{BD}&lt;/math&gt;'s length as &lt;math&gt;x&lt;/math&gt;. &lt;math&gt;\overline{CD}&lt;/math&gt;'s length must also be &lt;math&gt;x&lt;/math&gt; since &lt;math&gt;\angle BAD&lt;/math&gt; and &lt;math&gt;\angle DAC&lt;/math&gt; intercept arcs of equal length (because &lt;math&gt;\angle BAD=\angle DAC&lt;/math&gt;). Using [[Ptolemy's Theorem]], &lt;math&gt;7x+8x=9(AD)&lt;/math&gt;. The ratio is &lt;math&gt;\frac{5}{3}\implies\boxed{\text{(B)}}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> &lt;asy&gt;<br /> import graph;<br /> import geometry;<br /> import markers;<br /> <br /> unitsize(0.5 cm);<br /> <br /> pair A, B, C, D, E, I;<br /> <br /> A = (11/3,8*sqrt(5)/3);<br /> B = (0,0);<br /> C = (9,0);<br /> I = incenter(A,B,C);<br /> D = intersectionpoint(I--(I + 2*(I - A)), circumcircle(A,B,C));<br /> E = extension(A,D,B,C);<br /> <br /> draw(A--B--C--cycle);<br /> draw(circumcircle(A,B,C));<br /> draw(D--A);<br /> draw(D--B);<br /> draw(D--C);<br /> <br /> label(&quot;$A$&quot;, A, N);<br /> label(&quot;$B$&quot;, B, SW);<br /> label(&quot;$C$&quot;, C, SE);<br /> label(&quot;$D$&quot;, D, S);<br /> label(&quot;$E$&quot;, E, NE);<br /> <br /> markangle(radius = 20,B, A, C, marker(markinterval(2,stickframe(1,2mm),true)));<br /> markangle(radius = 20,B, C, D, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,D, B, C, marker(markinterval(1,stickframe(1,2mm),true)));<br /> markangle(radius = 20,C, B, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> markangle(radius = 20,C, D, A, marker(markinterval(1,stickframe(2,2mm),true)));<br /> &lt;/asy&gt;<br /> Let &lt;math&gt;E = \overline{BC}\cap \overline{AD}&lt;/math&gt;. Observe that &lt;math&gt;\angle ABC \cong \angle ADC&lt;/math&gt; because they subtend the same arc, &lt;math&gt;\overarc{AC}&lt;/math&gt;. Furthermore, &lt;math&gt;\angle BAE \cong \angle EAC&lt;/math&gt; because &lt;math&gt;\overline{AE}&lt;/math&gt; is an angle bisector, so &lt;math&gt;\triangle ABE \sim \triangle ADC&lt;/math&gt; by &lt;math&gt;\text{AA}&lt;/math&gt; similarity. Then &lt;math&gt;\dfrac{AD}{AB} = \dfrac{CD}{BE}&lt;/math&gt;. By the [[Angle Bisector Theorem]], &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{CE}&lt;/math&gt;, so &lt;math&gt;\dfrac{7}{BE} = \dfrac{8}{9-BE}&lt;/math&gt;. This in turn gives &lt;math&gt;BE = \frac{21}{5}&lt;/math&gt;. Plugging this into the similarity proportion gives: &lt;math&gt;\dfrac{AD}{7} = \dfrac{CD}{\tfrac{21}{5}} \implies \dfrac{AD}{CD} = {\dfrac{5}{3}} = \boxed{\text{(B)}}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2004|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=139936 1983 AIME Problems/Problem 1 2020-12-18T21:38:08Z <p>Coolmath2017: /* Solution 6 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt; and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt; and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.<br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression containing &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear that one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{\frac{1}{2}}&lt;/math&gt;. Raising both sides of &lt;math&gt;y^{40}=w&lt;/math&gt; to the &lt;math&gt;\frac{12}{40}&lt;/math&gt;th power gives &lt;math&gt;y^{12}=w^{\frac{3}{10}}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}w^{3/10}z^{12}=w&lt;/math&gt;. Simplifying, we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Since &lt;math&gt;\log_a b = \frac{1}{\log_b a}&lt;/math&gt;, the given conditions can be rewritten as &lt;math&gt;\log_w x = \frac{1}{24}&lt;/math&gt;, &lt;math&gt;\log_w y = \frac{1}{40}&lt;/math&gt;, and &lt;math&gt;\log_w xyz = \frac{1}{12}&lt;/math&gt;. Since &lt;math&gt;\log_a \frac{b}{c} = \log_a b - \log_a c&lt;/math&gt;, &lt;math&gt;\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}&lt;/math&gt;. Therefore, &lt;math&gt;\log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> If we convert all of the equations into exponential form, we receive &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The last equation can also be written as &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Also note that by multiplying the first two equations, we get, &lt;math&gt;x^{24}y^{40}= w^{2}&lt;/math&gt;. Taking the square root of this, we find that &lt;math&gt;x^{12}y^{20}=w&lt;/math&gt;. Recall, &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Thus, &lt;math&gt;z^{12}= y^{8}&lt;/math&gt;. Also recall, &lt;math&gt;y^{40}=w&lt;/math&gt;. Therefore, &lt;math&gt;z^{60}&lt;/math&gt; = &lt;math&gt;y^{40}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;. So, &lt;math&gt;\log_z w&lt;/math&gt; = &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> -Dhillonr25, Bobbob<br /> === Solution 6 ===<br /> Converting all of the logarithms to exponentials gives &lt;math&gt;x^{24} = w, y^{40} =w,&lt;/math&gt; and &lt;math&gt;x^{12}y^{12}z^{12}=w.&lt;/math&gt; <br /> Thus, we have &lt;math&gt;y^{40} = x^{24} \Rightarrow z^3=y^2.&lt;/math&gt; <br /> We are looking for &lt;math&gt;\log_z 3,&lt;/math&gt; which by substitution, is &lt;math&gt;\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=139935 1983 AIME Problems/Problem 1 2020-12-18T21:37:54Z <p>Coolmath2017: /* Solution 6 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt; and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt; and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.<br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression containing &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear that one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{\frac{1}{2}}&lt;/math&gt;. Raising both sides of &lt;math&gt;y^{40}=w&lt;/math&gt; to the &lt;math&gt;\frac{12}{40}&lt;/math&gt;th power gives &lt;math&gt;y^{12}=w^{\frac{3}{10}}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}w^{3/10}z^{12}=w&lt;/math&gt;. Simplifying, we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Since &lt;math&gt;\log_a b = \frac{1}{\log_b a}&lt;/math&gt;, the given conditions can be rewritten as &lt;math&gt;\log_w x = \frac{1}{24}&lt;/math&gt;, &lt;math&gt;\log_w y = \frac{1}{40}&lt;/math&gt;, and &lt;math&gt;\log_w xyz = \frac{1}{12}&lt;/math&gt;. Since &lt;math&gt;\log_a \frac{b}{c} = \log_a b - \log_a c&lt;/math&gt;, &lt;math&gt;\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}&lt;/math&gt;. Therefore, &lt;math&gt;\log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> If we convert all of the equations into exponential form, we receive &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The last equation can also be written as &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Also note that by multiplying the first two equations, we get, &lt;math&gt;x^{24}y^{40}= w^{2}&lt;/math&gt;. Taking the square root of this, we find that &lt;math&gt;x^{12}y^{20}=w&lt;/math&gt;. Recall, &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Thus, &lt;math&gt;z^{12}= y^{8}&lt;/math&gt;. Also recall, &lt;math&gt;y^{40}=w&lt;/math&gt;. Therefore, &lt;math&gt;z^{60}&lt;/math&gt; = &lt;math&gt;y^{40}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;. So, &lt;math&gt;\log_z w&lt;/math&gt; = &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> -Dhillonr25, Bobbob<br /> === Solution 6 ===<br /> Converting all of the logarithms to exponentials gives &lt;math&gt;x^{24} = w, y^{40} =w,&lt;/math&gt; and &lt;math&gt;x^{12}y^{12}z^{12}=w.&lt;/math&gt; <br /> Thus, we have &lt;math&gt;y^{40} = x^{24} \Rightarrow z^3=y^2.&lt;/math&gt; <br /> We are looking for &lt;math&gt;\log_z 3,&lt;/math&gt; which by substitution, is &lt;math&gt;\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} =\boxed{60}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=139934 1983 AIME Problems/Problem 1 2020-12-18T21:37:15Z <p>Coolmath2017: /* Solution 6 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt; and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt; and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.<br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression containing &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear that one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{\frac{1}{2}}&lt;/math&gt;. Raising both sides of &lt;math&gt;y^{40}=w&lt;/math&gt; to the &lt;math&gt;\frac{12}{40}&lt;/math&gt;th power gives &lt;math&gt;y^{12}=w^{\frac{3}{10}}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}w^{3/10}z^{12}=w&lt;/math&gt;. Simplifying, we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Since &lt;math&gt;\log_a b = \frac{1}{\log_b a}&lt;/math&gt;, the given conditions can be rewritten as &lt;math&gt;\log_w x = \frac{1}{24}&lt;/math&gt;, &lt;math&gt;\log_w y = \frac{1}{40}&lt;/math&gt;, and &lt;math&gt;\log_w xyz = \frac{1}{12}&lt;/math&gt;. Since &lt;math&gt;\log_a \frac{b}{c} = \log_a b - \log_a c&lt;/math&gt;, &lt;math&gt;\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}&lt;/math&gt;. Therefore, &lt;math&gt;\log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> If we convert all of the equations into exponential form, we receive &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The last equation can also be written as &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Also note that by multiplying the first two equations, we get, &lt;math&gt;x^{24}y^{40}= w^{2}&lt;/math&gt;. Taking the square root of this, we find that &lt;math&gt;x^{12}y^{20}=w&lt;/math&gt;. Recall, &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Thus, &lt;math&gt;z^{12}= y^{8}&lt;/math&gt;. Also recall, &lt;math&gt;y^{40}=w&lt;/math&gt;. Therefore, &lt;math&gt;z^{60}&lt;/math&gt; = &lt;math&gt;y^{40}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;. So, &lt;math&gt;\log_z w&lt;/math&gt; = &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> -Dhillonr25, Bobbob<br /> === Solution 6 ===<br /> Converting all of the logarithms to exponentials gives &lt;math&gt;x^{24} = w, y^{40} =w,&lt;/math&gt; and &lt;math&gt;x^{12}y^{12}z^{12}=w.&lt;/math&gt; <br /> Thus, we have &lt;math&gt;y^{40} = x^{24} \Rightarrow z^3=y^2.&lt;/math&gt; <br /> We are looking for &lt;math&gt;\log_z 3,&lt;/math&gt; which by substitution, is &lt;math&gt;\log_{y^{\frac{2}{3}}} y^{40} = 40 \div \frac{2}{3} = 40 \cdot \frac{3}{2} = \boxed{60}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=139933 1983 AIME Problems/Problem 1 2020-12-18T21:36:47Z <p>Coolmath2017: /* Solution 6 = */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt; and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt; and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.<br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression containing &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear that one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{\frac{1}{2}}&lt;/math&gt;. Raising both sides of &lt;math&gt;y^{40}=w&lt;/math&gt; to the &lt;math&gt;\frac{12}{40}&lt;/math&gt;th power gives &lt;math&gt;y^{12}=w^{\frac{3}{10}}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}w^{3/10}z^{12}=w&lt;/math&gt;. Simplifying, we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Since &lt;math&gt;\log_a b = \frac{1}{\log_b a}&lt;/math&gt;, the given conditions can be rewritten as &lt;math&gt;\log_w x = \frac{1}{24}&lt;/math&gt;, &lt;math&gt;\log_w y = \frac{1}{40}&lt;/math&gt;, and &lt;math&gt;\log_w xyz = \frac{1}{12}&lt;/math&gt;. Since &lt;math&gt;\log_a \frac{b}{c} = \log_a b - \log_a c&lt;/math&gt;, &lt;math&gt;\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}&lt;/math&gt;. Therefore, &lt;math&gt;\log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> If we convert all of the equations into exponential form, we receive &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The last equation can also be written as &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Also note that by multiplying the first two equations, we get, &lt;math&gt;x^{24}y^{40}= w^{2}&lt;/math&gt;. Taking the square root of this, we find that &lt;math&gt;x^{12}y^{20}=w&lt;/math&gt;. Recall, &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Thus, &lt;math&gt;z^{12}= y^{8}&lt;/math&gt;. Also recall, &lt;math&gt;y^{40}=w&lt;/math&gt;. Therefore, &lt;math&gt;z^{60}&lt;/math&gt; = &lt;math&gt;y^{40}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;. So, &lt;math&gt;\log_z w&lt;/math&gt; = &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> -Dhillonr25, Bobbob<br /> === Solution 6 ===<br /> Converting all of the logarithms to exponentials gives &lt;math&gt;x^{24} = w, y^{40} =w,&lt;/math&gt; and &lt;math&gt;x^{12}y^{12}z^{12}=w.&lt;/math&gt; <br /> Thus, we have &lt;math&gt;y^{40} = x^{24} \Rightarrow z^3=y^2.&lt;/math&gt; <br /> We are looking for &lt;math&gt;\log_z 3,&lt;/math&gt; which by substitution, is &lt;math&gt;\log_y^{\frac{2}{3}} y^{40} = 40 \div \frac{2}{3} = 40 \cdot \frac{3}{2} = \boxed{60}.&lt;/math&gt;<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1983_AIME_Problems/Problem_1&diff=139932 1983 AIME Problems/Problem 1 2020-12-18T21:29:29Z <p>Coolmath2017: /* Solution 5 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;x&lt;/math&gt;, &lt;math&gt;y&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt; all exceed &lt;math&gt;1&lt;/math&gt; and let &lt;math&gt;w&lt;/math&gt; be a positive number such that &lt;math&gt;\log_x w = 24&lt;/math&gt;, &lt;math&gt;\log_y w = 40&lt;/math&gt; and &lt;math&gt;\log_{xyz} w = 12&lt;/math&gt;. Find &lt;math&gt;\log_z w&lt;/math&gt;.<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> The [[logarithm]]ic notation doesn't tell us much, so we'll first convert everything to the equivalent exponential forms.<br /> <br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. If we now convert everything to a power of &lt;math&gt;120&lt;/math&gt;, it will be easy to isolate &lt;math&gt;z&lt;/math&gt; and &lt;math&gt;w&lt;/math&gt;.<br /> <br /> &lt;math&gt;x^{120}=w^5&lt;/math&gt;, &lt;math&gt;y^{120}=w^3&lt;/math&gt;, and &lt;math&gt;(xyz)^{120}=w^{10}&lt;/math&gt;.<br /> <br /> With some substitution, we get &lt;math&gt;w^5w^3z^{120}=w^{10}&lt;/math&gt; and &lt;math&gt;\log_zw=\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> First we'll convert everything to exponential form.<br /> &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The only expression containing &lt;math&gt;z&lt;/math&gt; is &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. It now becomes clear that one way to find &lt;math&gt;\log_z w&lt;/math&gt; is to find what &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; are in terms of &lt;math&gt;w&lt;/math&gt;. <br /> <br /> Taking the square root of the equation &lt;math&gt;x^{24}=w&lt;/math&gt; results in &lt;math&gt;x^{12}=w^{\frac{1}{2}}&lt;/math&gt;. Raising both sides of &lt;math&gt;y^{40}=w&lt;/math&gt; to the &lt;math&gt;\frac{12}{40}&lt;/math&gt;th power gives &lt;math&gt;y^{12}=w^{\frac{3}{10}}&lt;/math&gt;.<br /> <br /> Going back to &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;, we can substitute the &lt;math&gt;x^{12}&lt;/math&gt; and &lt;math&gt;y^{12}&lt;/math&gt; with &lt;math&gt;w^{1/2}&lt;/math&gt; and &lt;math&gt;w^{3/10}&lt;/math&gt;, respectively. We now have &lt;math&gt;w^{1/2}w^{3/10}z^{12}=w&lt;/math&gt;. Simplifying, we get &lt;math&gt;z^{60}=w&lt;/math&gt;.<br /> So our answer is &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 3 ===<br /> Applying the change of base formula,<br /> &lt;cmath&gt;\begin{align*} \log_x w = 24 &amp;\implies \frac{\log w}{\log x} = 24 \implies \frac{\log x}{\log w} = \frac 1 {24} \\<br /> \log_y w = 40 &amp;\implies \frac{\log w}{\log y} = 40 \implies \frac{\log y}{\log w} = \frac 1 {40} \\<br /> \log_{xyz} w = 12 &amp;\implies \frac{\log {w}}{\log {xyz}} = 12 \implies \frac{\log x +\log y + \log z}{\log w} = \frac 1 {12} \end{align*}&lt;/cmath&gt;<br /> Therefore, &lt;math&gt; \frac {\log z}{\log<br /> w} = \frac 1 {12} - \frac 1 {24} - \frac 1{40} = \frac 1 {60}&lt;/math&gt;. <br /> <br /> Hence, &lt;math&gt; \log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 4 ===<br /> Since &lt;math&gt;\log_a b = \frac{1}{\log_b a}&lt;/math&gt;, the given conditions can be rewritten as &lt;math&gt;\log_w x = \frac{1}{24}&lt;/math&gt;, &lt;math&gt;\log_w y = \frac{1}{40}&lt;/math&gt;, and &lt;math&gt;\log_w xyz = \frac{1}{12}&lt;/math&gt;. Since &lt;math&gt;\log_a \frac{b}{c} = \log_a b - \log_a c&lt;/math&gt;, &lt;math&gt;\log_w z = \log_w xyz - \log_w x - \log_w y = \frac{1}{12}-\frac{1}{24}-\frac{1}{40}=\frac{1}{60}&lt;/math&gt;. Therefore, &lt;math&gt;\log_z w = \boxed{060}&lt;/math&gt;.<br /> <br /> === Solution 5 ===<br /> <br /> If we convert all of the equations into exponential form, we receive &lt;math&gt;x^{24}=w&lt;/math&gt;, &lt;math&gt;y^{40}=w&lt;/math&gt;, and &lt;math&gt;(xyz)^{12}=w&lt;/math&gt;. The last equation can also be written as &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Also note that by multiplying the first two equations, we get, &lt;math&gt;x^{24}y^{40}= w^{2}&lt;/math&gt;. Taking the square root of this, we find that &lt;math&gt;x^{12}y^{20}=w&lt;/math&gt;. Recall, &lt;math&gt;x^{12}y^{12}z^{12}=w&lt;/math&gt;. Thus, &lt;math&gt;z^{12}= y^{8}&lt;/math&gt;. Also recall, &lt;math&gt;y^{40}=w&lt;/math&gt;. Therefore, &lt;math&gt;z^{60}&lt;/math&gt; = &lt;math&gt;y^{40}&lt;/math&gt; = &lt;math&gt;w&lt;/math&gt;. So, &lt;math&gt;\log_z w&lt;/math&gt; = &lt;math&gt;\boxed{060}&lt;/math&gt;.<br /> <br /> -Dhillonr25, Bobbob<br /> === Solution 6 ====<br /> <br /> == See Also ==<br /> {{AIME box|year=1983|before=First Question|num-a=2}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139756 1984 USAMO Problems/Problem 1 2020-12-16T02:54:30Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139755 1984 USAMO Problems/Problem 1 2020-12-16T02:54:06Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Call the roots of the polynomial &lt;math&gt;r_1, r_2, r_3,&lt;/math&gt; and &lt;math&gt;r_4.&lt;/math&gt; By Vieta's formula, we have &lt;cmath&gt;r_1 + r_2 + r_3 + r_4 = 18,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 +r_2r_4 + r_3r_4 = k,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2r_3 + r_2r_3r_4 + r_1r_2r_4 + r_1r_3r_4 = -200,&lt;/cmath&gt; and &lt;cmath&gt;r_1r_2r_3r_4 = -1984.&lt;/cmath&gt; From the problem statement, we know that &lt;math&gt;r_1r_2 = -32.&lt;/math&gt; Dividing &lt;math&gt;r_1r_2r_3r_4 = -1984&lt;/math&gt; by &lt;math&gt;r_1r_2 = -32&lt;/math&gt; gives &lt;math&gt;r_3r_4 = 62.&lt;/math&gt; Factoring the 3rd symmetric sum and substituting in the following values, we have &lt;math&gt;r_1r_2(r_3 + r_4) + r_3r_4(r_1 + r_2) = -200 \Rightarrow 62(r_1 + r_2) -32(r_3 + r_4) = -200.&lt;/math&gt; Let &lt;math&gt;r_1 + r_2 = x.&lt;/math&gt; From the first equation, we have &lt;math&gt;r_3+r_4 = 18-x.&lt;/math&gt; Thus, we have the equation &lt;math&gt;62(x) -32(18-x) = -200 \Rightarrow x=4.&lt;/math&gt; Substituting values and factoring the remaining values in the second symmetric sum gives &lt;math&gt;-32 + r_1(r_3 + r_4) + r_2(r_3 + r_4) +62 = k \Rightarrow -32 + (r_1+r_2)(r_3+r_4) + 62 = k \Rightarrow k = 86.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139754 1984 USAMO Problems/Problem 1 2020-12-16T02:53:55Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Call the roots of the polynomial &lt;math&gt;r_1, r_2, r_3,&lt;/math&gt; and &lt;math&gt;r_4.&lt;/math&gt; By Vieta's formula, we have &lt;cmath&gt;r_1 + r_2 + r_3 + r_4 = 18,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 +r_2r_4 + r_3r_4 = k,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2r_3 + r_2r_3r_4 + r_1r_2r_4 + r_1r_3r_4 = -200,&lt;/cmath&gt; and &lt;cmath&gt;r_1r_2r_3r_4 = -1984.&lt;/cmath&gt; From the problem statement, we know that &lt;math&gt;r_1r_2 = -32.&lt;/math&gt; Dividing &lt;math&gt;r_1r_2r_3r_4 = -1984&lt;/math&gt; by &lt;math&gt;r_1r_2 = -32&lt;/math&gt; gives &lt;math&gt;r_3r_4 = 62.&lt;/math&gt; Factoring the 3rd symmetric sum and substituting in the following values, we have &lt;math&gt;r_1r_2(r_3 + r_4) + r_3r_4(r_1 + r_2) = -200 \Rightarrow 62(r_1 + r_2) -32(r_3 + r_4) = -200.&lt;/math&gt; Let &lt;math&gt;r_1 + r_2 = x.&lt;/math&gt; From the first equation, we have &lt;math&gt;r_3+r_4 = 18-x.&lt;/math&gt; Thus, we have the equation &lt;math&gt;62(x) -32(18-x) = -200 \Rightarrow x=4.&lt;/math&gt; Substituting values and factoring the remaining values in the second symmetric sum gives &lt;math&gt;-32 + r_1(r_3 + r_4) + r_2(r_3 + r_4) +62 = k \Rightarrow -32 + (r_1+r_2)(r_3+r_4) + 62 = k \Rightarrow k = 86.&lt;/math&gt;<br /> ~coolmath2017<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139753 1984 USAMO Problems/Problem 1 2020-12-16T02:45:19Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> Call the roots of the polynomial &lt;math&gt;r_1, r_2, r_3,&lt;/math&gt; and &lt;math&gt;r_4.&lt;/math&gt; By Vieta's formula, we have &lt;cmath&gt;r_1 + r_2 + r_3 + r_4 = 18,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2 + r_1r_3 + r_1r_4 + r_2r_3 +r_2r_4 + r_3r_4 = k,&lt;/cmath&gt; &lt;cmath&gt;r_1r_2r_3 + r_2r_3r_4 + r_1r_2r_4 + r_1r_3r_4 = -200,&lt;/cmath&gt; and &lt;math&gt;r_1r_2r_3r_4 = -1984.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139752 1984 USAMO Problems/Problem 1 2020-12-16T02:41:55Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> === Solution 3 ===<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1984_USAMO_Problems/Problem_1&diff=139751 1984 USAMO Problems/Problem 1 2020-12-16T02:41:22Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> In the polynomial &lt;math&gt;x^4 - 18x^3 + kx^2 + 200x - 1984 = 0&lt;/math&gt;, the product of &lt;math&gt;2&lt;/math&gt; of its roots is &lt;math&gt;- 32&lt;/math&gt;. Find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> === Solution 1 ===<br /> <br /> <br /> Using Vieta's formulas, we have: <br /> <br /> &lt;cmath&gt;\begin{align*}a+b+c+d &amp;= 18,\\ ab+ac+ad+bc+bd+cd &amp;= k,\\ abc+abd+acd+bcd &amp;=-200,\\ abcd &amp;=-1984.\\ \end{align*}&lt;/cmath&gt; <br /> <br /> <br /> From the last of these equations, we see that &lt;math&gt;cd = \frac{abcd}{ab} = \frac{-1984}{-32} = 62&lt;/math&gt;. Thus, the second equation becomes &lt;math&gt;-32+ac+ad+bc+bd+62=k&lt;/math&gt;, and so &lt;math&gt;ac+ad+bc+bd=k-30&lt;/math&gt;. The key insight is now to factor the left-hand side as a product of two binomials: &lt;math&gt;(a+b)(c+d)=k-30&lt;/math&gt;, so that we now only need to determine &lt;math&gt;a+b&lt;/math&gt; and &lt;math&gt;c+d&lt;/math&gt; rather than all four of &lt;math&gt;a,b,c,d&lt;/math&gt;. <br /> <br /> Let &lt;math&gt;p=a+b&lt;/math&gt; and &lt;math&gt;q=c+d&lt;/math&gt;. Plugging our known values for &lt;math&gt;ab&lt;/math&gt; and &lt;math&gt;cd&lt;/math&gt; into the third Vieta equation, &lt;math&gt;-200 = abc+abd + acd + bcd = ab(c+d) + cd(a+b)&lt;/math&gt;, we have &lt;math&gt;-200 = -32(c+d) + 62(a+b) = 62p-32q&lt;/math&gt;. Moreover, the first Vieta equation, &lt;math&gt;a+b+c+d=18&lt;/math&gt;, gives &lt;math&gt;p+q=18&lt;/math&gt;. Thus we have two linear equations in &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt;, which we solve to obtain &lt;math&gt;p=4&lt;/math&gt; and &lt;math&gt;q=14&lt;/math&gt;. <br /> <br /> Therefore, we have &lt;math&gt;(\underbrace{a+b}_4)(\underbrace{c+d}_{14}) = k-30&lt;/math&gt;, yielding &lt;math&gt;k=4\cdot 14+30 = \boxed{86}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> We start as before: &lt;math&gt;ab=-32&lt;/math&gt; and &lt;math&gt;cd=62&lt;/math&gt;. We now observe that a and b must be the roots of a quadratic, &lt;math&gt;x^2+rx-32&lt;/math&gt;, where r is a constant (secretly, r is just -(a+b)=-p from Solution #1). Similarly, c and d must be the roots of a quadratic &lt;math&gt;x^2+sx+62&lt;/math&gt;. <br /> <br /> Now <br /> <br /> &lt;cmath&gt; \begin{align*}x^4-18x^3+kx^2+200x-1984 =&amp; (x^2+rx-32)(x^2+sx+62)\\ =&amp; x^4+(r+s)x^3+(62-32+rs)x^2\\<br /> &amp;+(62s-32r)x-1984.\end{align*} &lt;/cmath&gt;<br /> <br /> Equating the coefficients of &lt;math&gt;x^3&lt;/math&gt; and &lt;math&gt;x&lt;/math&gt; with their known values, we are left with essentially the same linear equations as in Solution #1, which we solve in the same way. Then we compute the coefficient of &lt;math&gt;x^2&lt;/math&gt; and get &lt;math&gt;k=\boxed{86}.&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{USAMO box|year=1984|before=First&lt;br&gt;Problem|num-a=2}}<br /> {{MAA Notice}}<br /> [[Category:Intermediate Algebra Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_19&diff=132987 2005 AMC 12A Problems/Problem 19 2020-09-02T02:06:59Z <p>Coolmath2017: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads &lt;tt&gt;002005&lt;/tt&gt;, how many miles has the car actually traveled?<br /> &lt;math&gt;<br /> (\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804<br /> &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> We find the number of numbers with a &lt;math&gt;4&lt;/math&gt; and subtract from &lt;math&gt;2005&lt;/math&gt;. Quick counting tells us that there are &lt;math&gt;200&lt;/math&gt; numbers with a 4 in the hundreds place, &lt;math&gt;200&lt;/math&gt; numbers with a 4 in the tens place, and &lt;math&gt;201&lt;/math&gt; numbers with a 4 in the units place (counting &lt;math&gt;2004&lt;/math&gt;). Now we apply the [[Principle of Inclusion-Exclusion]]. There are &lt;math&gt;20&lt;/math&gt; numbers with a 4 in the hundreds and in the tens, and &lt;math&gt;20&lt;/math&gt; for both the other two [[intersection]]s. The intersection of all three sets is just &lt;math&gt;2&lt;/math&gt;. So we get:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}&lt;/math&gt;&lt;/div&gt;<br /> <br /> ===Solution 2===<br /> Alternatively, consider that counting without the number &lt;math&gt;4&lt;/math&gt; is almost equivalent to counting in base &lt;math&gt;9&lt;/math&gt;; only, in base &lt;math&gt;9&lt;/math&gt;, the number &lt;math&gt;9&lt;/math&gt; is not counted. Since &lt;math&gt;4&lt;/math&gt; is skipped, the symbol &lt;math&gt;5&lt;/math&gt; represents &lt;math&gt;4&lt;/math&gt; miles of travel, and we have traveled &lt;math&gt;2004_9&lt;/math&gt; miles. By basic conversion, &lt;math&gt;2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Since any numbers containing one or more &lt;math&gt;4&lt;/math&gt;s were skipped, we need only to find the numbers that don't contain a &lt;math&gt;4&lt;/math&gt; at all. First we consider &lt;math&gt;1&lt;/math&gt; - &lt;math&gt;1999&lt;/math&gt;. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From &lt;math&gt;1&lt;/math&gt; - &lt;math&gt;1999&lt;/math&gt;, we have &lt;math&gt;2&lt;/math&gt; possibilities for the thousands place, and &lt;math&gt;9&lt;/math&gt; possibilities for the hundreds, tens, and ones places. This is &lt;math&gt;2 \cdot 9 \cdot 9 \cdot 9-1&lt;/math&gt; possibilities (because &lt;math&gt;0000&lt;/math&gt; doesn't count) or &lt;math&gt;1457&lt;/math&gt; numbers. From &lt;math&gt;2000&lt;/math&gt; - &lt;math&gt;2005&lt;/math&gt; there are &lt;math&gt;6&lt;/math&gt; numbers, &lt;math&gt;5&lt;/math&gt; of which don't contain a &lt;math&gt;4&lt;/math&gt;. Therefore the total is &lt;math&gt;1457 + 5&lt;/math&gt;, or &lt;math&gt;1462&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> <br /> We seek to find the amount of numbers that contain at least one &lt;math&gt;4,&lt;/math&gt; and subtract this number from &lt;math&gt;2005.&lt;/math&gt; <br /> <br /> We can simply apply casework to this problem.<br /> <br /> The amount of numbers with at least one &lt;math&gt;4&lt;/math&gt; that are one or two digit numbers are &lt;math&gt;4,14,24,34,40-49,54,\cdots,94&lt;/math&gt; which gives &lt;math&gt;19&lt;/math&gt; numbers.<br /> <br /> The amount of three digit numbers with at least one &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;8*19+100=252.&lt;/math&gt; <br /> <br /> The amount of four digit numbers with at least one &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;252+1+19=272&lt;/math&gt;<br /> <br /> This, our answer is &lt;math&gt;2005-19-252-272=162,&lt;/math&gt; or &lt;math&gt;\boxed{B}.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=18|num-a=20|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_19&diff=132986 2005 AMC 12A Problems/Problem 19 2020-09-02T02:06:31Z <p>Coolmath2017: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> A faulty car odometer proceeds from digit 3 to digit 5, always skipping the digit 4, regardless of position. If the odometer now reads &lt;tt&gt;002005&lt;/tt&gt;, how many miles has the car actually traveled?<br /> &lt;math&gt;<br /> (\mathrm {A}) \ 1404 \qquad (\mathrm {B}) \ 1462 \qquad (\mathrm {C})\ 1604 \qquad (\mathrm {D}) \ 1605 \qquad (\mathrm {E})\ 1804<br /> &lt;/math&gt;<br /> <br /> ==Solutions==<br /> <br /> === Solution 1===<br /> We find the number of numbers with a &lt;math&gt;4&lt;/math&gt; and subtract from &lt;math&gt;2005&lt;/math&gt;. Quick counting tells us that there are &lt;math&gt;200&lt;/math&gt; numbers with a 4 in the hundreds place, &lt;math&gt;200&lt;/math&gt; numbers with a 4 in the tens place, and &lt;math&gt;201&lt;/math&gt; numbers with a 4 in the units place (counting &lt;math&gt;2004&lt;/math&gt;). Now we apply the [[Principle of Inclusion-Exclusion]]. There are &lt;math&gt;20&lt;/math&gt; numbers with a 4 in the hundreds and in the tens, and &lt;math&gt;20&lt;/math&gt; for both the other two [[intersection]]s. The intersection of all three sets is just &lt;math&gt;2&lt;/math&gt;. So we get:<br /> <br /> &lt;div style=&quot;text-align:center;&quot;&gt;&lt;math&gt;2005-(200+200+201-20-20-20+2) = 1462 \Longrightarrow \mathrm{(B)}&lt;/math&gt;&lt;/div&gt;<br /> <br /> ===Solution 2===<br /> Alternatively, consider that counting without the number &lt;math&gt;4&lt;/math&gt; is almost equivalent to counting in base &lt;math&gt;9&lt;/math&gt;; only, in base &lt;math&gt;9&lt;/math&gt;, the number &lt;math&gt;9&lt;/math&gt; is not counted. Since &lt;math&gt;4&lt;/math&gt; is skipped, the symbol &lt;math&gt;5&lt;/math&gt; represents &lt;math&gt;4&lt;/math&gt; miles of travel, and we have traveled &lt;math&gt;2004_9&lt;/math&gt; miles. By basic conversion, &lt;math&gt;2004_9=9^3(2)+9^0(4)=729(2)+1(4)=1458+4=\boxed{1462}&lt;/math&gt;<br /> <br /> ===Solution 3===<br /> Since any numbers containing one or more &lt;math&gt;4&lt;/math&gt;s were skipped, we need only to find the numbers that don't contain a &lt;math&gt;4&lt;/math&gt; at all. First we consider &lt;math&gt;1&lt;/math&gt; - &lt;math&gt;1999&lt;/math&gt;. Single digits are simply four digit numbers with a zero in all but the ones place (this concept applies to double and triple digits numbers as well). From &lt;math&gt;1&lt;/math&gt; - &lt;math&gt;1999&lt;/math&gt;, we have &lt;math&gt;2&lt;/math&gt; possibilities for the thousands place, and &lt;math&gt;9&lt;/math&gt; possibilities for the hundreds, tens, and ones places. This is &lt;math&gt;2 \cdot 9 \cdot 9 \cdot 9-1&lt;/math&gt; possibilities (because &lt;math&gt;0000&lt;/math&gt; doesn't count) or &lt;math&gt;1457&lt;/math&gt; numbers. From &lt;math&gt;2000&lt;/math&gt; - &lt;math&gt;2005&lt;/math&gt; there are &lt;math&gt;6&lt;/math&gt; numbers, &lt;math&gt;5&lt;/math&gt; of which don't contain a &lt;math&gt;4&lt;/math&gt;. Therefore the total is &lt;math&gt;1457 + 5&lt;/math&gt;, or &lt;math&gt;1462&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;\boxed{\text{B}}&lt;/math&gt;.<br /> <br /> ===Solution 4===<br /> <br /> We seek to find the amount of numbers that contain at least one &lt;math&gt;4,&lt;/math&gt; and subtract this number from &lt;math&gt;2005.&lt;/math&gt; <br /> <br /> We can simply apply casework to this problem.<br /> <br /> The amount of numbers with at least one &lt;math&gt;4&lt;/math&gt; that are one or two digit numbers are &lt;math&gt;4,14,24,34,40-49,54,\cdots,94&lt;/math&gt; which gives &lt;math&gt;19&lt;/math&gt; numbers.<br /> <br /> The amount of three digit numbers with at least one &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;8*19+100=252.&lt;/math&gt; <br /> <br /> The amount of four digit numbers with at least one &lt;math&gt;4&lt;/math&gt; is &lt;math&gt;252+1+19=272&lt;/math&gt;<br /> <br /> This, our answer is &lt;math&gt;2005-19-252-272=162,&lt;/math&gt; or &lt;math&gt;B.&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=18|num-a=20|ab=A}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132907 2002 AMC 12B Problems/Problem 13 2020-09-01T01:24:21Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> == Solution 1 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132906 2002 AMC 12B Problems/Problem 13 2020-09-01T01:24:06Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132905 2002 AMC 12B Problems/Problem 13 2020-09-01T01:21:28Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> == Solution 1 ==<br /> <br /> Let the first term be &lt;math&gt;x&lt;/math&gt; and the common difference be &lt;math&gt;d.&lt;/math&gt; <br /> <br /> Thus, we know the sum of terms is &lt;math&gt;9(2x+17d).&lt;/math&gt;<br /> <br /> Because &lt;math&gt;9&lt;/math&gt; is already a perfect square, we know that &lt;math&gt;2x+17d&lt;/math&gt; must also be a perfect square.<br /> <br /> Since we want the smallest value of the sum, we know that &lt;math&gt;25=2x+17 \Rightarrow x=4&lt;/math&gt; and &lt;math&gt;d=1.&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;9*25 = \boxed{225}.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132904 2002 AMC 12B Problems/Problem 13 2020-09-01T01:20:49Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> == Solution 1 ==<br /> <br /> Let the first term be &lt;math&gt;x&lt;/math&gt; and the common difference be &lt;math&gt;d.&lt;/math&gt; <br /> <br /> Thus, we know the sum of terms is &lt;math&gt;9(2x+17d).&lt;/math&gt;<br /> <br /> Because &lt;math&gt;9&lt;/math&gt; is already a perfect square, we know that &lt;math&gt;2x+17d&lt;/math&gt; must also be a perfect square.<br /> <br /> Since we want the smallest value of the sum, we know that &lt;math&gt;25=2x+17 \rightarrow x=4&lt;/math&gt; and &lt;math&gt;d=1.&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;9*25 = \boxed{225}.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132903 2002 AMC 12B Problems/Problem 13 2020-09-01T01:20:21Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> == Solution 1 ==<br /> <br /> Let the first term be &lt;math&gt;x&lt;/math&gt; and the common difference be &lt;math&gt;d.&lt;/math&gt; <br /> <br /> Thus, we know the sum of terms is &lt;math&gt;9(2x+17d).&lt;/math&gt;<br /> <br /> Because &lt;math&gt;9&lt;/math&gt; is already a perfect square, we know that &lt;math&gt;2x+17d&lt;/math&gt; must also be a perfect square.<br /> <br /> Since we want the smallest value of the sum, we know that &lt;math&gt;25=2x+17 \rightarrow x=4&lt;/math&gt; and &lt;math&gt;d=1.&lt;/math&gt;<br /> <br /> Thus, the answer is &lt;math&gt;9*25 = \boxed{225}.&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2002_AMC_12B_Problems/Problem_13&diff=132902 2002 AMC 12B Problems/Problem 13 2020-09-01T01:17:23Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> The sum of &lt;math&gt;18&lt;/math&gt; consecutive positive integers is a [[perfect square]]. The smallest possible value of this sum is<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 169<br /> \qquad\mathrm{(B)}\ 225<br /> \qquad\mathrm{(C)}\ 289<br /> \qquad\mathrm{(D)}\ 361<br /> \qquad\mathrm{(E)}\ 441&lt;/math&gt;<br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> <br /> == Solution 2 ==<br /> Let &lt;math&gt;a, a+1, \ldots, a + 17&lt;/math&gt; be the consecutive positive integers. Their sum, &lt;math&gt;18a + \frac{17(18)}{2} = 9(2a+17)&lt;/math&gt;, is a perfect square. Since &lt;math&gt;9&lt;/math&gt; is a perfect square, it follows that &lt;math&gt;2a + 17&lt;/math&gt; is a perfect square. The smallest possible such perfect square is &lt;math&gt;25&lt;/math&gt; when &lt;math&gt;a = 4&lt;/math&gt;, and the sum is &lt;math&gt;225 \Rightarrow \mathrm{(B)}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> <br /> Notice that all five choices given are perfect squares.<br /> <br /> Let &lt;math&gt;a&lt;/math&gt; be the smallest number, we have &lt;cmath&gt;a+(a+1)+(a+2)+...+(a+17)=18a+\sum_{k=1}^{17}k=18a+153&lt;/cmath&gt;<br /> <br /> Subtract &lt;math&gt;153&lt;/math&gt; from each of the choices and then check its divisibility by &lt;math&gt;18&lt;/math&gt;, we have &lt;math&gt;225&lt;/math&gt; as the smallest possible sum. &lt;math&gt;\mathrm {(B)}&lt;/math&gt;<br /> <br /> ~ Nafer<br /> <br /> == See also ==<br /> {{AMC12 box|year=2002|ab=B|num-b=12|num-a=14}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=132580 2007 AMC 12B Problems/Problem 24 2020-08-26T23:14:19Z <p>Coolmath2017: /* Solution 5 (Similar to Solution 1) */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\text{gcd}(a,b)=1&lt;/math&gt; and &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must be a factor of &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for any integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Rewriting the expression over a common denominator yields &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt;. This expression must be equal to some integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm&lt;/math&gt;. Taking this &lt;math&gt;\pmod{a}&lt;/math&gt; yields &lt;math&gt;14b^2 \equiv 0\pmod{a}&lt;/math&gt;. Since &lt;math&gt;\gcd(a,b)=1&lt;/math&gt;, &lt;math&gt;14 \equiv 0\pmod{a}&lt;/math&gt;. This implies that &lt;math&gt;a|14&lt;/math&gt; so &lt;math&gt;a = 1, 2, 7, 14&lt;/math&gt;.<br /> <br /> <br /> We can then take &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{b}&lt;/math&gt; to get that &lt;math&gt;9 \equiv 0 \pmod{b}&lt;/math&gt;. Thus &lt;math&gt;b = 1, 3, 9&lt;/math&gt;.<br /> <br /> <br /> However, taking &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{3}&lt;/math&gt;, &lt;math&gt;b^2 \equiv 0\pmod{3}&lt;/math&gt; so &lt;math&gt;b&lt;/math&gt; cannot equal 1.<br /> <br /> <br /> Also, note that if &lt;math&gt;b = 9&lt;/math&gt;, &lt;math&gt;\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}&lt;/math&gt;. Since &lt;math&gt;a|14&lt;/math&gt;, &lt;math&gt;\frac{14}{a}&lt;/math&gt; will be an integer, but &lt;math&gt;\frac{a}{9}&lt;/math&gt; will not be an integer since none of the possible values of &lt;math&gt;a&lt;/math&gt; are multiples of 9. Thus, &lt;math&gt;b&lt;/math&gt; cannot equal 9.<br /> <br /> <br /> Thus, the only possible values of &lt;math&gt;b&lt;/math&gt; is 3, and &lt;math&gt;a&lt;/math&gt; can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Similar to Solution 1)==<br /> <br /> Rewriting &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; over a common denominator gives &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}.&lt;/math&gt;<br /> <br /> Thus, we have &lt;math&gt;9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.&lt;/math&gt;<br /> <br /> Next, we have &lt;math&gt;ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;a \in (1,2,7,4).&lt;/math&gt; <br /> <br /> Next, we have &lt;math&gt;b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;b \in (1,3,9).&lt;/math&gt;<br /> <br /> Now, we simply do casework on &lt;math&gt;b.&lt;/math&gt; <br /> <br /> Plugging in &lt;math&gt;b = 1,3&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; gives that there are &lt;math&gt;4&lt;/math&gt; total solutions for &lt;math&gt;(a,b).&lt;/math&gt; <br /> <br /> ~coolmath2017<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=132579 2007 AMC 12B Problems/Problem 24 2020-08-26T23:13:48Z <p>Coolmath2017: /* Solution 5 (Similar to Solution 1) */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\text{gcd}(a,b)=1&lt;/math&gt; and &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must be a factor of &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for any integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Rewriting the expression over a common denominator yields &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt;. This expression must be equal to some integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm&lt;/math&gt;. Taking this &lt;math&gt;\pmod{a}&lt;/math&gt; yields &lt;math&gt;14b^2 \equiv 0\pmod{a}&lt;/math&gt;. Since &lt;math&gt;\gcd(a,b)=1&lt;/math&gt;, &lt;math&gt;14 \equiv 0\pmod{a}&lt;/math&gt;. This implies that &lt;math&gt;a|14&lt;/math&gt; so &lt;math&gt;a = 1, 2, 7, 14&lt;/math&gt;.<br /> <br /> <br /> We can then take &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{b}&lt;/math&gt; to get that &lt;math&gt;9 \equiv 0 \pmod{b}&lt;/math&gt;. Thus &lt;math&gt;b = 1, 3, 9&lt;/math&gt;.<br /> <br /> <br /> However, taking &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{3}&lt;/math&gt;, &lt;math&gt;b^2 \equiv 0\pmod{3}&lt;/math&gt; so &lt;math&gt;b&lt;/math&gt; cannot equal 1.<br /> <br /> <br /> Also, note that if &lt;math&gt;b = 9&lt;/math&gt;, &lt;math&gt;\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}&lt;/math&gt;. Since &lt;math&gt;a|14&lt;/math&gt;, &lt;math&gt;\frac{14}{a}&lt;/math&gt; will be an integer, but &lt;math&gt;\frac{a}{9}&lt;/math&gt; will not be an integer since none of the possible values of &lt;math&gt;a&lt;/math&gt; are multiples of 9. Thus, &lt;math&gt;b&lt;/math&gt; cannot equal 9.<br /> <br /> <br /> Thus, the only possible values of &lt;math&gt;b&lt;/math&gt; is 3, and &lt;math&gt;a&lt;/math&gt; can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Similar to Solution 1)==<br /> <br /> Rewriting &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; over a common denominator gives &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}.&lt;/math&gt;<br /> <br /> Thus, we have &lt;math&gt;9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.&lt;/math&gt;<br /> <br /> Next, we have &lt;math&gt;ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;a \in (1,2,7,4).&lt;/math&gt; <br /> <br /> Next, we have &lt;math&gt;b \mid 9a^2 + 14b^2 \Rightarrow b \mid 9a^2 \Rightarrow b \mid 9.<br /> <br /> Thus, &lt;/math&gt;b \in (1,3,9).&lt;math&gt;<br /> <br /> Now, we simply do casework on &lt;/math&gt;b.&lt;math&gt; <br /> <br /> Plugging in &lt;/math&gt;b = 1,3&lt;math&gt; and &lt;/math&gt;9&lt;math&gt; gives that there are &lt;/math&gt;4&lt;math&gt; total solutions for &lt;/math&gt;(a,b).$<br /> <br /> ~coolmath2017<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=132578 2007 AMC 12B Problems/Problem 24 2020-08-26T23:10:30Z <p>Coolmath2017: /* Solution 5 (Similar to Solution 1) */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\text{gcd}(a,b)=1&lt;/math&gt; and &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must be a factor of &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for any integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Rewriting the expression over a common denominator yields &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt;. This expression must be equal to some integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm&lt;/math&gt;. Taking this &lt;math&gt;\pmod{a}&lt;/math&gt; yields &lt;math&gt;14b^2 \equiv 0\pmod{a}&lt;/math&gt;. Since &lt;math&gt;\gcd(a,b)=1&lt;/math&gt;, &lt;math&gt;14 \equiv 0\pmod{a}&lt;/math&gt;. This implies that &lt;math&gt;a|14&lt;/math&gt; so &lt;math&gt;a = 1, 2, 7, 14&lt;/math&gt;.<br /> <br /> <br /> We can then take &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{b}&lt;/math&gt; to get that &lt;math&gt;9 \equiv 0 \pmod{b}&lt;/math&gt;. Thus &lt;math&gt;b = 1, 3, 9&lt;/math&gt;.<br /> <br /> <br /> However, taking &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{3}&lt;/math&gt;, &lt;math&gt;b^2 \equiv 0\pmod{3}&lt;/math&gt; so &lt;math&gt;b&lt;/math&gt; cannot equal 1.<br /> <br /> <br /> Also, note that if &lt;math&gt;b = 9&lt;/math&gt;, &lt;math&gt;\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}&lt;/math&gt;. Since &lt;math&gt;a|14&lt;/math&gt;, &lt;math&gt;\frac{14}{a}&lt;/math&gt; will be an integer, but &lt;math&gt;\frac{a}{9}&lt;/math&gt; will not be an integer since none of the possible values of &lt;math&gt;a&lt;/math&gt; are multiples of 9. Thus, &lt;math&gt;b&lt;/math&gt; cannot equal 9.<br /> <br /> <br /> Thus, the only possible values of &lt;math&gt;b&lt;/math&gt; is 3, and &lt;math&gt;a&lt;/math&gt; can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Similar to Solution 1)==<br /> <br /> Rewriting &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; over a common denominator gives &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}.&lt;/math&gt;<br /> <br /> Thus, we have &lt;math&gt;9 \mid 9a^2 + 14b^2 \Rightarrow 3 \mid b.&lt;/math&gt;<br /> <br /> Next, we have &lt;math&gt;ab \mid 9a^2+14b^2 \Rightarrow ab \mid 14b^2 \Rightarrow a \mid 14b.&lt;/math&gt;<br /> <br /> Thus,<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=132577 2007 AMC 12B Problems/Problem 24 2020-08-26T23:07:25Z <p>Coolmath2017: /* Solution 4 */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\text{gcd}(a,b)=1&lt;/math&gt; and &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must be a factor of &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for any integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Rewriting the expression over a common denominator yields &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt;. This expression must be equal to some integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm&lt;/math&gt;. Taking this &lt;math&gt;\pmod{a}&lt;/math&gt; yields &lt;math&gt;14b^2 \equiv 0\pmod{a}&lt;/math&gt;. Since &lt;math&gt;\gcd(a,b)=1&lt;/math&gt;, &lt;math&gt;14 \equiv 0\pmod{a}&lt;/math&gt;. This implies that &lt;math&gt;a|14&lt;/math&gt; so &lt;math&gt;a = 1, 2, 7, 14&lt;/math&gt;.<br /> <br /> <br /> We can then take &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{b}&lt;/math&gt; to get that &lt;math&gt;9 \equiv 0 \pmod{b}&lt;/math&gt;. Thus &lt;math&gt;b = 1, 3, 9&lt;/math&gt;.<br /> <br /> <br /> However, taking &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{3}&lt;/math&gt;, &lt;math&gt;b^2 \equiv 0\pmod{3}&lt;/math&gt; so &lt;math&gt;b&lt;/math&gt; cannot equal 1.<br /> <br /> <br /> Also, note that if &lt;math&gt;b = 9&lt;/math&gt;, &lt;math&gt;\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}&lt;/math&gt;. Since &lt;math&gt;a|14&lt;/math&gt;, &lt;math&gt;\frac{14}{a}&lt;/math&gt; will be an integer, but &lt;math&gt;\frac{a}{9}&lt;/math&gt; will not be an integer since none of the possible values of &lt;math&gt;a&lt;/math&gt; are multiples of 9. Thus, &lt;math&gt;b&lt;/math&gt; cannot equal 9.<br /> <br /> <br /> Thus, the only possible values of &lt;math&gt;b&lt;/math&gt; is 3, and &lt;math&gt;a&lt;/math&gt; can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Similar to Solution 1)==<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_24&diff=132576 2007 AMC 12B Problems/Problem 24 2020-08-26T23:07:02Z <p>Coolmath2017: /* Solution 4 */</p> <hr /> <div>==Problem 24==<br /> How many pairs of positive integers &lt;math&gt;(a,b)&lt;/math&gt; are there such that &lt;math&gt;\text{gcd}(a,b)=1&lt;/math&gt; and &lt;math&gt;\frac{a}{b} + \frac{14b}{9a}&lt;/math&gt; is an integer?<br /> <br /> &lt;math&gt;\mathrm {(A)} 4\quad\mathrm {(B)} 6\quad\mathrm {(C)} 9\quad\mathrm {(D)} 12\quad\mathrm {(E)} \text{infinitely many}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Combining the fraction, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt; must be an integer.<br /> <br /> Since the denominator contains a factor of &lt;math&gt;9&lt;/math&gt;, &lt;math&gt;9 | 9a^2 + 14b^2 \quad\Longrightarrow\quad 9 | b^2 \quad\Longrightarrow\quad 3 | b&lt;/math&gt;<br /> <br /> Since &lt;math&gt;b = 3n&lt;/math&gt; for some positive integer &lt;math&gt;n&lt;/math&gt;, we can rewrite the fraction(divide by &lt;math&gt;9&lt;/math&gt; on both top and bottom) as &lt;math&gt;\frac{a^2 + 14n^2}{3an}&lt;/math&gt;<br /> <br /> Since the denominator now contains a factor of &lt;math&gt;n&lt;/math&gt;, we get &lt;math&gt;n | a^2 + 14n^2 \quad\Longrightarrow\quad n | a^2&lt;/math&gt;.<br /> <br /> But since &lt;math&gt;1=\gcd(a,b)=\gcd(a,3n)=\gcd(a,n)&lt;/math&gt;, we must have &lt;math&gt;n=1&lt;/math&gt;, and thus &lt;math&gt;b=3&lt;/math&gt;.<br /> <br /> For &lt;math&gt;b=3&lt;/math&gt; the original fraction simplifies to &lt;math&gt;\frac{a^2 + 14}{3a}&lt;/math&gt;.<br /> <br /> For that to be an integer, &lt;math&gt;a&lt;/math&gt; must be a factor of &lt;math&gt;14&lt;/math&gt;, and therefore we must have &lt;math&gt;a\in\{1,2,7,14\}&lt;/math&gt;. Each of these values does indeed yield an integer.<br /> <br /> Thus there are four solutions: &lt;math&gt;(1,3)&lt;/math&gt;, &lt;math&gt;(2,3)&lt;/math&gt;, &lt;math&gt;(7,3)&lt;/math&gt;, &lt;math&gt;(14,3)&lt;/math&gt; and the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> Let's assume that &lt;math&gt;\frac{a}{b} + \frac{14b}{9a} = m&lt;/math&gt; We get<br /> <br /> &lt;math&gt;9a^2 - 9mab + 14b^2 = 0&lt;/math&gt;<br /> <br /> Factoring this, we get &lt;math&gt;4&lt;/math&gt; equations-<br /> <br /> &lt;math&gt;(3a-2b)(3a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(3a-b)(3a-14b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-2b)(9a-7b) = 0&lt;/math&gt;<br /> <br /> &lt;math&gt;(a-b)(9a-14b) = 0&lt;/math&gt;<br /> <br /> (It's all negative, because if we had positive signs, &lt;math&gt;a&lt;/math&gt; would be the opposite sign of &lt;math&gt;b&lt;/math&gt;)<br /> <br /> Now we look at these, and see that-<br /> <br /> &lt;math&gt;3a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=b&lt;/math&gt;<br /> <br /> &lt;math&gt;3a=7b&lt;/math&gt; <br /> <br /> &lt;math&gt;3a=14b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=2b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=7b&lt;/math&gt;<br /> <br /> &lt;math&gt;a=b&lt;/math&gt; <br /> <br /> &lt;math&gt;9a=14b&lt;/math&gt;<br /> <br /> This gives us &lt;math&gt;8&lt;/math&gt; solutions, but we note that the middle term needs to give you back &lt;math&gt;9m&lt;/math&gt;.<br /> <br /> For example, in the case <br /> <br /> &lt;math&gt;(a-2b)(9a-7b)&lt;/math&gt;, the middle term is &lt;math&gt;-25ab&lt;/math&gt;, which is not equal by &lt;math&gt;-9m&lt;/math&gt; for any integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> Similar reason for the fourth equation. This eliminates the last four solutions out of the above eight listed, giving us 4 solutions total &lt;math&gt;\mathrm {(A)}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> <br /> Let &lt;math&gt;u = \frac{a}{b}&lt;/math&gt;. Then the given equation becomes &lt;math&gt;u + \frac{14}{9u} = \frac{9u^2 + 14}{9u}&lt;/math&gt;.<br /> <br /> Let's set this equal to some value, &lt;math&gt;k \Rightarrow \frac{9u^2 + 14}{9u} = k&lt;/math&gt;.<br /> <br /> Clearing the denominator and simplifying, we get a quadratic in terms of &lt;math&gt;u&lt;/math&gt;:<br /> <br /> &lt;math&gt;9u^2 - 9ku + 14 = 0 \Rightarrow u = \frac{9k \pm \sqrt{(9k)^2 - 504}}{18}&lt;/math&gt;<br /> <br /> Since &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are integers, &lt;math&gt;u&lt;/math&gt; is a rational number. This means that &lt;math&gt;\sqrt{(9k)^2 - 504}&lt;/math&gt; is an integer.<br /> <br /> Let &lt;math&gt;\sqrt{(9k)^2 - 504} = x&lt;/math&gt;. Squaring and rearranging yields:<br /> <br /> &lt;math&gt;(9k)^2 - x^2 = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;(9k+x)(9k-x) = 504&lt;/math&gt;.<br /> <br /> In order for both &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;a&lt;/math&gt; to be an integer, &lt;math&gt;9k + x&lt;/math&gt; and &lt;math&gt;9k - x&lt;/math&gt; must both be odd or even. (This is easily proven using modular arithmetic.) In the case of this problem, both must be even. Let &lt;math&gt;9k + x = 2m&lt;/math&gt; and &lt;math&gt;9k - x = 2n&lt;/math&gt;.<br /> <br /> Then:<br /> <br /> &lt;math&gt;2m \cdot 2n = 504&lt;/math&gt;<br /> <br /> &lt;math&gt;mn = 126&lt;/math&gt;.<br /> <br /> Factoring 126, we get &lt;math&gt;6&lt;/math&gt; pairs of numbers: &lt;math&gt;(1,126), (2,63), (3,42), (6,21), (7,18),&lt;/math&gt; and &lt;math&gt;(9,14)&lt;/math&gt;.<br /> <br /> Looking back at our equations for &lt;math&gt;m&lt;/math&gt; and &lt;math&gt;n&lt;/math&gt;, we can solve for &lt;math&gt;k = \frac{2m + 2n}{18} = \frac{m+n}{9}&lt;/math&gt;. Since &lt;math&gt;k&lt;/math&gt; is an integer, there are only &lt;math&gt;2&lt;/math&gt; pairs of &lt;math&gt;(m,n)&lt;/math&gt; that work: &lt;math&gt;(3,42)&lt;/math&gt; and &lt;math&gt;(6,21)&lt;/math&gt;. This means that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;k&lt;/math&gt; such that &lt;math&gt;u&lt;/math&gt; is an integer. But looking back at &lt;math&gt;u&lt;/math&gt; in terms of &lt;math&gt;k&lt;/math&gt;, we have &lt;math&gt;\pm&lt;/math&gt;, meaning that there are &lt;math&gt;2&lt;/math&gt; values of &lt;math&gt;u&lt;/math&gt; for every &lt;math&gt;k&lt;/math&gt;. Thus, the answer is &lt;math&gt;2 \cdot 2 = 4 \Rightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> <br /> Rewriting the expression over a common denominator yields &lt;math&gt;\frac{9a^2 + 14b^2}{9ab}&lt;/math&gt;. This expression must be equal to some integer &lt;math&gt;m&lt;/math&gt;.<br /> <br /> <br /> Thus, &lt;math&gt;\frac{9a^2 + 14b^2}{9ab} = m \rightarrow 9a^2 + 14b^2 = 9abm&lt;/math&gt;. Taking this &lt;math&gt;\pmod{a}&lt;/math&gt; yields &lt;math&gt;14b^2 \equiv 0\pmod{a}&lt;/math&gt;. Since &lt;math&gt;\gcd(a,b)=1&lt;/math&gt;, &lt;math&gt;14 \equiv 0\pmod{a}&lt;/math&gt;. This implies that &lt;math&gt;a|14&lt;/math&gt; so &lt;math&gt;a = 1, 2, 7, 14&lt;/math&gt;.<br /> <br /> <br /> We can then take &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{b}&lt;/math&gt; to get that &lt;math&gt;9 \equiv 0 \pmod{b}&lt;/math&gt;. Thus &lt;math&gt;b = 1, 3, 9&lt;/math&gt;.<br /> <br /> <br /> However, taking &lt;math&gt;9a^2 + 14b^2 = 9abm \pmod{3}&lt;/math&gt;, &lt;math&gt;b^2 \equiv 0\pmod{3}&lt;/math&gt; so &lt;math&gt;b&lt;/math&gt; cannot equal 1.<br /> <br /> <br /> Also, note that if &lt;math&gt;b = 9&lt;/math&gt;, &lt;math&gt;\frac{a}{b}+\frac{14b}{9a} = \frac{a}{9}+\frac{14}{a}&lt;/math&gt;. Since &lt;math&gt;a|14&lt;/math&gt;, &lt;math&gt;\frac{14}{a}&lt;/math&gt; will be an integer, but &lt;math&gt;\frac{a}{9}&lt;/math&gt; will not be an integer since none of the possible values of &lt;math&gt;a&lt;/math&gt; are multiples of 9. Thus, &lt;math&gt;b&lt;/math&gt; cannot equal 9.<br /> <br /> <br /> Thus, the only possible values of &lt;math&gt;b&lt;/math&gt; is 3, and &lt;math&gt;a&lt;/math&gt; can be 1, 2, 7, or 14. This yields 4 possible solutions, so the answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> ==Solution 5 (Similar to Solution 1)<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=23|num-a=25}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2019_AMC_10A_Problems/Problem_15&diff=132495 2019 AMC 10A Problems/Problem 15 2020-08-25T01:13:28Z <p>Coolmath2017: /* Solution 5 */</p> <hr /> <div>{{duplicate|[[2019 AMC 10A Problems|2019 AMC 10A #15]] and [[2019 AMC 12A Problems|2019 AMC 12A #9]]}}<br /> <br /> ==Problem==<br /> <br /> A sequence of numbers is defined recursively by &lt;math&gt;a_1 = 1&lt;/math&gt;, &lt;math&gt;a_2 = \frac{3}{7}&lt;/math&gt;, and<br /> &lt;cmath&gt;a_n=\frac{a_{n-2} \cdot a_{n-1}}{2a_{n-2} - a_{n-1}}&lt;/cmath&gt;for all &lt;math&gt;n \geq 3&lt;/math&gt; Then &lt;math&gt;a_{2019}&lt;/math&gt; can be written as &lt;math&gt;\frac{p}{q}&lt;/math&gt;, where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p+q ?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 2020 \qquad\textbf{(B) } 4039 \qquad\textbf{(C) } 6057 \qquad\textbf{(D) } 6061 \qquad\textbf{(E) } 8078&lt;/math&gt;<br /> <br /> ==Solution 1 (Induction)==<br /> <br /> Using the recursive formula, we find &lt;math&gt;a_3=\frac{3}{11}&lt;/math&gt;, &lt;math&gt;a_4=\frac{3}{15}&lt;/math&gt;, and so on. It appears that &lt;math&gt;a_n=\frac{3}{4n-1}&lt;/math&gt;, for all &lt;math&gt;n&lt;/math&gt;. Setting &lt;math&gt;n=2019&lt;/math&gt;, we find &lt;math&gt;a_{2019}=\frac{3}{8075}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(E) }8078}&lt;/math&gt;.<br /> <br /> To prove this formula, we use induction. We are given that &lt;math&gt;a_1=1&lt;/math&gt; and &lt;math&gt;a_2=\frac{3}{7}&lt;/math&gt;, which satisfy our formula. Now assume the formula holds true for all &lt;math&gt;n\le m&lt;/math&gt; for some positive integer &lt;math&gt;m&lt;/math&gt;. By our assumption, &lt;math&gt;a_{m-1}=\frac{3}{4m-5}&lt;/math&gt; and &lt;math&gt;a_m=\frac{3}{4m-1}&lt;/math&gt;. Using the recursive formula, &lt;cmath&gt;a_{m+1}=\frac{a_{m-1}\cdot a_m}{2a_{m-1}-a_m}=\frac{\frac{3}{4m-5}\cdot\frac{3}{4m-1}}{2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}}=\frac{\left(\frac{3}{4m-5}\cdot\frac{3}{4m-1}\right)(4m-5)(4m-1)}{\left(2\cdot\frac{3}{4m-5}-\frac{3}{4m-1}\right)(4m-5)(4m-1)}=\frac{9}{6(4m-1)-3(4m-5)}=\frac{3}{4(m+1)-1},&lt;/cmath&gt;<br /> so our induction is complete.<br /> <br /> ==Solution 2==<br /> <br /> Since we are interested in finding the sum of the numerator and the denominator, consider the sequence defined by &lt;math&gt;b_n = \frac{1}{a_n}&lt;/math&gt;.<br /> <br /> We have &lt;math&gt;\frac{1}{a_n} = \frac{2a_{n-2}-a_{n-1}}{a_{n-2} \cdot a_{n-1}}=\frac{2}{a_{n-1}}-\frac{1}{a_{n-2}}&lt;/math&gt;, so in other words, &lt;math&gt;b_n = 2b_{n-1}-b_{n-2}=3b_{n-2}-2b_{n-3}=4b_{n-3}-3b_{n-4}=\dots&lt;/math&gt;.<br /> <br /> By recursively following this pattern, we can see that &lt;math&gt;b_n=(n-1) \cdot b_2 - (n-2) \cdot b_1&lt;/math&gt;.<br /> <br /> By plugging in 2019, we thus find &lt;math&gt;b_{2019} = 2018 \cdot \frac{7}{3}-2017 = \frac{8075}{3}&lt;/math&gt;. Since the numerator and the denominator are relatively prime, the answer is &lt;math&gt;\boxed{\textbf{(E) } 8078}&lt;/math&gt;.<br /> <br /> -eric2020<br /> <br /> ==Solution 3==<br /> <br /> It seems reasonable to transform the equation into something else. Let &lt;math&gt;a_{n}=x&lt;/math&gt;, &lt;math&gt;a_{n-1}=y&lt;/math&gt;, and &lt;math&gt;a_{n-2}=z&lt;/math&gt;. Therefore, we have &lt;cmath&gt;x=\frac{zy}{2z-y}&lt;/cmath&gt;<br /> &lt;cmath&gt;2xz-xy=zy&lt;/cmath&gt;<br /> &lt;cmath&gt;2xz=y(x+z)&lt;/cmath&gt;<br /> &lt;cmath&gt;y=\frac{2xz}{x+z}&lt;/cmath&gt;<br /> Thus, &lt;math&gt;y&lt;/math&gt; is the harmonic mean of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;z&lt;/math&gt;. This implies &lt;math&gt;a_{n}&lt;/math&gt; is a harmonic sequence or equivalently &lt;math&gt;b_{n}=\frac{1}{a_{n}}&lt;/math&gt; is arithmetic. Now, we have &lt;math&gt;b_{1}=1&lt;/math&gt;, &lt;math&gt;b_{2}=\frac{7}{3}&lt;/math&gt;, &lt;math&gt;b_{3}=\frac{11}{3}&lt;/math&gt;, and so on. Since the common difference is &lt;math&gt;\frac{4}{3}&lt;/math&gt;, we can express &lt;math&gt;b_{n}&lt;/math&gt; explicitly as &lt;math&gt;b_{n}=\frac{4}{3}(n-1)+1&lt;/math&gt;. This gives &lt;math&gt;b_{2019}=\frac{4}{3}(2019-1)+1=\frac{8075}{3}&lt;/math&gt; which implies &lt;math&gt;a_{2019}=\frac{3}{8075}=\frac{p}{q}&lt;/math&gt;. &lt;math&gt;p+q=\boxed{\textbf{(E) } 8078}&lt;/math&gt;<br /> ~jakeg314<br /> <br /> ==Solution 4 (Not rigorous at all)==<br /> Noticing that there is clearly a pattern, but the formula for it is hidious, we first find the first few terms of the sequence to see if there is any pattern: &lt;math&gt;1, \frac{1}{7}, \frac{3}{11}, \frac{1}{5}, \frac{3}{19}, \frac{3}{23} ...&lt;/math&gt;<br /> <br /> Now, we notice that the numerator seems to be in a pattern: &lt;math&gt;1, 3, 3, 1, 3, 3, 1, 3, 3...&lt;/math&gt; Then, we notice that the only time the numerator is &lt;math&gt;1&lt;/math&gt; is when the index is a multiple of &lt;math&gt;4&lt;/math&gt;. Clearly, &lt;math&gt;2019&lt;/math&gt; is NOT a multiple of &lt;math&gt;4&lt;/math&gt;, so the numerator is &lt;math&gt;3&lt;/math&gt;. Then, using the positions of each term, we can come up with a simple formula for the denominator with &lt;math&gt;n&lt;/math&gt; as the position or index (This only applies for the numbers with numerator &lt;math&gt;3&lt;/math&gt;): &lt;math&gt;3n + (n - 1)&lt;/math&gt;.<br /> <br /> Plugging &lt;math&gt;n&lt;/math&gt; in for &lt;math&gt;2019&lt;/math&gt;, we get &lt;math&gt;8075&lt;/math&gt; for the denominator. Adding &lt;math&gt;3&lt;/math&gt; (The numerator) gives &lt;math&gt;\boxed{\textbf{(E) }8078}&lt;/math&gt;<br /> <br /> ~EricShi1685<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2019|ab=A|num-b=14|num-a=16}}<br /> {{AMC12 box|year=2019|ab=A|num-b=8|num-a=10}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1988_AHSME_Problems/Problem_8&diff=126226 1988 AHSME Problems/Problem 8 2020-06-22T21:50:20Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> If &lt;math&gt;\frac{b}{a} = 2&lt;/math&gt; and &lt;math&gt;\frac{c}{b} = 3&lt;/math&gt;, what is the ratio of &lt;math&gt;a + b&lt;/math&gt; to &lt;math&gt;b + c&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ \frac{1}{3}\qquad<br /> \textbf{(B)}\ \frac{3}{8}\qquad<br /> \textbf{(C)}\ \frac{3}{5}\qquad<br /> \textbf{(D)}\ \frac{2}{3}\qquad<br /> \textbf{(E)}\ \frac{3}{4} &lt;/math&gt; <br /> <br /> ==Solution 1 ==<br /> Since we are finding ratios, it would be helpful to put everything in terms of one variable. Since &lt;math&gt;b&lt;/math&gt; is in both equations, that would be a place to start. <br /> We manipulate the equations yielding &lt;math&gt;\frac{b}{2}=a&lt;/math&gt; and &lt;math&gt;c=3b&lt;/math&gt;. Since we are asked to find the ratio of &lt;math&gt;a+b&lt;/math&gt; to &lt;math&gt;b+c&lt;/math&gt;, we need to find &lt;math&gt;\frac{a+b}{b+c}&lt;/math&gt;. We found the &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;c&lt;/math&gt; in terms of &lt;math&gt;b&lt;/math&gt; so that means we can plug them in. We have: &lt;math&gt;\frac{\frac{b}{2}+b}{b+3b}=\frac{\frac{3}{2}b}{4b}=\frac{3}{8}&lt;/math&gt;. Thus the answer is &lt;math&gt;\frac{3}{8} \implies \boxed{\text{B}}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> WLOG, let &lt;math&gt;b=4, a=2, c=12.&lt;/math&gt;<br /> Thus, the answer is &lt;math&gt;\frac{4+2}{12+4}= \frac{3}{8}&lt;/math&gt;<br /> <br /> == See also ==<br /> {{AHSME box|year=1988|num-b=7|num-a=9}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_19&diff=126203 1990 AHSME Problems/Problem 19 2020-06-22T14:21:32Z <p>Coolmath2017: /* Solution 1 */</p> <hr /> <div>== Problem ==<br /> <br /> For how many integers &lt;math&gt;N&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1990&lt;/math&gt; is the improper fraction &lt;math&gt;\frac{N^2+7}{N+4}&lt;/math&gt; &lt;math&gt;\underline{not}&lt;/math&gt; in lowest terms?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 86\quad<br /> \text{(C) } 90\quad<br /> \text{(D) } 104\quad<br /> \text{(E) } 105&lt;/math&gt;<br /> <br /> == Solution ==<br /> What we want to know is for how many &lt;math&gt;n&lt;/math&gt; is &lt;cmath&gt;\gcd(n^2+7, n+4) &gt; 1.&lt;/cmath&gt; We start by setting &lt;cmath&gt;n+4 \equiv 0 \mod m&lt;/cmath&gt; for some arbitrary &lt;math&gt;m&lt;/math&gt;. This shows that &lt;math&gt;m&lt;/math&gt; evenly divides &lt;math&gt;n+4&lt;/math&gt;. Next we want to see under which conditions &lt;math&gt;m&lt;/math&gt; also divides &lt;math&gt;n^2 + 7&lt;/math&gt;. We know from the previous statement that &lt;cmath&gt;n \equiv -4 \mod m&lt;/cmath&gt; and thus &lt;cmath&gt;n^2 \equiv (-4)^2 \equiv 16 \mod m.&lt;/cmath&gt; Next we simply add &lt;math&gt;7&lt;/math&gt; to get &lt;cmath&gt;n^2 + 7 \equiv 23 \mod m.&lt;/cmath&gt; However, we also want &lt;cmath&gt;n^2 + 7 \equiv 0 \mod m&lt;/cmath&gt; which leads to &lt;cmath&gt;n^2 + 7\equiv 23 \equiv 0 \mod m&lt;/cmath&gt; from the previous statement. Since from that statement &lt;math&gt;23&lt;/math&gt; divides &lt;math&gt;m&lt;/math&gt; evenly, &lt;math&gt;m&lt;/math&gt; must be of the form &lt;math&gt;23x&lt;/math&gt;, for some arbitrary integer &lt;math&gt;x&lt;/math&gt;. After this, we can set &lt;cmath&gt;n+4=23x&lt;/cmath&gt; and &lt;cmath&gt;n=23x-4.&lt;/cmath&gt; Finally, we must find the largest &lt;math&gt;x&lt;/math&gt; such that &lt;cmath&gt;23x-4&lt;1990.&lt;/cmath&gt; This is a simple linear inequality for which the answer is &lt;math&gt;x=86&lt;/math&gt;, or &lt;math&gt;\fbox{B}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=18|num-a=20}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_19&diff=126202 1990 AHSME Problems/Problem 19 2020-06-22T14:21:24Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> For how many integers &lt;math&gt;N&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1990&lt;/math&gt; is the improper fraction &lt;math&gt;\frac{N^2+7}{N+4}&lt;/math&gt; &lt;math&gt;\underline{not}&lt;/math&gt; in lowest terms?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 86\quad<br /> \text{(C) } 90\quad<br /> \text{(D) } 104\quad<br /> \text{(E) } 105&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> What we want to know is for how many &lt;math&gt;n&lt;/math&gt; is &lt;cmath&gt;\gcd(n^2+7, n+4) &gt; 1.&lt;/cmath&gt; We start by setting &lt;cmath&gt;n+4 \equiv 0 \mod m&lt;/cmath&gt; for some arbitrary &lt;math&gt;m&lt;/math&gt;. This shows that &lt;math&gt;m&lt;/math&gt; evenly divides &lt;math&gt;n+4&lt;/math&gt;. Next we want to see under which conditions &lt;math&gt;m&lt;/math&gt; also divides &lt;math&gt;n^2 + 7&lt;/math&gt;. We know from the previous statement that &lt;cmath&gt;n \equiv -4 \mod m&lt;/cmath&gt; and thus &lt;cmath&gt;n^2 \equiv (-4)^2 \equiv 16 \mod m.&lt;/cmath&gt; Next we simply add &lt;math&gt;7&lt;/math&gt; to get &lt;cmath&gt;n^2 + 7 \equiv 23 \mod m.&lt;/cmath&gt; However, we also want &lt;cmath&gt;n^2 + 7 \equiv 0 \mod m&lt;/cmath&gt; which leads to &lt;cmath&gt;n^2 + 7\equiv 23 \equiv 0 \mod m&lt;/cmath&gt; from the previous statement. Since from that statement &lt;math&gt;23&lt;/math&gt; divides &lt;math&gt;m&lt;/math&gt; evenly, &lt;math&gt;m&lt;/math&gt; must be of the form &lt;math&gt;23x&lt;/math&gt;, for some arbitrary integer &lt;math&gt;x&lt;/math&gt;. After this, we can set &lt;cmath&gt;n+4=23x&lt;/cmath&gt; and &lt;cmath&gt;n=23x-4.&lt;/cmath&gt; Finally, we must find the largest &lt;math&gt;x&lt;/math&gt; such that &lt;cmath&gt;23x-4&lt;1990.&lt;/cmath&gt; This is a simple linear inequality for which the answer is &lt;math&gt;x=86&lt;/math&gt;, or &lt;math&gt;\fbox{B}&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=18|num-a=20}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_19&diff=126189 1990 AHSME Problems/Problem 19 2020-06-22T02:29:35Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> <br /> For how many integers &lt;math&gt;N&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1990&lt;/math&gt; is the improper fraction &lt;math&gt;\frac{N^2+7}{N+4}&lt;/math&gt; &lt;math&gt;\underline{not}&lt;/math&gt; in lowest terms?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 86\quad<br /> \text{(C) } 90\quad<br /> \text{(D) } 104\quad<br /> \text{(E) } 105&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> What we want to know is for how many &lt;math&gt;n&lt;/math&gt; is &lt;cmath&gt;\gcd(n^2+7, n+4) &gt; 1.&lt;/cmath&gt; We start by setting &lt;cmath&gt;n+4 \equiv 0 \mod m&lt;/cmath&gt; for some arbitrary &lt;math&gt;m&lt;/math&gt;. This shows that &lt;math&gt;m&lt;/math&gt; evenly divides &lt;math&gt;n+4&lt;/math&gt;. Next we want to see under which conditions &lt;math&gt;m&lt;/math&gt; also divides &lt;math&gt;n^2 + 7&lt;/math&gt;. We know from the previous statement that &lt;cmath&gt;n \equiv -4 \mod m&lt;/cmath&gt; and thus &lt;cmath&gt;n^2 \equiv (-4)^2 \equiv 16 \mod m.&lt;/cmath&gt; Next we simply add &lt;math&gt;7&lt;/math&gt; to get &lt;cmath&gt;n^2 + 7 \equiv 23 \mod m.&lt;/cmath&gt; However, we also want &lt;cmath&gt;n^2 + 7 \equiv 0 \mod m&lt;/cmath&gt; which leads to &lt;cmath&gt;n^2 + 7\equiv 23 \equiv 0 \mod m&lt;/cmath&gt; from the previous statement. Since from that statement &lt;math&gt;23&lt;/math&gt; divides &lt;math&gt;m&lt;/math&gt; evenly, &lt;math&gt;m&lt;/math&gt; must be of the form &lt;math&gt;23x&lt;/math&gt;, for some arbitrary integer &lt;math&gt;x&lt;/math&gt;. After this, we can set &lt;cmath&gt;n+4=23x&lt;/cmath&gt; and &lt;cmath&gt;n=23x-4.&lt;/cmath&gt; Finally, we must find the largest &lt;math&gt;x&lt;/math&gt; such that &lt;cmath&gt;23x-4&lt;1990.&lt;/cmath&gt; This is a simple linear inequality for which the answer is &lt;math&gt;x=86&lt;/math&gt;, or &lt;math&gt;\fbox{B}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Rearranging the expression &lt;math&gt;N^2 + 7,&lt;/math&gt; we get &lt;math&gt;N^2 + 7 = (N+4)(N-4) + 23.&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt; \frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.&lt;/math&gt;<br /> <br /> Thus, &lt;math&gt;N+4 \equiv 0 \mod 23.&lt;/math&gt; <br /> <br /> Hence, &lt;math&gt;N= 23x -4,&lt;/math&gt; for some arbitrary number &lt;math&gt;x.&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;x&lt;/math&gt; in &lt;math&gt;23x-4&lt;1990,&lt;/math&gt; we find the answer is &lt;math&gt; x= 86,&lt;/math&gt; or &lt;math&gt;\fbox{B}.&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=18|num-a=20}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=1990_AHSME_Problems/Problem_19&diff=126188 1990 AHSME Problems/Problem 19 2020-06-22T02:27:07Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> For how many integers &lt;math&gt;N&lt;/math&gt; between &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;1990&lt;/math&gt; is the improper fraction &lt;math&gt;\frac{N^2+7}{N+4}&lt;/math&gt; &lt;math&gt;\underline{not}&lt;/math&gt; in lowest terms?<br /> <br /> &lt;math&gt;\text{(A) } 0\quad<br /> \text{(B) } 86\quad<br /> \text{(C) } 90\quad<br /> \text{(D) } 104\quad<br /> \text{(E) } 105&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> What we want to know is for how many &lt;math&gt;n&lt;/math&gt; is &lt;cmath&gt;\gcd(n^2+7, n+4) &gt; 1.&lt;/cmath&gt; We start by setting &lt;cmath&gt;n+4 \equiv 0 \mod m&lt;/cmath&gt; for some arbitrary &lt;math&gt;m&lt;/math&gt;. This shows that &lt;math&gt;m&lt;/math&gt; evenly divides &lt;math&gt;n+4&lt;/math&gt;. Next we want to see under which conditions &lt;math&gt;m&lt;/math&gt; also divides &lt;math&gt;n^2 + 7&lt;/math&gt;. We know from the previous statement that &lt;cmath&gt;n \equiv -4 \mod m&lt;/cmath&gt; and thus &lt;cmath&gt;n^2 \equiv (-4)^2 \equiv 16 \mod m.&lt;/cmath&gt; Next we simply add &lt;math&gt;7&lt;/math&gt; to get &lt;cmath&gt;n^2 + 7 \equiv 23 \mod m.&lt;/cmath&gt; However, we also want &lt;cmath&gt;n^2 + 7 \equiv 0 \mod m&lt;/cmath&gt; which leads to &lt;cmath&gt;n^2 + 7\equiv 23 \equiv 0 \mod m&lt;/cmath&gt; from the previous statement. Since from that statement &lt;math&gt;23&lt;/math&gt; divides &lt;math&gt;m&lt;/math&gt; evenly, &lt;math&gt;m&lt;/math&gt; must be of the form &lt;math&gt;23x&lt;/math&gt;, for some arbitrary integer &lt;math&gt;x&lt;/math&gt;. After this, we can set &lt;cmath&gt;n+4=23x&lt;/cmath&gt; and &lt;cmath&gt;n=23x-4.&lt;/cmath&gt; Finally, we must find the largest &lt;math&gt;x&lt;/math&gt; such that &lt;cmath&gt;23x-4&lt;1990.&lt;/cmath&gt; This is a simple linear inequality for which the answer is &lt;math&gt;x=86&lt;/math&gt;, or &lt;math&gt;\fbox{B}&lt;/math&gt;.<br /> <br /> == Solution 2 ==<br /> <br /> Rearranging the expression &lt;math&gt;N^2 + 7, we get &lt;/math&gt;N^2 + 7 = (N+4)(N-4) + 23.&lt;math&gt;<br /> <br /> Thus, &lt;/math&gt; \frac{(N+4)(N-4) + 23}{N+4} = N - 4 + \frac{23}{N+4}.&lt;math&gt;<br /> <br /> Thus, &lt;/math&gt;N+4 \equiv 0 \mod 23.&lt;math&gt; <br /> <br /> Hence, &lt;/math&gt;N= 23x -4,&lt;math&gt; for some arbitrary number &lt;/math&gt;x.&lt;math&gt;<br /> <br /> Solving for &lt;/math&gt;x&lt;math&gt; in &lt;/math&gt;23x-4&lt;1990,&lt;math&gt; we find the answer is &lt;/math&gt; x= 86,&lt;math&gt; or &lt;/math&gt;\fbox{B}.$<br /> <br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AHSME box|year=1990|num-b=18|num-a=20}} <br /> <br /> [[Category: Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2020_AMC_10B_Problems/Problem_21&diff=125854 2020 AMC 10B Problems/Problem 21 2020-06-18T21:00:58Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>{{duplicate|[[2020 AMC 10B Problems|2020 AMC 10B #21]] and [[2020 AMC 12B Problems|2020 AMC 12B #18]]}}<br /> <br /> ==Problem==<br /> <br /> In square &lt;math&gt;ABCD&lt;/math&gt;, points &lt;math&gt;E&lt;/math&gt; and &lt;math&gt;H&lt;/math&gt; lie on &lt;math&gt;\overline{AB}&lt;/math&gt; and &lt;math&gt;\overline{DA}&lt;/math&gt;, respectively, so that &lt;math&gt;AE=AH.&lt;/math&gt; Points &lt;math&gt;F&lt;/math&gt; and &lt;math&gt;G&lt;/math&gt; lie on &lt;math&gt;\overline{BC}&lt;/math&gt; and &lt;math&gt;\overline{CD}&lt;/math&gt;, respectively, and points &lt;math&gt;I&lt;/math&gt; and &lt;math&gt;J&lt;/math&gt; lie on &lt;math&gt;\overline{EH}&lt;/math&gt; so that &lt;math&gt;\overline{FI} \perp \overline{EH}&lt;/math&gt; and &lt;math&gt;\overline{GJ} \perp \overline{EH}&lt;/math&gt;. See the figure below. Triangle &lt;math&gt;AEH&lt;/math&gt;, quadrilateral &lt;math&gt;BFIE&lt;/math&gt;, quadrilateral &lt;math&gt;DHJG&lt;/math&gt;, and pentagon &lt;math&gt;FCGJI&lt;/math&gt; each has area &lt;math&gt;1.&lt;/math&gt; What is &lt;math&gt;FI^2&lt;/math&gt;?<br /> &lt;asy&gt;<br /> real x=2sqrt(2);<br /> real y=2sqrt(16-8sqrt(2))-4+2sqrt(2);<br /> real z=2sqrt(8-4sqrt(2));<br /> pair A, B, C, D, E, F, G, H, I, J;<br /> A = (0,0);<br /> B = (4,0);<br /> C = (4,4);<br /> D = (0,4);<br /> E = (x,0);<br /> F = (4,y);<br /> G = (y,4);<br /> H = (0,x);<br /> I = F + z * dir(225);<br /> J = G + z * dir(225);<br /> <br /> draw(A--B--C--D--A);<br /> draw(H--E);<br /> draw(J--G^^F--I);<br /> draw(rightanglemark(G, J, I), linewidth(.5));<br /> draw(rightanglemark(F, I, E), linewidth(.5));<br /> <br /> dot(&quot;$A$&quot;, A, S);<br /> dot(&quot;$B$&quot;, B, S);<br /> dot(&quot;$C$&quot;, C, dir(90));<br /> dot(&quot;$D$&quot;, D, dir(90));<br /> dot(&quot;$E$&quot;, E, S);<br /> dot(&quot;$F$&quot;, F, dir(0));<br /> dot(&quot;$G$&quot;, G, N);<br /> dot(&quot;$H$&quot;, H, W);<br /> dot(&quot;$I$&quot;, I, SW);<br /> dot(&quot;$J$&quot;, J, SW);<br /> <br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } \frac{7}{3} \qquad \textbf{(B) } 8-4\sqrt2 \qquad \textbf{(C) } 1+\sqrt2 \qquad \textbf{(D) } \frac{7}{4}\sqrt2 \qquad \textbf{(E) } 2\sqrt2&lt;/math&gt;<br /> <br /> ==Solution==<br /> Since the total area is &lt;math&gt;4&lt;/math&gt;, the side length of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;. We see that since triangle &lt;math&gt;HAE&lt;/math&gt; is a right isosceles triangle with area 1, we can determine sides &lt;math&gt;HA&lt;/math&gt; and &lt;math&gt;AE&lt;/math&gt; both to be &lt;math&gt;\sqrt{2}&lt;/math&gt;. Now, consider extending &lt;math&gt;FB&lt;/math&gt; and &lt;math&gt;IE&lt;/math&gt; until they intersect. Let the point of intersection be &lt;math&gt;K&lt;/math&gt;. We note that &lt;math&gt;EBK&lt;/math&gt; is also a right isosceles triangle with side &lt;math&gt;2-\sqrt{2}&lt;/math&gt; and find it's area to be &lt;math&gt;3-2\sqrt{2}&lt;/math&gt;. Now, we notice that &lt;math&gt;FIK&lt;/math&gt; is also a right isosceles triangle and find it's area to be &lt;math&gt;\frac{1}{2}&lt;/math&gt;&lt;math&gt;FI^2&lt;/math&gt;. This is also equal to &lt;math&gt;1+3-2\sqrt{2}&lt;/math&gt; or &lt;math&gt;4-2\sqrt{2}&lt;/math&gt;. Since we are looking for &lt;math&gt;FI^2&lt;/math&gt;, we want two times this. That gives &lt;math&gt;\boxed{\textbf{(B)}\ 8-4\sqrt{2}}&lt;/math&gt;.~TLiu<br /> <br /> ==Solution 2==<br /> Since this is a geometry problem involving sides, and we know that &lt;math&gt;HE&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;, we can use our ruler and find the ratio between &lt;math&gt;FI&lt;/math&gt; and &lt;math&gt;HE&lt;/math&gt;. Measuring(on the booklet), we get that &lt;math&gt;HE&lt;/math&gt; is about &lt;math&gt;1.8&lt;/math&gt; inches and &lt;math&gt;FI&lt;/math&gt; is about &lt;math&gt;1.4&lt;/math&gt; inches. Thus, we can then multiply the length of &lt;math&gt;HE&lt;/math&gt; by the ratio of &lt;math&gt;\frac{1.4}{1.8},&lt;/math&gt; of which we then get &lt;math&gt;FI= \frac{14}{9}.&lt;/math&gt; We take the square of that and get &lt;math&gt;\frac{196}{81},&lt;/math&gt; and the closest answer to that is &lt;math&gt;\boxed{\textbf{(B)}\ 8-4\sqrt{2}}&lt;/math&gt;. ~Celloboy (Note that this is just a strategy I happened to use that worked. Do not press your luck with this strategy, for it was a lucky guess)<br /> <br /> ==Solution 3==<br /> Draw the auxiliary line &lt;math&gt;AC&lt;/math&gt;. Denote by &lt;math&gt;M&lt;/math&gt; the point it intersects with &lt;math&gt;HE&lt;/math&gt;, and by &lt;math&gt;N&lt;/math&gt; the point it intersects with &lt;math&gt;GF&lt;/math&gt;. Last, denote by &lt;math&gt;x&lt;/math&gt; the segment &lt;math&gt;FN&lt;/math&gt;, and by &lt;math&gt;y&lt;/math&gt; the segment &lt;math&gt;FI&lt;/math&gt;. We will find two equations for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt;, and then solve for &lt;math&gt;y^2&lt;/math&gt;. <br /> <br /> Since the overall area of &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;4 \;\; \Longrightarrow \;\; AB=2&lt;/math&gt;, and &lt;math&gt;AC=2\sqrt{2}&lt;/math&gt;. In addition, the area of &lt;math&gt;\bigtriangleup AME = \frac{1}{2} \;\; \Longrightarrow \;\; AM=1&lt;/math&gt;.<br /> <br /> The two equations for &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; are then:<br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; Length of &lt;math&gt;AC&lt;/math&gt;: &lt;math&gt;1+y+x = 2\sqrt{2} \;\; \Longrightarrow \;\; x = (2\sqrt{2}-1) - y&lt;/math&gt; <br /> <br /> &lt;math&gt;\bullet&lt;/math&gt; Area of CMIF: &lt;math&gt;\frac{1}{2}x^2+xy = \frac{1}{2} \;\; \Longrightarrow \;\; x(x+2y)=1&lt;/math&gt;.<br /> <br /> Substituting the first into the second, yields <br /> &lt;math&gt;\left[\left(2\sqrt{2}-1\right)-y\right]\cdot \left[\left(2\sqrt{2}-1\right)+y\right]=1&lt;/math&gt;<br /> <br /> Solving for &lt;math&gt;y^2&lt;/math&gt; gives &lt;math&gt;\boxed{\textbf{(B)}\ 8-4\sqrt{2}}&lt;/math&gt; ~DrB<br /> <br /> <br /> ==Solution 4==<br /> <br /> Plot a point &lt;math&gt;F'&lt;/math&gt; such that &lt;math&gt;F'I&lt;/math&gt; and &lt;math&gt;AB&lt;/math&gt; are parallel and extend line &lt;math&gt;FB&lt;/math&gt; to point &lt;math&gt;B'&lt;/math&gt; such that &lt;math&gt;FIB'F'&lt;/math&gt; forms a square. Extend line &lt;math&gt;AE&lt;/math&gt; to meet line &lt;math&gt;F'B'&lt;/math&gt; and point &lt;math&gt;E'&lt;/math&gt; is the intersection of the two. The area of this square is equivalent to &lt;math&gt;FI^2&lt;/math&gt;. We see that the area of square &lt;math&gt;ABCD&lt;/math&gt; is &lt;math&gt;4&lt;/math&gt;, meaning each side is of length 2. The area of the pentagon &lt;math&gt;EIFF'E'&lt;/math&gt; is &lt;math&gt;2&lt;/math&gt;. Length &lt;math&gt;AE=\sqrt{2}&lt;/math&gt;, thus &lt;math&gt;EB=2-\sqrt{2}&lt;/math&gt;. Triangle &lt;math&gt;EB'E'&lt;/math&gt; is isosceles, and the area of this triangle is &lt;math&gt;\frac{1}{2}(4-2\sqrt{2})(2-\sqrt{2})=6-4\sqrt{2}&lt;/math&gt;. Adding these two areas, we get &lt;cmath&gt;2+6-4\sqrt{2}=8-4\sqrt{2}\rightarrow \boxed{\mathrm{(B)}}&lt;/cmath&gt;. --OGBooger<br /> <br /> ==Solution 5 (HARD Calculation)==<br /> <br /> We can easily observe that the area of square &lt;math&gt;ABCD&lt;/math&gt; is 4 and its side length is 2 since all four regions that build up the square has area 1. <br /> Extend &lt;math&gt;FI&lt;/math&gt; and let the intersection with &lt;math&gt;AB&lt;/math&gt; be &lt;math&gt;K&lt;/math&gt;. Connect &lt;math&gt;AC&lt;/math&gt;, and let the intersection of &lt;math&gt;AC&lt;/math&gt; and &lt;math&gt;HE&lt;/math&gt; be &lt;math&gt;L&lt;/math&gt;.<br /> Notice that since the area of triangle &lt;math&gt;AEH&lt;/math&gt; is 1 and &lt;math&gt;AE=AH&lt;/math&gt; , &lt;math&gt;AE=AH=\sqrt{2}&lt;/math&gt;, therefore &lt;math&gt;BE=HD=2-\sqrt{2}&lt;/math&gt;.<br /> Let &lt;math&gt;CG=CF=m&lt;/math&gt;, then &lt;math&gt;BF=DG=2-m&lt;/math&gt;.<br /> Also notice that &lt;math&gt;KB=2-m&lt;/math&gt;, thus &lt;math&gt;KE=KB-BE=2-m-(2-\sqrt{2})=\sqrt{2}-m&lt;/math&gt;.<br /> Now use the condition that the area of quadrilateral &lt;math&gt;BFIE&lt;/math&gt; is 1, we can set up the following equation: <br /> &lt;math&gt;\frac{1}{2}(2-m)^2-\frac{1}{4}(\sqrt{2}-m)^2=1&lt;/math&gt;<br /> We solve the equation and yield &lt;math&gt;m=\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}&lt;/math&gt;.<br /> Now notice that<br /> &lt;math&gt;FI=AC-AL=2\sqrt{2}-1-\frac{\sqrt{2}}{2}*\frac{8-2\sqrt{2}-\sqrt{64-32\sqrt{2}}}{2}&lt;/math&gt;<br /> &lt;math&gt;=2\sqrt{2}-1-\frac{8\sqrt{2}-4-\sqrt{128-64\sqrt2}}{4}&lt;/math&gt;<br /> &lt;math&gt;=\frac{\sqrt{128-64\sqrt{2}}}{4}&lt;/math&gt;.<br /> Hence &lt;math&gt;FI^2=\frac{128-64\sqrt{2}}{16}=8-4\sqrt{2}&lt;/math&gt;. -HarryW<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2020|ab=B|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2020|ab=B|num-b=17|num-a=19}}<br /> <br /> {{MAA Notice}}<br /> <br /> <br /> <br /> <br /> <br /> <br /> <br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, &lt;math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;/math&gt;x_1=r&lt;math&gt;, then &lt;/math&gt;x_2=(-1+i√3)/2&lt;math&gt; r, &lt;/math&gt;x_3=((-1+i√3)/2)^2&lt;math&gt; &lt;/math&gt;r=(-1-i√3)/2&lt;math&gt; &lt;/math&gt;r&lt;math&gt;, &lt;/math&gt;x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li<br /> ==Solution==<br /> <br /> <br /> Let &lt;math&gt;x_1=r&lt;/math&gt;, then &lt;math&gt;x_2=(-1+i√3)/2&lt;/math&gt; r, &lt;math&gt;x_3=((-1+i√3)/2)^2&lt;/math&gt; &lt;math&gt;r=(-1-i√3)/2&lt;/math&gt; &lt;math&gt;r&lt;/math&gt;, $x_4=((-1+i√3)/2)^3 &lt;cmath&gt;&lt;/cmath&gt;r which means x_4 will be back to x_1<br /> Now we have 3 different roots of the polynomial, x_(1 ) 〖,x〗_2 and x_3. next we gonna prove that all 5 roots of the polynomial must be chosen from those 3 roots. Let us assume that there has one root x_4=p which is different from the three roots we already know, then there must be another two roots, x_5=〖((-1+i√3)/2)〗^2 p=(-1-i√3)/2 p and x_6=〖((-1+i√3)/2)〗^3 p=p, different from all known roots. So we got 6 different roots for the polynomial, which is impossible. Therefore the assumption of different root is wrong.<br /> The polynomial then can be written like f(x)=〖(x-x_1)〗^m 〖(x-x_2)〗^n 〖(x-x_3)〗^q,m,n,q are non-negative integers and m+n+q=5. Since a,b,c and d are real numbers, n must be equal to q. Therefore (m,n,q) can only be (1,2,2) or (3,1,1), so the answer is (C) 2<br /> <br /> ~Yelong_Li</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=122085 2005 AMC 12A Problems/Problem 22 2020-05-06T02:33:24Z <p>Coolmath2017: /* Problem */</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 112. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Its surface area is &lt;cmath&gt;2lw+2lh+2wh=384,&lt;/cmath&gt; *I fixed a typo here*<br /> and the sum of all its edges is &lt;cmath&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}.&lt;/cmath&gt;<br /> Notice that &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,&lt;/cmath&gt; so the diameter is<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.&lt;/cmath&gt; The radius is half of the diameter, so<br /> &lt;cmath&gt;r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> As in the previous solution, we have that &lt;math&gt;2lw+2lh+2wh=384&lt;/math&gt; and &lt;math&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28&lt;/math&gt;, and the diameter of the sphere is the space diagonal of the prism, &lt;math&gt;\sqrt{l^2 + w^2 + h^2}&lt;/math&gt;.<br /> <br /> <br /> Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that &lt;math&gt;h=0&lt;/math&gt;. (This essentially means that we have an infinitesimally thin box.) We now have that &lt;math&gt;2lw = 384&lt;/math&gt; and &lt;math&gt;l + w = 28&lt;/math&gt;, and we are solving for &lt;math&gt;\sqrt{l^2 + w^2}&lt;/math&gt;. Because &lt;cmath&gt;(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,&lt;/cmath&gt; this means that &lt;cmath&gt;l^2 + w^2 = 28^2 - 384 = 400,&lt;/cmath&gt; so the space diagonal is &lt;math&gt;\sqrt{400} = 20&lt;/math&gt;. Since the diameter of the sphere is &lt;math&gt;20&lt;/math&gt;, the radius is &lt;math&gt;\boxed{\textbf{(B) } 10}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2005_AMC_12A_Problems/Problem_22&diff=122084 2005 AMC 12A Problems/Problem 22 2020-05-06T02:33:08Z <p>Coolmath2017: /* Problem */</p> <hr /> <div>== Problem ==<br /> A rectangular box &lt;math&gt; P &lt;/math&gt; is [[inscribe]]d in a [[sphere]] of [[radius]] &lt;math&gt;r&lt;/math&gt;. The [[surface area]] of &lt;math&gt;P&lt;/math&gt; is 384, and the sum of the lengths of its 12 edges is 28. What is &lt;math&gt;r&lt;/math&gt;?<br /> <br /> &lt;math&gt;\mathrm{(A)}\ 8\qquad \mathrm{(B)}\ 10\qquad \mathrm{(C)}\ 12\qquad \mathrm{(D)}\ 14\qquad \mathrm{(E)}\ 16&lt;/math&gt;<br /> <br /> == Solution==<br /> <br /> Box P has dimensions &lt;math&gt;l&lt;/math&gt;, &lt;math&gt;w&lt;/math&gt;, and &lt;math&gt;h&lt;/math&gt;. <br /> Its surface area is &lt;cmath&gt;2lw+2lh+2wh=384,&lt;/cmath&gt; *I fixed a typo here*<br /> and the sum of all its edges is &lt;cmath&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28.&lt;/cmath&gt;<br /> <br /> The diameter of the sphere is the space diagonal of the prism, which is &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2}.&lt;/cmath&gt;<br /> Notice that &lt;cmath&gt;(l + w + h)^2 - (2lw + 2lh + 2wh) = l^2 + w^2 + h^2 = 784 - 384 = 400,&lt;/cmath&gt; so the diameter is<br /> &lt;cmath&gt;\sqrt{l^2 + w^2 +h^2} = \sqrt{400} = 20.&lt;/cmath&gt; The radius is half of the diameter, so<br /> &lt;cmath&gt;r=\frac{20}{2} = \boxed{\textbf{(B)} 10}.&lt;/cmath&gt;<br /> <br /> == Solution 2 ==<br /> As in the previous solution, we have that &lt;math&gt;2lw+2lh+2wh=384&lt;/math&gt; and &lt;math&gt;l + w + h = \dfrac{4l+4w+4h}{4} = \dfrac{112}{4} = 28&lt;/math&gt;, and the diameter of the sphere is the space diagonal of the prism, &lt;math&gt;\sqrt{l^2 + w^2 + h^2}&lt;/math&gt;.<br /> <br /> <br /> Now, since this is competition math, we only need to find the space diagonal of any one box that fits the requirements, so assume that &lt;math&gt;h=0&lt;/math&gt;. (This essentially means that we have an infinitesimally thin box.) We now have that &lt;math&gt;2lw = 384&lt;/math&gt; and &lt;math&gt;l + w = 28&lt;/math&gt;, and we are solving for &lt;math&gt;\sqrt{l^2 + w^2}&lt;/math&gt;. Because &lt;cmath&gt;(l + w)^2 - 2lw = l^2 + 2lw + w^2 - 2lw = l^2 + w^2,&lt;/cmath&gt; this means that &lt;cmath&gt;l^2 + w^2 = 28^2 - 384 = 400,&lt;/cmath&gt; so the space diagonal is &lt;math&gt;\sqrt{400} = 20&lt;/math&gt;. Since the diameter of the sphere is &lt;math&gt;20&lt;/math&gt;, the radius is &lt;math&gt;\boxed{\textbf{(B) } 10}&lt;/math&gt;. ~[[User:emerald_block|emerald_block]]<br /> <br /> == See also ==<br /> {{AMC12 box|year=2005|num-b=21|num-a=23|ab=A}}<br /> <br /> [[Category:Intermediate Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121971 2006 AMC 12A Problems/Problem 6 2020-05-03T23:37:57Z <p>Coolmath2017: /* Solution 2 (Cheap) */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2 (Cheap) ==<br /> Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or &lt;math&gt;4&lt;/math&gt;. Next, we plug the answer choices in to see which one works. Trying &lt;math&gt;A&lt;/math&gt;, we get the area of one hexagon is &lt;math&gt;72&lt;/math&gt; , as desired, so the answer is &lt;math&gt;A&lt;/math&gt; .<br /> <br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121970 2006 AMC 12A Problems/Problem 6 2020-05-03T23:37:48Z <p>Coolmath2017: /* Solution 2 (Cheap) */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2 (Cheap) ==<br /> Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or &lt;math&gt;4&lt;/math&gt;. Next, we plug the answer choices in to see which one works. Trying &lt;math&gt;A&lt;/math&gt;, we get the area of one hexagon is &lt;math&gt;72&lt;/math&gt; , as desired, so the answer is &lt;math&gt;A&lt;/math&gt; .<br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121969 2006 AMC 12A Problems/Problem 6 2020-05-03T23:37:28Z <p>Coolmath2017: /* Solution 2 (Cheap) */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2 (Cheap) ==<br /> Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or &lt;math&gt;4&lt;/math&gt;. Next, we plug the answer choices in to see which one works. Trying &lt;math&gt;A&lt;/math&gt;, we get the area of one hexagon is &lt;math&gt;72&lt;/math&gt; , as desired, so the answer is &lt;math&gt;A&lt;/math&gt; .<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121968 2006 AMC 12A Problems/Problem 6 2020-05-03T23:36:05Z <p>Coolmath2017: /* Solution 2 (Cheap) */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2 (Cheap) ==<br /> Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or &lt;math&gt;4&lt;/math&gt;. Next, we plug the answer choices in to see which one works. Trying &lt;math&gt;A&lt;/math&gt;, we get the area of one hexagon is &lt;math&gt;72&lt;/math&gt;, as desired, so the answer is \Longrightarrow \mathrm{(A)}$<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121967 2006 AMC 12A Problems/Problem 6 2020-05-03T23:35:12Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2 (Cheap) ==<br /> Because the two hexagons are congruent, we know that the perpendicular line to A is half of BC, or &lt;math&gt;4&lt;/math&gt;. Next, we plug the answer choices in to see which one works.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2006_AMC_12A_Problems/Problem_6&diff=121966 2006 AMC 12A Problems/Problem 6 2020-05-03T23:32:16Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>{{duplicate|[[2006 AMC 12A Problems|2006 AMC 12A #6]] and [[2006 AMC 10A Problems/Problem 7|2006 AMC 10A #7]]}}<br /> == Problem ==<br /> The &lt;math&gt;8\times18&lt;/math&gt; [[rectangle]] &lt;math&gt;ABCD&lt;/math&gt; is cut into two congruent hexagons, as shown, in such a way that the two hexagons can be repositioned without overlap to form a square. What is &lt;math&gt;y&lt;/math&gt;? &lt;!-- [[Image:2006 AMC 12A Problem 6.png]] --&gt;<br /> &lt;asy&gt;<br /> unitsize(3mm);<br /> defaultpen(fontsize(10pt)+linewidth(.8pt));<br /> dotfactor=4;<br /> draw((0,4)--(18,4)--(18,-4)--(0,-4)--cycle);<br /> draw((6,4)--(6,0)--(12,0)--(12,-4));<br /> label(&quot;$A$&quot;,(0,4),NW);<br /> label(&quot;$B$&quot;,(18,4),NE);<br /> label(&quot;$C$&quot;,(18,-4),SE);<br /> label(&quot;$D$&quot;,(0,-4),SW);<br /> label(&quot;$y$&quot;,(3,4),S);<br /> label(&quot;$y$&quot;,(15,-4),N);<br /> label(&quot;$18$&quot;,(9,4),N);<br /> label(&quot;$18$&quot;,(9,-4),S);<br /> label(&quot;$8$&quot;,(0,0),W);<br /> label(&quot;$8$&quot;,(18,0),E);<br /> dot((0,4));<br /> dot((18,4));<br /> dot((18,-4));<br /> dot((0,-4));&lt;/asy&gt;<br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 7\qquad\mathrm{(C)}\ 8\qquad\mathrm{(D)}\ 9\qquad\mathrm{(E)}\ 10&lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> Since the two [[hexagon]]s are going to be repositioned to form a [[square (geometry) | square]] without overlap, the [[area]] will remain the same. The rectangle's area is &lt;math&gt;18\cdot8=144&lt;/math&gt;. This means the square will have four sides of length 12. The only way to do this is shown below.&lt;br&gt;<br /> <br /> &lt;asy&gt;<br /> size(175);<br /> pair A,B,C,D,E,F,G,H;<br /> A=(0,8);<br /> B=(12,12);<br /> C=(12,4);<br /> D=(0,0);<br /> E=(0,12);<br /> F=(12,0);<br /> G=(6,4);<br /> H=(6,8);<br /> draw(A--E--B--C--G--H--A--D--F--C);<br /> label(&quot;$A$&quot;,A,W); label(&quot;$B$&quot;,B,NE); label(&quot;$C$&quot;,(12.6,4)); label(&quot;$D$&quot;,D,SW);<br /> label(&quot;$12$&quot;,E--B,N); label(&quot;$12$&quot;,D--F,S); <br /> label(&quot;$4$&quot;,E--A,W); label(&quot;$4$&quot;,(12.4,-1.75),E);<br /> label(&quot;$8$&quot;,A--D,W); label(&quot;$8$&quot;,(12.4,4),E);<br /> label(&quot;$y$&quot;,A--H,S); label(&quot;$y$&quot;,G--C,N);<br /> &lt;/asy&gt;<br /> <br /> As you can see from the diagram, the [[line segment]] denoted as &lt;math&gt;y&lt;/math&gt; is half the length of the side of the square, which leads to &lt;math&gt; y = \frac{12}{2} = 6 \Longrightarrow \mathrm{(A)}&lt;/math&gt;.<br /> <br /> == Solution 2==<br /> Because the two hexagons are congruent, we know that<br /> <br /> == See also ==<br /> {{AMC12 box|year=2006|ab=A|num-b=5|num-a=7}}<br /> {{AMC10 box|year=2006|ab=A|num-b=6|num-a=8}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_4&diff=121889 2007 AMC 12B Problems/Problem 4 2020-05-01T23:31:55Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>==Problem==<br /> At Frank's Fruit Market, 3 bananas cost as much as 2 apples, and 6 apples cost as much as 4 oranges. How many oranges cost as much as 18 bananas?<br /> <br /> ==Solution==<br /> <br /> 18 bananas cost the same as 12 apples, and 12 apples cost the same as 8 oranges, so 18 bananas cost the same as &lt;math&gt;8 \Rightarrow \mathrm {(B)}&lt;/math&gt; oranges.<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2007|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2007_AMC_12B_Problems/Problem_3&diff=121888 2007 AMC 12B Problems/Problem 3 2020-05-01T23:31:33Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>==Problem==<br /> The point &lt;math&gt;O&lt;/math&gt; is the center of the circle circumscribed about triangle &lt;math&gt;ABC&lt;/math&gt;, with &lt;math&gt;\angle BOC = 120^{\circ}&lt;/math&gt; and &lt;math&gt;\angle AOB = 140^{\circ}&lt;/math&gt;, as shown. What is the degree measure of &lt;math&gt;\angle ABC&lt;/math&gt;?<br /> <br /> [[Image:2007_12B_AMC-3.png]]<br /> <br /> &lt;math&gt;\mathrm {(A)} 35 \qquad \mathrm {(B)} 40 \qquad \mathrm {(C)} 45 \qquad \mathrm {(D)} 50 \qquad \mathrm {(E)} 60&lt;/math&gt;<br /> ==Solution==<br /> Since triangles &lt;math&gt;ABO&lt;/math&gt; and &lt;math&gt;BOC&lt;/math&gt; are isosceles, &lt;math&gt;\angle ABO=20^o&lt;/math&gt; and &lt;math&gt;\angle OBC=30^o&lt;/math&gt;. Therefore, &lt;math&gt;\angle ABC=50^o&lt;/math&gt;, or $\mathim{(D)}.<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2007|ab=B|num-b=2|num-a=4}}<br /> <br /> [[Category:Introductory Algebra Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_7&diff=121832 2008 AMC 12B Problems/Problem 7 2020-04-29T22:25:36Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>==Problem 7==<br /> For real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, define &lt;math&gt;a\textdollar b = (a - b)^2&lt;/math&gt;. What is &lt;math&gt;(x - y)^2\textdollar(y - x)^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy&lt;/math&gt;<br /> <br /> ==Solution 1 ==<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (y-x)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (x-y)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;^2&lt;/math&gt;<br /> <br /> &lt;math&gt;0 \Rightarrow \textbf{(A)}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> WLOG, let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; both be &lt;math&gt;0&lt;/math&gt;. Thus, &lt;math&gt;(0-0)^2&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;0 \Rightarrow \textbf{(A)}&lt;/math&gt;<br /> <br /> ~coolmath2017<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_7&diff=121831 2008 AMC 12B Problems/Problem 7 2020-04-29T22:25:21Z <p>Coolmath2017: /* Solution 2 */</p> <hr /> <div>==Problem 7==<br /> For real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, define &lt;math&gt;a\textdollar b = (a - b)^2&lt;/math&gt;. What is &lt;math&gt;(x - y)^2\textdollar(y - x)^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy&lt;/math&gt;<br /> <br /> ==Solution 1 ==<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (y-x)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (x-y)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;^2&lt;/math&gt;<br /> <br /> &lt;math&gt;0 \Rightarrow \textbf{(A)}&lt;/math&gt;<br /> <br /> == Solution 2==<br /> WLOG, let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; both be &lt;math&gt;0&lt;/math&gt;. Thus, &lt;math&gt;(0-0)^2&lt;/math&gt; &lt;math&gt;=&lt;/math&gt; &lt;math&gt;0 \Rightarrow \textbf{(A)}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_12B_Problems/Problem_7&diff=121830 2008 AMC 12B Problems/Problem 7 2020-04-29T22:23:47Z <p>Coolmath2017: /* Solution */</p> <hr /> <div>==Problem 7==<br /> For real numbers &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt;, define &lt;math&gt;a\textdollar b = (a - b)^2&lt;/math&gt;. What is &lt;math&gt;(x - y)^2\textdollar(y - x)^2&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A)}\ 0 \qquad \textbf{(B)}\ x^2 + y^2 \qquad \textbf{(C)}\ 2x^2 \qquad \textbf{(D)}\ 2y^2 \qquad \textbf{(E)}\ 4xy&lt;/math&gt;<br /> <br /> ==Solution 1 ==<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (y-x)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;\left[ (x-y)^2 - (x-y)^2 \right]^2&lt;/math&gt;<br /> <br /> &lt;math&gt;^2&lt;/math&gt;<br /> <br /> &lt;math&gt;0 \Rightarrow \textbf{(A)}&lt;/math&gt;<br /> <br /> == Solution 2 ==<br /> WLOG, let &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;y&lt;/math&gt; both be 0. Thus,&lt;math&gt;0^2&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2008|ab=B|num-b=6|num-a=8}}<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_9&diff=121766 2009 AMC 12B Problems/Problem 9 2020-04-28T01:19:41Z <p>Coolmath2017: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has vertices &lt;math&gt;A = (3,0)&lt;/math&gt;, &lt;math&gt;B = (0,3)&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, where &lt;math&gt;C&lt;/math&gt; is on the line &lt;math&gt;x + y = 7&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad<br /> \mathrm{(B)}\ 8\qquad<br /> \mathrm{(C)}\ 10\qquad<br /> \mathrm{(D)}\ 12\qquad<br /> \mathrm{(E)}\ 14&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> Because the line &lt;math&gt;x + y = 7&lt;/math&gt; is parallel to &lt;math&gt;\overline {AB}&lt;/math&gt;, the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is independent of the location of &lt;math&gt;C&lt;/math&gt; on that line. Therefore it may be assumed that &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;(7,0)&lt;/math&gt;. In that case the triangle has base &lt;math&gt;AC = 4&lt;/math&gt; and altitude &lt;math&gt;3&lt;/math&gt;, so its area is &lt;math&gt;\frac 12 \cdot 4 \cdot 3 = \boxed {6}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> The base of the triangle is &lt;math&gt;AB = \sqrt{3^2 + 3^2} = 3\sqrt 2&lt;/math&gt;. Its altitude is the distance between the point &lt;math&gt;A&lt;/math&gt; and the parallel line &lt;math&gt;x + y = 7&lt;/math&gt;, which is &lt;math&gt;\frac 4{\sqrt 2} = 2\sqrt 2&lt;/math&gt;. Therefore its area is &lt;math&gt;\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(3,0), B=(0,3);<br /> draw ( (-1,0) -- (9,0), dashed );<br /> draw ( (0,-1) -- (0,9), dashed );<br /> dot(A); dot(B); draw(A--B);<br /> draw ( (-1,8) -- (8,-1) );<br /> label( &quot;$A$&quot;, A, S );<br /> label( &quot;$B$&quot;, B, W );<br /> label( &quot;$3$&quot;, A--(0,0), S );<br /> label( &quot;$3$&quot;, B--(0,0), W );<br /> label( &quot;$x+y=7$&quot;, (8,-1), SE );<br /> pair C = intersectionpoint(A--(10,7),(7,0)--(0,7));<br /> draw( A--C, dashed );<br /> draw(rightanglemark(A,C,(7,0)));<br /> draw(rightanglemark(C,A,B));<br /> label( &quot;$4$&quot;, A--(7,0), S );<br /> label( &quot;$3\sqrt 2$&quot;, 0.67*B+0.33*A, NE );<br /> label( &quot;$\frac 4{\sqrt 2}$&quot;, A--C, NW );<br /> label( &quot;$\frac 4{\sqrt 2}$&quot;, C--(7,0), NE );<br /> &lt;/asy&gt;<br /> <br /> === Solution 3 ===<br /> By Shoelace, our area is:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.&lt;/cmath&gt;<br /> We know &lt;math&gt;x+y=7&lt;/math&gt; so we get:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |9-21|=\boxed 6&lt;/cmath&gt;<br /> <br /> === Solution 4 ===<br /> WLOG, let the coordinates of &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;(3,4)&lt;/math&gt; , or any coordinate, for that matter. Applying the shoelace formula, we get the area as &lt;math&gt;\boxed 6&lt;/math&gt;.<br /> <br /> ~coolmath2017<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Coolmath2017 https://artofproblemsolving.com/wiki/index.php?title=2009_AMC_12B_Problems/Problem_9&diff=121765 2009 AMC 12B Problems/Problem 9 2020-04-28T01:18:46Z <p>Coolmath2017: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> Triangle &lt;math&gt;ABC&lt;/math&gt; has vertices &lt;math&gt;A = (3,0)&lt;/math&gt;, &lt;math&gt;B = (0,3)&lt;/math&gt;, and &lt;math&gt;C&lt;/math&gt;, where &lt;math&gt;C&lt;/math&gt; is on the line &lt;math&gt;x + y = 7&lt;/math&gt;. What is the area of &lt;math&gt;\triangle ABC&lt;/math&gt;?<br /> <br /> <br /> &lt;math&gt;\mathrm{(A)}\ 6\qquad<br /> \mathrm{(B)}\ 8\qquad<br /> \mathrm{(C)}\ 10\qquad<br /> \mathrm{(D)}\ 12\qquad<br /> \mathrm{(E)}\ 14&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> === Solution 1 ===<br /> Because the line &lt;math&gt;x + y = 7&lt;/math&gt; is parallel to &lt;math&gt;\overline {AB}&lt;/math&gt;, the area of &lt;math&gt;\triangle ABC&lt;/math&gt; is independent of the location of &lt;math&gt;C&lt;/math&gt; on that line. Therefore it may be assumed that &lt;math&gt;C&lt;/math&gt; is &lt;math&gt;(7,0)&lt;/math&gt;. In that case the triangle has base &lt;math&gt;AC = 4&lt;/math&gt; and altitude &lt;math&gt;3&lt;/math&gt;, so its area is &lt;math&gt;\frac 12 \cdot 4 \cdot 3 = \boxed {6}&lt;/math&gt;.<br /> <br /> === Solution 2 ===<br /> The base of the triangle is &lt;math&gt;AB = \sqrt{3^2 + 3^2} = 3\sqrt 2&lt;/math&gt;. Its altitude is the distance between the point &lt;math&gt;A&lt;/math&gt; and the parallel line &lt;math&gt;x + y = 7&lt;/math&gt;, which is &lt;math&gt;\frac 4{\sqrt 2} = 2\sqrt 2&lt;/math&gt;. Therefore its area is &lt;math&gt;\frac 12 \cdot 3\sqrt 2 \cdot 2\sqrt 2 = \boxed{6}&lt;/math&gt;. The answer is &lt;math&gt;\mathrm{(A)}&lt;/math&gt;.<br /> <br /> &lt;asy&gt;<br /> unitsize(0.75cm);<br /> defaultpen(0.8);<br /> pair A=(3,0), B=(0,3);<br /> draw ( (-1,0) -- (9,0), dashed );<br /> draw ( (0,-1) -- (0,9), dashed );<br /> dot(A); dot(B); draw(A--B);<br /> draw ( (-1,8) -- (8,-1) );<br /> label( &quot;$A$&quot;, A, S );<br /> label( &quot;$B$&quot;, B, W );<br /> label( &quot;$3$&quot;, A--(0,0), S );<br /> label( &quot;$3$&quot;, B--(0,0), W );<br /> label( &quot;$x+y=7$&quot;, (8,-1), SE );<br /> pair C = intersectionpoint(A--(10,7),(7,0)--(0,7));<br /> draw( A--C, dashed );<br /> draw(rightanglemark(A,C,(7,0)));<br /> draw(rightanglemark(C,A,B));<br /> label( &quot;$4$&quot;, A--(7,0), S );<br /> label( &quot;$3\sqrt 2$&quot;, 0.67*B+0.33*A, NE );<br /> label( &quot;$\frac 4{\sqrt 2}$&quot;, A--C, NW );<br /> label( &quot;$\frac 4{\sqrt 2}\$&quot;, C--(7,0), NE );<br /> &lt;/asy&gt;<br /> <br /> === Solution 3 ===<br /> By Shoelace, our area is:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |(3\cdot 3 + 0 \cdot y+ 0 \cdot x)-(0*0+3(x+y))|.&lt;/cmath&gt;<br /> We know &lt;math&gt;x+y=7&lt;/math&gt; so we get:<br /> &lt;cmath&gt;\frac {1}{2} \cdot |9-21|=\boxed 6&lt;/cmath&gt;<br /> <br /> === Solution 4 ===<br /> WLOG, let the coordinates of &lt;math&gt;C&lt;/math&gt; be &lt;math&gt;(3,4)&lt;/math&gt; , or any coordinate, for that matter. Applying the shoelace formula, we get the area as &lt;math&gt;\boxed 6&lt;/math&gt;.<br /> <br /> == See also ==<br /> {{AMC12 box|year=2009|ab=B|num-b=8|num-a=10}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:Area Problems]]<br /> {{MAA Notice}}</div> Coolmath2017