https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Kundusne000&feedformat=atomAoPS Wiki - User contributions [en]2024-03-28T11:41:31ZUser contributionsMediaWiki 1.31.1https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_11&diff=1963322003 AIME II Problems/Problem 112023-08-01T23:47:34Z<p>Kundusne000: /* Solution 3 */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
<br />
We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math><br />
<br />
Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</math><br />
<br />
From the third equation, we get <br />
<math>CN=\frac{168} {25}.</math><br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta BCN,</math> we have <br />
<br />
<math>BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.</math> <br />
<br />
Thus, <br />
<math>MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.</math> <br />
<br />
In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get <br />
<math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math><br />
<br />
Thus, <br />
<math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.</math><br />
<br />
Hence, the answer is <math>527+11+40=\boxed{578}.</math><br />
<br />
~ minor edits by kundusne000<br />
<br />
===Solution 2===<br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta AMD</math>, we get <math>DM=\frac{5\sqrt{11}} {2}</math>. Since <math>ABC</math> is a right triangle, <math>M</math> is the circumcenter and thus, <math>CM=\frac{25} {2}</math>. We let <math>\angle CMD=\theta</math>. By the [[Law of Cosines]],<br />
<br />
<math>24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).</math><br />
<br />
It follows that <math>\sin \theta = \frac{527} {625}</math>. Thus,<br />
<math>[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}</math>.<br />
<br />
===Solution 3===<br />
<br />
Suppose <math>ABC</math> is plotted on the [[cartesian plane]] with <math>C</math> at <math>(0,0)</math>, <math>A</math> at <math>(0,7)</math>, and <math>B</math> at <math>(24,0)</math>.<br />
Then <math>M</math> is at <math>(12,3.5)</math>. Since <math>\Delta ABD</math> is isosceles, <math>MD</math> is perpendicular to <math>AM</math>, and since <math>AM=12.5</math> and <math>AD=15, MD=2.5\sqrt{11}</math>.<br />
The slope of <math>AM</math> is <math>-\frac{7}{24}</math> so the slope of <math>MD</math> is <math>\frac{24}{7}</math>.<br />
Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem.<br />
So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},\, 3.5-2.4\sqrt{11})</math>.<br />
Since we know the coordinates of each of the vertices of <math>\Delta CMD</math>, we can apply the [[Shoelace Theorem]] to find the area of <math>\Delta CMD, \frac{527 \sqrt{11}}{40}</math>.<br />
<br />
~ minor edit by kundusne000<br />
<br />
===Solution 4===<br />
<br />
Let <math>E</math> be the intersection of lines <math>BC</math> and <math>DM</math>. Since triangles <math>\Delta CME</math> and <math>\Delta CMD</math> share a side and height, the area of <math>\Delta CDM</math> is equal to <math>\frac{DM}{EM}</math> times the area of <math>\Delta CME</math>.<br />
By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solving yields <math>EM=\frac{175}{48}</math>. Using the same method but for <math>EB</math> yields <math>EB=\frac{625}{48}</math>. As in previous solutions, by the [[Pythagorean Theorem]], <math>DM=\frac{5\sqrt{11}}{2}</math>. So, <math>\frac{DM}{EM}=\frac{24\sqrt{11}}{35}</math>.<br />
Now, since we know both <math>CB</math> and <math>EB</math>, we can find that <math>CE=\frac{527}{48}</math>. The height of <math>\Delta CME</math> is the length from point <math>M</math> to <math>CB</math>. Since <math>M</math> is the midpoint of <math>AB</math>, the height is just <math>\frac{1}{2}\cdot7=\frac{7}{2}</math>. Using this, we can find that the area of <math>\Delta CMD</math> is <math>\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}</math>, giving our answer of <math>\boxed{578}</math>.<br />
<br />
Solution by someonenumber011.<br />
<br />
== See also ==<br />
Video Solution from Khan Academy: https://www.youtube.com/watch?v=smtrrefmC40<br />
{{AIME box|year=2003|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_11&diff=1963312003 AIME II Problems/Problem 112023-08-01T23:46:06Z<p>Kundusne000: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
<br />
We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math><br />
<br />
Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</math><br />
<br />
From the third equation, we get <br />
<math>CN=\frac{168} {25}.</math><br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta BCN,</math> we have <br />
<br />
<math>BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.</math> <br />
<br />
Thus, <br />
<math>MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.</math> <br />
<br />
In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get <br />
<math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math><br />
<br />
Thus, <br />
<math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.</math><br />
<br />
Hence, the answer is <math>527+11+40=\boxed{578}.</math><br />
<br />
~ minor edits by kundusne000<br />
<br />
===Solution 2===<br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta AMD</math>, we get <math>DM=\frac{5\sqrt{11}} {2}</math>. Since <math>ABC</math> is a right triangle, <math>M</math> is the circumcenter and thus, <math>CM=\frac{25} {2}</math>. We let <math>\angle CMD=\theta</math>. By the [[Law of Cosines]],<br />
<br />
<math>24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).</math><br />
<br />
It follows that <math>\sin \theta = \frac{527} {625}</math>. Thus,<br />
<math>[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}</math>.<br />
<br />
===Solution 3===<br />
<br />
Suppose <math>ABC</math> is plotted on the [[cartesian plane]] with <math>C</math> at <math>(0,0)</math>, <math>A</math> at <math>(0,7)</math>, and <math>B</math> at <math>(24,0)</math>.<br />
Then <math>M</math> is at <math>(12,3.5)</math>. Since <math>\Delta ABD</math> is isosceles, <math>MD</math> is perpendicular to <math>AM</math>, and since <math>AM=12.5</math> and <math>AD=15, MD=2.5\sqrt{11}</math>.<br />
The slope of <math>AM</math> is <math>-\frac{7}{24}</math> so the slope of <math>MD</math> is <math>\frac{24}{7}</math>.<br />
Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem.<br />
So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},\, 2.5-2.4\sqrt{11})</math>.<br />
Since we know the coordinates of each of the vertices of <math>\Delta CMD</math>, we can apply the [[Shoelace Theorem]] to find the area of <math>\Delta CMD, \frac{527 \sqrt{11}}{40}</math>.<br />
<br />
<br />
===Solution 4===<br />
<br />
Let <math>E</math> be the intersection of lines <math>BC</math> and <math>DM</math>. Since triangles <math>\Delta CME</math> and <math>\Delta CMD</math> share a side and height, the area of <math>\Delta CDM</math> is equal to <math>\frac{DM}{EM}</math> times the area of <math>\Delta CME</math>.<br />
By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solving yields <math>EM=\frac{175}{48}</math>. Using the same method but for <math>EB</math> yields <math>EB=\frac{625}{48}</math>. As in previous solutions, by the [[Pythagorean Theorem]], <math>DM=\frac{5\sqrt{11}}{2}</math>. So, <math>\frac{DM}{EM}=\frac{24\sqrt{11}}{35}</math>.<br />
Now, since we know both <math>CB</math> and <math>EB</math>, we can find that <math>CE=\frac{527}{48}</math>. The height of <math>\Delta CME</math> is the length from point <math>M</math> to <math>CB</math>. Since <math>M</math> is the midpoint of <math>AB</math>, the height is just <math>\frac{1}{2}\cdot7=\frac{7}{2}</math>. Using this, we can find that the area of <math>\Delta CMD</math> is <math>\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}</math>, giving our answer of <math>\boxed{578}</math>.<br />
<br />
Solution by someonenumber011.<br />
<br />
== See also ==<br />
Video Solution from Khan Academy: https://www.youtube.com/watch?v=smtrrefmC40<br />
{{AIME box|year=2003|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2003_AIME_II_Problems/Problem_11&diff=1963302003 AIME II Problems/Problem 112023-08-01T23:45:15Z<p>Kundusne000: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Triangle <math>ABC</math> is a right triangle with <math>AC = 7,</math> <math>BC = 24,</math> and right angle at <math>C.</math> Point <math>M</math> is the midpoint of <math>AB,</math> and <math>D</math> is on the same side of line <math>AB</math> as <math>C</math> so that <math>AD = BD = 15.</math> Given that the area of triangle <math>CDM</math> may be expressed as <math>\frac {m\sqrt {n}}{p},</math> where <math>m,</math> <math>n,</math> and <math>p</math> are positive integers, <math>m</math> and <math>p</math> are relatively prime, and <math>n</math> is not divisible by the square of any prime, find <math>m + n + p.</math><br />
<br />
== Solution ==<br />
===Solution 1===<br />
<br />
We use the [[Pythagorean Theorem]] on <math>ABC</math> to determine that <math>AB=25.</math><br />
<br />
Let <math>N</math> be the orthogonal projection from <math>C</math> to <math>AB.</math> Thus, <math>[CDM]=\frac{(DM)(MN)} {2}</math>, <math>MN=BN-BM</math>, and <math>[ABC]=\frac{24 \cdot 7} {2} =\frac{25 \cdot (CN)} {2}.</math><br />
<br />
From the third equation, we get <br />
<math>CN=\frac{168} {25}.</math><br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta ACN,</math> we have <br />
<br />
<math>BN=\sqrt{\left(\frac{24 \cdot 25} {25}\right)^2-\left(\frac{24 \cdot 7} {25}\right)^2}=\frac{24} {25}\sqrt{25^2-7^2}=\frac{576} {25}.</math> <br />
<br />
Thus, <br />
<math>MN=\frac{576} {25}-\frac{25} {2}=\frac{527} {50}.</math> <br />
<br />
In <math>\Delta ADM</math>, we use the [[Pythagorean Theorem]] to get <br />
<math>DM=\sqrt{15^2-\left(\frac{25} {2}\right)^2}=\frac{5} {2} \sqrt{11}.</math><br />
<br />
Thus, <br />
<math>[CDM]=\frac{527 \cdot 5\sqrt{11}} {50 \cdot 2 \cdot 2}= \frac{527\sqrt{11}} {40}.</math><br />
<br />
Hence, the answer is <math>527+11+40=\boxed{578}.</math><br />
<br />
===Solution 2===<br />
<br />
By the [[Pythagorean Theorem]] in <math>\Delta AMD</math>, we get <math>DM=\frac{5\sqrt{11}} {2}</math>. Since <math>ABC</math> is a right triangle, <math>M</math> is the circumcenter and thus, <math>CM=\frac{25} {2}</math>. We let <math>\angle CMD=\theta</math>. By the [[Law of Cosines]],<br />
<br />
<math>24^2 = 2 \cdot (12.5)^2-2 \cdot (12.5)^2 * \cos (90+\theta).</math><br />
<br />
It follows that <math>\sin \theta = \frac{527} {625}</math>. Thus,<br />
<math>[CMD]=\frac{1} {2} (12.5) \left(\frac{5\sqrt{11}} {2}\right)\left(\frac{527} {625}\right)=\frac{527\sqrt{11}} {40}</math>.<br />
<br />
===Solution 3===<br />
<br />
Suppose <math>ABC</math> is plotted on the [[cartesian plane]] with <math>C</math> at <math>(0,0)</math>, <math>A</math> at <math>(0,7)</math>, and <math>B</math> at <math>(24,0)</math>.<br />
Then <math>M</math> is at <math>(12,3.5)</math>. Since <math>\Delta ABD</math> is isosceles, <math>MD</math> is perpendicular to <math>AM</math>, and since <math>AM=12.5</math> and <math>AD=15, MD=2.5\sqrt{11}</math>.<br />
The slope of <math>AM</math> is <math>-\frac{7}{24}</math> so the slope of <math>MD</math> is <math>\frac{24}{7}</math>.<br />
Draw a vertical line through <math>M</math> and a horizontal line through <math>D</math>. Suppose these two lines meet at <math>X</math>. then <math>MX=\frac{24}{7}DX</math> so <math>MD=\frac{25}{7}DX=\frac{25}{24}MD</math> by the pythagorean theorem.<br />
So <math>MX=2.4\sqrt{11}</math> and <math>DX=.7\sqrt{11}</math> so the coordinates of D are <math>(12-.7\sqrt{11},\, 2.5-2.4\sqrt{11})</math>.<br />
Since we know the coordinates of each of the vertices of <math>\Delta CMD</math>, we can apply the [[Shoelace Theorem]] to find the area of <math>\Delta CMD, \frac{527 \sqrt{11}}{40}</math>.<br />
<br />
<br />
===Solution 4===<br />
<br />
Let <math>E</math> be the intersection of lines <math>BC</math> and <math>DM</math>. Since triangles <math>\Delta CME</math> and <math>\Delta CMD</math> share a side and height, the area of <math>\Delta CDM</math> is equal to <math>\frac{DM}{EM}</math> times the area of <math>\Delta CME</math>.<br />
By AA similarity, <math>\Delta EMB</math> is similar to <math>\Delta ACB</math>, <math>\frac{EM}{AC}=\frac{MB}{CB}</math>. Solving yields <math>EM=\frac{175}{48}</math>. Using the same method but for <math>EB</math> yields <math>EB=\frac{625}{48}</math>. As in previous solutions, by the [[Pythagorean Theorem]], <math>DM=\frac{5\sqrt{11}}{2}</math>. So, <math>\frac{DM}{EM}=\frac{24\sqrt{11}}{35}</math>.<br />
Now, since we know both <math>CB</math> and <math>EB</math>, we can find that <math>CE=\frac{527}{48}</math>. The height of <math>\Delta CME</math> is the length from point <math>M</math> to <math>CB</math>. Since <math>M</math> is the midpoint of <math>AB</math>, the height is just <math>\frac{1}{2}\cdot7=\frac{7}{2}</math>. Using this, we can find that the area of <math>\Delta CMD</math> is <math>\frac{1}{2}\cdot\frac{7}{2}\cdot\frac{527}{48}\cdot\frac{24\sqrt{11}}{35}=\frac{527\sqrt{11}}{40}</math>, giving our answer of <math>\boxed{578}</math>.<br />
<br />
Solution by someonenumber011.<br />
<br />
== See also ==<br />
Video Solution from Khan Academy: https://www.youtube.com/watch?v=smtrrefmC40<br />
{{AIME box|year=2003|n=II|num-b=10|num-a=12}}<br />
<br />
[[Category: Intermediate Geometry Problems]]<br />
{{MAA Notice}}</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_12&diff=1851192020 AIME I Problems/Problem 122022-12-29T02:41:00Z<p>Kundusne000: /* = Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be the least positive integer for which <math>149^n-2^n</math> is divisible by <math>3^3\cdot5^5\cdot7^7.</math> Find the number of positive integer divisors of <math>n.</math><br />
<br />
= Solution 1= <br />
As usual, denote <math>v_p(n)</math> the highest power of prime <math>p</math> that divides <math>n</math>.<br />
Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>. <br />
<br />
<br />
Now, setting <math>n = 4c</math> (necessitated by <math>149^n \equiv 2^n \pmod 5</math> in order to set up LTE), we see <cmath>v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})</cmath> and since <math>149^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4c}-2^{4c})=1+v_5(c)</math> meaning that we have that by LTE, <math>5^4 | c</math> and <math>4 \cdot 5^4</math> divides <math>n</math>. <br />
<br />
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.<br />
<br />
~kevinmathz<br />
<br />
clarified by another user <br />
<br />
notation note from another user<br />
<br />
== Solution 2 (Simpler, just basic mods and Fermat's theorem)==<br />
Note that for all <math>n</math>, <math>149^n - 2^n</math> is divisible by <math>149-2 = 147</math> by difference of <math>n</math>th powers. That is <math>3\cdot7^2</math>, so now we can clearly see that the smallest <math>n</math> to make the expression divisible by <math>3^3</math> is just <math>3^2</math>. Similarly, we can reason that the smallest <math>n</math> to make the expression divisible by <math>7^7</math> is just <math>7^5</math>. <br />
<br />
Finally, for <math>5^5</math>, take <math>\pmod {5}</math> and <math>\pmod {25}</math> of each quantity (They happen to both be <math>-1</math> and <math>2</math> respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum <math>n</math> for divisibility by <math>5</math> is <math>4</math>, and other values are factors of <math>4</math>. Testing all of them(just <math>1</math>,<math>2</math>,<math>4</math> using mods-not too bad), <math>4</math> is indeed the smallest value to make the expression divisible by <math>5</math>, and this clearly is NOT divisible by <math>25</math>.<br />
Therefore, the smallest <math>n</math> to make this expression divisible by <math>5^5</math> is <math>2^2 \cdot 5^4</math>.<br />
<br />
Calculating the LCM of all these, one gets <math>2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Using the factor counting formula,<br />
the answer is <math>3\cdot3\cdot5\cdot6</math> = <math>\boxed{270}</math>.<br />
<br />
~Solution by thanosaops<br />
<br />
~formatted by MY-2<br />
<br />
~also formatted by pandyhu2001<br />
<br />
== Solution 3 (Elementary and Thorough) ==<br />
As usual, denote <math>v_p(n)</math> the highest power of prime <math>p</math> that divides <math>n</math>. For divisibility by <math>3^3</math>, notice that <math>v_3(149^3 - 2^3) = 2</math> as <math>149^3 - 2^3 =</math> <math>(147)(149^2 + 2\cdot149 + 2^2)</math>, and upon checking mods, <math>149^2 + 2\cdot149 + 2^2</math> is divisible by <math>3</math> but not <math>9</math>. In addition, <math>149^9 - 2^9</math> is divisible by <math>3^3</math> because <math>149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)</math>, and the rightmost factor equates to <math>1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}</math>. In fact, <math>n = 9 = 3^2</math> is the least possible choice to ensure divisibility by <math>3^3</math> because if <math>n = a \cdot 3^b</math>, with <math>3 \nmid a</math> and <math>b < 2</math>, we write <cmath>149^{a \cdot 3^b} - 2^{a \cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).</cmath> Then, the rightmost factor is equivalent to <math>\pm a \pmod{3} \not\equiv 0 \pmod{3}</math>, and <math>v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3</math>.<br />
<br />
For divisibility by <math>7^7</math>, we'll induct, claiming that <math>v_7(149^{7^k} - 2^{7^k}) = k + 2</math> for whole numbers <math>k</math>. The base case is clear. Then, <cmath>v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).</cmath> By the induction hypothesis, <math>v_7(149^{7^k} - 2^{7^k}) = k + 2</math>. Then, notice that <cmath>S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.</cmath> This tells us that <math>S(k)</math> is divisible by <math>7</math>, but not <math>49</math> so that <math>v_7\left(S(k)\right) = 1</math>, completing our induction. We can verify that <math>7^5</math> is the least choice of <math>n</math> to ensure divisibility by <math>7^7</math> by arguing similarly to the <math>3^3</math> case. <br />
<br />
Finally, for <math>5^5</math>, we take the powers of <math>149</math> and <math>2</math> in mod <math>5</math> and mod <math>25</math>. Writing out these mods, we have that <math>149^n \equiv 2^n \pmod{5}</math> if and only if <math>4 | n</math>, in which <math>149^n \equiv 2^n \equiv 1 \pmod{5}</math>. So here we claim that <math>v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1</math> and perform yet another induction. The base case is true: <math>5 | 149^4 - 2^4</math>, but <math>149^4 - 2^4 \equiv 1 - 16 \pmod{25}</math>. Now then, assuming the induction statement to hold for some <math>k</math>, <cmath>v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).</cmath> Note that <math>S'(k) = 149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}</math> equates to <math>S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}</math> in both mod <math>5</math> and mod <math>25</math>. We notice that <math>S''(k) \equiv 0 \pmod{5}</math>. Writing out the powers of <math>2</math> mod <math>25</math>, we have <math>S''(0) \equiv 5 \pmod{25}</math>. Also <math>2^n \equiv 1 \pmod{25}</math> when <math>n</math> is a multiple of <math>20</math>. Hence for <math>k > 0</math>, <math>S''(k) \equiv 5 \mod{25}</math>. Thus, <math>v_5\left(S'(k)\right) = 1</math>, completing our induction. Applying the same argument from the previous two cases, <math>4\cdot5^4</math> is the least choice to ensure divisibility by <math>5^5</math>.<br />
<br />
Our answer is the number of divisors of <math>\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. It is <math>(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}</math>.<br />
<br />
~hnkevin42<br />
<br />
==Solution 4 (Official MAA)==<br />
Analyze each prime power separately.<br />
Start with the case of <math>3^3</math>. By the Binomial Theorem,<br />
<cmath>\begin{align*}<br />
149^n - 2^n &= (147+2)^n - 2^n \\<br />
&= \binom n1 \cdot 147 \cdot 2^{n-1}<br />
+ \binom n2 \cdot 147^2 \cdot 2^{n-2}\\<br />
&\qquad+ \binom n3 \cdot 147^3 \cdot 2^{n-3}<br />
+ \cdots.<br />
\end{align*}</cmath>Because <math>147</math> is divisible by <math>3</math>, all terms after the first two are divisible by <math>3^3</math>, and the exponent of <math>3</math> in the first term is less than that in the second term. Hence it is necessary and sufficient that <math>3^3 \mid 147n</math>, that is, <math>3^2 \mid n</math>.<br />
For the <math>7^7</math> case, consider the same expansion as in the previous case. Because <math>147</math> is divisible by <math>49 = 7^2</math>, all terms after the first three are divisible by <math>7^7</math>, and the exponent of <math>7</math> in the first term is less than that in the second and third term. Hence it is necessary and sufficient that <math>7^7 \mid 147n</math>, that is, <math>7^5 \mid n</math>.<br />
For the <math>5^5</math> case, working modulo <math>5</math> gives <math>149^n - 2^n \equiv 4^n - 2^n = 2^n(2^n-1) \pmod 5</math>, so it must be that <math>4 \mid n</math>. Let <math>n = 4m</math>, and let <math>c = 149^4 - 2^4 = (149^2-2^2)(149^2+2^2) = 147 \cdot 151 \cdot 22205</math>. Note that <math>\frac c5</math> is an integer not divisible by <math>5</math>. Expand by the Binomial Theorem again to get<br />
<cmath>\begin{align*}<br />
(149^4)^m - (2^4)^m &= (c+16)^m - (16)^m \\<br />
&= \binom m1 \cdot c \cdot 16^{m-1}<br />
+ \binom m2 \cdot c^2 \cdot 16^{m-2} \\<br />
&\qquad+ \binom m3 \cdot c^3 \cdot 16^{m-3}<br />
+ \binom m4 \cdot c^4 \cdot 16^{m-4}<br />
+ \cdots.<br />
\end{align*}</cmath>All terms after the first four are divisible by <math>5^5</math>, and the exponent of <math>5</math> in the first term is less than that in the second, third, or fourth term. Hence it is necessary and sufficient that <math>5^5 \mid cm</math>. Thus <math>5^4 \mid m</math>, and it follows that <math>4 \cdot 5^4 \mid n</math>.<br />
Therefore the least <math>n</math> is <math>3^2 \cdot (2^2 \cdot 5^4) \cdot 7^5</math>. The requested number of divisors is <math>(1+2)(1+2)(1+4)(1+5) = 270</math>.<br />
<br />
The results of the above cases can be generalized using the following lemma.<br />
<br />
Lifting the Exponent Lemma: Let <math>p</math> be an odd prime, and let <math>a</math> and <math>b</math> be integers relatively prime to <math>p</math> such that <math>p \mid (a-b)</math>. Let <math>n</math> be a positive integer. Then the number of factors of <math>p</math> that divide <math>a^n - b^n</math> is equal to the number of factors of <math>p</math> that divide <math>a-b</math> plus the number of factors of <math>p</math> that divide <math>n</math>.<br />
<br />
== Video Solution ==<br />
https://www.youtube.com/watch?v=WyXBnSv-R34<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Intermediate Number Theory Problems]]</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2020_AIME_I_Problems/Problem_12&diff=1851182020 AIME I Problems/Problem 122022-12-29T02:40:25Z<p>Kundusne000: /* Solution 1 */</p>
<hr />
<div>== Problem ==<br />
Let <math>n</math> be the least positive integer for which <math>149^n-2^n</math> is divisible by <math>3^3\cdot5^5\cdot7^7.</math> Find the number of positive integer divisors of <math>n.</math><br />
<br />
== Solution 1= <br />
As usual, denote <math>v_p(n)</math> the highest power of prime <math>p</math> that divides <math>n</math>.<br />
Lifting the Exponent shows that <cmath>v_3(149^n-2^n) = v_3(n) + v_3(147) = v_3(n)+1</cmath> so thus, <math>3^2</math> divides <math>n</math>. It also shows that <cmath>v_7(149^n-2^n) = v_7(n) + v_7(147) = v_7(n)+2</cmath> so thus, <math>7^5</math> divides <math>n</math>. <br />
<br />
<br />
Now, setting <math>n = 4c</math> (necessitated by <math>149^n \equiv 2^n \pmod 5</math> in order to set up LTE), we see <cmath>v_5(149^{4c}-2^{4c}) = v_5(149^{4c}-16^{c})</cmath> and since <math>149^{4} \equiv 1 \pmod{25}</math> and <math>16^1 \equiv 16 \pmod{25}</math> then <math>v_5(149^{4c}-2^{4c})=1+v_5(c)</math> meaning that we have that by LTE, <math>5^4 | c</math> and <math>4 \cdot 5^4</math> divides <math>n</math>. <br />
<br />
Since <math>3^2</math>, <math>7^5</math> and <math>4\cdot 5^4</math> all divide <math>n</math>, the smallest value of <math>n</math> working is their LCM, also <math>3^2 \cdot 7^5 \cdot 4 \cdot 5^4 = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Thus the number of divisors is <math>(2+1)(2+1)(4+1)(5+1) = \boxed{270}</math>.<br />
<br />
~kevinmathz<br />
<br />
clarified by another user <br />
<br />
notation note from another user<br />
<br />
== Solution 2 (Simpler, just basic mods and Fermat's theorem)==<br />
Note that for all <math>n</math>, <math>149^n - 2^n</math> is divisible by <math>149-2 = 147</math> by difference of <math>n</math>th powers. That is <math>3\cdot7^2</math>, so now we can clearly see that the smallest <math>n</math> to make the expression divisible by <math>3^3</math> is just <math>3^2</math>. Similarly, we can reason that the smallest <math>n</math> to make the expression divisible by <math>7^7</math> is just <math>7^5</math>. <br />
<br />
Finally, for <math>5^5</math>, take <math>\pmod {5}</math> and <math>\pmod {25}</math> of each quantity (They happen to both be <math>-1</math> and <math>2</math> respectively, so you only need to compute once). One knows from Fermat's theorem that the maximum possible minimum <math>n</math> for divisibility by <math>5</math> is <math>4</math>, and other values are factors of <math>4</math>. Testing all of them(just <math>1</math>,<math>2</math>,<math>4</math> using mods-not too bad), <math>4</math> is indeed the smallest value to make the expression divisible by <math>5</math>, and this clearly is NOT divisible by <math>25</math>.<br />
Therefore, the smallest <math>n</math> to make this expression divisible by <math>5^5</math> is <math>2^2 \cdot 5^4</math>.<br />
<br />
Calculating the LCM of all these, one gets <math>2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. Using the factor counting formula,<br />
the answer is <math>3\cdot3\cdot5\cdot6</math> = <math>\boxed{270}</math>.<br />
<br />
~Solution by thanosaops<br />
<br />
~formatted by MY-2<br />
<br />
~also formatted by pandyhu2001<br />
<br />
== Solution 3 (Elementary and Thorough) ==<br />
As usual, denote <math>v_p(n)</math> the highest power of prime <math>p</math> that divides <math>n</math>. For divisibility by <math>3^3</math>, notice that <math>v_3(149^3 - 2^3) = 2</math> as <math>149^3 - 2^3 =</math> <math>(147)(149^2 + 2\cdot149 + 2^2)</math>, and upon checking mods, <math>149^2 + 2\cdot149 + 2^2</math> is divisible by <math>3</math> but not <math>9</math>. In addition, <math>149^9 - 2^9</math> is divisible by <math>3^3</math> because <math>149^9 - 2^9 = (149^3 - 2^3)(149^6 + 149^3\cdot2^3 + 2^6)</math>, and the rightmost factor equates to <math>1 + 1 + 1 \pmod{3} \equiv 0 \pmod{3}</math>. In fact, <math>n = 9 = 3^2</math> is the least possible choice to ensure divisibility by <math>3^3</math> because if <math>n = a \cdot 3^b</math>, with <math>3 \nmid a</math> and <math>b < 2</math>, we write <cmath>149^{a \cdot 3^b} - 2^{a \cdot 3^b} = (149^{3^b} - 2^{3^b})(149^{3^b(a - 1)} + 149^{3^b(a - 2)}\cdot2^{3^b}+\cdots2^{3^b(a - 1)}).</cmath> Then, the rightmost factor is equivalent to <math>\pm a \pmod{3} \not\equiv 0 \pmod{3}</math>, and <math>v_3(149^{3^b} - 2^{3^b}) = b + 1 < 3</math>.<br />
<br />
For divisibility by <math>7^7</math>, we'll induct, claiming that <math>v_7(149^{7^k} - 2^{7^k}) = k + 2</math> for whole numbers <math>k</math>. The base case is clear. Then, <cmath>v_7(149^{7^{k+1}} - 2^{7^{k+1}}) = v_7(149^{7^k} - 2^{7^k}) + v_7(149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k}).</cmath> By the induction hypothesis, <math>v_7(149^{7^k} - 2^{7^k}) = k + 2</math>. Then, notice that <cmath>S(k) = 149^{6\cdot7^k} + 2^{7^k}\cdot149^{5\cdot7^k} + \cdots + 2^{5\cdot7^k}\cdot149^{7^k} + 2^{6\cdot7^k} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{7} \equiv 7 \cdot 2^{6\cdot7^k}\pmod{49}.</cmath> This tells us that <math>S(k)</math> is divisible by <math>7</math>, but not <math>49</math> so that <math>v_7\left(S(k)\right) = 1</math>, completing our induction. We can verify that <math>7^5</math> is the least choice of <math>n</math> to ensure divisibility by <math>7^7</math> by arguing similarly to the <math>3^3</math> case. <br />
<br />
Finally, for <math>5^5</math>, we take the powers of <math>149</math> and <math>2</math> in mod <math>5</math> and mod <math>25</math>. Writing out these mods, we have that <math>149^n \equiv 2^n \pmod{5}</math> if and only if <math>4 | n</math>, in which <math>149^n \equiv 2^n \equiv 1 \pmod{5}</math>. So here we claim that <math>v_5(149^{4\cdot5^k} - 2^{4\cdot5^k}) = k + 1</math> and perform yet another induction. The base case is true: <math>5 | 149^4 - 2^4</math>, but <math>149^4 - 2^4 \equiv 1 - 16 \pmod{25}</math>. Now then, assuming the induction statement to hold for some <math>k</math>, <cmath>v_5(149^{4\cdot5^{k+1}} - 2^{4\cdot5^{k+1}}) = (k+1) + v_5(149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}).</cmath> Note that <math>S'(k) = 149^{4\cdot4\cdot5^k}+2^{4\cdot5^k}\cdot149^{3\cdot4\cdot5^k}+\cdots+2^{3\cdot4\cdot5^k}\cdot149^{4\cdot5^k}+2^{4\cdot4\cdot5^k}</math> equates to <math>S''(k) = 1 + 2^{4\cdot5^k} + \cdots + 2^{16\cdot5^k}</math> in both mod <math>5</math> and mod <math>25</math>. We notice that <math>S''(k) \equiv 0 \pmod{5}</math>. Writing out the powers of <math>2</math> mod <math>25</math>, we have <math>S''(0) \equiv 5 \pmod{25}</math>. Also <math>2^n \equiv 1 \pmod{25}</math> when <math>n</math> is a multiple of <math>20</math>. Hence for <math>k > 0</math>, <math>S''(k) \equiv 5 \mod{25}</math>. Thus, <math>v_5\left(S'(k)\right) = 1</math>, completing our induction. Applying the same argument from the previous two cases, <math>4\cdot5^4</math> is the least choice to ensure divisibility by <math>5^5</math>.<br />
<br />
Our answer is the number of divisors of <math>\text{lcm}(3^2, 7^5, 2^2\cdot5^4) = 2^2 \cdot 3^2 \cdot 5^4 \cdot 7^5</math>. It is <math>(2 + 1)(2 + 1)(4 + 1)(5 + 1) = \boxed{270}</math>.<br />
<br />
~hnkevin42<br />
<br />
==Solution 4 (Official MAA)==<br />
Analyze each prime power separately.<br />
Start with the case of <math>3^3</math>. By the Binomial Theorem,<br />
<cmath>\begin{align*}<br />
149^n - 2^n &= (147+2)^n - 2^n \\<br />
&= \binom n1 \cdot 147 \cdot 2^{n-1}<br />
+ \binom n2 \cdot 147^2 \cdot 2^{n-2}\\<br />
&\qquad+ \binom n3 \cdot 147^3 \cdot 2^{n-3}<br />
+ \cdots.<br />
\end{align*}</cmath>Because <math>147</math> is divisible by <math>3</math>, all terms after the first two are divisible by <math>3^3</math>, and the exponent of <math>3</math> in the first term is less than that in the second term. Hence it is necessary and sufficient that <math>3^3 \mid 147n</math>, that is, <math>3^2 \mid n</math>.<br />
For the <math>7^7</math> case, consider the same expansion as in the previous case. Because <math>147</math> is divisible by <math>49 = 7^2</math>, all terms after the first three are divisible by <math>7^7</math>, and the exponent of <math>7</math> in the first term is less than that in the second and third term. Hence it is necessary and sufficient that <math>7^7 \mid 147n</math>, that is, <math>7^5 \mid n</math>.<br />
For the <math>5^5</math> case, working modulo <math>5</math> gives <math>149^n - 2^n \equiv 4^n - 2^n = 2^n(2^n-1) \pmod 5</math>, so it must be that <math>4 \mid n</math>. Let <math>n = 4m</math>, and let <math>c = 149^4 - 2^4 = (149^2-2^2)(149^2+2^2) = 147 \cdot 151 \cdot 22205</math>. Note that <math>\frac c5</math> is an integer not divisible by <math>5</math>. Expand by the Binomial Theorem again to get<br />
<cmath>\begin{align*}<br />
(149^4)^m - (2^4)^m &= (c+16)^m - (16)^m \\<br />
&= \binom m1 \cdot c \cdot 16^{m-1}<br />
+ \binom m2 \cdot c^2 \cdot 16^{m-2} \\<br />
&\qquad+ \binom m3 \cdot c^3 \cdot 16^{m-3}<br />
+ \binom m4 \cdot c^4 \cdot 16^{m-4}<br />
+ \cdots.<br />
\end{align*}</cmath>All terms after the first four are divisible by <math>5^5</math>, and the exponent of <math>5</math> in the first term is less than that in the second, third, or fourth term. Hence it is necessary and sufficient that <math>5^5 \mid cm</math>. Thus <math>5^4 \mid m</math>, and it follows that <math>4 \cdot 5^4 \mid n</math>.<br />
Therefore the least <math>n</math> is <math>3^2 \cdot (2^2 \cdot 5^4) \cdot 7^5</math>. The requested number of divisors is <math>(1+2)(1+2)(1+4)(1+5) = 270</math>.<br />
<br />
The results of the above cases can be generalized using the following lemma.<br />
<br />
Lifting the Exponent Lemma: Let <math>p</math> be an odd prime, and let <math>a</math> and <math>b</math> be integers relatively prime to <math>p</math> such that <math>p \mid (a-b)</math>. Let <math>n</math> be a positive integer. Then the number of factors of <math>p</math> that divide <math>a^n - b^n</math> is equal to the number of factors of <math>p</math> that divide <math>a-b</math> plus the number of factors of <math>p</math> that divide <math>n</math>.<br />
<br />
== Video Solution ==<br />
https://www.youtube.com/watch?v=WyXBnSv-R34<br />
<br />
==See Also==<br />
<br />
{{AIME box|year=2020|n=I|num-b=11|num-a=13}}<br />
{{MAA Notice}}<br />
<br />
[[Category: Intermediate Number Theory Problems]]</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_23&diff=1593832018 AMC 10B Problems/Problem 232021-08-01T22:28:20Z<p>Kundusne000: </p>
<hr />
<div>==Problem==<br />
<br />
How many ordered pairs <math>(a, b)</math> of positive integers satisfy the equation <br />
<cmath>a\cdot b + 63 = 20\cdot \text{lcm}(a, b) + 12\cdot\text{gcd}(a,b),</cmath><br />
where <math>\text{gcd}(a,b)</math> denotes the greatest common divisor of <math>a</math> and <math>b</math>, and <math>\text{lcm}(a,b)</math> denotes their least common multiple?<br />
<br />
<math>\textbf{(A)} \text{ 0} \qquad \textbf{(B)} \text{ 2} \qquad \textbf{(C)} \text{ 4} \qquad \textbf{(D)} \text{ 6} \qquad \textbf{(E)} \text{ 8}</math><br />
<br />
<br />
==Solution==<br />
Let <math>x = \text{lcm}(a, b)</math>, and <math>y = \text{gcd}(a, b)</math>. Therefore, <math>a\cdot b = lcm(a, b)\cdot gcd(a, b) = x\cdot y</math>. Thus, the equation becomes<br />
<br />
<cmath>x\cdot y + 63 = 20x + 12y</cmath><br />
<cmath>x\cdot y - 20x - 12y + 63 = 0</cmath><br />
<br />
Using [[Simon's Favorite Factoring Trick]], we rewrite this equation as<br />
<br />
<cmath>(x - 12)(y - 20) - 240 + 63 = 0</cmath><br />
<cmath>(x - 12)(y - 20) = 177</cmath><br />
<br />
Since <math>177 = 3\cdot 59</math> and <math>x > y</math>, we have <math>x - 12 = 59</math> and <math>y - 20 = 3</math>, or <math>x - 12 = 177</math> and <math>y - 20 = 1</math>. This gives us the solutions <math>(71, 23)</math> and <math>(189, 21)</math>. Since the <math>\text{GCD}</math> must be a divisor of the <math>\text{LCM}</math>, the first pair does not work. Assume <math>a>b</math>. We must have <math>a = 21 \cdot 9</math> and <math>b = 21</math>, and we could then have <math>a<b</math>, so there are <math>\boxed{2}</math> solutions.<br />
(awesomeag)<br />
<br />
Edited by IronicNinja, Firebolt360, and mprincess0229~<br />
<br />
== Video Solution ==<br />
https://youtu.be/zfChnbMGLVQ?t=494<br />
<br />
~ pi_is_3.14<br />
<br />
==Video Solution==<br />
<br />
https://www.youtube.com/watch?v=JWGHYUeOx-k<br />
<br />
==See Also==<br />
{{AMC10 box|year=2018|ab=B|num-b=22|num-a=24}}<br />
{{MAA Notice}}</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=User:Palaashgang&diff=157144User:Palaashgang2021-07-01T15:13:27Z<p>Kundusne000: /* User Count */</p>
<hr />
<div><br><br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#dddddd"><br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="107px"><math>e^{\pi\cdot i}+28+1</math></font></center><br />
</div><br />
<div style="border:2px solid black; background:#cccccc;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">About Me</div></font>==<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black"><br />
PalaashGang is 11 years old.<br><br />
<br />
PalaashGang is attempting the AMC 8 2020. Edit : He got a 20<br> <br />
<br />
PalaashGang is attempting the CEMC Gauss 8 2021. Edit : He got a 142<br> <br />
<br />
PalaashGang is good at Skating<br />
<br />
PalaashGang is a National Indian Skater who took part in the SGFI National Championships 2019<br />
<br />
PalaashGang loves to read and play FĂĽr Elise on the piano.<br />
<br />
PalaashGang knows HTML, CSS, JS, Java and Scratch<br />
<br />
PalaashGang has completed 10000+ Problems on Alcumus<br />
<br />
PalaashGang is the CEO and founder if XiAMST and Euclidean Math Circle.<br />
<br />
PalaashGang is an active learner and volunteer on schoolhouse.world<br />
<br />
PalaashGang gets AMC 10s between 130-140 and AMC 12s between 120-135<br />
<br />
παλάς is PalaashGang's AoPS Wiki Signature <br />
</font></div><br />
</div><br />
<div style="border:2px solid black; background:#bbbbbb;-webkit-border-radius: 10px; align:center"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">Goals</div></font>==<br />
<div style="margin-left: 10px; margin-right: 10px; margin-bottom:10px">A User Count of 300<br />
<br />
AIME Qual<br />
<br />
Get 5 or more on AIME Mocks<br />
<br />
Represent India in Skating<br />
<br />
Play "Flight of the Bumblebee" on the Piano<br />
<br />
Be on the Alcumus Hall of Fame<br />
</div><br />
</div></div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=User:Bissue&diff=156550User:Bissue2021-06-21T17:13:16Z<p>Kundusne000: /* User Count */</p>
<hr />
<div>Here is where I'll fully "publish" mock tests I've written. So far there's only one, but there will definitely be more to come.<br />
<br />
I'll also add a user count here. It won't change whether the action occurs, but instead how I feel about the action.<br />
<br />
(am clout chaser)<br />
<br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Increase the user count below if this is your first time discovering this page.</font></div><br />
<br />
<center><font size="100px"><br />
<br />
<font color="#FF0000">3<br>RedFireTruck was here</font><br />
<br />
</font><br />
centslordm was here; pigeon orz</center><br />
</div><br />
<br />
<br />
__NOTOC__<div style="border:2px solid black; -webkit-border-radius: 10px; background:#F0F2F3"><br />
<br />
==<font color="black" style="font-family: ITC Avant Garde Gothic Std, Verdana"><div style="margin-left:10px">User Count</div></font>==<br />
<br />
<div style="margin-left: 10px; margin-bottom:10px"><font color="black">Increase the user count below if this is your first time discovering this page <b>and</b> you have used these problems as practice.</font></div><br />
<br />
<center><font size="100px"><br />
<br />
0<br />
<br />
</font></center><br />
</div><br />
---------------------<br />
<br />
==Apocalyptic AMC 8 (2020)==<br />
This contest took place between September 8th and November 6th. All problems were written by me <b>(bissue)</b> and testsolved by <b>nikenissan, bobthegod78, knightime1010, truffle, cw357,</b> and <b>ApraTrip.</b><br />
<br />
The mock was separated into two sections: The AMC 8 and The Tiebreakers.<br />
<br />
Standard AMC 8 rules were used. All participants had 40 minutes to complete as many of the 25 problems as they could. Correct answers were worth 1 point each, while incorrect or blank answers were worth 0 points each.<br />
<br />
The Tiebreakers were used to break ties between participants with the same score on the AMC 8. The rules were the same as those used for the ARML tiebreaker. For more information, see the original post in the AoPS Mock Contests Forum here:<br />
<br />
https://artofproblemsolving.com/community/c594864h2255517<br />
<br />
Full statistics and discussion threads can be found using the link above as well.<br />
<br />
==Problem 1==<br />
<br />
To walk up a single floor in her eighteen floor apartment building, Sarah needs to take nine steps up a flight of stairs. If Sarah starts on Floor <math>3</math> and walks up <math>100</math> steps, she would end up on the flight of stairs connecting which two floors?<br />
<br />
<math>\textbf{(A)} ~ \mbox{11 and 12} \qquad \textbf{(B)} ~ \mbox{12 and 13} \qquad \textbf{(C)} ~ \mbox{13 and 14} \qquad \textbf{(D)} ~ \mbox{14 and 15} \qquad \textbf{(E)} ~ \mbox{15 and 16}</math><br />
<br />
==Problem 2==<br />
<br />
Abby, Barb, and Carlos each have <math>35</math>, <math>42</math>, and <math>31</math> trading cards respectively. If they share their trading cards equally between each other, how many more trading cards would Carlos have than before?<br />
<br />
<math>\textbf{(A)} ~ 4 \qquad \textbf{(B)} ~ 5 \qquad \textbf{(C)} ~ 6 \qquad \textbf{(D)} ~ 9 \qquad \textbf{(E)} ~ 11</math><br />
<br />
==Problem 3==<br />
<br />
In triangle <math>ABC</math> the measure of angle <math>\angle A</math> is the average of the measures of angles <math>\angle B</math> and <math>\angle C</math>. What is the measure of angle <math>\angle A</math>?<br />
<br />
<math>\textbf{(A)} ~ 45^{\circ} \qquad \textbf{(B)} ~ 60^{\circ} \qquad \textbf{(C)} ~ 75^{\circ} \qquad \textbf{(D)} ~ 90^{\circ} \qquad \textbf{(E)} ~ 120^{\circ}</math><br />
<br />
==Problem 4==<br />
<br />
A spruce tree grows by <math>25</math> feet, increasing its height by <math>25 \%</math>. If the tree grows for a second time by <math>25</math> feet, by what percent would its height increase?<br />
<br />
<math>\textbf{(A)} ~ 5 \% \qquad \textbf{(B)} ~ 15 \% \qquad \textbf{(C)} ~ 20 \% \qquad \textbf{(D)} ~ 25 \% \qquad \textbf{(E)} ~ 30 \% </math><br />
<br />
==Problem 5==<br />
<br />
Find the sum of the digits of <math>\dfrac{5 \times 10^{2020}}{2}</math>.<br />
<br />
<math>\textbf{(A)} ~ 1 \qquad \textbf{(B)} ~ 2 \qquad \textbf{(C)} ~ 5 \qquad \textbf{(D)} ~ 7 \qquad \textbf{(E)} ~ 8</math><br />
<br />
==Problem 6==<br />
<br />
Square <math>B</math> with side length three is attached to a side of square <math>A</math> with side length four, as shown in the figure below. Find the area of the shaded region.<br />
<asy><br />
size(150);<br />
draw((0, 0)--(4, 0)--(4, 4)--(0, 4)--cycle);<br />
draw((4, 1)--(7, 1)--(7, 4)--(4, 4)--cycle);<br />
filldraw((0, 0)--(4, 0)--(4, 2.285714)--cycle, grey);<br />
filldraw((4, 1)--(7, 1)--(7, 4)--(4, 2.285714)--cycle, grey);<br />
label("A", (2, 2));<br />
label("B", (5.5, 2.5));<br />
</asy><br />
<math>\textbf{(A)} ~ 10 \qquad \textbf{(B)} ~ 10 \frac{1}{2} \qquad \textbf{(C)} ~ 11 \qquad \textbf{(D)} ~ 14 \qquad \textbf{(E)} ~ 14 \frac{1}{2}</math><br />
<br />
==Problem 7==<br />
<br />
When expressed as a decimal rounded to the nearest ten-thousandth, what is the value of <math>\dfrac{125+3}{125 \times 3}</math>?<br />
<br />
<math>\textbf{(A)} ~ 0.3412 \qquad \textbf{(B)} ~ 0.3413 \qquad \textbf{(C)} ~ 0.3414 \qquad \textbf{(D)} ~ 0.3415 \qquad \textbf{(E)} ~ 0.3416</math><br />
<br />
==Problem 8==<br />
<br />
What is the value of<br />
<cmath>(1+2+3)-(2+3+4)+(3+4+5)-\cdots -(98+99+100)?</cmath><br />
<math>\textbf{(A)} ~ -150 \qquad \textbf{(B)} ~ -147 \qquad \textbf{(C)} ~ -144 \qquad \textbf{(D)} ~ 147 \qquad \textbf{(E)} ~ 150</math><br />
<br />
==Problem 9==<br />
<br />
Kayla writes down the first <math>N</math> positive integers. What is the sum of all possible values of <math>N</math> if Kayla writes five multiples of <math>13</math> and six multiples of <math>12</math>?<br />
<br />
<math>\textbf{(A)} ~ 447 \qquad \textbf{(B)} ~ 453 \qquad \textbf{(C)} ~ 518 \qquad \textbf{(D)} ~ 525 \qquad \textbf{(E)} ~ 548</math><br />
<br />
==Problem 10==<br />
<br />
In Murphy's seventh grade homeroom, <math>\frac{7}{12}</math> of the students like tennis, <math>\frac{2}{3}</math> of the students like badminton, and <math>\frac{1}{12}</math> of the students like neither. What is the minimum possible number of students who like both tennis and badminton?<br />
<br />
<math>\textbf{(A)} ~ 1 \qquad \textbf{(B)} ~ 2 \qquad \textbf{(C)} ~ 3 \qquad \textbf{(D)} ~ 4 \qquad \textbf{(E)} ~ 6</math><br />
<br />
==Problem 11==<br />
<br />
For how many values of <math>N</math> does there exist a regular <math>N</math> sided polygon whose vertices all lie on the vertices of a regular <math>24</math> sided polygon?<br />
<br />
<math>\textbf{(A)} ~ 6 \qquad \textbf{(B)} ~ 7 \qquad \textbf{(C)} ~ 8 \qquad \textbf{(D)} ~ 9 \qquad \textbf{(E)} ~ 10</math><br />
<br />
==Problem 12==<br />
<br />
Quadrilateral <math>WXYZ</math> has its vertices on the sides of rectangle <math>ABCD</math> with <math>AB=7</math> and <math>BC=5</math>, as shown below. What is the area of quadrilateral <math>WXYZ</math>?<br />
<asy><br />
size(150);<br />
draw((0, 0)--(7, 0)--(7, 5)--(0, 5)--cycle);<br />
label("A", (0, 0), SW);<br />
label("B", (7, 0), SE);<br />
label("C", (7, 5), NE);<br />
label("D", (0, 5), NW);<br />
filldraw((0, 1)--(4, 0)--(7, 3)--(4, 5)--cycle, grey);<br />
label("W", (0, 1), W);<br />
label("X", (4, 0), S);<br />
label("Y", (7, 3), E);<br />
label("Z", (4, 5), N);<br />
label("4", (2, -0.5));<br />
label("3", (5.5, -0.5));<br />
label("4", (2, 5.5));<br />
label("3", (5.5, 5.5));<br />
</asy><br />
<math>\textbf{(A)} ~ 15 \dfrac{1}{2} \qquad \textbf{(B)} ~ 16 \qquad \textbf{(C)} ~ 16 \dfrac{1}{2} \qquad \textbf{(D)} ~ 17 \qquad \textbf{(E)} ~ 17 \dfrac{1}{2}</math><br />
<br />
==Problem 13==<br />
<br />
To drive to the supermarket, Mable drives for <math>m</math> miles, then drives <math>12</math> miles per hour faster for the remaining <math>\frac{4}{3}m</math> miles. The amount of time Mable spent driving at each of the two speeds was equal. What was Mable's average speed during her drive to the supermarket, in miles per hour?<br />
<br />
<math>\textbf{(A)} ~ \dfrac{81}{2} \qquad \textbf{(B)} ~ \dfrac{288}{7} \qquad \textbf{(C)} ~ 42 \qquad \textbf{(D)} ~ \dfrac{300}{7} \qquad \textbf{(E)} ~ 50</math><br />
<br />
==Problem 14==<br />
<br />
Six circles of radius one are cut out of the rectangle below. What is the area of the shaded region?<br />
<asy><br />
size(150);<br />
filldraw((0, 0)--(6, 0)--(6, 4)--(0, 4)--cycle, grey);<br />
filldraw(circle((1, 1), 1), white);<br />
filldraw(circle((3, 1), 1), white);<br />
filldraw(circle((5, 1), 1), white);<br />
filldraw(circle((1, 3), 1), white);<br />
filldraw(circle((3, 3), 1), white);<br />
filldraw(circle((5, 3), 1), white);<br />
</asy><br />
<math>\textbf{(A)} ~ 20-6\pi \qquad \textbf{(B)} ~ 24-6\pi \qquad \textbf{(C)} ~ 28-6\pi \qquad \textbf{(D)} ~ 30-6\pi \qquad \textbf{(E)} ~ 32-6\pi</math><br />
<br />
==Problem 15==<br />
<br />
One metronome beeps at a steady rate of <math>72</math> beeps per minute, while another metronome beeps at a steady rate of <math>96</math> beeps per minute. If both metronomes beep at the same time once, how long will it take, in seconds, until they first beep at the same time again?<br />
<br />
<math>\textbf{(A)} ~ 2 \dfrac{1}{2} \qquad \textbf{(B)} ~ 5 \qquad \textbf{(C)} ~ 10 \qquad \textbf{(D)} ~ 18 \qquad \textbf{(E)} ~ 24</math><br />
<br />
==Problem 16==<br />
<br />
A square with side length two is placed on a table, forming a <math>30</math> degree angle with the table's surface. How much higher is the top vertex of the square than the table?<br />
<asy><br />
size(150);<br />
draw((0, 0)--(0.882, 0.4714)--(0.4106, 1.3534)--(-0.4714, 0.882)--cycle);<br />
draw((-0.5, 0)--(1, 0), linewidth(3));<br />
draw((-0.75, 1.3534)--(-0.65, 1.3534));<br />
draw((-0.7, 1.3534)--(-0.7, 0));<br />
draw((-0.75, 0)--(-0.65, 0));<br />
</asy><br />
<math>\textbf{(A)} ~ \dfrac{5}{2} \qquad \textbf{(B)} ~ \sqrt{3}+1 \qquad \textbf{(C)} ~ \dfrac{4\sqrt{3}}{3} \qquad \textbf{(D)} ~ 3 \qquad \textbf{(E)} ~ \dfrac{3\sqrt{3}}{2}+1</math><br />
<br />
==Problem 17==<br />
<br />
Kurtis' school schedule is made up of four classes, followed by lunch, followed by three more classes. In how many ways can Kurtis arrange his schedule if two of his classes (Reading and Writing) must occur one immediately after the other?<br />
<br />
<math>\textbf{(A)} ~ 600 \qquad \textbf{(B)} ~ 840 \qquad \textbf{(C)} ~ 1200 \qquad \textbf{(D)} ~ 1440 \qquad \textbf{(E)} ~ 1680</math><br />
<br />
==Problem 18==<br />
<br />
When the number <math>25</math> is added to a list of numbers with total sum <math>S</math>, the average of all the numbers increases by one. What is the sum of the digits of the greatest possible value of <math>S</math>?<br />
<br />
<math>\textbf{(A)} ~ 6 \qquad \textbf{(B)} ~ 7 \qquad \textbf{(C)} ~ 8 \qquad \textbf{(D)} ~ 9 \qquad \textbf{(E)} ~ 12</math><br />
<br />
==Problem 19==<br />
<br />
A magician randomly picks a three digit positive integer to put into her hat and pulls out the same number with its digits in reverse order. (For example <math>496</math> would become <math>694</math> and <math>720</math> would become <math>27</math>.) What is the probability the magician pulls out a multiple of <math>22</math>?<br />
<br />
<math>\textbf{(A)} ~ \dfrac{1}{15} \qquad \textbf{(B)} ~ \dfrac{1}{18} \qquad \textbf{(C)} ~ \dfrac{1}{20} \qquad \textbf{(D)} ~ \dfrac{1}{25} \qquad \textbf{(E)} ~ \dfrac{1}{30}</math><br />
<br />
==Problem 20==<br />
<br />
Tyrone has three books to read in six days. He reads one-half of a single book every day. In how many ways can he finish all the books if he may not read the same book two days in a row?<br />
<br />
<math>\textbf{(A)} ~ 12 \qquad \textbf{(B)} ~ 18 \qquad \textbf{(C)} ~ 24 \qquad \textbf{(D)} ~ 30 \qquad \textbf{(E)} ~ 36</math><br />
<br />
==Problem 21==<br />
<br />
There exists a circle that is tangent to <math>\overline{AB}</math> and <math>\overline{BC}</math> at <math>A</math> and <math>C</math>, respectively. If <math>AB=BC=13</math> and <math>AC=10</math>, what is the radius of the circle?<br />
<asy><br />
size(150);<br />
draw((-5, 0)--(5, 0)--(0, -12)--cycle);<br />
draw(circle((0, 2.08333), 5.41666));<br />
label("A", (-5, 0), W);<br />
label("C", (5, 0), E);<br />
label("B", (0, -12), S);<br />
label("13", (-2.7, -6), W);<br />
label("13", (2.7, -6), E);<br />
label("10", (0, 0.2), N);<br />
</asy><br />
<math>\textbf{(A)} ~ \dfrac{60}{13} \qquad \textbf{(B)} ~ 5 \qquad \textbf{(C)} ~ \dfrac{26}{5} \qquad \textbf{(D)} ~ \dfrac{65}{12} \qquad \textbf{(E)} ~ \dfrac{156}{25}</math><br />
<br />
==Problem 22==<br />
<br />
For each of the distinct sets of numbers containing only positive integers between <math>1</math> and <math>9</math> inclusive, Jordan writes the sum of the numbers in that set. What is the sum of the numbers Jordan writes?<br />
<br />
<math>\textbf{(A)} ~ 11520 \qquad \textbf{(B)} ~ 11565 \qquad \textbf{(C)} ~ 11610 \qquad \textbf{(D)} ~ 11655 \qquad \textbf{(E)} ~ 11700</math><br />
<br />
==Problem 23==<br />
<br />
In rectangle <math>ABCD</math>, the perpendicular from <math>B</math> to diagonal <math>\overline{AC}</math> bisects segment <math>\overline{CD}</math>. Which of the following is closest to <math>\frac{AB}{BC}</math>?<br />
<br />
<math>\textbf{(A)} ~ \dfrac{5}{4} \qquad \textbf{(B)} ~ \dfrac{4}{3} \qquad \textbf{(C)} ~ \dfrac{7}{5} \qquad \textbf{(D)} ~ \dfrac{3}{2} \qquad \textbf{(E)} ~ \dfrac{8}{5}</math><br />
<br />
==Problem 24==<br />
<br />
How many ordered triples of positive integers <math>(a, b, c)</math> satisfy <math>\text{gcd}(a, b, c)=20</math> and <math>\text{lcm}(a, b, c)=240</math>?<br />
<br />
<math>\textbf{(A)} ~ 18 \qquad \textbf{(B)} ~ 24 \qquad \textbf{(C)} ~ 36 \qquad \textbf{(D)} ~ 54 \qquad \textbf{(E)} ~ 72 </math><br />
<br />
==Problem 25==<br />
<br />
Cheyanne rolls two standard six sided dice, then repeatedly rerolls all dice which show an odd number and stops as soon as all dice show an even number. What is the probability Cheyanne stops after exactly four rounds of rerolling?<br />
<br />
<math>\textbf{(A)} ~ \dfrac{61}{1024} \qquad \textbf{(B)} ~ \dfrac{1}{16} \qquad \textbf{(C)} ~ \dfrac{67}{1024} \qquad \textbf{(D)} ~ \dfrac{9}{128} \qquad \textbf{(E)} ~ \dfrac{29}{256}</math><br />
<br />
------------------<br />
==Tiebreaker Problem 1==<br />
<br />
A whiteboard has positive real numbers <math>1</math> and <math>m</math> written on it. Every second, if the numbers <math>x</math> and <math>y</math> are on the whiteboard, a ghost will replace those numbers with <math>|x^2-y^2|</math> and <math>2xy</math>. The ghost stops once one number on the whiteboard is <math>m</math> times the other. For how many positive real numbers <math>m</math> does the ghost stop after exactly <math>16</math> seconds?<br />
<br />
==Tiebreaker Problem 2==<br />
<br />
The perpendicular bisectors of triangle <math>ABC</math> can be described in the coordinate plane as lines <math>y=0</math>, <math>y=x</math>, and <math>y=\sqrt{3}x</math>. Given that triangle <math>ABC</math> has circumradius <math>1</math>, find its area.<br />
<br />
==Tiebreaker Problem 3==<br />
<br />
The diagram below is constructed by attaching an equilateral triangle, a square, a regular pentagon, and a regular hexagon together. Compute the measure of the obtuse angle formed by the three red vertices.<br />
<asy><br />
import graph; size(10cm); <br />
real labelscalefactor = 0.5; /* changes label-to-point distance */<br />
pen dps = linewidth(0.7) + fontsize(10); defaultpen(dps); /* default pen style */ <br />
pen dotstyle = black; /* point style */ <br />
real xmin = -29.36, xmax = -9.8, ymin = 4.78, ymax = 17.66; /* image dimensions */<br />
<br />
draw((-22,12)--(-19,12)--(-20.5,14.598076211353318)--cycle, linewidth(2)); <br />
draw((-20.5,14.598076211353318)--(-19,12)--(-16.401923788646684,13.5)--(-17.90192378864668,16.098076211353316)--cycle, linewidth(2)); <br />
draw((-16.401923788646684,13.5)--(-19,12)--(-18.376264927546725,9.065557197798585)--(-15.392699241441907,8.751971807995622)--(-14.172489312214504,11.492608180923423)--cycle, linewidth(2)); <br />
draw((-18.376264927546725,9.065557197798585)--(-19,12)--(-21.85316954888546,12.927050983124847)--(-24.082604025317647,10.919659164048271)--(-23.45886895286437,7.985216361846854)--(-20.60569940397891,7.058165378722009)--cycle, linewidth(2)); <br />
Label laxis; laxis.p = fontsize(10); <br />
string blank(real x) {return "";} <br />
xaxis(xmin, xmax, Ticks(laxis, blank, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); <br />
yaxis(ymin, ymax, Ticks(laxis, blank, Step = 1, Size = 2, NoZero),EndArrow(6), above = true); /* draws axes; NoZero hides '0' label */ <br />
/* draw figures */<br />
draw((-22,12)--(-19,12), linewidth(2)); <br />
draw((-19,12)--(-20.5,14.598076211353318), linewidth(2)); <br />
draw((-20.5,14.598076211353318)--(-22,12), linewidth(2)); <br />
draw((-20.5,14.598076211353318)--(-19,12), linewidth(2)); <br />
draw((-19,12)--(-16.401923788646684,13.5), linewidth(2)); <br />
draw((-16.401923788646684,13.5)--(-17.90192378864668,16.098076211353316), linewidth(2)); <br />
draw((-17.90192378864668,16.098076211353316)--(-20.5,14.598076211353318), linewidth(2)); <br />
draw((-16.401923788646684,13.5)--(-19,12), linewidth(2)); <br />
draw((-19,12)--(-18.376264927546725,9.065557197798585), linewidth(2)); <br />
draw((-18.376264927546725,9.065557197798585)--(-15.392699241441907,8.751971807995622), linewidth(2)); <br />
draw((-15.392699241441907,8.751971807995622)--(-14.172489312214504,11.492608180923423), linewidth(2)); <br />
draw((-14.172489312214504,11.492608180923423)--(-16.401923788646684,13.5), linewidth(2)); <br />
draw((-18.376264927546725,9.065557197798585)--(-19,12), linewidth(2)); <br />
draw((-19,12)--(-21.85316954888546,12.927050983124847), linewidth(2)); <br />
draw((-21.85316954888546,12.927050983124847)--(-24.082604025317647,10.919659164048271), linewidth(2)); <br />
draw((-24.082604025317647,10.919659164048271)--(-23.45886895286437,7.985216361846854), linewidth(2)); <br />
draw((-23.45886895286437,7.985216361846854)--(-20.60569940397891,7.058165378722009), linewidth(2)); <br />
draw((-20.60569940397891,7.058165378722009)--(-18.376264927546725,9.065557197798585), linewidth(2)); <br />
/* dots and labels */<br />
dot((-22,12),red); <br />
dot((-19,12),red); <br />
dot((-21.526119073220972,12.820785841919117),linewidth(4pt) + dotstyle+red); <br />
clip((xmin,ymin)--(xmin,ymax)--(xmax,ymax)--(xmax,ymin)--cycle); <br />
</asy><br />
------------------------------<br />
==Answer Key==<br />
<br />
AMC 8: DBBCD / CBBAD / AECBA / BCDDD / DACEA<br />
<br />
Tiebreakers: (<math>65280</math>, <math>\dfrac{3-\sqrt{3}}{4}</math>, <math>102</math>)</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=2008_AMC_10A_Problems&diff=1565452008 AMC 10A Problems2021-06-21T16:25:13Z<p>Kundusne000: /* Problem 17 */</p>
<hr />
<div>{{AMC10 Problems|year=2008|ab=A}}<br />
==Problem 1==<br />
A bakery owner turns on his doughnut machine at <math>\text{8:30}\ {\small\text{AM}}</math>. At <math>\text{11:10}\ {\small\text{AM}}</math> the machine has completed one third of the day's job. At what time will the doughnut machine complete the job?<br />
<br />
<math>\mathrm{(A)}\ \text{1:50}\ {\small\text{PM}}\qquad\mathrm{(B)}\ \text{3:00}\ {\small\text{PM}}\qquad\mathrm{(C)}\ \text{3:30}\ {\small\text{PM}}\qquad\mathrm{(D)}\ \text{4:30}\ {\small\text{PM}}\qquad\mathrm{(E)}\ \text{5:50}\ {\small\text{PM}}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 1|Solution]]<br />
<br />
==Problem 2==<br />
A square is drawn inside a rectangle. The ratio of the width of the rectangle to a side of the square is <math>2:1</math>. The ratio of the rectangle's length to its width is <math>2:1</math>. What percent of the rectangle's area is in the square?<br />
<br />
<math>\mathrm{(A)}\ 12.5\qquad\mathrm{(B)}\ 25\qquad\mathrm{(C)}\ 50\qquad\mathrm{(D)}\ 75\qquad\mathrm{(E)}\ 87.5</math><br />
<br />
[[2008 AMC 10A Problems/Problem 2|Solution]]<br />
<br />
==Problem 3==<br />
For the positive integer <math>n</math>, let <math>\langle n\rangle</math> denote the sum of all the positive divisors of <math>n</math> with the exception of <math>n</math> itself. For example, <math>\langle 4\rangle=1+2=3</math> and <math>\langle 12 \rangle =1+2+3+4+6=16</math>. What is <math>\langle\langle\langle 6\rangle\rangle\rangle</math>?<br />
<br />
<math>\mathrm{(A)}\ 6\qquad\mathrm{(B)}\ 12\qquad\mathrm{(C)}\ 24\qquad\mathrm{(D)}\ 32\qquad\mathrm{(E)}\ 36</math><br />
<br />
[[2008 AMC 10A Problems/Problem 3|Solution]]<br />
<br />
==Problem 4==<br />
Suppose that <math>\tfrac{2}{3}</math> of <math>10</math> bananas are worth as much as <math>8</math> oranges. How many oranges are worth as much as <math>\tfrac{1}{2}</math> of <math>5</math> bananas?<br />
<br />
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ \frac{5}{2}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{7}{2}\qquad\mathrm{(E)}\ 4</math><br />
<br />
[[2008 AMC 10A Problems/Problem 4|Solution]]<br />
<br />
==Problem 5==<br />
Which of the following is equal to the product<br />
<cmath>\frac{8}{4}\cdot\frac{12}{8}\cdot\frac{16}{12}\cdot\cdots\cdot\frac{4n+4}{4n}\cdot\cdots\cdot\frac{2008}{2004}?</cmath><br />
<br />
<math>\mathrm{(A)}\ 251\qquad\mathrm{(B)}\ 502\qquad\mathrm{(C)}\ 1004\qquad\mathrm{(D)}\ 2008\qquad\mathrm{(E)}\ 4016</math><br />
<br />
[[2008 AMC 10A Problems/Problem 5|Solution]]<br />
<br />
==Problem 6==<br />
A triathlete competes in a triathlon in which the swimming, biking, and running segments are all of the same length. The triathlete swims at a rate of 3 kilometers per hour, bikes at a rate of 20 kilometers per hour, and runs at a rate of 10 kilometers per hour. Which of the following is closest to the triathlete's average speed, in kilometers per hour, for the entire race?<br />
<br />
<math>\mathrm{(A)}\ 3\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 5\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 7</math><br />
<br />
[[2008 AMC 10A Problems/Problem 6|Solution]]<br />
<br />
==Problem 7==<br />
The fraction<br />
<cmath>\frac{\left(3^{2008}\right)^2-\left(3^{2006}\right)^2}{\left(3^{2007}\right)^2-\left(3^{2005}\right)^2}</cmath><br />
simplifies to which of the following?<br />
<br />
<math>\mathrm{(A)}\ 1\qquad\mathrm{(B)}\ \frac{9}{4}\qquad\mathrm{(C)}\ 3\qquad\mathrm{(D)}\ \frac{9}{2}\qquad\mathrm{(E)}\ 9</math><br />
<br />
[[2008 AMC 10A Problems/Problem 7|Solution]]<br />
<br />
==Problem 8==<br />
Heather compares the price of a new computer at two different stores. Store <math>A</math> offers <math>15\%</math> off the sticker price followed by a <math>\textdollar90</math> rebate, and store <math>B</math> offers <math>25\%</math> off the same sticker price with no rebate. Heather saves <math>\textdollar15</math> by buying the computer at store <math>A</math> instead of store <math>B</math>. What is the sticker price of the computer, in dollars?<br />
<br />
<math>\mathrm{(A)}\ 750\qquad\mathrm{(B)}\ 900\qquad\mathrm{(C)}\ 1000\qquad\mathrm{(D)}\ 1050\qquad\mathrm{(E)}\ 1500</math><br />
<br />
<br />
[[2008 AMC 10A Problems/Problem 8|Solution]]<br />
<br />
==Problem 9==<br />
Suppose that<br />
<cmath>\frac{2x}{3}-\frac{x}{6}</cmath><br />
is an integer. Which of the following statements must be true about <math>x</math>?<br />
<br />
<math>\mathrm{(A)}\ \text{It is negative.}\\\qquad\mathrm{(B)}\ \text{It is even, but not necessarily a multiple of 3.}\\\qquad\mathrm{(C)}\ \text{It is a multiple of 3, but not necessarily even.}\\\qquad\mathrm{(D)}\ \text{It is a multiple of 6, but not necessarily a multiple of 12.}\\\qquad\mathrm{(E)}\ \text{It is a multiple of 12.}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 9|Solution]]<br />
<br />
==Problem 10==<br />
Each of the sides of a square <math>S_1</math> with area <math>16</math> is bisected, and a smaller square <math>S_2</math> is constructed using the bisection points as vertices. The same process is carried out on <math>S_2</math> to construct an even smaller square <math>S_3</math>. What is the area of <math>S_3</math>?<br />
<br />
<math>\mathrm{(A)}\ \frac{1}{2}\qquad\mathrm{(B)}\ 1\qquad\mathrm{(C)}\ 2\qquad\mathrm{(D)}\ 3\qquad\mathrm{(E)}\ 4</math><br />
<br />
[[2008 AMC 10A Problems/Problem 10|Solution]]<br />
<br />
==Problem 11==<br />
While Steve and LeRoy are fishing 1 mile from shore, their boat springs a leak, and water comes in at a constant rate of 10 gallons per minute. The boat will sink if it takes in more than 30 gallons of water. Steve starts rowing toward the shore at a constant rate of 4 miles per hour while LeRoy bails water out of the boat. What is the slowest rate, in gallons per minute, at which LeRoy can bail if they are to reach the shore without sinking?<br />
<br />
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 4\qquad\mathrm{(C)}\ 6\qquad\mathrm{(D)}\ 8\qquad\mathrm{(E)}\ 10</math><br />
<br />
[[2008 AMC 10A Problems/Problem 11|Solution]]<br />
<br />
==Problem 12==<br />
In a collection of red, blue, and green marbles, there are <math>25\%</math> more red marbles than blue marbles, and there are <math>60\%</math> more green marbles than red marbles. Suppose that there are <math>r</math> red marbles. What is the total number of marbles in the collection?<br />
<math>\mathrm{(A)}\ 2.85r\qquad\mathrm{(B)}\ 3r\qquad\mathrm{(C)}\ 3.4r\qquad\mathrm{(D)}\ 3.85r\qquad\mathrm{(E)}\ 4.25r</math><br />
<br />
[[2008 AMC 10A Problems/Problem 12|Solution]]<br />
<br />
==Problem 13==<br />
Doug can paint a room in <math>5</math> hours. Dave can paint the same room in <math>7</math> hours. Doug and Dave paint the room together and take a one-hour break for lunch. Let <math>t</math> be the total time, in hours, required for them to complete the job working together, including lunch. Which of the following equations is satisfied by <math>t</math>?<br />
<br />
<math>\mathrm{(A)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t+1\right)=1\qquad\mathrm{(B)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t+1=1\qquad\mathrm{(C)}\ \left(\frac{1}{5}+\frac{1}{7}\right)t=1\\\mathrm{(D)}\ \left(\frac{1}{5}+\frac{1}{7}\right)\left(t-1\right)=1\qquad\mathrm{(E)}\ \left(5+7\right)t=1</math><br />
<br />
[[2008 AMC 10A Problems/Problem 13|Solution]]<br />
<br />
==Problem 14==<br />
Older television screens have an aspect ratio of <math>4: 3</math>. That is, the ratio of the width to the height is <math>4: 3</math>. The aspect ratio of many movies is not <math>4: 3</math>, so they are sometimes shown on a television screen by "letterboxing" - darkening strips of equal height at the top and bottom of the screen, as shown. Suppose a movie has an aspect ratio of <math>2: 1</math> and is shown on an older television screen with a <math>27</math>-inch diagonal. What is the height, in inches, of each darkened strip?<br />
<asy>unitsize(1mm);<br />
filldraw((0,0)--(21.6,0)--(21.6,2.7)--(0,2.7)--cycle,grey,black);<br />
filldraw((0,13.5)--(21.6,13.5)--(21.6,16.2)--(0,16.2)--cycle,grey,black);<br />
draw((0,0)--(21.6,0)--(21.6,16.2)--(0,16.2)--cycle);</asy><br />
<math>\mathrm{(A)}\ 2\qquad\mathrm{(B)}\ 2.25\qquad\mathrm{(C)}\ 2.5\qquad\mathrm{(D)}\ 2.7\qquad\mathrm{(E)}\ 3</math><br />
<br />
[[2008 AMC 10A Problems/Problem 14|Solution]]<br />
<br />
==Problem 15==<br />
Yesterday Han drove 1 hour longer than Ian at an average speed 5 miles per hour faster than Ian. Jan drove 2 hours longer than Ian at an average speed 10 miles per hour faster than Ian. Han drove 70 miles more than Ian. How many more miles did Jan drive than Ian?<br />
<br />
<math>\mathrm{(A)}\ 120\qquad\mathrm{(B)}\ 130\qquad\mathrm{(C)}\ 140\qquad\mathrm{(D)}\ 150\qquad\mathrm{(E)}\ 160</math><br />
<br />
[[2008 AMC 10A Problems/Problem 15|Solution]]<br />
<br />
==Problem 16==<br />
Points <math>A</math> and <math>B</math> lie on a circle centered at <math>O</math>, and <math>\angle AOB = 60^\circ</math>. A second circle is internally tangent to the first and tangent to both <math>\overline{OA}</math> and <math>\overline{OB}</math>. What is the ratio of the area of the smaller circle to that of the larger circle?<br />
<br />
<math>\mathrm{(A)}\ \frac{1}{16}\qquad\mathrm{(B)}\ \frac{1}{9}\qquad\mathrm{(C)}\ \frac{1}{8}\qquad\mathrm{(D)}\ \frac{1}{6}\qquad\mathrm{(E)}\ \frac{1}{4}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 16|Solution]]<br />
<br />
==Problem 17==<br />
An equilateral triangle has side length 6. What is the area of the region containing all points that are outside the triangle but not more than 3 units from a point on the triangle?<br />
<br />
<math>\mathrm{(A)}\ 36+24\sqrt{3}\qquad\mathrm{(B)}\ 54+9\pi\qquad\mathrm{(C)}\ 54+18\sqrt{3}+6\pi\qquad\mathrm{(D)}\ \left(2\sqrt{3}+3\right)^2\pi\\\mathrm{(E)}\ 9\left(\sqrt{3}+1\right)^2\pi</math><br />
<br />
[[2008 AMC 10A Problems/Problem 17|Solution]]<br />
<br />
==Problem 18==<br />
A right triangle has perimeter 32 and area 20. What is the length of its hypotenuse?<br />
<br />
<math>\mathrm{(A)}\ \frac{57}{4}\qquad\mathrm{(B)}\ \frac{59}{4}\qquad\mathrm{(C)}\ \frac{61}{4}\qquad\mathrm{(D)}\ \frac{63}{4}\qquad\mathrm{(E)}\ \frac{65}{4}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 18|Solution]]<br />
<br />
==Problem 19==<br />
Rectangle <math>PQRS</math> lies in a plane with <math>PQ=RS=2</math> and <math>QR=SP=6</math>. The rectangle is rotated <math>90^\circ</math> clockwise about <math>R</math>, then rotated <math>90^\circ</math> clockwise about the point <math>S</math> moved to after the first rotation. What is the length of the path traveled by point <math>P</math>?<br />
<br />
<math>\mathrm{(A)}\ \left(2\sqrt{3}+\sqrt{5}\right)\pi\qquad\mathrm{(B)}\ 6\pi\qquad\mathrm{(C)}\ \left(3+\sqrt{10}\right)\pi\qquad\mathrm{(D)}\ \left(\sqrt{3}+2\sqrt{5}\right)\pi\\\mathrm{(E)}\ 2\sqrt{10}\pi</math><br />
<br />
[[2008 AMC 10A Problems/Problem 19|Solution]]<br />
<br />
==Problem 20==<br />
Trapezoid <math>ABCD</math> has bases <math>\overline{AB}</math> and <math>\overline{CD}</math> and diagonals intersecting at <math>K</math>. Suppose that <math>AB = 9</math>, <math>DC = 12</math>, and the area of <math>\triangle AKD</math> is <math>24</math>. What is the area of trapezoid <math>ABCD</math>?<br />
<br />
<math>\mathrm{(A)}\ 92\qquad\mathrm{(B)}\ 94\qquad\mathrm{(C)}\ 96\qquad\mathrm{(D)}\ 98 \qquad\mathrm{(E)}\ 100</math><br />
<br />
[[2008 AMC 10A Problems/Problem 20|Solution]]<br />
<br />
==Problem 21==<br />
A cube with side length <math>1</math> is sliced by a plane that passes through two diagonally opposite vertices <math>A</math> and <math>C</math> and the midpoints <math>B</math> and <math>D</math> of two opposite edges not containing <math>A</math> or <math>C</math>, as shown. What is the area of quadrilateral <math>ABCD</math>?<br />
<br />
<asy><br />
size(4cm);<br />
import three;<br />
unitsize(3cm);<br />
defaultpen(fontsize(8)+linewidth(0.7));<br />
currentprojection=obliqueX;<br />
<br />
draw((0.5,0,0)--(0,0,0)--(0,0,1)--(0,0,0)--(0,1,0),linetype("4 4"));<br />
draw((0.5,0,1)--(0,0,1)--(0,1,1)--(0.5,1,1)--(0.5,0,1)--(0.5,0,0)--(0.5,1,0)--(0.5,1,1));<br />
draw((0.5,1,0)--(0,1,0)--(0,1,1));<br />
dot((0.5,0,0));<br />
label("$A$",(0.5,0,0),WSW);<br />
dot((0,1,1));<br />
label("$C$",(0,1,1),NE);<br />
dot((0.5,1,0.5));<br />
label("$D$",(0.5,1,0.5),ESE);<br />
dot((0,0,0.5));<br />
label("$B$",(0,0,0.5),NW);</asy><br />
<br />
<math>\mathrm{(A)}\ \frac{\sqrt{6}}{2}\qquad\mathrm{(B)}\ \frac{5}{4}\qquad\mathrm{(C)}\ \sqrt{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 21|Solution]]<br />
<br />
==Problem 22==<br />
Jacob uses the following procedure to write down a sequence of numbers. First he chooses the first term to be 6. To generate each succeeding term, he flips a fair coin. If it comes up heads, he doubles the previous term and subtracts 1. If it comes up tails, he takes half of the previous term and subtracts 1. What is the probability that the fourth term in Jacob's sequence is an integer?<br />
<br />
<math>\mathrm{(A)}\ \frac{1}{6}\qquad\mathrm{(B)}\ \frac{1}{3}\qquad\mathrm{(C)}\ \frac{1}{2}\qquad\mathrm{(D)}\ \frac{5}{8}\qquad\mathrm{(E)}\ \frac{3}{4}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 22|Solution]]<br />
<br />
==Problem 23==<br />
Two subsets of the set <math>S=\lbrace a,b,c,d,e\rbrace</math> are to be chosen so that their union is <math>S</math> and their intersection contains exactly two elements. In how many ways can this be done, assuming that the order in which the subsets are chosen does not matter?<br />
<br />
<math>\mathrm{(A)}\ 20\qquad\mathrm{(B)}\ 40\qquad\mathrm{(C)}\ 60\qquad\mathrm{(D)}\ 160\qquad\mathrm{(E)}\ 320</math><br />
<br />
[[2008 AMC 10A Problems/Problem 23|Solution]]<br />
<br />
==Problem 24==<br />
Let <math>k={2008}^{2}+{2}^{2008}</math>. What is the units digit of <math>k^2+2^k</math>?<br />
<br />
<math>\mathrm{(A)}\ 0\qquad\mathrm{(B)}\ 2\qquad\mathrm{(C)}\ 4\qquad\mathrm{(D)}\ 6\qquad\mathrm{(E)}\ 8</math><br />
<br />
[[2008 AMC 10A Problems/Problem 24|Solution]]<br />
<br />
==Problem 25==<br />
A round table has radius <math>4</math>. Six rectangular place mats are placed on the table. Each place mat has width <math>1</math> and length <math>x</math> as shown. They are positioned so that each mat has two corners on the edge of the table, these two corners being end points of the same side of length <math>x</math>. Further, the mats are positioned so that the inner corners each touch an inner corner of an adjacent mat. What is <math>x</math>?<br />
<br />
<asy>unitsize(4mm);<br />
defaultpen(linewidth(.8)+fontsize(8));<br />
draw(Circle((0,0),4));<br />
path mat=(-2.687,-1.5513)--(-2.687,1.5513)--(-3.687,1.5513)--(-3.687,-1.5513)--cycle;<br />
draw(mat);<br />
draw(rotate(60)*mat);<br />
draw(rotate(120)*mat);<br />
draw(rotate(180)*mat);<br />
draw(rotate(240)*mat);<br />
draw(rotate(300)*mat);<br />
label("\(x\)",(-1.55,2.1),E);<br />
label("\(1\)",(-0.5,3.8),S);</asy><br />
<br />
<math>\mathrm{(A)}\ 2\sqrt{5}-\sqrt{3}\qquad\mathrm{(B)}\ 3\qquad\mathrm{(C)}\ \frac{3\sqrt{7}-\sqrt{3}}{2}\qquad\mathrm{(D)}\ 2\sqrt{3}\qquad\mathrm{(E)}\ \frac{5+2\sqrt{3}}{2}</math><br />
<br />
[[2008 AMC 10A Problems/Problem 25|Solution]]<br />
<br />
==See also==<br />
{{AMC10 box|year=2008|ab=A|before=[[2007 AMC 10B Problems]]|after=[[2008 AMC 10B Problems]]}}<br />
*[[AMC 10 Problems and Solutions]]<br />
*[[AMC Problems and Solutions]]<br />
{{MAA Notice}}</div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=User:Flamekhoemberish&diff=156399User:Flamekhoemberish2021-06-19T18:40:34Z<p>Kundusne000: /* User Count */</p>
<hr />
<div>==User Count==<br />
If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">2</font></center><br />
<br />
==Introduction==<br />
Hi Everyone,<br />
<br />
I am Flame Kho. I am currently working in England and quite fluent in both English and Mandarin.<br />
<br />
I have a broad range of interests: mathematics, Cover Chess, classic Chinese songs, travels, etc. I also enjoy creating certificates for chess winners. Visit this page for my interests and artworks. :)<br />
<br />
Below is my upcoming avatar, which I will change two weeks from now.<br />
<br />
[[File:FlameKhoAvatar.jpeg|center|300px]]<br />
~FlameKhoEmberish<br />
<br />
==Mathematics==<br />
I am an amateur problem-solver. I am so proud of my contribution to [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 2021 AMC 10A Problem 25] (Solution 4).<br />
<br />
==Cover Chess==<br />
Cover Chess is a variant of Xiangqi (Chinese Chess), where all pieces are covered initially. The game has a considerable amount of luck--even amateur players have a chance of beating masters.<br />
<br />
I managed many Cover Chess tournaments and participated in certificate-making. See my artworks below.<br />
<br />
==Guess Songs==<br />
I am a fan of Chinese classic music. I recognize a lot of songs. If you want to challenge me on guessing songs, I will always be here! :)<br />
<br />
==Artworks==<br />
In this section, I will show you my artworks across different areas. I am the original author for all of the artworks below.<br />
<br />
Hope you enjoy them! :)<br />
<br />
===Certificates===<br />
====Cover Chess====<br />
<center>[[File:Cover Chess Certificate 1.jpeg|450px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 2.jpeg|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 3.jpeg|450px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 7.jpeg|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 8.jpeg|center|450px]]</center><p><br />
<center>[[File:Cover Chess Certificate 4.jpeg|300px]]&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 5.jpeg|300px]]&nbsp;&nbsp;&nbsp;&nbsp;[[File:Cover Chess Certificate 6.jpeg|300px]]</center><br />
<br />
====Guess Songs====<br />
[[File:GuessSongs.jpeg|center|300px]]<br />
<br />
===Posters===<br />
====Valentine Speed Chess Tournament====<br />
<center>[[File:Season 2.jpeg|300px]][[File:Season 3.jpeg|300px]][[File:Season 4.jpeg|300px]]</center><br />
<center>[[File:Season 5.jpeg|300px]][[File:Season 6.jpeg|300px]][[File:Season 7.jpeg|300px]]</center><br />
<center>[[File:Season 8.jpeg|300px]][[File:Season 9.jpeg|300px]][[File:Season 10.jpeg|300px]]</center><br />
<br />
====Master Challenges====<br />
[[File:Master 1.jpeg|center|600px]]<p><br />
[[File:Master 2.png|center|600px]]<p><br />
[[File:Master 3.jpeg|center|600px]]<br />
<br />
====Other Chess Tournaments====<br />
<center>[[File:Chess Poster 1.jpeg|400px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Chess Poster 2.jpeg|400px]]</center><p><br />
<center>[[File:Chess Poster 3.jpeg|400px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Chess Poster 4.jpeg|400px]]</center><br />
<br />
===Photographs===<br />
[[File:Church.jpeg|center]]<p><br />
[[File:Outdoors.jpeg|center]]<br />
<br />
===Drawings===<br />
[[File:Drawing.jpeg|center|300px]]<br />
<br />
==Cartoons==<br />
[[File:Cartoon.jpeg|center]]<br />
<br />
==Stickers==<br />
Finally, I am a huge fan of stickers. Below are my favorite ones. As you can tell, I love cats in particular.<br />
<center>[[File:Coming.png|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Hahahahaha.png|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Bad Words.jpg|250px]]</center><p><br />
<center>[[File:Yes.gif|250px]]&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;[[File:Fighting.gif|250px]]</center></div>Kundusne000https://artofproblemsolving.com/wiki/index.php?title=User:MRENTHUSIASM&diff=156200User:MRENTHUSIASM2021-06-17T16:39:54Z<p>Kundusne000: /* User Count */</p>
<hr />
<div>==User Count==<br />
If this is your first time visiting this page, edit it by incrementing the user count below by one.</font></div><br />
<center><font size="100px">4</font></center><br />
<br />
==Introduction==<br />
Hi Everyone,<br />
<br />
I am a hedgehog who likes Message Board halping, funny pictures, emoji wars, and AoPS Wiki contributions. Keep these fun things coming!<br />
<br />
This hedgehog represents my attitude towards math--enthusiastic and victorious!<br />
[[File:Victorious-Hedgehog.png|center]]<br />
And here is my personified self-portrait:<br />
[[File:Enthusiastic.gif|center]]<br />
~MRENTHUSIASM<br />
<br />
==AoPS Bio==<br />
Jerry graduated from Northeastern University in 2018 as a mathematics and computer science combined major. He has ten years of experience working with students. During his last two years in high school, he served as the math club president and prepared lively lectures on contest math. As a result, he inspired many kids to get interested in math and participate in math competitions (such as AMC and ARML). For the last eight years, he has worked with Math League to write math contests for students from grades 4 through 12 in the USA. Starting in 2019, he has served as an instructor and a Message Board Halper for many AoPS online classes and truly enjoyed that experience! In his mind, nothing is more rewarding than educating tomorrow’s bright minds. In his spare time, he loves contributing to the AoPS Wiki and playing chess and all sorts of board games, especially Scotland Yard and Samurai.<br />
<br />
==Contributions==<br />
Below are my contributions to the AMC/AIME Problems' Solutions in the AoPS Wiki. I understand that in this communal Wiki, we all have to collaborate and make compromises. <i><b>In any of my solutions, if you find flaws and/or want major revisions, please contact me via my [https://artofproblemsolving.com/wiki/index.php/User_talk:MRENTHUSIASM user talk page] or the private messaging system before you take action. I am sure that we can work things out.</b></i><br />
<br />
Thank you for your cooperation, and hope you enjoy reading my solutions. :)<br />
<br />
~MRENTHUSIASM<br />
<br />
===1984 AIME===<br />
* [https://artofproblemsolving.com/wiki/index.php/1984_AIME_Problems/Problem_10 Problem 10] (Solution 3)<br />
<br />
===1985 AIME===<br />
* [https://artofproblemsolving.com/wiki/index.php/1985_AIME_Problems/Problem_12 Problem 12] (Solution 7)<br />
<br />
===1986 AIME===<br />
* [https://artofproblemsolving.com/wiki/index.php/1986_AIME_Problems/Problem_2 Problem 2] (Solutions 1, 2)<br />
<br />
===2009 AMC 10B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_1 Problem 1] (Solutions 1, 2, 3)<br />
<br />
===2009 AMC 12B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2009_AMC_12B_Problems/Problem_1 Problem 1]: same as [https://artofproblemsolving.com/wiki/index.php/2009_AMC_10B_Problems/Problem_1 2009 AMC 10B Problem 1] (Solutions 1, 2, 3)<br />
<br />
===2020 AMC 8===<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_1 Problem 1] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_8_Problems/Problem_3 Problem 3] (Solution 3)<br />
<br />
===2020 AMC 10A===<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_3 Problem 3] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_5 Problem 5] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_11 Problem 11] (Solution 5)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_17 Problem 17] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_24 Problem 24] (Solution 11)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_25 Problem 25] (Solution 3)<br />
<br />
===2020 AMC 10B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_3 Problem 3] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_4 Problem 4] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_8 Problem 8] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_13 Problem 13] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_15 Problem 15] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_23 Problem 23] (Solution 5)<br />
<br />
===2020 AMC 12A===<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_1 Problem 1] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_6 Problem 6] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_8 Problem 8]: same as [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_11 2020 AMC 10A Problem 11] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_9 Problem 9] (Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_10 Problem 10] (Solution 5)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_15 Problem 15] (Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_23 Problem 23]: same as [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10A_Problems/Problem_25 2020 AMC 10A Problem 25] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_24 Problem 24] (Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12A_Problems/Problem_25 Problem 25] (Solution 1, Remark)<br />
<br />
===2020 AMC 12B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_3 Problem 3]: same as [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_3 2020 AMC 10B Problem 3] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_4 Problem 4]: same as [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_4 2020 AMC 10B Problem 4] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_5 Problem 5] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_6 Problem 6] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_10 Problem 10] (Diagram, Solution 7)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_12 Problem 12] (Diagram)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_13 Problem 13] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2020_AMC_12B_Problems/Problem_19 Problem 19]: same as [https://artofproblemsolving.com/wiki/index.php/2020_AMC_10B_Problems/Problem_23 2020 AMC 10B Problem 23] (Solution 5)<br />
<br />
===2020 AIME I===<br />
<br />
===2020 AIME II===<br />
<br />
===2021 AMC 10A===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_1 Problem 1] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_2 Problem 2] (Solutions 2, 3, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 Problem 3] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_4 Problem 4] (Solutions 1, 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_5 Problem 5] (Solutions 1, 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_6 Problem 6] (Solutions 1, 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_7 Problem 7] (Solutions 2, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9 Problem 9] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_10 Problem 10] (Solution 7)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_11 Problem 11] (Solutions 1, 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12 Problem 12] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_13 Problem 13] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 Problem 17] (Diagram, Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_18 Problem 18] (Solutions 1, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_20 Problem 20] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_22 Problem 22] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_23 Problem 23] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_24 Problem 24] (Diagram, Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_25 Problem 25] (Solutions 2, 4)<br />
<br />
===2021 AMC 10B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 Problem 6] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_7 Problem 7] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_8 Problem 8] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_16 Problem 16] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_17 Problem 17] (Solution 3)<br />
<br />
===2021 AMC 12A===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_2 Problem 2] (Solutions 2, 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_3 Problem 3]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_3 2021 AMC 10A Problem 3] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_4 Problem 4]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_7 2021 AMC 10A Problem 7] (Solutions 2, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_6 Problem 6] (Solutions 2, 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_7 Problem 7]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_9 2021 AMC 10A Problem 9] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_9 Problem 9]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_10 2021 AMC 10A Problem 10] (Solution 7)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_10 Problem 10]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_12 2021 AMC 10A Problem 12] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_11 Problem 11] (Solutions 2, 3, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_13 Problem 13] (Solutions 2, 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_14 Problem 14] (Solutions 2, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_15 Problem 15] (Solutions 3, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_17 Problem 17]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_17 2021 AMC 10A Problem 17] (Diagram, Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_18 Problem 18]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_18 2021 AMC 10A Problem 18] (Solutions 1, 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_19 Problem 19] (Solutions 2, 3, Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_21 Problem 21] (Solution 2, Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_22 Problem 22] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_23 Problem 23]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10A_Problems/Problem_23 2021 AMC 10A Problem 23] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_24 Problem 24] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12A_Problems/Problem_25 Problem 25] (Solution 1)<br />
<br />
===2021 AMC 12B===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_4 Problem 4]: same as [https://artofproblemsolving.com/wiki/index.php/2021_AMC_10B_Problems/Problem_6 2021 AMC 10B Problem 6] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_11 Problem 11] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_20 Problem 20] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_21 Problem 21] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AMC_12B_Problems/Problem_23 Problem 23] (Solution 4)<br />
<br />
===2021 AIME I===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_1 Problem 1] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_2 Problem 2] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_3 Problem 3] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_7 Problem 7] (Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_8 Problem 8] (Solution 3, Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_9 Problem 9] (Diagram, Solution 6)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_10 Problem 10] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_I_Problems/Problem_11 Problem 11] (Solution 4)<br />
<br />
===2021 AIME II===<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_1 Problem 1] (Solution 3, Remark)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_2 Problem 2] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_3 Problem 3] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_4 Problem 4] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_5 Problem 5] (Solutions 2, 5)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_6 Problem 6] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_7 Problem 7] (Solution 4)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_8 Problem 8] (Solution 1)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_9 Problem 9] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_10 Problem 10] (Diagram, Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_11 Problem 11] (Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_12 Problem 12] (Diagram, Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_13 Problem 13] (Solution 3)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_14 Problem 14] (Diagram, Solution 2)<br />
* [https://artofproblemsolving.com/wiki/index.php/2021_AIME_II_Problems/Problem_15 Problem 15] (Solution 2)</div>Kundusne000