https://artofproblemsolving.com/wiki/api.php?action=feedcontributions&user=Olutosinfires&feedformat=atom AoPS Wiki - User contributions [en] 2021-08-05T11:40:41Z User contributions MediaWiki 1.31.1 https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_17&diff=105202 2018 AMC 12B Problems/Problem 17 2019-04-09T02:51:40Z <p>Olutosinfires: /* Solution 9 (Quick inspection) */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We claim that, between any two fractions &lt;math&gt;a/b&lt;/math&gt; and &lt;math&gt;c/d&lt;/math&gt;, if &lt;math&gt;bc-ad=1&lt;/math&gt;, the fraction with smallest denominator between them is &lt;math&gt;\frac{a+c}{b+d}&lt;/math&gt;. To prove this, we see that<br /> <br /> &lt;cmath&gt;\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},&lt;/cmath&gt;<br /> which reduces to &lt;math&gt;q\geq b+d&lt;/math&gt;. We can easily find that &lt;math&gt;p=a+c&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 2 (requires justification)==<br /> <br /> Assume that the difference &lt;math&gt;\frac{p}{q} - \frac{5}{9}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{9q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;9p - 5q = 1&lt;/math&gt;<br /> <br /> Also assume that the difference &lt;math&gt;\frac{4}{7} - \frac{p}{q}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{7q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;4q - 7p = 1&lt;/math&gt;<br /> <br /> Solving the system of equations yields &lt;math&gt;q=16&lt;/math&gt; and &lt;math&gt;p=9&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Cross-multiply the inequality to get &lt;cmath&gt;35q &lt; 63p &lt; 36q.&lt;/cmath&gt;<br /> <br /> Then,<br /> &lt;cmath&gt;0 &lt; 63p-35q &lt; q,&lt;/cmath&gt;<br /> &lt;cmath&gt;0 &lt; 7(9p-5q) &lt; q.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; are integers, &lt;math&gt;9p-5q&lt;/math&gt; is an integer. To minimize &lt;math&gt;q&lt;/math&gt;, start from &lt;math&gt;9p-5q=1&lt;/math&gt;, which gives &lt;math&gt;p=\frac{5q+1}{9}&lt;/math&gt;. This limits &lt;math&gt;q&lt;/math&gt; to be greater than &lt;math&gt;7&lt;/math&gt;, so test values of &lt;math&gt;q&lt;/math&gt; starting from &lt;math&gt;q=8&lt;/math&gt;. However, &lt;math&gt;q=8&lt;/math&gt; to &lt;math&gt;q=14&lt;/math&gt; do not give integer values of &lt;math&gt;p&lt;/math&gt;. <br /> <br /> Once &lt;math&gt;q&gt;14&lt;/math&gt;, it is possible for &lt;math&gt;9p-5q&lt;/math&gt; to be equal to &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; could also be equal to &lt;math&gt;\frac{5q+2}{9}.&lt;/math&gt; The next value, &lt;math&gt;q=15&lt;/math&gt;, is not a solution, but &lt;math&gt;q=16&lt;/math&gt; gives &lt;math&gt;p=\frac{5\cdot 16 + 1}{9} = 9&lt;/math&gt;. Thus, the smallest possible value of &lt;math&gt;q&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Graph the regions &lt;math&gt;y &gt; \frac{5}{9}x&lt;/math&gt; and &lt;math&gt;y &lt; \frac{4}{7}x&lt;/math&gt;. Note that the lattice point &lt;math&gt;(16,9)&lt;/math&gt; is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is &lt;math&gt;\frac{9}{16}&lt;/math&gt; and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;. <br /> <br /> Remark: This also gives an intuitive geometric proof of the mediant using vectors.<br /> <br /> ==Solution 5 (Using answer choices to prove mediant)==<br /> As the other solutions do, the mediant &lt;math&gt;=\frac{9}{16}&lt;/math&gt; is between the two fractions, with a difference of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;. Suppose that the answer was not &lt;math&gt;A&lt;/math&gt;, then the answer must be &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; as otherwise &lt;math&gt;p&lt;/math&gt; would be negative. Then, the possible fractions with lower denominator would be &lt;math&gt;\frac{k-11}{k}&lt;/math&gt; for &lt;math&gt;k=12,13,14,15&lt;/math&gt; and &lt;math&gt;\frac{k-13}{k}&lt;/math&gt; for &lt;math&gt;k=14,15,&lt;/math&gt; which are clearly not anywhere close to &lt;math&gt;\frac{4}{7}\approx 0.6&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> Inverting the given inequality we get <br /> &lt;cmath&gt;\frac{7}{4} &lt; \frac{q}{p} &lt; \frac{9}{5}&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;35p &lt; 20q &lt; 36p&lt;/cmath&gt;<br /> <br /> We can now substitute &lt;math&gt;q = p + k&lt;/math&gt;. Note we need to find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;35p &lt; 20p + 20k &lt; 36p&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;15p &lt; 20k &lt; 16p&lt;/cmath&gt;<br /> <br /> Cleary &lt;math&gt;p&lt;/math&gt; is greater than &lt;math&gt;k&lt;/math&gt;. We will now substitute &lt;math&gt;p = k + x&lt;/math&gt; to get<br /> <br /> &lt;cmath&gt;15k + 15x &lt; 20k &lt; 16k + 16x&lt;/cmath&gt;<br /> <br /> The inequality &lt;math&gt;15k + 15x &lt; 20k&lt;/math&gt; simplifies to &lt;math&gt;3x &lt; k&lt;/math&gt;.<br /> The inequality &lt;math&gt;20k &lt; 16k + 16x&lt;/math&gt; simplifies to &lt;math&gt;k &lt; 4x&lt;/math&gt;.<br /> Combining the two we get<br /> &lt;cmath&gt;3x &lt; k &lt; 4x&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are integers, the smallest values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; that satisfy the above equation are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; respectively. Substituting these back in, we arrive with an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Start with &lt;math&gt;\frac{5}{9}&lt;/math&gt;. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to &lt;math&gt;\frac{5}{9}&lt;/math&gt;, add &lt;math&gt;1&lt;/math&gt; to the numerator; otherwise, if it is greater than or equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;, add one &lt;math&gt;1&lt;/math&gt; to the denominator. We have:<br /> <br /> &lt;cmath&gt;\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;16 - 9 = \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Because q and p are positive integers with &lt;math&gt;p&lt;q&lt;/math&gt;, we can let &lt;math&gt;q=p+k&lt;/math&gt; where &lt;math&gt;k\in{\mathbb{Z}}&lt;/math&gt;. Now, the problem condition reduces to<br /> <br /> &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt;<br /> <br /> Our first inequality is &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;/math&gt; which gives us &lt;math&gt;5p+5k&lt;9p\implies \frac{5}{4}k&lt;p&lt;/math&gt;.<br /> <br /> Our second inequality is &lt;math&gt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt; which gives us &lt;math&gt;7p&lt;4p+4k\implies p&lt;\frac{4}{3}k&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;\frac{5}{4}k&lt;p&lt;\frac{4}{3}k\implies 15k&lt;12p&lt;16k&lt;/math&gt;.<br /> <br /> It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.<br /> <br /> Now, simple casework through the answer choices of the problem reveals that &lt;math&gt;q-p=p+k-p=k\implies k\ge{\boxed{7}}&lt;/math&gt;.<br /> <br /> == Solution 9 (Quick inspection) ==<br /> Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.<br /> The interval can also be written as &lt;math&gt;0.5556&lt;x&lt;0.5714&lt;/math&gt;. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range. <br /> <br /> The denominators to be considered are &lt;math&gt;9,10,11,12...&lt;/math&gt;. We check &lt;math&gt;\frac{6}{10}, \frac{6}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}&lt;/math&gt;. At this point we know that we've got our fraction and our answer is &lt;math&gt;16-9=\boxed{\textbf{A } 7}&lt;/math&gt;<br /> <br /> The inspection was made faster by considering the fact that &lt;math&gt;\frac{a+1}{b+1}&gt;\frac{a}{b}&lt;/math&gt;. <br /> <br /> So, once a fraction was gotten which was greater than &lt;math&gt;\frac{4}{7}&lt;/math&gt; we jump to the next denominator. <br /> <br /> We then make sure we consider fractions with higher positive difference between the denominator and numerator. And we also do not forget that the numerator must be greater than half of the denominator.<br /> <br /> (&lt;math&gt;\frac{8}{14}&lt;/math&gt; was obviously skipped because it is equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;.)<br /> <br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_17&diff=105201 2018 AMC 12B Problems/Problem 17 2019-04-09T02:48:45Z <p>Olutosinfires: /* Solution 9 (Quick inspection) */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We claim that, between any two fractions &lt;math&gt;a/b&lt;/math&gt; and &lt;math&gt;c/d&lt;/math&gt;, if &lt;math&gt;bc-ad=1&lt;/math&gt;, the fraction with smallest denominator between them is &lt;math&gt;\frac{a+c}{b+d}&lt;/math&gt;. To prove this, we see that<br /> <br /> &lt;cmath&gt;\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},&lt;/cmath&gt;<br /> which reduces to &lt;math&gt;q\geq b+d&lt;/math&gt;. We can easily find that &lt;math&gt;p=a+c&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 2 (requires justification)==<br /> <br /> Assume that the difference &lt;math&gt;\frac{p}{q} - \frac{5}{9}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{9q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;9p - 5q = 1&lt;/math&gt;<br /> <br /> Also assume that the difference &lt;math&gt;\frac{4}{7} - \frac{p}{q}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{7q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;4q - 7p = 1&lt;/math&gt;<br /> <br /> Solving the system of equations yields &lt;math&gt;q=16&lt;/math&gt; and &lt;math&gt;p=9&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Cross-multiply the inequality to get &lt;cmath&gt;35q &lt; 63p &lt; 36q.&lt;/cmath&gt;<br /> <br /> Then,<br /> &lt;cmath&gt;0 &lt; 63p-35q &lt; q,&lt;/cmath&gt;<br /> &lt;cmath&gt;0 &lt; 7(9p-5q) &lt; q.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; are integers, &lt;math&gt;9p-5q&lt;/math&gt; is an integer. To minimize &lt;math&gt;q&lt;/math&gt;, start from &lt;math&gt;9p-5q=1&lt;/math&gt;, which gives &lt;math&gt;p=\frac{5q+1}{9}&lt;/math&gt;. This limits &lt;math&gt;q&lt;/math&gt; to be greater than &lt;math&gt;7&lt;/math&gt;, so test values of &lt;math&gt;q&lt;/math&gt; starting from &lt;math&gt;q=8&lt;/math&gt;. However, &lt;math&gt;q=8&lt;/math&gt; to &lt;math&gt;q=14&lt;/math&gt; do not give integer values of &lt;math&gt;p&lt;/math&gt;. <br /> <br /> Once &lt;math&gt;q&gt;14&lt;/math&gt;, it is possible for &lt;math&gt;9p-5q&lt;/math&gt; to be equal to &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; could also be equal to &lt;math&gt;\frac{5q+2}{9}.&lt;/math&gt; The next value, &lt;math&gt;q=15&lt;/math&gt;, is not a solution, but &lt;math&gt;q=16&lt;/math&gt; gives &lt;math&gt;p=\frac{5\cdot 16 + 1}{9} = 9&lt;/math&gt;. Thus, the smallest possible value of &lt;math&gt;q&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Graph the regions &lt;math&gt;y &gt; \frac{5}{9}x&lt;/math&gt; and &lt;math&gt;y &lt; \frac{4}{7}x&lt;/math&gt;. Note that the lattice point &lt;math&gt;(16,9)&lt;/math&gt; is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is &lt;math&gt;\frac{9}{16}&lt;/math&gt; and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;. <br /> <br /> Remark: This also gives an intuitive geometric proof of the mediant using vectors.<br /> <br /> ==Solution 5 (Using answer choices to prove mediant)==<br /> As the other solutions do, the mediant &lt;math&gt;=\frac{9}{16}&lt;/math&gt; is between the two fractions, with a difference of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;. Suppose that the answer was not &lt;math&gt;A&lt;/math&gt;, then the answer must be &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; as otherwise &lt;math&gt;p&lt;/math&gt; would be negative. Then, the possible fractions with lower denominator would be &lt;math&gt;\frac{k-11}{k}&lt;/math&gt; for &lt;math&gt;k=12,13,14,15&lt;/math&gt; and &lt;math&gt;\frac{k-13}{k}&lt;/math&gt; for &lt;math&gt;k=14,15,&lt;/math&gt; which are clearly not anywhere close to &lt;math&gt;\frac{4}{7}\approx 0.6&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> Inverting the given inequality we get <br /> &lt;cmath&gt;\frac{7}{4} &lt; \frac{q}{p} &lt; \frac{9}{5}&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;35p &lt; 20q &lt; 36p&lt;/cmath&gt;<br /> <br /> We can now substitute &lt;math&gt;q = p + k&lt;/math&gt;. Note we need to find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;35p &lt; 20p + 20k &lt; 36p&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;15p &lt; 20k &lt; 16p&lt;/cmath&gt;<br /> <br /> Cleary &lt;math&gt;p&lt;/math&gt; is greater than &lt;math&gt;k&lt;/math&gt;. We will now substitute &lt;math&gt;p = k + x&lt;/math&gt; to get<br /> <br /> &lt;cmath&gt;15k + 15x &lt; 20k &lt; 16k + 16x&lt;/cmath&gt;<br /> <br /> The inequality &lt;math&gt;15k + 15x &lt; 20k&lt;/math&gt; simplifies to &lt;math&gt;3x &lt; k&lt;/math&gt;.<br /> The inequality &lt;math&gt;20k &lt; 16k + 16x&lt;/math&gt; simplifies to &lt;math&gt;k &lt; 4x&lt;/math&gt;.<br /> Combining the two we get<br /> &lt;cmath&gt;3x &lt; k &lt; 4x&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are integers, the smallest values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; that satisfy the above equation are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; respectively. Substituting these back in, we arrive with an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Start with &lt;math&gt;\frac{5}{9}&lt;/math&gt;. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to &lt;math&gt;\frac{5}{9}&lt;/math&gt;, add &lt;math&gt;1&lt;/math&gt; to the numerator; otherwise, if it is greater than or equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;, add one &lt;math&gt;1&lt;/math&gt; to the denominator. We have:<br /> <br /> &lt;cmath&gt;\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;16 - 9 = \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Because q and p are positive integers with &lt;math&gt;p&lt;q&lt;/math&gt;, we can let &lt;math&gt;q=p+k&lt;/math&gt; where &lt;math&gt;k\in{\mathbb{Z}}&lt;/math&gt;. Now, the problem condition reduces to<br /> <br /> &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt;<br /> <br /> Our first inequality is &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;/math&gt; which gives us &lt;math&gt;5p+5k&lt;9p\implies \frac{5}{4}k&lt;p&lt;/math&gt;.<br /> <br /> Our second inequality is &lt;math&gt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt; which gives us &lt;math&gt;7p&lt;4p+4k\implies p&lt;\frac{4}{3}k&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;\frac{5}{4}k&lt;p&lt;\frac{4}{3}k\implies 15k&lt;12p&lt;16k&lt;/math&gt;.<br /> <br /> It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.<br /> <br /> Now, simple casework through the answer choices of the problem reveals that &lt;math&gt;q-p=p+k-p=k\implies k\ge{\boxed{7}}&lt;/math&gt;.<br /> <br /> == Solution 9 (Quick inspection) ==<br /> Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.<br /> The interval can also be written as &lt;math&gt;0.5556&lt;x&lt;0.5714&lt;/math&gt;. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range. <br /> <br /> The denominators to be considered are &lt;math&gt;9,10,11,12...&lt;/math&gt;. We check &lt;math&gt;\frac{6}{10}, \frac{6}{11}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}&lt;/math&gt;. At this point we know that we've got our fraction and our answer is &lt;math&gt;16-9=\boxed{\textbf{A } 7}&lt;/math&gt;<br /> <br /> The inspection was made faster by considering the fact that &lt;math&gt;\frac{a+1}{b+1}&gt;\frac{a}{b}&lt;/math&gt;. <br /> <br /> So, once a fraction was gotten which was greater than &lt;math&gt;\frac{4}{7}&lt;/math&gt; we jump to the next denominator. <br /> <br /> We then make sure we consider fractions with higher positive difference between the denominator and numerator. And we also do not forget that the numerator must be greater than half of the denominator.<br /> <br /> (&lt;math&gt;\frac{8}{14}&lt;/math&gt; was obviously skipped because it is equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;.)<br /> <br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_17&diff=105200 2018 AMC 12B Problems/Problem 17 2019-04-09T02:48:15Z <p>Olutosinfires: /* Solution 9 (Quick inspection) */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We claim that, between any two fractions &lt;math&gt;a/b&lt;/math&gt; and &lt;math&gt;c/d&lt;/math&gt;, if &lt;math&gt;bc-ad=1&lt;/math&gt;, the fraction with smallest denominator between them is &lt;math&gt;\frac{a+c}{b+d}&lt;/math&gt;. To prove this, we see that<br /> <br /> &lt;cmath&gt;\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},&lt;/cmath&gt;<br /> which reduces to &lt;math&gt;q\geq b+d&lt;/math&gt;. We can easily find that &lt;math&gt;p=a+c&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 2 (requires justification)==<br /> <br /> Assume that the difference &lt;math&gt;\frac{p}{q} - \frac{5}{9}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{9q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;9p - 5q = 1&lt;/math&gt;<br /> <br /> Also assume that the difference &lt;math&gt;\frac{4}{7} - \frac{p}{q}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{7q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;4q - 7p = 1&lt;/math&gt;<br /> <br /> Solving the system of equations yields &lt;math&gt;q=16&lt;/math&gt; and &lt;math&gt;p=9&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Cross-multiply the inequality to get &lt;cmath&gt;35q &lt; 63p &lt; 36q.&lt;/cmath&gt;<br /> <br /> Then,<br /> &lt;cmath&gt;0 &lt; 63p-35q &lt; q,&lt;/cmath&gt;<br /> &lt;cmath&gt;0 &lt; 7(9p-5q) &lt; q.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; are integers, &lt;math&gt;9p-5q&lt;/math&gt; is an integer. To minimize &lt;math&gt;q&lt;/math&gt;, start from &lt;math&gt;9p-5q=1&lt;/math&gt;, which gives &lt;math&gt;p=\frac{5q+1}{9}&lt;/math&gt;. This limits &lt;math&gt;q&lt;/math&gt; to be greater than &lt;math&gt;7&lt;/math&gt;, so test values of &lt;math&gt;q&lt;/math&gt; starting from &lt;math&gt;q=8&lt;/math&gt;. However, &lt;math&gt;q=8&lt;/math&gt; to &lt;math&gt;q=14&lt;/math&gt; do not give integer values of &lt;math&gt;p&lt;/math&gt;. <br /> <br /> Once &lt;math&gt;q&gt;14&lt;/math&gt;, it is possible for &lt;math&gt;9p-5q&lt;/math&gt; to be equal to &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; could also be equal to &lt;math&gt;\frac{5q+2}{9}.&lt;/math&gt; The next value, &lt;math&gt;q=15&lt;/math&gt;, is not a solution, but &lt;math&gt;q=16&lt;/math&gt; gives &lt;math&gt;p=\frac{5\cdot 16 + 1}{9} = 9&lt;/math&gt;. Thus, the smallest possible value of &lt;math&gt;q&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Graph the regions &lt;math&gt;y &gt; \frac{5}{9}x&lt;/math&gt; and &lt;math&gt;y &lt; \frac{4}{7}x&lt;/math&gt;. Note that the lattice point &lt;math&gt;(16,9)&lt;/math&gt; is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is &lt;math&gt;\frac{9}{16}&lt;/math&gt; and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;. <br /> <br /> Remark: This also gives an intuitive geometric proof of the mediant using vectors.<br /> <br /> ==Solution 5 (Using answer choices to prove mediant)==<br /> As the other solutions do, the mediant &lt;math&gt;=\frac{9}{16}&lt;/math&gt; is between the two fractions, with a difference of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;. Suppose that the answer was not &lt;math&gt;A&lt;/math&gt;, then the answer must be &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; as otherwise &lt;math&gt;p&lt;/math&gt; would be negative. Then, the possible fractions with lower denominator would be &lt;math&gt;\frac{k-11}{k}&lt;/math&gt; for &lt;math&gt;k=12,13,14,15&lt;/math&gt; and &lt;math&gt;\frac{k-13}{k}&lt;/math&gt; for &lt;math&gt;k=14,15,&lt;/math&gt; which are clearly not anywhere close to &lt;math&gt;\frac{4}{7}\approx 0.6&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> Inverting the given inequality we get <br /> &lt;cmath&gt;\frac{7}{4} &lt; \frac{q}{p} &lt; \frac{9}{5}&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;35p &lt; 20q &lt; 36p&lt;/cmath&gt;<br /> <br /> We can now substitute &lt;math&gt;q = p + k&lt;/math&gt;. Note we need to find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;35p &lt; 20p + 20k &lt; 36p&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;15p &lt; 20k &lt; 16p&lt;/cmath&gt;<br /> <br /> Cleary &lt;math&gt;p&lt;/math&gt; is greater than &lt;math&gt;k&lt;/math&gt;. We will now substitute &lt;math&gt;p = k + x&lt;/math&gt; to get<br /> <br /> &lt;cmath&gt;15k + 15x &lt; 20k &lt; 16k + 16x&lt;/cmath&gt;<br /> <br /> The inequality &lt;math&gt;15k + 15x &lt; 20k&lt;/math&gt; simplifies to &lt;math&gt;3x &lt; k&lt;/math&gt;.<br /> The inequality &lt;math&gt;20k &lt; 16k + 16x&lt;/math&gt; simplifies to &lt;math&gt;k &lt; 4x&lt;/math&gt;.<br /> Combining the two we get<br /> &lt;cmath&gt;3x &lt; k &lt; 4x&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are integers, the smallest values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; that satisfy the above equation are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; respectively. Substituting these back in, we arrive with an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Start with &lt;math&gt;\frac{5}{9}&lt;/math&gt;. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to &lt;math&gt;\frac{5}{9}&lt;/math&gt;, add &lt;math&gt;1&lt;/math&gt; to the numerator; otherwise, if it is greater than or equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;, add one &lt;math&gt;1&lt;/math&gt; to the denominator. We have:<br /> <br /> &lt;cmath&gt;\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;16 - 9 = \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Because q and p are positive integers with &lt;math&gt;p&lt;q&lt;/math&gt;, we can let &lt;math&gt;q=p+k&lt;/math&gt; where &lt;math&gt;k\in{\mathbb{Z}}&lt;/math&gt;. Now, the problem condition reduces to<br /> <br /> &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt;<br /> <br /> Our first inequality is &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;/math&gt; which gives us &lt;math&gt;5p+5k&lt;9p\implies \frac{5}{4}k&lt;p&lt;/math&gt;.<br /> <br /> Our second inequality is &lt;math&gt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt; which gives us &lt;math&gt;7p&lt;4p+4k\implies p&lt;\frac{4}{3}k&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;\frac{5}{4}k&lt;p&lt;\frac{4}{3}k\implies 15k&lt;12p&lt;16k&lt;/math&gt;.<br /> <br /> It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.<br /> <br /> Now, simple casework through the answer choices of the problem reveals that &lt;math&gt;q-p=p+k-p=k\implies k\ge{\boxed{7}}&lt;/math&gt;.<br /> <br /> == Solution 9 (Quick inspection) ==<br /> Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.<br /> The interval can also be written as &lt;math&gt;0.5556&lt;x&lt;0.5714&lt;/math&gt;. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range. <br /> <br /> The denominators to be considered are &lt;math&gt;9,10,11,12...&lt;/math&gt;. We check &lt;math&gt;\frac{6}{10}, \frac{6}{11}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}&lt;/math&gt;. At this point we know that we've got our fraction and our answer is &lt;math&gt;16-9=\boxed{\textbf{A } 7}&lt;/math&gt;<br /> <br /> The inspection was made faster by considering the fact that &lt;math&gt;\frac{a+1}{b+1}&gt;\frac{a}{b}&lt;/math&gt;. <br /> <br /> So, once a fraction was gotten which was greater than &lt;math&gt;\frac{4}{7}&lt;/math&gt; we jump to the next denominator. <br /> <br /> We then make sure we consider fractions with higher positive difference between the denominator and numerator. And we also do not forget that the numerator must be greater than half of the denominator.<br /> <br /> &lt;math&gt;\frac{8}{14}&lt;/math&gt; was obviously skipped because it is equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;.<br /> <br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_17&diff=105199 2018 AMC 12B Problems/Problem 17 2019-04-09T02:44:49Z <p>Olutosinfires: /* Solution 9 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We claim that, between any two fractions &lt;math&gt;a/b&lt;/math&gt; and &lt;math&gt;c/d&lt;/math&gt;, if &lt;math&gt;bc-ad=1&lt;/math&gt;, the fraction with smallest denominator between them is &lt;math&gt;\frac{a+c}{b+d}&lt;/math&gt;. To prove this, we see that<br /> <br /> &lt;cmath&gt;\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},&lt;/cmath&gt;<br /> which reduces to &lt;math&gt;q\geq b+d&lt;/math&gt;. We can easily find that &lt;math&gt;p=a+c&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 2 (requires justification)==<br /> <br /> Assume that the difference &lt;math&gt;\frac{p}{q} - \frac{5}{9}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{9q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;9p - 5q = 1&lt;/math&gt;<br /> <br /> Also assume that the difference &lt;math&gt;\frac{4}{7} - \frac{p}{q}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{7q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;4q - 7p = 1&lt;/math&gt;<br /> <br /> Solving the system of equations yields &lt;math&gt;q=16&lt;/math&gt; and &lt;math&gt;p=9&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Cross-multiply the inequality to get &lt;cmath&gt;35q &lt; 63p &lt; 36q.&lt;/cmath&gt;<br /> <br /> Then,<br /> &lt;cmath&gt;0 &lt; 63p-35q &lt; q,&lt;/cmath&gt;<br /> &lt;cmath&gt;0 &lt; 7(9p-5q) &lt; q.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; are integers, &lt;math&gt;9p-5q&lt;/math&gt; is an integer. To minimize &lt;math&gt;q&lt;/math&gt;, start from &lt;math&gt;9p-5q=1&lt;/math&gt;, which gives &lt;math&gt;p=\frac{5q+1}{9}&lt;/math&gt;. This limits &lt;math&gt;q&lt;/math&gt; to be greater than &lt;math&gt;7&lt;/math&gt;, so test values of &lt;math&gt;q&lt;/math&gt; starting from &lt;math&gt;q=8&lt;/math&gt;. However, &lt;math&gt;q=8&lt;/math&gt; to &lt;math&gt;q=14&lt;/math&gt; do not give integer values of &lt;math&gt;p&lt;/math&gt;. <br /> <br /> Once &lt;math&gt;q&gt;14&lt;/math&gt;, it is possible for &lt;math&gt;9p-5q&lt;/math&gt; to be equal to &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; could also be equal to &lt;math&gt;\frac{5q+2}{9}.&lt;/math&gt; The next value, &lt;math&gt;q=15&lt;/math&gt;, is not a solution, but &lt;math&gt;q=16&lt;/math&gt; gives &lt;math&gt;p=\frac{5\cdot 16 + 1}{9} = 9&lt;/math&gt;. Thus, the smallest possible value of &lt;math&gt;q&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Graph the regions &lt;math&gt;y &gt; \frac{5}{9}x&lt;/math&gt; and &lt;math&gt;y &lt; \frac{4}{7}x&lt;/math&gt;. Note that the lattice point &lt;math&gt;(16,9)&lt;/math&gt; is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is &lt;math&gt;\frac{9}{16}&lt;/math&gt; and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;. <br /> <br /> Remark: This also gives an intuitive geometric proof of the mediant using vectors.<br /> <br /> ==Solution 5 (Using answer choices to prove mediant)==<br /> As the other solutions do, the mediant &lt;math&gt;=\frac{9}{16}&lt;/math&gt; is between the two fractions, with a difference of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;. Suppose that the answer was not &lt;math&gt;A&lt;/math&gt;, then the answer must be &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; as otherwise &lt;math&gt;p&lt;/math&gt; would be negative. Then, the possible fractions with lower denominator would be &lt;math&gt;\frac{k-11}{k}&lt;/math&gt; for &lt;math&gt;k=12,13,14,15&lt;/math&gt; and &lt;math&gt;\frac{k-13}{k}&lt;/math&gt; for &lt;math&gt;k=14,15,&lt;/math&gt; which are clearly not anywhere close to &lt;math&gt;\frac{4}{7}\approx 0.6&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> Inverting the given inequality we get <br /> &lt;cmath&gt;\frac{7}{4} &lt; \frac{q}{p} &lt; \frac{9}{5}&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;35p &lt; 20q &lt; 36p&lt;/cmath&gt;<br /> <br /> We can now substitute &lt;math&gt;q = p + k&lt;/math&gt;. Note we need to find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;35p &lt; 20p + 20k &lt; 36p&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;15p &lt; 20k &lt; 16p&lt;/cmath&gt;<br /> <br /> Cleary &lt;math&gt;p&lt;/math&gt; is greater than &lt;math&gt;k&lt;/math&gt;. We will now substitute &lt;math&gt;p = k + x&lt;/math&gt; to get<br /> <br /> &lt;cmath&gt;15k + 15x &lt; 20k &lt; 16k + 16x&lt;/cmath&gt;<br /> <br /> The inequality &lt;math&gt;15k + 15x &lt; 20k&lt;/math&gt; simplifies to &lt;math&gt;3x &lt; k&lt;/math&gt;.<br /> The inequality &lt;math&gt;20k &lt; 16k + 16x&lt;/math&gt; simplifies to &lt;math&gt;k &lt; 4x&lt;/math&gt;.<br /> Combining the two we get<br /> &lt;cmath&gt;3x &lt; k &lt; 4x&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are integers, the smallest values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; that satisfy the above equation are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; respectively. Substituting these back in, we arrive with an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Start with &lt;math&gt;\frac{5}{9}&lt;/math&gt;. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to &lt;math&gt;\frac{5}{9}&lt;/math&gt;, add &lt;math&gt;1&lt;/math&gt; to the numerator; otherwise, if it is greater than or equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;, add one &lt;math&gt;1&lt;/math&gt; to the denominator. We have:<br /> <br /> &lt;cmath&gt;\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;16 - 9 = \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Because q and p are positive integers with &lt;math&gt;p&lt;q&lt;/math&gt;, we can let &lt;math&gt;q=p+k&lt;/math&gt; where &lt;math&gt;k\in{\mathbb{Z}}&lt;/math&gt;. Now, the problem condition reduces to<br /> <br /> &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt;<br /> <br /> Our first inequality is &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;/math&gt; which gives us &lt;math&gt;5p+5k&lt;9p\implies \frac{5}{4}k&lt;p&lt;/math&gt;.<br /> <br /> Our second inequality is &lt;math&gt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt; which gives us &lt;math&gt;7p&lt;4p+4k\implies p&lt;\frac{4}{3}k&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;\frac{5}{4}k&lt;p&lt;\frac{4}{3}k\implies 15k&lt;12p&lt;16k&lt;/math&gt;.<br /> <br /> It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.<br /> <br /> Now, simple casework through the answer choices of the problem reveals that &lt;math&gt;q-p=p+k-p=k\implies k\ge{\boxed{7}}&lt;/math&gt;.<br /> <br /> == Solution 9 (Quick inspection) ==<br /> Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.<br /> The interval can also be written as &lt;math&gt;0.5556&lt;x&lt;0.5714&lt;/math&gt;. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range. <br /> <br /> The denominators to be considered are &lt;math&gt;9,10,11,12...&lt;/math&gt;. We check &lt;math&gt;\frac{6}{10}, \frac{6}{11}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}&lt;/math&gt;. At this point we know that we've got our fraction and our answer is &lt;math&gt;16-9=\boxed{\textbf{A } 7}&lt;/math&gt;<br /> <br /> The inspection was made faster by considering the fact that &lt;math&gt;\frac{a+1}{b+1}&gt;\frac{a}{b}&lt;/math&gt;. So, once a fraction was gotten which was greater than &lt;math&gt;\frac{4}{7}&lt;/math&gt; we jump to the next denominator and also make sure we consider fractions with higher positive difference between the denominator and numerator, making sure also that the numerator is greater than half of the denominator.<br /> <br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_17&diff=105198 2018 AMC 12B Problems/Problem 17 2019-04-09T02:37:02Z <p>Olutosinfires: /* Solution 8 */</p> <hr /> <div>== Problem ==<br /> <br /> Let &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; be positive integers such that &lt;cmath&gt;\frac{5}{9} &lt; \frac{p}{q} &lt; \frac{4}{7}&lt;/cmath&gt;and &lt;math&gt;q&lt;/math&gt; is as small as possible. What is &lt;math&gt;q-p&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 7 \qquad \textbf{(B) } 11 \qquad \textbf{(C) } 13 \qquad \textbf{(D) } 17 \qquad \textbf{(E) } 19 &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> We claim that, between any two fractions &lt;math&gt;a/b&lt;/math&gt; and &lt;math&gt;c/d&lt;/math&gt;, if &lt;math&gt;bc-ad=1&lt;/math&gt;, the fraction with smallest denominator between them is &lt;math&gt;\frac{a+c}{b+d}&lt;/math&gt;. To prove this, we see that<br /> <br /> &lt;cmath&gt;\frac{1}{bd}=\frac{c}{d}-\frac{a}{b}=\left(\frac{c}{d}-\frac{p}{q}\right)+\left(\frac{p}{q}-\frac{a}{b}\right) \geq \frac{1}{dq}+\frac{1}{bq},&lt;/cmath&gt;<br /> which reduces to &lt;math&gt;q\geq b+d&lt;/math&gt;. We can easily find that &lt;math&gt;p=a+c&lt;/math&gt;, giving an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 2 (requires justification)==<br /> <br /> Assume that the difference &lt;math&gt;\frac{p}{q} - \frac{5}{9}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{9q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;9p - 5q = 1&lt;/math&gt;<br /> <br /> Also assume that the difference &lt;math&gt;\frac{4}{7} - \frac{p}{q}&lt;/math&gt; results in a fraction of the form &lt;math&gt;\frac{1}{7q}&lt;/math&gt;. Then,<br /> <br /> &lt;math&gt;4q - 7p = 1&lt;/math&gt;<br /> <br /> Solving the system of equations yields &lt;math&gt;q=16&lt;/math&gt; and &lt;math&gt;p=9&lt;/math&gt;. Therefore, the answer is &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> Cross-multiply the inequality to get &lt;cmath&gt;35q &lt; 63p &lt; 36q.&lt;/cmath&gt;<br /> <br /> Then,<br /> &lt;cmath&gt;0 &lt; 63p-35q &lt; q,&lt;/cmath&gt;<br /> &lt;cmath&gt;0 &lt; 7(9p-5q) &lt; q.&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;p&lt;/math&gt;, &lt;math&gt;q&lt;/math&gt; are integers, &lt;math&gt;9p-5q&lt;/math&gt; is an integer. To minimize &lt;math&gt;q&lt;/math&gt;, start from &lt;math&gt;9p-5q=1&lt;/math&gt;, which gives &lt;math&gt;p=\frac{5q+1}{9}&lt;/math&gt;. This limits &lt;math&gt;q&lt;/math&gt; to be greater than &lt;math&gt;7&lt;/math&gt;, so test values of &lt;math&gt;q&lt;/math&gt; starting from &lt;math&gt;q=8&lt;/math&gt;. However, &lt;math&gt;q=8&lt;/math&gt; to &lt;math&gt;q=14&lt;/math&gt; do not give integer values of &lt;math&gt;p&lt;/math&gt;. <br /> <br /> Once &lt;math&gt;q&gt;14&lt;/math&gt;, it is possible for &lt;math&gt;9p-5q&lt;/math&gt; to be equal to &lt;math&gt;2&lt;/math&gt;, so &lt;math&gt;p&lt;/math&gt; could also be equal to &lt;math&gt;\frac{5q+2}{9}.&lt;/math&gt; The next value, &lt;math&gt;q=15&lt;/math&gt;, is not a solution, but &lt;math&gt;q=16&lt;/math&gt; gives &lt;math&gt;p=\frac{5\cdot 16 + 1}{9} = 9&lt;/math&gt;. Thus, the smallest possible value of &lt;math&gt;q&lt;/math&gt; is &lt;math&gt;16&lt;/math&gt;, and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Graph the regions &lt;math&gt;y &gt; \frac{5}{9}x&lt;/math&gt; and &lt;math&gt;y &lt; \frac{4}{7}x&lt;/math&gt;. Note that the lattice point &lt;math&gt;(16,9)&lt;/math&gt; is the smallest magnitude one which appears within the region bounded by the two graphs. Thus, our fraction is &lt;math&gt;\frac{9}{16}&lt;/math&gt; and the answer is &lt;math&gt;16-9= \boxed{\textbf{(A)}\ 7}&lt;/math&gt;. <br /> <br /> Remark: This also gives an intuitive geometric proof of the mediant using vectors.<br /> <br /> ==Solution 5 (Using answer choices to prove mediant)==<br /> As the other solutions do, the mediant &lt;math&gt;=\frac{9}{16}&lt;/math&gt; is between the two fractions, with a difference of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;. Suppose that the answer was not &lt;math&gt;A&lt;/math&gt;, then the answer must be &lt;math&gt;B&lt;/math&gt; or &lt;math&gt;C&lt;/math&gt; as otherwise &lt;math&gt;p&lt;/math&gt; would be negative. Then, the possible fractions with lower denominator would be &lt;math&gt;\frac{k-11}{k}&lt;/math&gt; for &lt;math&gt;k=12,13,14,15&lt;/math&gt; and &lt;math&gt;\frac{k-13}{k}&lt;/math&gt; for &lt;math&gt;k=14,15,&lt;/math&gt; which are clearly not anywhere close to &lt;math&gt;\frac{4}{7}\approx 0.6&lt;/math&gt;<br /> <br /> ==Solution 6==<br /> <br /> Inverting the given inequality we get <br /> &lt;cmath&gt;\frac{7}{4} &lt; \frac{q}{p} &lt; \frac{9}{5}&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;35p &lt; 20q &lt; 36p&lt;/cmath&gt;<br /> <br /> We can now substitute &lt;math&gt;q = p + k&lt;/math&gt;. Note we need to find &lt;math&gt;k&lt;/math&gt;.<br /> <br /> &lt;cmath&gt;35p &lt; 20p + 20k &lt; 36p&lt;/cmath&gt;<br /> <br /> which simplifies to &lt;cmath&gt;15p &lt; 20k &lt; 16p&lt;/cmath&gt;<br /> <br /> Cleary &lt;math&gt;p&lt;/math&gt; is greater than &lt;math&gt;k&lt;/math&gt;. We will now substitute &lt;math&gt;p = k + x&lt;/math&gt; to get<br /> <br /> &lt;cmath&gt;15k + 15x &lt; 20k &lt; 16k + 16x&lt;/cmath&gt;<br /> <br /> The inequality &lt;math&gt;15k + 15x &lt; 20k&lt;/math&gt; simplifies to &lt;math&gt;3x &lt; k&lt;/math&gt;.<br /> The inequality &lt;math&gt;20k &lt; 16k + 16x&lt;/math&gt; simplifies to &lt;math&gt;k &lt; 4x&lt;/math&gt;.<br /> Combining the two we get<br /> &lt;cmath&gt;3x &lt; k &lt; 4x&lt;/cmath&gt;<br /> <br /> Since &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; are integers, the smallest values of &lt;math&gt;x&lt;/math&gt; and &lt;math&gt;k&lt;/math&gt; that satisfy the above equation are &lt;math&gt;2&lt;/math&gt; and &lt;math&gt;7&lt;/math&gt; respectively. Substituting these back in, we arrive with an answer of &lt;math&gt;\boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 7==<br /> Start with &lt;math&gt;\frac{5}{9}&lt;/math&gt;. Repeat the following process until you arrive at the answer: if the fraction is less than or equal to &lt;math&gt;\frac{5}{9}&lt;/math&gt;, add &lt;math&gt;1&lt;/math&gt; to the numerator; otherwise, if it is greater than or equal to &lt;math&gt;\frac{4}{7}&lt;/math&gt;, add one &lt;math&gt;1&lt;/math&gt; to the denominator. We have:<br /> <br /> &lt;cmath&gt;\frac{5}{9}, \frac{6}{9}, \frac{6}{10}, \frac{6}{11}, \frac{7}{11}, \frac{7}{12}, \frac{7}{13}, \frac{8}{13}, \frac{8}{14}, \frac{8}{15}, \frac{9}{15}, \frac{9}{16}&lt;/cmath&gt;<br /> <br /> &lt;math&gt;16 - 9 = \boxed{\textbf{(A)}\ 7}&lt;/math&gt;.<br /> <br /> ==Solution 8==<br /> Because q and p are positive integers with &lt;math&gt;p&lt;q&lt;/math&gt;, we can let &lt;math&gt;q=p+k&lt;/math&gt; where &lt;math&gt;k\in{\mathbb{Z}}&lt;/math&gt;. Now, the problem condition reduces to<br /> <br /> &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt;<br /> <br /> Our first inequality is &lt;math&gt;\frac{5}{9}&lt;\frac{p}{p+k}&lt;/math&gt; which gives us &lt;math&gt;5p+5k&lt;9p\implies \frac{5}{4}k&lt;p&lt;/math&gt;.<br /> <br /> Our second inequality is &lt;math&gt;\frac{p}{p+k}&lt;\frac{4}{7}&lt;/math&gt; which gives us &lt;math&gt;7p&lt;4p+4k\implies p&lt;\frac{4}{3}k&lt;/math&gt;.<br /> <br /> Hence, &lt;math&gt;\frac{5}{4}k&lt;p&lt;\frac{4}{3}k\implies 15k&lt;12p&lt;16k&lt;/math&gt;.<br /> <br /> It is clear that we are aiming to find the least positive integer value of k such that there is at least one value of p that satisfies the inequality.<br /> <br /> Now, simple casework through the answer choices of the problem reveals that &lt;math&gt;q-p=p+k-p=k\implies k\ge{\boxed{7}}&lt;/math&gt;.<br /> <br /> == Solution 9 ==<br /> Checking possible fractions within the interval can get us to the answer, but only if we do it with more skill.<br /> The interval can also be written as &lt;math&gt;0.5556&lt;x&lt;0.5714&lt;/math&gt;. This represents fraction with the numerator a little bit more than half the denominator. Every fraction we consider must not exceed this range. <br /> <br /> The denominators to be considered are &lt;math&gt;9,10,11,12...&lt;/math&gt;. We check &lt;math&gt;\frac{6}{10}, \frac{6}{11}, \frac{7}{13}, \frac{8}{15}, \frac{9}{16}&lt;/math&gt;. At this point we know that we've got our fraction and our answer is &lt;math&gt;16-9=\boxed{textbf{A } 7}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=16|num-a=18}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_16&diff=105197 2018 AMC 12B Problems/Problem 16 2019-04-09T02:05:50Z <p>Olutosinfires: /* Solution */</p> <hr /> <div>== Problem ==<br /> <br /> The solutions to the equation &lt;math&gt;(z+6)^8=81&lt;/math&gt; are connected in the complex plane to form a convex regular polygon, three of whose vertices are labeled &lt;math&gt;A,B,&lt;/math&gt; and &lt;math&gt;C&lt;/math&gt;. What is the least possible area of &lt;math&gt;\triangle ABC?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } \frac{1}{6}\sqrt{6} \qquad \textbf{(B) } \frac{3}{2}\sqrt{2}-\frac{3}{2} \qquad \textbf{(C) } 2\sqrt3-3\sqrt2 \qquad \textbf{(D) } \frac{1}{2}\sqrt{2} \qquad \textbf{(E) } \sqrt 3-1&lt;/math&gt;<br /> <br /> == Solution ==<br /> <br /> The answer is the same if we consider &lt;math&gt;z^8=81.&lt;/math&gt; Now we just need to find the area of the triangle bounded by &lt;math&gt;\sqrt 3i, \sqrt 3,&lt;/math&gt; and &lt;math&gt;\frac{\sqrt 3}{\sqrt 2}+\frac{\sqrt 3}{\sqrt 2}i.&lt;/math&gt; This is just &lt;math&gt;\boxed{\textbf{B.}}&lt;/math&gt;<br /> <br /> == Solution 2 (Understanding the polygon) ==<br /> <br /> The polygon formed will be a regular octagon since there are &lt;math&gt;8&lt;/math&gt; roots of &lt;math&gt;z^8=81&lt;/math&gt;. By normal math computation, we can figure out that two roots of &lt;math&gt;z^8=81&lt;/math&gt; are &lt;math&gt;\sqrt{3}&lt;/math&gt; and &lt;math&gt;-\sqrt{3}&lt;/math&gt;. These will lie on the real axis of the plane. Since it's a regular polygon, there has to be points on the vertical plane also which will be &lt;math&gt;\sqrt{3}i&lt;/math&gt; and &lt;math&gt;-\sqrt{3}i&lt;/math&gt;.<br /> <br /> Clearly, the rest of the points will lie in each quadrant. The next thing is to get their coordinates (note that to answer this question, we do not need all the coordinates, only 3 consecutive ones are needed).<br /> <br /> The circumcircle of the octagon will have the equation &lt;math&gt;i^2+r^2=3&lt;/math&gt;. The coordinates of the point in the first quadrant will be equal in magnitude and both positive, so &lt;math&gt;i=r&lt;/math&gt;. Solving gives &lt;math&gt;i=r=\frac{\sqrt{3}}{\sqrt{2}}&lt;/math&gt; (meaning that the root represented is &lt;math&gt;\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i&lt;/math&gt;).<br /> <br /> This way we can deduce the values of the &lt;math&gt;8&lt;/math&gt; roots of the equation to be &lt;math&gt;\sqrt{3},-\sqrt{3},-\sqrt{3}i,\sqrt{3}i,\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,\frac{\sqrt{3}}{\sqrt{2}}-\frac{\sqrt{3}}{\sqrt{2}}i,-\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i&lt;/math&gt;.<br /> <br /> To get the area, &lt;math&gt;3&lt;/math&gt; consecutive points such as &lt;math&gt;\sqrt{3},&lt;/math&gt; &lt;math&gt;\frac{\sqrt{3}}{\sqrt{2}}+\frac{\sqrt{3}}{\sqrt{2}}i,&lt;/math&gt; and &lt;math&gt;\sqrt{3}i&lt;/math&gt; can be used. The area can be computed using different methods like using the shoelace formula, or subtracting areas to find the area.<br /> The answer you get is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> (This method is not actually as long as it seems if you understand what you're doing while doing it. Also calculations can be made a little easier by solving using &lt;math&gt;x^8=1&lt;/math&gt; and multiplying your answer by &lt;math&gt;\sqrt{3}&lt;/math&gt;).<br /> <br /> <br /> ~OlutosinNGA<br /> <br /> == See Also ==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]<br /> [[Category:Intermediate Geometry Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=105194 2018 AMC 12B Problems/Problem 15 2019-04-09T00:34:06Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br /> <br /> &lt;math&gt;\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 &lt;/math&gt;<br /> <br /> == Solution 1 (For Dummies) ==<br /> Analyze that the three-digit integers divisible by &lt;math&gt;3&lt;/math&gt; start from &lt;math&gt;102&lt;/math&gt;. In the &lt;math&gt;200&lt;/math&gt;'s, it starts from &lt;math&gt;201&lt;/math&gt;. In the &lt;math&gt;300&lt;/math&gt;'s, it starts from &lt;math&gt;300&lt;/math&gt;. We see that the units digits is &lt;math&gt;0, 1, &lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Write out the 1- and 2-digit multiples of &lt;math&gt;3&lt;/math&gt; starting from &lt;math&gt;0, 1,&lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt; Count up the ones that meet the conditions. Then, add up and multiply by &lt;math&gt;3&lt;/math&gt;, since there are three sets of three from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9.&lt;/math&gt; Then, subtract the amount that started from &lt;math&gt;0&lt;/math&gt;, since the &lt;math&gt;300&lt;/math&gt;'s ll contain the digit &lt;math&gt;3&lt;/math&gt;.<br /> <br /> We get: &lt;cmath&gt;3(12+12+12)-12.&lt;/cmath&gt;<br /> <br /> This gives us: &lt;cmath&gt;\boxed{\textbf{(A) } 96}.&lt;/cmath&gt;<br /> <br /> == Solution 2==<br /> <br /> There are &lt;math&gt;4&lt;/math&gt; choices for the last digit (&lt;math&gt;1, 5, 7, 9&lt;/math&gt;), and &lt;math&gt;8&lt;/math&gt; choices for the first digit (exclude &lt;math&gt;0&lt;/math&gt;). We know what the second digit mod &lt;math&gt;3&lt;/math&gt; is, so there are &lt;math&gt;3&lt;/math&gt; choices for it (pick from one of the sets &lt;math&gt;\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}&lt;/math&gt;). The answer is &lt;math&gt;4\cdot 8 \cdot 3 = \boxed{96}&lt;/math&gt; (Plasma_Vortex)<br /> <br /> == Solution 3==<br /> <br /> Consider the number of &lt;math&gt;2&lt;/math&gt;-digit numbers that do not contain the digit &lt;math&gt;3&lt;/math&gt;, which is &lt;math&gt;90-18=72&lt;/math&gt;. For any of these &lt;math&gt;2&lt;/math&gt;-digit numbers, we can append &lt;math&gt;1,5,7,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; to reach a desirable &lt;math&gt;3&lt;/math&gt;-digit number. However, &lt;math&gt;1 \equiv 7 \equiv 1&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt;, and thus we need to count any &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\equiv 2&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt; twice. There are &lt;math&gt;(98-11)/3+1=30&lt;/math&gt; total such numbers that have remainder &lt;math&gt;2&lt;/math&gt;, but &lt;math&gt;6&lt;/math&gt; of them &lt;math&gt;(23,32,35,38,53,83)&lt;/math&gt; contain &lt;math&gt;3&lt;/math&gt;, so the number we want is &lt;math&gt;30-6=24&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;72+24= \boxed{96}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that this isn't a great solution, but a more practical one to achieve the answer.<br /> <br /> Notice that there are &lt;math&gt;300&lt;/math&gt; numbers that have &lt;math&gt;3&lt;/math&gt; digits and are divisible by &lt;math&gt;3&lt;/math&gt; (from &lt;math&gt;102&lt;/math&gt; to &lt;math&gt;999&lt;/math&gt;). Now one by one apply the restrictions. <br /> <br /> The restriction for only odd numbers would mean that half the numbers are taken out &lt;math&gt;\Rightarrow 300*\frac{1}{2} = 150&lt;/math&gt;. <br /> <br /> Next, apply the restriction of no &lt;math&gt;3&lt;/math&gt;s. For the units digit, that would mean multiplying by &lt;math&gt;\frac{4}{5}&lt;/math&gt; (remember that now you only have odd numbers to choose from). <br /> <br /> For the tens that would mean multiplying by &lt;math&gt;\frac{9}{10}&lt;/math&gt;, and for the hundreds that would mean multiplying by &lt;math&gt;\frac{8}{9}&lt;/math&gt; (because you cant have 0 here). <br /> <br /> Thus, we get &lt;math&gt;150*\frac{4}{5}*\frac{9}{10}*\frac{8}{9}=96&lt;/math&gt;, which is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> Sol by IronicNinja~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> [[Category:Introductory Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12B_Problems/Problem_15&diff=105193 2018 AMC 12B Problems/Problem 15 2019-04-09T00:32:41Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> How many odd positive 3-digit integers are divisible by 3 but do not contain the digit 3?<br /> <br /> &lt;math&gt;\textbf{(A) } 96 \qquad \textbf{(B) } 97 \qquad \textbf{(C) } 98 \qquad \textbf{(D) } 102 \qquad \textbf{(E) } 120 &lt;/math&gt;<br /> <br /> == Solution 1 (For Dummies) ==<br /> Analyze that the three-digit integers divisible by &lt;math&gt;3&lt;/math&gt; start from &lt;math&gt;102&lt;/math&gt;. In the &lt;math&gt;200&lt;/math&gt;'s, it starts from &lt;math&gt;201&lt;/math&gt;. In the &lt;math&gt;300&lt;/math&gt;'s, it starts from &lt;math&gt;300&lt;/math&gt;. We see that the units digits is &lt;math&gt;0, 1, &lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt;<br /> <br /> Write out the 1- and 2-digit multiples of &lt;math&gt;3&lt;/math&gt; starting from &lt;math&gt;0, 1,&lt;/math&gt; and &lt;math&gt;2.&lt;/math&gt; Count up the ones that meet the conditions. Then, add up and multiply by &lt;math&gt;3&lt;/math&gt;, since there are three sets of three from &lt;math&gt;1&lt;/math&gt; to &lt;math&gt;9.&lt;/math&gt; Then, subtract the amount that started from &lt;math&gt;0&lt;/math&gt;, since the &lt;math&gt;300&lt;/math&gt;'s ll contain the digit &lt;math&gt;3&lt;/math&gt;.<br /> <br /> We get: &lt;cmath&gt;3(12+12+12)-12.&lt;/cmath&gt;<br /> <br /> This gives us: &lt;cmath&gt;\boxed{\textbf{(A) } 96}.&lt;/cmath&gt;<br /> <br /> == Solution 2==<br /> <br /> There are &lt;math&gt;4&lt;/math&gt; choices for the last digit (&lt;math&gt;1, 5, 7, 9&lt;/math&gt;), and &lt;math&gt;8&lt;/math&gt; choices for the first digit (exclude &lt;math&gt;0&lt;/math&gt;). We know what the second digit mod &lt;math&gt;3&lt;/math&gt; is, so there are &lt;math&gt;3&lt;/math&gt; choices for it (pick from one of the sets &lt;math&gt;\{0, 6, 9\},\{1, 4, 7\}, \{2, 5, 8\}&lt;/math&gt;). The answer is &lt;math&gt;4\cdot 8 \cdot 3 = \boxed{96}&lt;/math&gt; (Plasma_Vortex)<br /> <br /> == Solution 3==<br /> <br /> Consider the number of &lt;math&gt;2&lt;/math&gt;-digit numbers that do not contain the digit &lt;math&gt;3&lt;/math&gt;, which is &lt;math&gt;90-18=72&lt;/math&gt;. For any of these &lt;math&gt;2&lt;/math&gt;-digit numbers, we can append &lt;math&gt;1,5,7,&lt;/math&gt; or &lt;math&gt;9&lt;/math&gt; to reach a desirable &lt;math&gt;3&lt;/math&gt;-digit number. However, &lt;math&gt;1 \equiv 7 \equiv 1&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt;, and thus we need to count any &lt;math&gt;2&lt;/math&gt;-digit number &lt;math&gt;\equiv 2&lt;/math&gt; &lt;math&gt;(mod&lt;/math&gt; &lt;math&gt;3)&lt;/math&gt; twice. There are &lt;math&gt;(98-11)/3+1=30&lt;/math&gt; total such numbers that have remainder &lt;math&gt;2&lt;/math&gt;, but &lt;math&gt;6&lt;/math&gt; of them &lt;math&gt;(23,32,35,38,53,83)&lt;/math&gt; contain &lt;math&gt;3&lt;/math&gt;, so the number we want is &lt;math&gt;30-6=24&lt;/math&gt;. Therefore, the final answer is &lt;math&gt;72+24= \boxed{96}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that this isn't a great solution, but a more practical one to achieve the answer.<br /> <br /> Notice that there are &lt;math&gt;300&lt;/math&gt; numbers that have &lt;math&gt;3&lt;/math&gt; digits and are divisible by &lt;math&gt;3&lt;/math&gt; (from &lt;math&gt;102&lt;/math&gt; to &lt;math&gt;999&lt;/math&gt;). Now one by one apply the restrictions. <br /> <br /> The restriction for only odd numbers would mean that half the numbers are taken out &lt;math&gt;\Rightarrow 300*\frac{1}{2} = 150&lt;/math&gt;. <br /> <br /> Next, apply the restriction of no &lt;math&gt;3&lt;/math&gt;s. For the units digit, that would mean multiplying by &lt;math&gt;\frac{4}{5}&lt;/math&gt; (remember that now you only have odd numbers to choose from). <br /> <br /> For the tens that would mean multiplying by &lt;math&gt;\frac{9}{10}&lt;/math&gt;, and for the hundreds that would mean multiplying by &lt;math&gt;\frac{8}{9}&lt;/math&gt; (because you cant have 0 here). <br /> <br /> Thus, we get &lt;math&gt;150*\frac{4}{5}=120*\frac{9}{10}=108*\frac{8}{9}=96&lt;/math&gt;, which is &lt;math&gt;\boxed{A}&lt;/math&gt;.<br /> <br /> Sol by IronicNinja~<br /> <br /> ==See Also==<br /> <br /> {{AMC12 box|year=2018|ab=B|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Combinatorics Problems]]<br /> [[Category:Introductory Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_13&diff=105168 2018 AMC 10B Problems/Problem 13 2019-04-07T19:33:34Z <p>Olutosinfires: /* Solution 3 */</p> <hr /> <div>==Problem==<br /> How many of the first &lt;math&gt;2018&lt;/math&gt; numbers in the sequence &lt;math&gt;101, 1001, 10001, 100001, \dots&lt;/math&gt; are divisible by &lt;math&gt;101&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }253 \qquad<br /> \textbf{(B) }504 \qquad<br /> \textbf{(C) }505 \qquad<br /> \textbf{(D) }506 \qquad<br /> \textbf{(E) }1009 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> The number &lt;math&gt;10^n+1&lt;/math&gt; is divisible by 101 if and only if &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt;. We note that &lt;math&gt;(10,10^2,10^3,10^4)\equiv (10,-1,-10,1)\pmod{101}&lt;/math&gt;, so the powers of 10 are 4-periodic mod 101. It follows that &lt;math&gt;10^n\equiv -1\pmod{101}&lt;/math&gt; if and only if &lt;math&gt;n\equiv 2\pmod 4&lt;/math&gt;.<br /> <br /> In the given list, &lt;math&gt;10^2+1,10^3+1,10^4+1,\dots,10^{2019}+1&lt;/math&gt;, the desired exponents are &lt;math&gt;2,6,10,\dots,2018&lt;/math&gt;, and there are &lt;math&gt;2020/4=\boxed{\textbf{(C) } 505}&lt;/math&gt; numbers in that list.<br /> <br /> ==Solution 2==<br /> Note that &lt;math&gt;10^{2k}+1&lt;/math&gt; for some odd &lt;math&gt;k&lt;/math&gt; will suffice &lt;math&gt;\mod {101}&lt;/math&gt;. Each &lt;math&gt;2k \in \{2,6,10,\dots,2018\}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;<br /> (AOPS12142015)<br /> <br /> ==Solution 3==<br /> If we divide each number by &lt;math&gt;101&lt;/math&gt;, we see a pattern occuring in every 4 numbers. &lt;math&gt;101, 1000001, 10000000001, \dots&lt;/math&gt;. We divide &lt;math&gt;2018&lt;/math&gt; by &lt;math&gt;4&lt;/math&gt; to get &lt;math&gt;504&lt;/math&gt; with &lt;math&gt;2&lt;/math&gt; left over. Looking at our pattern of four numbers from above, the first number is divisible by &lt;math&gt;101&lt;/math&gt;. This means that the first of the &lt;math&gt;2&lt;/math&gt; left over will be divisible by &lt;math&gt;101&lt;/math&gt;, so our answer is &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> <br /> ==Solution 4==<br /> Note that &lt;math&gt;909&lt;/math&gt; is divisible by &lt;math&gt;101&lt;/math&gt;, and thus &lt;math&gt;9999&lt;/math&gt; is too. We know that &lt;math&gt;101&lt;/math&gt; is divisible and &lt;math&gt;1001&lt;/math&gt; isn't so let us start from &lt;math&gt;10001&lt;/math&gt;. We subtract &lt;math&gt;9999&lt;/math&gt; to get 2. Likewise from &lt;math&gt;100001&lt;/math&gt; we subtract, but we instead subtract &lt;math&gt;9999&lt;/math&gt; times &lt;math&gt;10&lt;/math&gt; or &lt;math&gt;99990&lt;/math&gt; to get &lt;math&gt;11&lt;/math&gt;. We do it again and multiply the 9's by &lt;math&gt;10&lt;/math&gt; to get &lt;math&gt;101&lt;/math&gt;. Following the same knowledge, we can use mod &lt;math&gt;101&lt;/math&gt; to finish the problem. The sequence will just be subtracting 1, multiplying by 10, then adding 1. Thus the sequence is &lt;math&gt;{0,-9,-99 ( 2),11, 0, ...}&lt;/math&gt;. Thus it repeats every four. Consider the sequence after the 1st term and we have 2017 numbers. Divide &lt;math&gt;2017&lt;/math&gt; by four to get &lt;math&gt;504&lt;/math&gt; remainder &lt;math&gt;1&lt;/math&gt;. Thus the answer is &lt;math&gt;504&lt;/math&gt; plus the 1st term or &lt;math&gt;\boxed{\textbf{(C) } 505}&lt;/math&gt;.<br /> <br /> -googleghosh<br /> <br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=12|num-a=14}}<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105111 2018 AMC 10B Problems/Problem 10 2019-04-03T15:51:15Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> <br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt; Area of &lt;math&gt;BCHE = BC * BE = \sqrt{13}&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 *&lt;/math&gt; Area of &lt;math&gt;FBE / BE&lt;/math&gt; (since Area &lt;math&gt;= \frac{1}{2}bh&lt;/math&gt;).<br /> <br /> Area of &lt;math&gt;FBE = \frac{1}{2} * FB * FE = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}&lt;/math&gt;<br /> <br /> Volume of pyramid &lt;math&gt;=\frac{1}{3} * \sqrt{13} * \frac{6}{\sqrt{13}} = 2&lt;/math&gt;<br /> <br /> Answer is &lt;math&gt;\boxed{\textbf{E } 2}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105110 2018 AMC 10B Problems/Problem 10 2019-04-03T15:49:38Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> <br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt; Area of &lt;math&gt;BCHE = BC * BE = \sqrt{13}&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * Area of FBE / BE&lt;/math&gt; (since &lt;math&gt;Area = \frac{1}{2}bh&lt;/math&gt;).<br /> <br /> &lt;math&gt;Area of FBE = \frac{1}{2} * FB * FE = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}&lt;/math&gt;<br /> <br /> Volume of pyramid &lt;math&gt;=\frac{1}{3} * \sqrt{13} * \frac{6}{\sqrt{13}} = 2&lt;/math&gt;<br /> <br /> Answer is &lt;math&gt;\boxed{\textbf{E } 2}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105109 2018 AMC 10B Problems/Problem 10 2019-04-03T15:48:34Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> <br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt; Area of &lt;math&gt;BCHE = BC * BE = \sqrt{13}&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * Area of FBE / BE&lt;/math&gt; (since &lt;math&gt;Area = \frac{1}{2}bh&lt;/math&gt;).<br /> <br /> &lt;math&gt;Area of FBE = \frac{1}{2} * FB * FE = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}&lt;/math&gt;<br /> <br /> Volume of pyramid &lt;math&gt;=\frac{1}{3} * \sqrt{13} * \frac{6}{\sqrt{13}} = 2&lt;/math&gt;<br /> <br /> Answer is &lt;math&gt;\boxed{\textbf{E } 2}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105108 2018 AMC 10B Problems/Problem 10 2019-04-03T15:37:57Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> <br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt; Area of &lt;math&gt;BCHE = BC * BE = \sqrt{13}&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * Area of FBE / BE&lt;/math&gt; (since &lt;math&gt;Area = \frac{1}{2}bh&lt;/math&gt;).<br /> <br /> &lt;math&gt;Area of FBE = \frac{1}{2} * FB * FE = 3&lt;/math&gt;<br /> <br /> &lt;math&gt;h = 2 * 3 / \sqrt{13} = \frac{6}{\sqrt{13}}&lt;/math&gt;<br /> <br /> Volume of pyramid &lt;math&gt;=\frac{\sqrt{13} * \frac{6}{\sqrt{13}}{3} = 2&lt;/math&gt;<br /> <br /> Answer is &lt;math&gt;\boxed{E 2}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105107 2018 AMC 10B Problems/Problem 10 2019-04-03T15:19:27Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> <br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105106 2018 AMC 10B Problems/Problem 10 2019-04-03T15:18:53Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; drawn from &lt;math&gt;F&lt;/math&gt; to &lt;math&gt;BE&lt;/math&gt;.<br /> &lt;math&gt;BE=\sqrt{BF^2 + EF^2}=\sqrt{13}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105105 2018 AMC 10B Problems/Problem 10 2019-04-03T15:17:59Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;FBE&lt;/math&gt; .<br /> $BE=\sqrt{BF^2 + EF^2}=\sqrt{13}<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105104 2018 AMC 10B Problems/Problem 10 2019-04-03T15:13:25Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, &lt;math&gt;BCHE&lt;/math&gt;, and is equal to &lt;math&gt;BC * BE&lt;/math&gt;. The height, &lt;math&gt;h&lt;/math&gt;, is equal to the height of triangle &lt;math&gt;&lt;/math&gt;.<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_10&diff=105103 2018 AMC 10B Problems/Problem 10 2019-04-03T15:11:22Z <p>Olutosinfires: /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> In the rectangular parallelepiped shown, &lt;math&gt;AB&lt;/math&gt; = &lt;math&gt;3&lt;/math&gt;, &lt;math&gt;BC&lt;/math&gt; = &lt;math&gt;1&lt;/math&gt;, and &lt;math&gt;CG&lt;/math&gt; = &lt;math&gt;2&lt;/math&gt;. Point &lt;math&gt;M&lt;/math&gt; is the midpoint of &lt;math&gt;\overline{FG}&lt;/math&gt;. What is the volume of the rectangular pyramid with base &lt;math&gt;BCHE&lt;/math&gt; and apex &lt;math&gt;M&lt;/math&gt;?<br /> <br /> <br /> &lt;asy&gt;<br /> size(250);<br /> defaultpen(fontsize(10pt));<br /> pair A =origin;<br /> pair B = (4.75,0);<br /> pair E1=(0,3);<br /> pair F = (4.75,3);<br /> pair G = (5.95,4.2);<br /> pair C = (5.95,1.2);<br /> pair D = (1.2,1.2);<br /> pair H= (1.2,4.2);<br /> pair M = ((4.75+5.95)/2,3.6);<br /> draw(E1--M--H--E1--A--B--E1--F--B--M--C--G--H);<br /> draw(B--C);<br /> draw(F--G);<br /> draw(A--D--H--C--D,dashed);<br /> label(&quot;$A$&quot;,A,SW);<br /> label(&quot;$B$&quot;,B,SE);<br /> label(&quot;$C$&quot;,C,E);<br /> label(&quot;$D$&quot;,D,W);<br /> label(&quot;$E$&quot;,E1,W);<br /> label(&quot;$F$&quot;,F,SW);<br /> label(&quot;$G$&quot;,G,NE);<br /> label(&quot;$H$&quot;,H,NW);<br /> label(&quot;$M$&quot;,M,N);<br /> dot(A);<br /> dot(B);<br /> dot(E1);<br /> dot(F);<br /> dot(G);<br /> dot(C);<br /> dot(D);<br /> dot(H);<br /> dot(M);<br /> label(&quot;3&quot;,A/2+B/2,S);<br /> label(&quot;2&quot;,C/2+G/2,E);<br /> label(&quot;1&quot;,C/2+B/2,SE);<br /> &lt;/asy&gt;<br /> <br /> &lt;math&gt;\textbf{(A) } 1 \qquad \textbf{(B) } \frac{4}{3} \qquad \textbf{(C) } \frac{3}{2} \qquad \textbf{(D) } \frac{5}{3} \qquad \textbf{(E) } 2&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> Consider the cross-sectional plane and label its area &lt;math&gt;b&lt;/math&gt;. Note that the volume of the triangular prism that encloses the pyramid is &lt;math&gt;\frac{bh}{2}=3&lt;/math&gt;, and we want the rectangular pyramid that shares the base and height with the triangular prism. The volume of the pyramid is &lt;math&gt;\frac{bh}{3}&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{2}&lt;/math&gt;. (AOPS12142015)<br /> <br /> ==Solution 2==<br /> We can start by finding the total volume of the parallelepiped. It is &lt;math&gt;2 \cdot 3 \cdot 1 = 6&lt;/math&gt;, because a rectangular parallelepiped is a rectangular prism.<br /> <br /> Next, we can consider the wedge-shaped section made when the plane &lt;math&gt;BCHE&lt;/math&gt; cuts the figure. We can find the volume of the triangular pyramid with base EFB and apex M. The area of EFB is &lt;math&gt;\frac{1}{2} \cdot 2 \cdot 3 = 3&lt;/math&gt;. Since BC is given to be &lt;math&gt;1&lt;/math&gt;, we have that FM is &lt;math&gt;\frac{1}{2}&lt;/math&gt;. Using the formula for the volume of a triangular pyramid, we have &lt;math&gt;V = \frac{1}{3} \cdot \frac{1}{2} \cdot 3 = \frac{1}{2}&lt;/math&gt;. Also, since the triangular pyramid with base HGC and apex M has the exact same dimensions, it has volume &lt;math&gt;\frac{1}{2}&lt;/math&gt; as well.<br /> <br /> The original wedge we considered in the last step has volume &lt;math&gt;3&lt;/math&gt;, because it is half of the volume of the parallelepiped. We can subtract out the parts we found to have &lt;math&gt;3 - \frac{1}{2} \cdot 2 = 2&lt;/math&gt;. Thus, the volume of the figure we are trying to find is &lt;math&gt;2&lt;/math&gt;. This means that the correct answer choice is &lt;math&gt;\boxed{E}&lt;/math&gt;.<br /> <br /> Written by: Archimedes15<br /> <br /> NOTE: For those who think that it isn't a rectangular prism, please read the problem. It says &quot;rectangular parallelepiped.&quot; If a parallelepiped is such that all of the faces are rectangles, it is a rectangular prism.<br /> <br /> ==Solution 3==<br /> If you look carefully, you will see that on the either side of the pyramid in question, there are two congruent tetrahedra. The volume of one is &lt;math&gt;\frac{1}{3}Bh&lt;/math&gt;, with its base being half of one of the rectangular prism's faces and its height being half of one of the edges, so its volume is &lt;math&gt;\frac{1}{3} (3 \times 2/2 \times \frac{1}{2})=\frac{1}{2}&lt;/math&gt;. We can obtain the answer by subtracting twice this value from the diagonal half prism, or<br /> &lt;math&gt;(\frac{1}{2} \times 3 \times 2 \times 1) - (2 \times \frac{1}{2})= &lt;/math&gt; &lt;math&gt;\boxed{2}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> You can calculate the volume of the rectangular pyramid by using the formula, &lt;math&gt;\frac{Ah}{3}&lt;/math&gt;. &lt;math&gt;A&lt;/math&gt; is the area of the base, , and<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=9|num-a=11}}<br /> <br /> [[Category:Introductory Geometry Problems]]<br /> [[Category:3D Geometry Problems]]<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10B_Problems/Problem_4&diff=101986 2018 AMC 10B Problems/Problem 4 2019-02-13T07:32:07Z <p>Olutosinfires: /* Solution 2 */ and /* Solution 3 */</p> <hr /> <div>== Problem ==<br /> <br /> A three-dimensional rectangular box with dimensions &lt;math&gt;X&lt;/math&gt;, &lt;math&gt;Y&lt;/math&gt;, and &lt;math&gt;Z&lt;/math&gt; has faces whose surface areas are &lt;math&gt;24&lt;/math&gt;, &lt;math&gt;24&lt;/math&gt;, &lt;math&gt;48&lt;/math&gt;, &lt;math&gt;48&lt;/math&gt;, &lt;math&gt;72&lt;/math&gt;, and &lt;math&gt;72&lt;/math&gt; square units. What is &lt;math&gt;X&lt;/math&gt; + &lt;math&gt;Y&lt;/math&gt; + &lt;math&gt;Z&lt;/math&gt;?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }18 \qquad<br /> \textbf{(B) }22 \qquad<br /> \textbf{(C) }24 \qquad<br /> \textbf{(D) }30 \qquad<br /> \textbf{(E) }36 \qquad<br /> &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Let &lt;math&gt;X&lt;/math&gt; be the length of the shortest dimension and &lt;math&gt;Z&lt;/math&gt; be the length of the longest dimension. Thus, &lt;math&gt;XY = 24&lt;/math&gt;, &lt;math&gt;YZ = 72&lt;/math&gt;, and &lt;math&gt;XZ = 48&lt;/math&gt;.<br /> Divide the first two equations to get &lt;math&gt;\frac{Z}{X} = 3&lt;/math&gt;. Then, multiply by the last equation to get &lt;math&gt;Z^2 = 144&lt;/math&gt;, giving &lt;math&gt;Z = 12&lt;/math&gt;. Following, &lt;math&gt;X = 4&lt;/math&gt; and &lt;math&gt;Y = 6&lt;/math&gt;.<br /> <br /> The final answer is &lt;math&gt;4 + 6 + 12 = 22&lt;/math&gt;. &lt;math&gt;\boxed{B}&lt;/math&gt;<br /> <br /> ==Solution 2==<br /> If you find the GCD of &lt;math&gt;24&lt;/math&gt;, &lt;math&gt;48&lt;/math&gt;, and &lt;math&gt;72&lt;/math&gt; you get your first number, &lt;math&gt;12&lt;/math&gt;. After this, do &lt;math&gt;48 \div 12&lt;/math&gt; and &lt;math&gt;72 \div 12&lt;/math&gt; to get &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;6&lt;/math&gt;, the other 2 numbers. When you add up your &lt;math&gt;3&lt;/math&gt; numbers, you get &lt;math&gt;22&lt;/math&gt; which is &lt;math&gt;\boxed{B}&lt;/math&gt;.<br /> <br /> ==Solution 3==<br /> Without a loss of generality, &lt;math&gt;XY=24&lt;/math&gt;, &lt;math&gt;XZ=48&lt;/math&gt;, &lt;math&gt;YZ=72&lt;/math&gt;.<br /> Multiplying the three gives &lt;math&gt;(XYZ)^2=2^{10}\cdot3^4&lt;/math&gt; &lt;math&gt;\Rightarrow&lt;/math&gt; &lt;math&gt;XYZ=2^5\cdot3^2&lt;/math&gt;. &lt;math&gt;\frac{XYZ}{XY}=12&lt;/math&gt;<br /> &lt;math&gt;\frac{XYZ}{XZ}=6&lt;/math&gt; and &lt;math&gt;\frac{XYZ}{YZ}=4&lt;/math&gt;<br /> <br /> &lt;math&gt;\therefore X+Y+Z=4+6+12=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{B } 22}&lt;/math&gt;<br /> <br /> ==See Also==<br /> <br /> {{AMC10 box|year=2018|ab=B|num-b=3|num-a=5}}<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_21&diff=100839 2018 AMC 10A Problems/Problem 21 2019-01-25T10:56:04Z <p>Olutosinfires: /* Solution 6 (Calculus Needed) */</p> <hr /> <div>== Problem ==<br /> <br /> Which of the following describes the set of values of &lt;math&gt;a&lt;/math&gt; for which the curves &lt;math&gt;x^2+y^2=a^2&lt;/math&gt; and &lt;math&gt;y=x^2-a&lt;/math&gt; in the real &lt;math&gt;xy&lt;/math&gt;-plane intersect at exactly &lt;math&gt;3&lt;/math&gt; points?<br /> <br /> &lt;math&gt;<br /> \textbf{(A) }a=\frac14 \qquad<br /> \textbf{(B) }\frac14 &lt; a &lt; \frac12 \qquad<br /> \textbf{(C) }a&gt;\frac14 \qquad<br /> \textbf{(D) }a=\frac12 \qquad<br /> \textbf{(E) }a&gt;\frac12 \qquad<br /> &lt;/math&gt;<br /> <br /> == Solution 1 ==<br /> <br /> Substituting &lt;math&gt;y=x^2-a&lt;/math&gt; into &lt;math&gt;x^2+y^2=a^2&lt;/math&gt;, we get<br /> &lt;cmath&gt;<br /> x^2+(x^2-a)^2=a^2 \implies x^2+x^4-2ax^2=0 \implies x^2(x^2-(2a-1))=0<br /> &lt;/cmath&gt;<br /> Since this is a quartic, there are 4 total roots (counting multiplicity). We see that &lt;math&gt;x=0&lt;/math&gt; always at least one intersection at &lt;math&gt;(0,-a)&lt;/math&gt; (and is in fact a double root). <br /> <br /> The other two intersection points have &lt;math&gt;x&lt;/math&gt; coordinates &lt;math&gt;\pm\sqrt{2a-1}&lt;/math&gt;. We must have &lt;math&gt;2a-1&gt; 0,&lt;/math&gt; otherwise we are in the case where the parabola lies entirely above the circle (tangent to it at the point &lt;math&gt;(0,a)&lt;/math&gt;). This only results in a single intersection point in the real coordinate plane. Thus, we see &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> (projecteulerlover)<br /> <br /> == Solution 2 ==<br /> <br /> &lt;asy&gt;<br /> Label f; <br /> f.p=fontsize(6);<br /> xaxis(-2,2,Ticks(f, 0.2)); <br /> yaxis(-2,2,Ticks(f, 0.2)); <br /> real g(real x) <br /> { <br /> return x^2-1; <br /> } <br /> draw(graph(g, 1.7, -1.7));<br /> real h(real x) <br /> { <br /> return sqrt(1-x^2); <br /> } <br /> draw(graph(h, 1, -1));<br /> real j(real x) <br /> { <br /> return -sqrt(1-x^2); <br /> } <br /> draw(graph(j, 1, -1));<br /> &lt;/asy&gt;<br /> <br /> Looking at a graph, it is obvious that the two curves intersect at (0, -a). We also see that if the parabola goes 'in' the circle. Then, by going out of it (as it will), it will intersect five times, an impossibility. Thus we only look for cases where the parabola becomes externally tangent to the circle. We have &lt;math&gt;x^2 - a = -\sqrt{a^2 - x^2}&lt;/math&gt;. Squaring both sides and solving yields &lt;math&gt;x^4 - (2a - 1)x^2 = 0&lt;/math&gt;. Since x = 0 is already accounted for, we only need to find 1 solution for &lt;math&gt;x^2 = 2a - 1&lt;/math&gt;, where the right hand side portion is obviously increasing. Since a = 1/2 begets x = 0 (an overcount), we have &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; is the right answer.<br /> <br /> Solution by JohnHankock<br /> <br /> == Solution 3 ==<br /> <br /> This describes a unit parabola, with a circle centered on the axis of symmetry and tangent to the vertex. As the curvature of the unit parabola at the vertex is 2, the radius of the circle that matches it has a radius of &lt;math&gt;\frac{1}{2}&lt;/math&gt;. This circle is tangent to an infinitesimally close pair of points, one on each side. Therefore, it is tangent to only 1 point. When a larger circle is used, it is tangent to 3 points because the points on either side are now separated from the vertex. Therefore, &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; is correct.<br /> <br /> &lt;math&gt;QED \blacksquare&lt;/math&gt;<br /> <br /> == Solution 4 ==<br /> <br /> Notice, the equations are of that of a circle of radius a centered at the origin and a parabola translated down by a units. They always intersect at the point &lt;math&gt;(0, a)&lt;/math&gt;, and they have symmetry across the y-axis, thus, for them to intersect at exactly 3 points, it suffices to find the y solution. <br /> <br /> First, rewrite the second equation to &lt;math&gt;y=x^2-a\implies x^2=y+a&lt;/math&gt;<br /> And substitute into the first equation: &lt;math&gt;y+a+y^2=a^2&lt;/math&gt; <br /> Since we're only interested in seeing the interval in which a can exist, we find the discriminant: &lt;math&gt;1-4a+4a^2&lt;/math&gt;. This value must not be less than 0 (It is the square root part of the quadratic formula). To find when it is 0, we find the roots: <br /> &lt;cmath&gt;4a^2-4a+1=0 \implies a=\frac{4\pm\sqrt{16-16}}{8}=\frac{1}{2}&lt;/cmath&gt;<br /> Since &lt;math&gt;\lim_{a\to \infty}(4a^2-4a+1)=\infty&lt;/math&gt;, our range is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> Solution by ktong<br /> <br /> == Solution 5 (Cheating with Answer Choices) ==<br /> Simply plug in &lt;math&gt;a = 0, \frac{1}{2}, \frac{1}{4}, 1&lt;/math&gt; and sole the systems. (This sholdn't take too long.) And then realize that only &lt;math&gt;a=1&lt;/math&gt; yields three real solutions for &lt;math&gt;x&lt;/math&gt;, so we are done and the answer is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt;.<br /> <br /> ~ ccx09<br /> <br /> == Solution 6 (Calculus Needed) ==<br /> <br /> In order to solve for the values of &lt;math&gt;a&lt;/math&gt;, we need to just count multiplicities of the roots when the equations are set equal to each other: in other words, take the derivative. We know that &lt;math&gt;\sqrt{a^2 - x^2} = x^2 - a&lt;/math&gt;. Now, we take square of both sides, and rearrange to obtain &lt;math&gt;x^4 - (2a - 1)x^2 = 0&lt;/math&gt;. Now, we may take the second derivative of the equation to obtain &lt;math&gt;6x^2 - (2a - 1) = 0&lt;/math&gt;. Now, we must take discriminant. Since we need the roots of that equation to be real and not repetitive (otherwise they would not intersect each other at three points), the discriminant must be greater than zero. Thus,<br /> <br /> &lt;math&gt;<br /> b^2 - 4ac &gt; 0 \rightarrow 0 - 4(6)(-(2a - 1)) &gt; 0 \rightarrow a &gt; \frac{1}{2}<br /> &lt;/math&gt;<br /> The answer is &lt;math&gt;\boxed{\textbf{(E) }a&gt;\frac12}&lt;/math&gt; and we are done. <br /> <br /> ~awesome1st ~OlutosinNGA&lt;math&gt;edit&lt;/math&gt;<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=20|num-a=22}}<br /> {{AMC12 box|year=2018|ab=A|num-b=15|num-a=17}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Algebra Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_20&diff=100836 2018 AMC 10A Problems/Problem 20 2019-01-25T09:10:00Z <p>Olutosinfires: /* Solution 3 */</p> <hr /> <div>A scanning code consists of a &lt;math&gt;7 \times 7&lt;/math&gt; grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of &lt;math&gt;49&lt;/math&gt; squares. A scanning code is called &lt;math&gt;\textit{symmetric}&lt;/math&gt; if its look does not change when the entire square is rotated by a multiple of &lt;math&gt;90 ^{\circ}&lt;/math&gt; counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw a &lt;math&gt;7 \times 7&lt;/math&gt; square.<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|c|c|c|c|}<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> G &amp; D &amp; B &amp; A &amp; B &amp; D &amp; G \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> \end{tabular} &lt;/math&gt;<br /> <br /> Start from the center and label all protruding cells symmetrically. (Note that &quot;I&quot; is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)<br /> <br /> More specifically, since there are &lt;math&gt;4&lt;/math&gt; given lines of symmetry (&lt;math&gt;2&lt;/math&gt; diagonals, &lt;math&gt;1&lt;/math&gt; vertical, &lt;math&gt;1&lt;/math&gt; horizontal) and they split the plot into &lt;math&gt;8&lt;/math&gt; equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has &lt;math&gt;10&lt;/math&gt; distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose &lt;math&gt;2^{10}=1024&lt;/math&gt; but then subtract the &lt;math&gt;2&lt;/math&gt; cases where all are white or all are black. That leaves us with &lt;math&gt;\boxed{(B)}, 1022&lt;/math&gt;. ∎<br /> <br /> There are only ten squares we get to actually choose, and two independent choices for each, for a total of &lt;math&gt;2^{10} = 1024&lt;/math&gt; codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of &lt;math&gt;\fbox{\textbf{(B)} \text{ 1022}}&lt;/math&gt;.<br /> ~Nosysnow<br /> <br /> <br /> Note that this problem is very similar to the 1996 AIME Problem 7.<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt;<br /> size(100pt);<br /> draw((1,0)--(8,0),linewidth(0.5));<br /> draw((1,2)--(6,2),linewidth(0.5));<br /> draw((1,4)--(4,4),linewidth(0.5));<br /> draw((1,6)--(2,6),linewidth(0.5));<br /> draw((2,6)--(2,0),linewidth(0.5));<br /> draw((4,4)--(4,0),linewidth(0.5));<br /> draw((6,2)--(6,0),linewidth(0.5));<br /> draw((1,0)--(1,7),dashed+linewidth(0.5));<br /> draw((1,7)--(8,0),dashed+linewidth(0.5));<br /> &lt;/asy&gt;<br /> <br /> Imagine folding the scanning code along its lines of symmetry. There will be &lt;math&gt;10&lt;/math&gt; regions which you have control over coloring. Since we must subtract off &lt;math&gt;2&lt;/math&gt; cases for the all-black and all-white cases, the answer is &lt;math&gt;2^{10}-2=\boxed{\textbf{(B) } 1022.}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> This &lt;math&gt;7 \times 7&lt;/math&gt; square drawn in &lt;math&gt;Solution 1&lt;/math&gt; satisfies the conditions given in the problem. Calculating the number of ways of coloring it will solve the problem.<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|c|c|c|c|}<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> G &amp; D &amp; B &amp; A &amp; B &amp; D &amp; G \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> \end{tabular} &lt;/math&gt;<br /> <br /> In the grid, 10 letters are used: &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt;, and &lt;math&gt;K&lt;/math&gt;. Each of the letters must have its own color, either white or black. This means, for example, all &lt;math&gt;K&lt;/math&gt;'s must have the same color for the grid to be symmetrical.<br /> <br /> So there are &lt;math&gt;2^{10}&lt;/math&gt; ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is &lt;math&gt;2^{10}-2=1024-2=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) } 1022}&lt;/math&gt;.<br /> <br /> ~OlutosinNGA<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_20&diff=100835 2018 AMC 10A Problems/Problem 20 2019-01-25T09:06:35Z <p>Olutosinfires: /* Solution 2 */ and /* Solution 3 */</p> <hr /> <div>A scanning code consists of a &lt;math&gt;7 \times 7&lt;/math&gt; grid of squares, with some of its squares colored black and the rest colored white. There must be at least one square of each color in this grid of &lt;math&gt;49&lt;/math&gt; squares. A scanning code is called &lt;math&gt;\textit{symmetric}&lt;/math&gt; if its look does not change when the entire square is rotated by a multiple of &lt;math&gt;90 ^{\circ}&lt;/math&gt; counterclockwise around its center, nor when it is reflected across a line joining opposite corners or a line joining midpoints of opposite sides. What is the total number of possible symmetric scanning codes?<br /> <br /> &lt;math&gt;\textbf{(A)} \text{ 510} \qquad \textbf{(B)} \text{ 1022} \qquad \textbf{(C)} \text{ 8190} \qquad \textbf{(D)} \text{ 8192} \qquad \textbf{(E)} \text{ 65,534}&lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> Draw a &lt;math&gt;7 \times 7&lt;/math&gt; square.<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|c|c|c|c|}<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> G &amp; D &amp; B &amp; A &amp; B &amp; D &amp; G \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> \end{tabular} &lt;/math&gt;<br /> <br /> Start from the center and label all protruding cells symmetrically. (Note that &quot;I&quot; is left out of this labelling, so there are only 10 labels, not 11, as ending in K would suggest!)<br /> <br /> More specifically, since there are &lt;math&gt;4&lt;/math&gt; given lines of symmetry (&lt;math&gt;2&lt;/math&gt; diagonals, &lt;math&gt;1&lt;/math&gt; vertical, &lt;math&gt;1&lt;/math&gt; horizontal) and they split the plot into &lt;math&gt;8&lt;/math&gt; equivalent sections, we can take just one-eighth and study it in particular. Each of these sections has &lt;math&gt;10&lt;/math&gt; distinct sub-squares, whether partially or in full. So since each can be colored either white or black, we choose &lt;math&gt;2^{10}=1024&lt;/math&gt; but then subtract the &lt;math&gt;2&lt;/math&gt; cases where all are white or all are black. That leaves us with &lt;math&gt;\boxed{(B)}, 1022&lt;/math&gt;. ∎<br /> <br /> There are only ten squares we get to actually choose, and two independent choices for each, for a total of &lt;math&gt;2^{10} = 1024&lt;/math&gt; codes. Two codes must be subtracted (due to the rule that there must be at least one square of each color) for an answer of &lt;math&gt;\fbox{\textbf{(B)} \text{ 1022}}&lt;/math&gt;.<br /> ~Nosysnow<br /> <br /> <br /> Note that this problem is very similar to the 1996 AIME Problem 7.<br /> <br /> ==Solution 2==<br /> <br /> &lt;asy&gt;<br /> size(100pt);<br /> draw((1,0)--(8,0),linewidth(0.5));<br /> draw((1,2)--(6,2),linewidth(0.5));<br /> draw((1,4)--(4,4),linewidth(0.5));<br /> draw((1,6)--(2,6),linewidth(0.5));<br /> draw((2,6)--(2,0),linewidth(0.5));<br /> draw((4,4)--(4,0),linewidth(0.5));<br /> draw((6,2)--(6,0),linewidth(0.5));<br /> draw((1,0)--(1,7),dashed+linewidth(0.5));<br /> draw((1,7)--(8,0),dashed+linewidth(0.5));<br /> &lt;/asy&gt;<br /> <br /> Imagine folding the scanning code along its lines of symmetry. There will be &lt;math&gt;10&lt;/math&gt; regions which you have control over coloring. Since we must subtract off &lt;math&gt;2&lt;/math&gt; cases for the all-black and all-white cases, the answer is &lt;math&gt;2^{10}-2=\boxed{\textbf{(B) } 1022.}&lt;/math&gt;<br /> <br /> ==Solution 3==<br /> This &lt;math&gt;7 \times 7&lt;/math&gt; square drawn in &lt;math&gt;Solution 1&lt;/math&gt; satisfies the conditions given in the problem. The number of ways of coloring it determines will solve the problem.<br /> <br /> &lt;math&gt; \begin{tabular}{|c|c|c|c|c|c|c|}<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> G &amp; D &amp; B &amp; A &amp; B &amp; D &amp; G \\<br /> \hline<br /> H &amp; E &amp; C &amp; B &amp; C &amp; E &amp; H \\<br /> \hline<br /> J &amp; F &amp; E &amp; D &amp; E &amp; F &amp; J \\<br /> \hline<br /> K &amp; J &amp; H &amp; G &amp; H &amp; J &amp; K \\<br /> \hline<br /> \end{tabular} &lt;/math&gt;<br /> <br /> In the grid, 10 letters are used: &lt;math&gt;A&lt;/math&gt;, &lt;math&gt;B&lt;/math&gt;, &lt;math&gt;C&lt;/math&gt;, &lt;math&gt;D&lt;/math&gt;, &lt;math&gt;E&lt;/math&gt;, &lt;math&gt;F&lt;/math&gt;, &lt;math&gt;G&lt;/math&gt;, &lt;math&gt;H&lt;/math&gt;, &lt;math&gt;J&lt;/math&gt;, and &lt;math&gt;K&lt;/math&gt;. Each of the letters must have its own color, either white or black. This means, for example, all &lt;math&gt;K&lt;/math&gt;'s must have the same color for the grid to be symmetrical.<br /> <br /> So there are &lt;math&gt;2^{10}&lt;/math&gt; ways to color the grid, including a completely black grid and a completely white grid. Since the grid must contain at least one square with each color, the number of ways is &lt;math&gt;2^{10}-2=1024-2=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(B) } 1022}&lt;/math&gt;.<br /> <br /> ~OlutosinNGA<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=19|num-a=21}}<br /> {{AMC12 box|year=2018|ab=|num-b=14|num-a=16}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Combinatorics Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_14&diff=100834 2018 AMC 12A Problems/Problem 14 2019-01-25T08:24:57Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>==Problem==<br /> <br /> The solutions to the equation &lt;math&gt;\log_{3x} 4 = \log_{2x} 8&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is a positive real number other than &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; or &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;, can be written as &lt;math&gt;\tfrac {p}{q}&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p + q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad <br /> \textbf{(B) } 13 \qquad <br /> \textbf{(C) } 17 \qquad <br /> \textbf{(D) } 31 \qquad <br /> \textbf{(E) } 35 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> Base switch to log 2 and you have &lt;math&gt;\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}&lt;/math&gt; .<br /> <br /> &lt;math&gt;\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}&lt;/math&gt;<br /> <br /> &lt;math&gt;2*\log_2 2x = 3*\log_2 3x&lt;/math&gt;<br /> <br /> Then &lt;math&gt;\log_2 (2x)^2 = \log_2 (3x)^3&lt;/math&gt;. so &lt;math&gt;4x^2=27x^3&lt;/math&gt; and we have &lt;math&gt;x=\frac{4}{27}&lt;/math&gt; leading to &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt; (jeremylu)<br /> <br /> ==Solution 2==<br /> If you multiply both sides by &lt;math&gt;\log_2 (3x)&lt;/math&gt;<br /> <br /> then it should come out to &lt;math&gt;\log_2 (3x)&lt;/math&gt; * &lt;math&gt;\log_{3x} (4)&lt;/math&gt; = &lt;math&gt;\log_2 {3x}&lt;/math&gt; * &lt;math&gt;\log_{2x} (8)&lt;/math&gt;<br /> <br /> that then becomes &lt;math&gt;\log_2 (4)&lt;/math&gt; * &lt;math&gt;\log_{3x} (3x)&lt;/math&gt; = &lt;math&gt;\log_2 (8)&lt;/math&gt; * &lt;math&gt;\log_{2x} (3x)&lt;/math&gt;<br /> <br /> which simplifies to &lt;math&gt;2*1 = 3\log_{2x} (3x)&lt;/math&gt;<br /> <br /> so now &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\log_{2x} (3x)&lt;/math&gt; putting in exponent form gets<br /> <br /> &lt;math&gt;(2x)^2&lt;/math&gt; = &lt;math&gt;(3x)^3&lt;/math&gt;<br /> <br /> so &lt;math&gt;4x^2&lt;/math&gt; = &lt;math&gt;27x^3&lt;/math&gt;<br /> <br /> dividing yields &lt;math&gt;x = 4/27&lt;/math&gt; and <br /> <br /> &lt;math&gt;4+27 =&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt;<br /> - Pikachu13307<br /> <br /> ==Solution 3==<br /> <br /> We can convert both &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; into &lt;math&gt;2^2&lt;/math&gt; and &lt;math&gt;2^3&lt;/math&gt;, respectively, giving:<br /> <br /> &lt;math&gt;2\log_{3x} (2) = 3\log_{2x} (2)&lt;/math&gt;<br /> <br /> Converting the bases of the right side, we get &lt;math&gt;\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\ln 2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}&lt;/math&gt;<br /> <br /> Which simplifies to<br /> <br /> &lt;math&gt;2\ln (2x) = 3\ln (3x)&lt;/math&gt;<br /> <br /> Log expansion allows us to see that<br /> <br /> &lt;math&gt;2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)&lt;/math&gt;, which then simplifies to<br /> <br /> &lt;math&gt;\ln (x) = 2\ln 2 - 3\ln 3&lt;/math&gt;<br /> <br /> Thus,<br /> <br /> &lt;math&gt;x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}&lt;/math&gt;<br /> <br /> And<br /> <br /> &lt;math&gt;x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}&lt;/math&gt; <br /> <br /> -lepetitmoulin<br /> <br /> ==Solution 4==<br /> &lt;math&gt;\log_{3x} 4=\log_{2x} 8&lt;/math&gt; is the same as &lt;math&gt;2\log_{3x} 2=3\log_{2x} 2&lt;/math&gt;<br /> <br /> Using Reciprocal law, we get &lt;math&gt;\log_{(3x)^\frac{1}{2}} 2=\log_{(2x)^\frac{1}{3}} 2&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow (3x)^\frac{1}{2}=(2x)^\frac{1}{3}&lt;/math&gt; &lt;math&gt;\Rightarrow 27x^3=4x^2&lt;/math&gt; &lt;math&gt;\Rightarrow \frac{x^3}{x^2}=\frac{4}{27}=x&lt;/math&gt;<br /> <br /> &lt;math&gt;\therefore \frac{p}{q}=\frac{4}{27}&lt;/math&gt; &lt;math&gt;\Rightarrow p+q=4+27=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } 31}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_14&diff=100833 2018 AMC 12A Problems/Problem 14 2019-01-25T08:22:00Z <p>Olutosinfires: /* Solution 3 */ and /* Solution */</p> <hr /> <div>==Problem==<br /> <br /> The solutions to the equation &lt;math&gt;\log_{3x} 4 = \log_{2x} 8&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is a positive real number other than &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; or &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;, can be written as &lt;math&gt;\tfrac {p}{q}&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p + q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad <br /> \textbf{(B) } 13 \qquad <br /> \textbf{(C) } 17 \qquad <br /> \textbf{(D) } 31 \qquad <br /> \textbf{(E) } 35 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> Base switch to log 2 and you have &lt;math&gt;\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}&lt;/math&gt; .<br /> <br /> &lt;math&gt;\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}&lt;/math&gt;<br /> <br /> &lt;math&gt;2*\log_2 2x = 3*\log_2 3x&lt;/math&gt;<br /> <br /> Then &lt;math&gt;\log_2 (2x)^2 = \log_2 (3x)^3&lt;/math&gt;. so &lt;math&gt;4x^2=27x^3&lt;/math&gt; and we have &lt;math&gt;x=\frac{4}{27}&lt;/math&gt; leading to &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt; (jeremylu)<br /> <br /> ==Solution 2==<br /> If you multiply both sides by &lt;math&gt;\log_2 (3x)&lt;/math&gt;<br /> <br /> then it should come out to &lt;math&gt;\log_2 (3x)&lt;/math&gt; * &lt;math&gt;\log_{3x} (4)&lt;/math&gt; = &lt;math&gt;\log_2 {3x}&lt;/math&gt; * &lt;math&gt;\log_{2x} (8)&lt;/math&gt;<br /> <br /> that then becomes &lt;math&gt;\log_2 (4)&lt;/math&gt; * &lt;math&gt;\log_{3x} (3x)&lt;/math&gt; = &lt;math&gt;\log_2 (8)&lt;/math&gt; * &lt;math&gt;\log_{2x} (3x)&lt;/math&gt;<br /> <br /> which simplifies to &lt;math&gt;2*1 = 3\log_{2x} (3x)&lt;/math&gt;<br /> <br /> so now &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\log_{2x} (3x)&lt;/math&gt; putting in exponent form gets<br /> <br /> &lt;math&gt;(2x)^2&lt;/math&gt; = &lt;math&gt;(3x)^3&lt;/math&gt;<br /> <br /> so &lt;math&gt;4x^2&lt;/math&gt; = &lt;math&gt;27x^3&lt;/math&gt;<br /> <br /> dividing yields &lt;math&gt;x = 4/27&lt;/math&gt; and <br /> <br /> &lt;math&gt;4+27 =&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt;<br /> - Pikachu13307<br /> <br /> ==Solution 3==<br /> <br /> We can convert both &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; into &lt;math&gt;2^2&lt;/math&gt; and &lt;math&gt;2^3&lt;/math&gt;, respectively, giving:<br /> <br /> &lt;math&gt;2\log_{3x} (2) = 3\log_{2x} (2)&lt;/math&gt;<br /> <br /> Converting the bases of the right side, we get &lt;math&gt;\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\ln 2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}&lt;/math&gt;<br /> <br /> Which simplifies to<br /> <br /> &lt;math&gt;2\ln (2x) = 3\ln (3x)&lt;/math&gt;<br /> <br /> Log expansion allows us to see that<br /> <br /> &lt;math&gt;2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)&lt;/math&gt;, which then simplifies to<br /> <br /> &lt;math&gt;\ln (x) = 2\ln 2 - 3\ln 3&lt;/math&gt;<br /> <br /> Thus,<br /> <br /> &lt;math&gt;x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}&lt;/math&gt;<br /> <br /> And<br /> <br /> &lt;math&gt;x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}&lt;/math&gt; <br /> <br /> -lepetitmoulin<br /> <br /> ==Solution 4==<br /> &lt;math&gt;\log_{3x} 4=\log_{2x} 8&lt;/math&gt; is the same as &lt;math&gt;2\log_{3x} 2=3\log_{2x} 2&lt;/math&gt;<br /> <br /> Using Reciprocal law, we get &lt;math&gt;\log_{(3x)^\frac{1}{2}} 2=\log_{(2x)^\frac{1}{3}} 2&lt;/math&gt;<br /> <br /> &lt;math&gt;\Rightarrow (3x)^\frac{1}{2}=(2x)^\frac{1}{3}&lt;/math&gt; &lt;math&gt;\Rightarrow 27x^3=4x^2&lt;/math&gt; &lt;math&gt;\Rightarrow \frac{x^3}{x^2}=\frac{4}{27}=x&lt;/math&gt;<br /> <br /> &lt;math&gt;\therefore \frac{p}{q}=\frac{4}{27}&lt;/math&gt; &lt;math&gt;\Rightarrow p+q=4+27=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } 31}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_12A_Problems/Problem_14&diff=100832 2018 AMC 12A Problems/Problem 14 2019-01-25T07:55:14Z <p>Olutosinfires: /* Solution 2 */</p> <hr /> <div>==Problem==<br /> <br /> The solutions to the equation &lt;math&gt;\log_{3x} 4 = \log_{2x} 8&lt;/math&gt;, where &lt;math&gt;x&lt;/math&gt; is a positive real number other than &lt;math&gt;\tfrac{1}{3}&lt;/math&gt; or &lt;math&gt;\tfrac{1}{2}&lt;/math&gt;, can be written as &lt;math&gt;\tfrac {p}{q}&lt;/math&gt; where &lt;math&gt;p&lt;/math&gt; and &lt;math&gt;q&lt;/math&gt; are relatively prime positive integers. What is &lt;math&gt;p + q&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 5 \qquad <br /> \textbf{(B) } 13 \qquad <br /> \textbf{(C) } 17 \qquad <br /> \textbf{(D) } 31 \qquad <br /> \textbf{(E) } 35 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> <br /> Base switch to log 2 and you have &lt;math&gt;\frac{\log_2 4}{\log_2 3x} = \frac{\log_2 8}{\log_2 2x}&lt;/math&gt; .<br /> <br /> &lt;math&gt;\frac{2}{\log_2 3x} = \frac{3}{\log_2 2x}&lt;/math&gt;<br /> <br /> &lt;math&gt;2*\log_2 2x = 3*\log_2 3x&lt;/math&gt;<br /> <br /> Then &lt;math&gt;\log_2 (2x)^2 = \log_2 (3x)^3&lt;/math&gt;. so &lt;math&gt;4x^2=27x^3&lt;/math&gt; and we have &lt;math&gt;x=\frac{4}{27}&lt;/math&gt; leading to &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt; (jeremylu)<br /> <br /> ==Solution 2==<br /> If you multiply both sides by &lt;math&gt;\log_2 (3x)&lt;/math&gt;<br /> <br /> then it should come out to &lt;math&gt;\log_2 (3x)&lt;/math&gt; * &lt;math&gt;\log_{3x} (4)&lt;/math&gt; = &lt;math&gt;\log_2 {3x}&lt;/math&gt; * &lt;math&gt;\log_{2x} (8)&lt;/math&gt;<br /> <br /> that then becomes &lt;math&gt;\log_2 (4)&lt;/math&gt; * &lt;math&gt;\log_{3x} (3x)&lt;/math&gt; = &lt;math&gt;\log_2 (8)&lt;/math&gt; * &lt;math&gt;\log_{2x} (3x)&lt;/math&gt;<br /> <br /> which simplifies to &lt;math&gt;2*1 = 3\log_{2x} (3x)&lt;/math&gt;<br /> <br /> so now &lt;math&gt;\frac{2}{3}&lt;/math&gt; = &lt;math&gt;\log_{2x} (3x)&lt;/math&gt; putting in exponent form gets<br /> <br /> &lt;math&gt;(2x)^2&lt;/math&gt; = &lt;math&gt;(3x)^3&lt;/math&gt;<br /> <br /> so &lt;math&gt;4x^2&lt;/math&gt; = &lt;math&gt;27x^3&lt;/math&gt;<br /> <br /> dividing yields &lt;math&gt;x = 4/27&lt;/math&gt; and <br /> <br /> &lt;math&gt;4+27 =&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D)}31}&lt;/math&gt;<br /> - Pikachu13307<br /> <br /> ==Solution 3==<br /> <br /> We can convert both &lt;math&gt;4&lt;/math&gt; and &lt;math&gt;8&lt;/math&gt; into &lt;math&gt;2^2&lt;/math&gt; and &lt;math&gt;2^3&lt;/math&gt;, respectively, giving:<br /> <br /> &lt;math&gt;2\log_{3x} (2) = 3\log_{2x} (2)&lt;/math&gt;<br /> <br /> Converting the bases of the right side, we get &lt;math&gt;\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}&lt;/math&gt;<br /> <br /> &lt;math&gt;\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)&lt;/math&gt;<br /> <br /> Dividing both sides by &lt;math&gt;\ln 2&lt;/math&gt;, we get<br /> <br /> &lt;math&gt;\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}&lt;/math&gt;<br /> <br /> Which simplifies to<br /> <br /> &lt;math&gt;2\ln (2x) = 3\ln (3x)&lt;/math&gt;<br /> <br /> Log expansion allows us to see that<br /> <br /> &lt;math&gt;2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)&lt;/math&gt;, which then simplifies to<br /> <br /> &lt;math&gt;\ln (x) = 2\ln 2 - 3\ln 3&lt;/math&gt;<br /> <br /> Thus,<br /> <br /> &lt;math&gt;x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}&lt;/math&gt;<br /> <br /> And<br /> <br /> &lt;math&gt;x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{\textbf{(D)}31}&lt;/math&gt; <br /> <br /> -lepetitmoulin<br /> <br /> ==See Also==<br /> {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}}<br /> {{MAA Notice}}</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=100831 2018 AMC 10A Problems/Problem 18 2019-01-25T07:46:01Z <p>Olutosinfires: /* Solution 7 */</p> <hr /> <div>==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{3281}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Solution 5==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 6==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{(D)3281}&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 7==<br /> To get the number of integers, we can get the highest positive integer that can be represented using &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;.<br /> <br /> Note that the least nonnegative integer that can be represented is &lt;math&gt;0&lt;/math&gt;, when all &lt;math&gt;a_i=0&lt;/math&gt;. The highest number will be the number when all &lt;math&gt;a_i=1&lt;/math&gt;. That will be &lt;cmath&gt;3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}&lt;/cmath&gt; &lt;cmath&gt;=3280&lt;/cmath&gt;<br /> <br /> Therefore, there are &lt;math&gt;3280&lt;/math&gt; positive integers and &lt;math&gt;(3280+1)&lt;/math&gt; nonnegative integers (while including &lt;math&gt;0&lt;/math&gt;) that can be represented. Our answer is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=100830 2018 AMC 10A Problems/Problem 18 2019-01-25T07:44:02Z <p>Olutosinfires: /* Solution 7 */</p> <hr /> <div>==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{3281}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Solution 5==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 6==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{(D)3281}&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 7==<br /> To get the number of integers, we can get the highest positive integer that can be represented using &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;.<br /> <br /> Note that the least nonnegative integer that can be represented is &lt;math&gt;0&lt;/math&gt;, when all &lt;math&gt;a_i=0&lt;/math&gt;. The highest number will be the number when all &lt;math&gt;a_i=1&lt;/math&gt;. That will be &lt;cmath&gt;3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}&lt;/cmath&gt; &lt;cmath&gt;=3280&lt;/cmath&gt;<br /> <br /> Therefore, there are &lt;math&gt;(3280+1)&lt;/math&gt; nonnegative integers that can be represented. Our answer is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=100829 2018 AMC 10A Problems/Problem 18 2019-01-25T07:42:38Z <p>Olutosinfires: /* Solution 7 */</p> <hr /> <div>==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{3281}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Solution 5==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 6==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{(D)3281}&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 7==<br /> To get the number of integers, we can get the highest positive integer that can be represented using &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;.<br /> <br /> Note that the least nonnegative integer that can be represented is &lt;math&gt;0&lt;/math&gt;, when all &lt;math&gt;a_i=0&lt;/math&gt;. The highest number will be the number when all &lt;math&gt;a_i=1&lt;/math&gt;. That will be &lt;cmath&gt;3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}&lt;/cmath&gt; &lt;cmath&gt;=3280&lt;/cmath&gt;<br /> <br /> Therefore, there are &lt;math&gt;(3280+1)&lt;/math&gt; nonnegative integers that can be represented. Our answer is &lt;math&gt;\boxed{\textbf{(D) } 3281}&lt;/math&gt;<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=100828 2018 AMC 10A Problems/Problem 18 2019-01-25T07:40:02Z <p>Olutosinfires: /* Solution 7 */</p> <hr /> <div>==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{3281}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Solution 5==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 6==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{(D)3281}&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 7==<br /> To get the number of integers, we can get the highest positive integer that can be represented using &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;.<br /> <br /> Note that the least nonnegative integer that can be represented is &lt;math&gt;0&lt;/math&gt;, when all &lt;math&gt;a_i=0&lt;/math&gt;. The highest number will be the number when all &lt;math&gt;a_i=1&lt;/math&gt;. That will be &lt;cmath&gt;3^7+3^6+3^5+3^4+3^3+3^2+3^1+3^0=\frac{3^8-1}{3-1}&lt;/cmath&gt; &lt;cmath&gt;=3280&lt;/cmath&gt;<br /> <br /> Therefore, there are &lt;math&gt;(3280+1)&lt;/math&gt; nonnegative integers that can be represented. Our answer is$\boxed{}<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_18&diff=100826 2018 AMC 10A Problems/Problem 18 2019-01-25T07:27:14Z <p>Olutosinfires: /* Solution 6 */ and /* Solution 7 */</p> <hr /> <div>==Problem==<br /> How many nonnegative integers can be written in the form &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt;<br /> where &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; for &lt;math&gt;0\le i \le 7&lt;/math&gt;?<br /> <br /> &lt;math&gt;\textbf{(A) } 512 \qquad <br /> \textbf{(B) } 729 \qquad <br /> \textbf{(C) } 1094 \qquad <br /> \textbf{(D) } 3281 \qquad <br /> \textbf{(E) } 59,048 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> This looks like balanced ternary, in which all the integers with absolute values less than &lt;math&gt;\frac{3^n}{2}&lt;/math&gt; are represented in &lt;math&gt;n&lt;/math&gt; digits. There are 8 digits. Plugging in 8 into the formula for the balanced ternary gives a maximum bound of &lt;math&gt;|x|=3280.5&lt;/math&gt;, which means there are 3280 positive integers, 0, and 3280 negative integers. Since we want all nonnegative integers, there are &lt;math&gt;3280+1=\boxed{3281}&lt;/math&gt; integers or &lt;math&gt;\boxed{D}&lt;/math&gt;.<br /> <br /> ==Solution 2==<br /> <br /> Note that all numbers formed from this sum are either positive, negative or zero. The number of positive numbers formed by this sum is equal to the number of negative numbers formed by this sum, because of symmetry. There is only one way to achieve a sum of zero, if all &lt;math&gt;a_i=0&lt;/math&gt;. The total number of ways to pick &lt;math&gt;a_i&lt;/math&gt; from &lt;math&gt;i=0, 1, 2, 3, ... 7&lt;/math&gt; is &lt;math&gt;3^8=6561&lt;/math&gt;. &lt;math&gt;\frac{6561-1}{2}=3280&lt;/math&gt; gives the number of possible negative integers. The question asks for the number of nonnegative integers, so subtracting from the total gives &lt;math&gt;6561-3280=\boxed{3281}&lt;/math&gt;. (RegularHexagon)<br /> <br /> ==Solution 3==<br /> Note that the number of total possibilities (ignoring the conditions set by the problem) is &lt;math&gt;3^8=6561&lt;/math&gt;. So, E is clearly unrealistic. <br /> <br /> Note that if &lt;math&gt;a_7&lt;/math&gt; is 1, then it's impossible for &lt;cmath&gt;a_7\cdot3^7+a_6\cdot3^6+a_5\cdot3^5+a_4\cdot3^4+a_3\cdot3^3+a_2\cdot3^2+a_1\cdot3^1+a_0\cdot3^0,&lt;/cmath&gt; to be negative. Therefore, if &lt;math&gt;a_7&lt;/math&gt; is 1, there are &lt;math&gt;3^7=2187&lt;/math&gt; possibilities. (We also must convince ourselves that these &lt;math&gt;2187&lt;/math&gt; different sets of coefficients must necessarily yield &lt;math&gt;2187&lt;/math&gt; different integer results.)<br /> <br /> As A, B, and C are all less than 2187, the answer must be &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;<br /> <br /> ==Solution 4==<br /> Note that we can do some simple casework:<br /> If &lt;math&gt;a_7=1&lt;/math&gt;, then we can choose anything for the other 7 variables, so this give us &lt;math&gt;3^7&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt; and &lt;math&gt;a_6=1&lt;/math&gt;, then we can choose anything for the other 6 variables, giving us &lt;math&gt;3^6&lt;/math&gt;.<br /> If &lt;math&gt;a_7=0&lt;/math&gt;, &lt;math&gt;a_6=0&lt;/math&gt;, and &lt;math&gt;a_5=1&lt;/math&gt;, then we have &lt;math&gt;3^5&lt;/math&gt;.<br /> Continuing in this vein, we have &lt;math&gt;3^7+3^6+\cdots+3^1+3^0&lt;/math&gt; ways to choose the variables' values, except we have to add 1<br /> because we haven't counted the case where all variables are 0. So our total sum is &lt;math&gt;\boxed{(D) 3281}&lt;/math&gt;.<br /> Note that we have counted all possibilities, because the largest positive positive power of 3 must be greater than or equal to the largest negative positive power of 3, for the number to be nonnegative.<br /> <br /> ==Solution 5==<br /> The key is to realize that this question is basically taking place in &lt;math&gt;a\in\{0,1,2\}&lt;/math&gt; if each value of &lt;math&gt;a&lt;/math&gt; was increased by &lt;math&gt;1&lt;/math&gt;, essentially making it into base &lt;math&gt;3&lt;/math&gt;. Then the range would be from &lt;math&gt;0\cdot3^7+&lt;/math&gt; &lt;math&gt;0\cdot3^6+&lt;/math&gt; &lt;math&gt;0\cdot3^5+&lt;/math&gt; &lt;math&gt;0\cdot3^4+&lt;/math&gt; &lt;math&gt;0\cdot3^3+&lt;/math&gt; &lt;math&gt;0\cdot3^2+&lt;/math&gt; &lt;math&gt;0\cdot3^1+&lt;/math&gt; &lt;math&gt;0\cdot3^0=&lt;/math&gt; &lt;math&gt;0&lt;/math&gt; to &lt;math&gt;2\cdot3^7+&lt;/math&gt; &lt;math&gt;2\cdot3^6+&lt;/math&gt; &lt;math&gt;2\cdot3^5+&lt;/math&gt; &lt;math&gt;2\cdot3^4+&lt;/math&gt; &lt;math&gt;2\cdot3^3+&lt;/math&gt; &lt;math&gt;2\cdot3^2+&lt;/math&gt; &lt;math&gt;2\cdot3^1+&lt;/math&gt; &lt;math&gt;2\cdot3^0=&lt;/math&gt; &lt;math&gt;3^8-1=&lt;/math&gt; &lt;math&gt;6561-1=&lt;/math&gt; &lt;math&gt;6560&lt;/math&gt;, yielding &lt;math&gt;6561&lt;/math&gt; different values. Since the distribution for all &lt;math&gt;a_i\in \{-1,0,1\}&lt;/math&gt; the question originally gave is symmetrical, we retain the &lt;math&gt;3280&lt;/math&gt; positive integers and one &lt;math&gt;0&lt;/math&gt; but discard the &lt;math&gt;3280&lt;/math&gt; negative integers. Thus, we are left with the answer, &lt;math&gt;\boxed{\textbf{(D)} 3281}\qquad&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 6==<br /> First, set &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq1&lt;/math&gt;. The range would be the integers for which &lt;math&gt;[-1,1]&lt;/math&gt;. If &lt;math&gt;a_i=0&lt;/math&gt; for all &lt;math&gt;i\geq2&lt;/math&gt;, our set expands to include all integers such that &lt;math&gt;-4\leq\mathbb{Z}\leq4&lt;/math&gt;. Similarly, when &lt;math&gt;i\geq3&lt;/math&gt; we get &lt;math&gt;-13\leq\mathbb{Z}\leq13&lt;/math&gt;, and when &lt;math&gt;i\geq4&lt;/math&gt; the range is &lt;math&gt;-40\leq\mathbb{Z}\leq40&lt;/math&gt;. The pattern continues until we reach &lt;math&gt;i=7&lt;/math&gt;, where &lt;math&gt;-3280\leq\mathbb{Z}\leq3280&lt;/math&gt;. Because we are only looking for positive integers, we filter out all &lt;math&gt;\mathbb{Z}&lt;0&lt;/math&gt;, leaving us with all integers between &lt;math&gt;0\leq\mathbb{Z}\leq3280&lt;/math&gt;, inclusive. The answer becomes &lt;math&gt;\boxed{(D)3281}&lt;/math&gt;. ∎ --anna0kear<br /> <br /> ==Solution 7==<br /> To get the number of integers, we can get the highest positive integer that can be represented using<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=17|num-a=19}}<br /> {{AMC12 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_17&diff=100813 2018 AMC 10A Problems/Problem 17 2019-01-24T14:20:07Z <p>Olutosinfires: /* Solution 2 */</p> <hr /> <div>== Problem ==<br /> Let &lt;math&gt;S&lt;/math&gt; be a set of 6 integers taken from &lt;math&gt;\{1,2,\dots,12\}&lt;/math&gt; with the property that if &lt;math&gt;a&lt;/math&gt; and &lt;math&gt;b&lt;/math&gt; are elements of &lt;math&gt;S&lt;/math&gt; with &lt;math&gt;a&lt;b&lt;/math&gt;, then &lt;math&gt;b&lt;/math&gt; is not a multiple of &lt;math&gt;a&lt;/math&gt;. What is the least possible value of an element in &lt;math&gt;S?&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{(A)}\ 2\qquad\textbf{(B)}\ 3\qquad\textbf{(C)}\ 4\qquad\textbf{(D)}\ 5\qquad\textbf{(E)}\ 7&lt;/math&gt;<br /> <br /> == Solution ==<br /> If we start with &lt;math&gt;1&lt;/math&gt;, we can include nothing else, so that won't work.<br /> <br /> If we start with &lt;math&gt;2&lt;/math&gt;, we would have to include every odd number except &lt;math&gt;1&lt;/math&gt; to fill out the set, but then &lt;math&gt;3&lt;/math&gt; and &lt;math&gt;9&lt;/math&gt; would violate the rule, so that won't work.<br /> <br /> Experimentation with &lt;math&gt;3&lt;/math&gt; shows it's likewise impossible. You can include &lt;math&gt;7&lt;/math&gt;, &lt;math&gt;11&lt;/math&gt;, and either &lt;math&gt;5&lt;/math&gt; or &lt;math&gt;10&lt;/math&gt; (which are always safe). But after adding either &lt;math&gt;4&lt;/math&gt; or &lt;math&gt;8&lt;/math&gt; we have no more places to go.<br /> <br /> Finally, starting with &lt;math&gt;4&lt;/math&gt;, we find that the sequence &lt;math&gt;4,5,6,7,9,11&lt;/math&gt; works, giving us &lt;math&gt;\boxed{\textbf{(C)} \text{ 4}}&lt;/math&gt;.<br /> (Random_Guy)<br /> <br /> ==Solution 2==<br /> We know that all the odd numbers (except 1) can be used.<br /> <br /> &lt;math&gt;3, 5, 7, 9, 11&lt;/math&gt;<br /> <br /> Now we have 7 to choose from for the last number (out of &lt;math&gt;1, 2, 4, 6, 8, 10, 12&lt;/math&gt;). We can eliminate 1, 2, 10, and 12, and we have &lt;math&gt;4, 6, 8&lt;/math&gt; to choose from. But wait, 9 is a multiple of 3! Now we have to take out either 3 or 9 from the list. If we take out &lt;math&gt;9&lt;/math&gt;, none of the numbers would work, but if we take out &lt;math&gt;3&lt;/math&gt;, we get:<br /> <br /> &lt;math&gt;4, 5, 6, 7, 9, 11&lt;/math&gt;<br /> <br /> So the least number is &lt;math&gt;4&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C)} \text{ 4}}&lt;/math&gt;.<br /> <br /> -Baolan<br /> <br /> ==Solution 3==<br /> We can get the multiples for the numbers in the original set with multiples in the same original set<br /> <br /> &lt;math&gt;1: all&lt;/math&gt; &lt;math&gt;numbers&lt;/math&gt; &lt;math&gt;within&lt;/math&gt; &lt;math&gt;range&lt;/math&gt;<br /> <br /> &lt;math&gt;2: 4,6,8,10,12&lt;/math&gt;<br /> <br /> &lt;math&gt;3: 6,9,12&lt;/math&gt;<br /> <br /> &lt;math&gt;4: 8,12&lt;/math&gt;<br /> <br /> &lt;math&gt;5: 10&lt;/math&gt;<br /> <br /> &lt;math&gt;6: 12&lt;/math&gt;<br /> <br /> It will be safe to start with 5 or 6 since they have the smallest number of multiples as listed above, but since the question asks for the least, it will be better to try others.<br /> <br /> Trying &lt;math&gt;4&lt;/math&gt;, we can get &lt;math&gt;4,5,6,7,9,11&lt;/math&gt;. So &lt;math&gt;4&lt;/math&gt; works.<br /> Trying &lt;math&gt;3&lt;/math&gt; won't work, so the least is &lt;math&gt;4&lt;/math&gt;. This means the answer is &lt;math&gt;\boxed{\textbf{(C) } 4}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also ==<br /> {{AMC10 box|year=2018|ab=A|num-b=16|num-a=18}}<br /> {{AMC12 box|year=2018|ab=A|num-b=11|num-a=13}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Number Theory Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=100812 2018 AMC 10A Problems/Problem 13 2019-01-24T13:48:17Z <p>Olutosinfires: /* Solution 4 */</p> <hr /> <div>== Problem ==<br /> <br /> A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First, we need to realize that the crease line is just the perpendicular bisector of side &lt;math&gt;AB&lt;/math&gt;, the hypotenuse of right triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;. Draw this line and call the intersection point with &lt;math&gt;AC&lt;/math&gt; as &lt;math&gt;E&lt;/math&gt;. Now, &lt;math&gt;\triangle ACB&lt;/math&gt; is similar to &lt;math&gt;\triangle ADE&lt;/math&gt; by &lt;math&gt;AA&lt;/math&gt; similarity. Setting up the ratios, we find that<br /> &lt;cmath&gt;\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> ===Note===<br /> In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, because &lt;math&gt;A&lt;/math&gt; must be reflected onto &lt;math&gt;B&lt;/math&gt;. (by pulusona)<br /> <br /> ==Solution 2==<br /> <br /> Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than &lt;math&gt;\frac{7}{4}&lt;/math&gt; units and somewhat less than &lt;math&gt;2&lt;/math&gt; units. The only answer choice in range is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> This is pretty much a cop-out, but it's allowed in the rules technically.<br /> ==Solution 3==<br /> <br /> Since &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle, we can see that the slope of line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;\frac{BC}{AC}&lt;/math&gt; = &lt;math&gt;\frac{3}{4}&lt;/math&gt;. We know that if we fold &lt;math&gt;\triangle ABC&lt;/math&gt; so that point &lt;math&gt;A&lt;/math&gt; meets point &lt;math&gt;B&lt;/math&gt; the crease line will be perpendicular to &lt;math&gt;AB&lt;/math&gt; and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is &lt;math&gt;-\frac{4}{3}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> pen dotstyle = black;<br /> <br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> <br /> dot((0,0),dotstyle); <br /> label(&quot;$A$&quot;, (0.03153837092244126,0.07822624343603715), SW); <br /> dot((4,0),dotstyle); <br /> label(&quot;$C$&quot;, (4.028913881471271,0.07822624343603715), SE); <br /> dot((4,3),dotstyle); <br /> label(&quot;$B$&quot;, (4.028913881471271,3.078221223847919), NE); <br /> dot((2,1.5),dotstyle);<br /> label(&quot;$D$&quot;, (2.0341528211973956,1.578223733641978), SE); <br /> dot((2,0),dotstyle);<br /> label(&quot;$E$&quot;, (2.0341528211973956,0.07822624343603715), NE); <br /> dot((3.1249518689638927,0),dotstyle);<br /> label(&quot;$F$&quot;, (3.1571875913515854,0.07822624343603715), NE); <br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> &lt;/asy&gt;<br /> Let us call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;, the midpoint of &lt;math&gt;AC&lt;/math&gt; point &lt;math&gt;E&lt;/math&gt;, and the crease line &lt;math&gt;DF&lt;/math&gt;. We know that &lt;math&gt;DE&lt;/math&gt; is parallel to &lt;math&gt;AC&lt;/math&gt; and that &lt;math&gt;DE&lt;/math&gt;'s length is &lt;math&gt;\frac{AC}{2}=\frac{3}{2}&lt;/math&gt;. Using our slope calculation from earlier, we can see that&lt;math&gt;-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}&lt;/math&gt;. With this information, we can solve for &lt;math&gt;EF&lt;/math&gt;: <br /> &lt;cmath&gt;-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.&lt;/cmath&gt;<br /> We can then use the Pythagorean Theorem to find &lt;math&gt;DF&lt;/math&gt;.<br /> &lt;cmath&gt;\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;\propto M o o n R a b b i t&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 4==<br /> Make use of the diagram in Solution 3. It can be deduced that &lt;math&gt;AF=BF&lt;/math&gt;. Let &lt;math&gt;DF=x&lt;/math&gt;. In &lt;math&gt;\triangle ADF&lt;/math&gt;, &lt;math&gt;AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}&lt;/math&gt;. Then &lt;math&gt;FC&lt;/math&gt; also would be &lt;math&gt;4-\sqrt{x^2+2.5^2}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle BCF&lt;/math&gt;, &lt;math&gt;BF^2=FC^2+BC^2 \Rightarrow (\sqrt{x^2+2.5^2})^2=(4-\sqrt{x^2+2.5^2})^2+3^2&lt;/math&gt;. After some quick math, we get &lt;math&gt;\sqrt{x^2+2.5^2}=\frac{25}{8}&lt;/math&gt;.<br /> Solving for &lt;math&gt;x&lt;/math&gt; will give &lt;math&gt;x=\frac{15}{8}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\therefore&lt;/math&gt; &lt;math&gt;DF=x=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> ~OlutosinNGA<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_13&diff=100811 2018 AMC 10A Problems/Problem 13 2019-01-24T13:45:07Z <p>Olutosinfires: /* Solution 3 */ and /* Solution 4*/</p> <hr /> <div>== Problem ==<br /> <br /> A paper triangle with sides of lengths 3,4, and 5 inches, as shown, is folded so that point &lt;math&gt;A&lt;/math&gt; falls on point &lt;math&gt;B&lt;/math&gt;. What is the length in inches of the crease?<br /> &lt;asy&gt;<br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> label(&quot;$A$&quot;, (0,0), SW);<br /> label(&quot;$B$&quot;, (4,3), NE);<br /> label(&quot;$C$&quot;, (4,0), SE);<br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> &lt;/asy&gt;<br /> &lt;math&gt;\textbf{(A) } 1+\frac12 \sqrt2 \qquad \textbf{(B) } \sqrt3 \qquad \textbf{(C) } \frac74 \qquad \textbf{(D) } \frac{15}{8} \qquad \textbf{(E) } 2 &lt;/math&gt;<br /> <br /> ==Solution 1==<br /> <br /> First, we need to realize that the crease line is just the perpendicular bisector of side &lt;math&gt;AB&lt;/math&gt;, the hypotenuse of right triangle &lt;math&gt;\triangle ABC&lt;/math&gt;. Call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;. Draw this line and call the intersection point with &lt;math&gt;AC&lt;/math&gt; as &lt;math&gt;E&lt;/math&gt;. Now, &lt;math&gt;\triangle ACB&lt;/math&gt; is similar to &lt;math&gt;\triangle ADE&lt;/math&gt; by &lt;math&gt;AA&lt;/math&gt; similarity. Setting up the ratios, we find that<br /> &lt;cmath&gt;\frac{BC}{AC}=\frac{DE}{AD} \Rightarrow \frac{3}{4}=\frac{DE}{\frac{5}{2}} \Rightarrow DE=\frac{15}{8}.&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> ~Nivek<br /> ===Note===<br /> In general, whenever we are asked to make a crease, think about that crease as a line of reflection over which the diagram is reflected. This is why the crease must be the perpendicular bisector of &lt;math&gt;AB&lt;/math&gt;, because &lt;math&gt;A&lt;/math&gt; must be reflected onto &lt;math&gt;B&lt;/math&gt;. (by pulusona)<br /> <br /> ==Solution 2==<br /> <br /> Use the ruler and graph paper you brought to quickly draw a 3-4-5 triangle of any scale (don't trust the diagram in the booklet). Very carefully fold the acute vertices together and make a crease. Measure the crease with the ruler. If you were reasonably careful, you should see that it measures somewhat more than &lt;math&gt;\frac{7}{4}&lt;/math&gt; units and somewhat less than &lt;math&gt;2&lt;/math&gt; units. The only answer choice in range is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> This is pretty much a cop-out, but it's allowed in the rules technically.<br /> ==Solution 3==<br /> <br /> Since &lt;math&gt;\triangle ABC&lt;/math&gt; is a right triangle, we can see that the slope of line &lt;math&gt;AB&lt;/math&gt; is &lt;math&gt;\frac{BC}{AC}&lt;/math&gt; = &lt;math&gt;\frac{3}{4}&lt;/math&gt;. We know that if we fold &lt;math&gt;\triangle ABC&lt;/math&gt; so that point &lt;math&gt;A&lt;/math&gt; meets point &lt;math&gt;B&lt;/math&gt; the crease line will be perpendicular to &lt;math&gt;AB&lt;/math&gt; and we also know that the slopes of perpendicular lines are negative reciprocals of one another. Then, we can see that the slope of our crease line is &lt;math&gt;-\frac{4}{3}&lt;/math&gt;.<br /> &lt;asy&gt;<br /> pen dotstyle = black;<br /> <br /> draw((0,0)--(4,0)--(4,3)--(0,0));<br /> fill(origin--(0,0)--(4,3)--(4,0)--cycle, gray);<br /> <br /> dot((0,0),dotstyle); <br /> label(&quot;$A$&quot;, (0.03153837092244126,0.07822624343603715), SW); <br /> dot((4,0),dotstyle); <br /> label(&quot;$C$&quot;, (4.028913881471271,0.07822624343603715), SE); <br /> dot((4,3),dotstyle); <br /> label(&quot;$B$&quot;, (4.028913881471271,3.078221223847919), NE); <br /> dot((2,1.5),dotstyle);<br /> label(&quot;$D$&quot;, (2.0341528211973956,1.578223733641978), SE); <br /> dot((2,0),dotstyle);<br /> label(&quot;$E$&quot;, (2.0341528211973956,0.07822624343603715), NE); <br /> dot((3.1249518689638927,0),dotstyle);<br /> label(&quot;$F$&quot;, (3.1571875913515854,0.07822624343603715), NE); <br /> <br /> label(&quot;$4$&quot;, (2,0), S);<br /> label(&quot;$3$&quot;, (4,1.5), E);<br /> label(&quot;$5$&quot;, (2,1.5), NW);<br /> &lt;/asy&gt;<br /> Let us call the midpoint of &lt;math&gt;AB&lt;/math&gt; point &lt;math&gt;D&lt;/math&gt;, the midpoint of &lt;math&gt;AC&lt;/math&gt; point &lt;math&gt;E&lt;/math&gt;, and the crease line &lt;math&gt;DF&lt;/math&gt;. We know that &lt;math&gt;DE&lt;/math&gt; is parallel to &lt;math&gt;AC&lt;/math&gt; and that &lt;math&gt;DE&lt;/math&gt;'s length is &lt;math&gt;\frac{AC}{2}=\frac{3}{2}&lt;/math&gt;. Using our slope calculation from earlier, we can see that&lt;math&gt;-\frac{DE}{EF}=-\frac{\frac{3}{2}}{EF}=-\frac{4}{3}&lt;/math&gt;. With this information, we can solve for &lt;math&gt;EF&lt;/math&gt;: <br /> &lt;cmath&gt;-4EF=(-\frac{3}{2})(3) \Rightarrow -4EF=-\frac{9}{2} \Rightarrow 4EF=\frac{9}{2} \Rightarrow EF=\frac{9}{8}.&lt;/cmath&gt;<br /> We can then use the Pythagorean Theorem to find &lt;math&gt;DF&lt;/math&gt;.<br /> &lt;cmath&gt;\frac{3}{2}^2+\frac{9}{8}^2=DF^2 \Rightarrow \frac{9}{4}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8}{4\cdot2\cdot8}+\frac{9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9\cdot2\cdot8+9\cdot9}{8\cdot8}=DF^2 \Rightarrow \frac{9(2\cdot8+9)}{8\cdot8}=DF^2&lt;/cmath&gt;<br /> &lt;cmath&gt;\Rightarrow DF=\sqrt{\frac{9(2\cdot8+9)}{8\cdot8}} \Rightarrow DF=\frac{3\cdot5}{8} \Rightarrow DF=\frac{15}{8}&lt;/cmath&gt;<br /> Thus, our answer is &lt;math&gt;\boxed{\textbf{D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> <br /> &lt;math&gt;\propto M o o n R a b b i t&lt;/math&gt;<br /> <br /> <br /> <br /> ==Solution 4==<br /> Make use of the diagram in Solution 3. It can be deduced that &lt;math&gt;AF=BF&lt;/math&gt;. Let &lt;math&gt;DF=x&lt;/math&gt;. In &lt;math&gt;\triangle ADF&lt;/math&gt;, &lt;math&gt;AF^2=x^2+2.5^2 \Rightarrow AF=\sqrt{x^2+2.5^2}&lt;/math&gt;. Then &lt;math&gt;FC&lt;/math&gt; also would be &lt;math&gt;4-\sqrt{x^2+2.5^2}&lt;/math&gt;.<br /> <br /> In &lt;math&gt;\triangle BCF&lt;/math&gt;, &lt;math&gt;BF^2=FC^2+BC^2 \Rightarrow (\sqrt{x^2+2.5^2})^2=(4-\sqrt{x^2+2.5^2})^2+3^2&lt;/math&gt;. After some quick math, we get &lt;math&gt;\sqrt{x^2+2.5^2}=\frac{25}{8}&lt;/math&gt;.<br /> Solving for &lt;math&gt;x&lt;/math&gt; will give &lt;math&gt;x=\frac{15}{8}&lt;/math&gt;.<br /> <br /> &lt;math&gt;\therefore&lt;/math&gt; &lt;math&gt;DF=x=&lt;/math&gt; &lt;math&gt;\boxed{\textbf{(D) } \frac{15}{8}}&lt;/math&gt;.<br /> <br /> == See Also ==<br /> <br /> {{AMC10 box|year=2018|ab=A|num-b=12|num-a=14}}<br /> {{AMC12 box|year=2018|ab=A|num-b=10|num-a=12}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Introductory Geometry Problems]]</div> Olutosinfires https://artofproblemsolving.com/wiki/index.php?title=2018_AMC_10A_Problems/Problem_12&diff=100791 2018 AMC 10A Problems/Problem 12 2019-01-23T23:07:39Z <p>Olutosinfires: /* Solutions */</p> <hr /> <div>How many ordered pairs of real numbers &lt;math&gt;(x,y)&lt;/math&gt; satisfy the following system of equations?<br /> &lt;cmath&gt;x+3y=3&lt;/cmath&gt; <br /> &lt;cmath&gt;\big||x|-|y|\big|=1&lt;/cmath&gt;<br /> &lt;math&gt;\textbf{(A) } 1 \qquad <br /> \textbf{(B) } 2 \qquad <br /> \textbf{(C) } 3 \qquad <br /> \textbf{(D) } 4 \qquad <br /> \textbf{(E) } 8 &lt;/math&gt;<br /> <br /> ==Solutions==<br /> ===Solution 1===<br /> We can solve this by graphing the equations. The second equation looks challenging to graph, but start by graphing it in the first quadrant only (which is easy since the inner absolute value signs can be ignored), then simply reflect that graph into the other quadrants.<br /> <br /> The graph looks something like this:<br /> &lt;asy&gt;<br /> draw((-3,0)--(3,0), Arrows);<br /> draw((0,-3)--(0,3), Arrows);<br /> draw((2,3)--(0,1)--(-2,3), blue);<br /> draw((-3,2)--(-1,0)--(-3,-2), blue);<br /> draw((-2,-3)--(0,-1)--(2,-3), blue);<br /> draw((3,-2)--(1,0)--(3,2), blue);<br /> draw((-3,2)--(3,0), red);<br /> dot((-3,2));<br /> dot((3/2,1/2));<br /> dot((0,1));<br /> &lt;/asy&gt;<br /> Now, it becomes clear that there are &lt;math&gt;\boxed{\textbf{(C) } 3}&lt;/math&gt; intersection points. (pinetree1) BOI<br /> <br /> ===Solution 2===<br /> &lt;math&gt;x+3y=3&lt;/math&gt; can be rewritten to &lt;math&gt;x=3-3y&lt;/math&gt;. Substituting &lt;math&gt;3-3y&lt;/math&gt; for &lt;math&gt;x&lt;/math&gt; in the second equation will give &lt;math&gt;||3-3y|-y|=1&lt;/math&gt;. Splitting this question into casework for the ranges of &lt;math&gt;y&lt;/math&gt; will give us the total number of solutions. <br /> <br /> &lt;math&gt;\textbf{Case 1:}&lt;/math&gt; &lt;math&gt;y&gt;1&lt;/math&gt;:<br /> &lt;math&gt;3-3y&lt;/math&gt; will be negative so &lt;math&gt;|3-3y| = 3y-3.&lt;/math&gt;<br /> &lt;math&gt;|3y-3-y| = |2y-3| = 1&lt;/math&gt;<br /> Subcase 1: &lt;math&gt;y&gt;\frac{3}{2}&lt;/math&gt;<br /> &lt;math&gt;2y-3&lt;/math&gt; is positive so &lt;math&gt;2y-3 = 1&lt;/math&gt; and &lt;math&gt;y = 2&lt;/math&gt; and &lt;math&gt;x = 3-3(2) = -3&lt;/math&gt;<br /> Subcase 2: &lt;math&gt;1&lt;y&lt;\frac{3}{2}&lt;/math&gt;<br /> &lt;math&gt;2y-3&lt;/math&gt; is negative so &lt;math&gt;|2y-3| = 3-2y = 1&lt;/math&gt;. &lt;math&gt;2y = 2&lt;/math&gt; and so there are no solutions (&lt;math&gt;y&lt;/math&gt; can't equal to &lt;math&gt;1&lt;/math&gt;)<br /> <br /> &lt;math&gt;\textbf{Case 2:}&lt;/math&gt; &lt;math&gt;y = 1&lt;/math&gt;:<br /> It is fairly clear that &lt;math&gt;x = 0.&lt;/math&gt;<br /> <br /> &lt;math&gt;\textbf{Case 3:}&lt;/math&gt; &lt;math&gt;y&lt;1&lt;/math&gt;:<br /> &lt;math&gt;3-3y&lt;/math&gt; will be positive so &lt;math&gt;|3-3y-y| = |3-4y| = 1&lt;/math&gt;<br /> Subcase 1: &lt;math&gt;y&gt;\frac{4}{3}&lt;/math&gt;<br /> &lt;math&gt;3-4y&lt;/math&gt; will be negative so &lt;math&gt;4y-3 = 1&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;4y = 4&lt;/math&gt;. There are no solutions (again, &lt;math&gt;y&lt;/math&gt; can't equal to &lt;math&gt;1&lt;/math&gt;)<br /> Subcase 2: &lt;math&gt;y&lt;\frac{4}{3}&lt;/math&gt;<br /> &lt;math&gt;3-4y&lt;/math&gt; will be positive so &lt;math&gt;3-4y = 1&lt;/math&gt; &lt;math&gt;\rightarrow&lt;/math&gt; &lt;math&gt;4y = 2&lt;/math&gt;. &lt;math&gt;y = \frac{1}{2}&lt;/math&gt; and &lt;math&gt;x = \frac{3}{2}&lt;/math&gt;.<br /> Thus, the solutions are: &lt;math&gt;(-3,2), (0,1), \left(\frac{3}{2},\frac{1}{2} \right)&lt;/math&gt;, and the answer is &lt;math&gt;\boxed{\textbf{(C) } 3}&lt;/math&gt;.<br /> &lt;math&gt;\text{\LaTeX}&lt;/math&gt; edit by pretzel, very minor &lt;math&gt;\text{\LaTeX}&lt;/math&gt; edits by Bryanli, very very minor &lt;math&gt;\text{\LaTeX}&lt;/math&gt; edit by ssb02<br /> <br /> ===Solution 3===<br /> Note that &lt;math&gt;||x| - |y||&lt;/math&gt; can take on either of four values: &lt;math&gt;x + y&lt;/math&gt;, &lt;math&gt;x - y&lt;/math&gt;, &lt;math&gt;-x + y&lt;/math&gt;, &lt;math&gt;-x -y&lt;/math&gt;. <br /> Solving the equations (by elimination, either adding the two equations or subtracting),<br /> we obtain the three solutions: &lt;math&gt;(0, 1)&lt;/math&gt;, &lt;math&gt;(-3,2)&lt;/math&gt;, &lt;math&gt;(1.5, 0.5)&lt;/math&gt; so the answer is &lt;math&gt;\boxed{\textbf{(C) } 3}&lt;/math&gt;. One of those equations overlap into &lt;math&gt;(0, 1)&lt;/math&gt; so there's only 3 solutions.<br /> <br /> ~trumpeter, ccx09<br /> ~minor edit, XxHalo711<br /> <br /> ===Solution 4===<br /> Just as in solution &lt;math&gt;2&lt;/math&gt;, we derive the equation &lt;math&gt;||3-3y|-|y||=1&lt;/math&gt;. If we remove the absolute values, the equation collapses into four different possible values. &lt;math&gt;3-2y&lt;/math&gt;, &lt;math&gt;3-4y&lt;/math&gt;, &lt;math&gt;2y-3&lt;/math&gt;, and &lt;math&gt;4y-3&lt;/math&gt;, each equal to either &lt;math&gt;1&lt;/math&gt; or &lt;math&gt;-1&lt;/math&gt;. Remember that if &lt;math&gt;P-Q=a&lt;/math&gt;, then &lt;math&gt;Q-P=-a&lt;/math&gt;. Because we have already taken &lt;math&gt;1&lt;/math&gt; and &lt;math&gt;-1&lt;/math&gt; into account, we can eliminate one of the conjugates of each pair, namely &lt;math&gt;3-2y&lt;/math&gt; and &lt;math&gt;2y-3&lt;/math&gt;, and &lt;math&gt;3-4y&lt;/math&gt; and &lt;math&gt;4y-3&lt;/math&gt;. Find the values of &lt;math&gt;y&lt;/math&gt; when &lt;math&gt;3-2y=1&lt;/math&gt;, &lt;math&gt;3-2y=-1&lt;/math&gt;, &lt;math&gt;3-4y=1&lt;/math&gt; and &lt;math&gt;3-4y=-1&lt;/math&gt;. We see that &lt;math&gt;3-2y=1&lt;/math&gt; and &lt;math&gt;3-4y=-1&lt;/math&gt; give us the same value for &lt;math&gt;y&lt;/math&gt;, so the answer is &lt;math&gt;\boxed{\textbf{(C) } 3}&lt;/math&gt;<br /> <br /> ~Zeric Hang<br /> <br /> ===Solution 5===<br /> Just as in solution &lt;math&gt;2&lt;/math&gt;, we derive the equation &lt;math&gt;x=3-3y&lt;/math&gt;. Squaring both sides in the second equation gives &lt;math&gt;x^2+y^2-2|xy|=1&lt;/math&gt;. Putting &lt;math&gt;x=3-3y&lt;/math&gt; and doing a little calculation gives &lt;math&gt;10y^2-18y+9-2|3y-3y^2|=1&lt;/math&gt;. From here we know that &lt;math&gt;3y-3y^2&lt;/math&gt; is either positive or negative.<br /> <br /> When positive, we get &lt;math&gt;2y^2-3y+1=0&lt;/math&gt; and then, &lt;math&gt;y=1/2&lt;/math&gt; or &lt;math&gt;y=1&lt;/math&gt;.<br /> When negative, we get &lt;math&gt;y^2-3y+2=0&lt;/math&gt; and then, &lt;math&gt;y=2&lt;/math&gt; or &lt;math&gt;y=1&lt;/math&gt;. Clearly, there are &lt;math&gt;3&lt;/math&gt; different pairs of values and that gives us &lt;math&gt;\boxed{\textbf{(C) } 3}&lt;/math&gt;<br /> <br /> ~OlutosinNGA<br /> <br /> ==See Also==<br /> {{AMC10 box|year=2018|ab=A|num-b=11|num-a=13}}<br /> {{AMC12 box|year=2018|ab=A|num-b=9|num-a=11}}<br /> {{MAA Notice}}<br /> <br /> [[Category:Intermediate Algebra Problems]]</div> Olutosinfires