Difference between revisions of "Mock AIME I 2015 Problems/Problem 9"

(Created page with "Since <math>a</math> is a multiple of <math>b</math>, let <math>a=kb</math>.<math>\newline</math> We can rewrite the first and second conditions as:<math>\newline</math> (a) <...")
 
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Since <math>a</math> is a multiple of <math>b</math>, let <math>a=kb</math>.<math>\newline</math>
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Since <math>a</math> is a multiple of <math>b</math>, let <math>a=kb</math>.
We can rewrite the first and second conditions as:<math>\newline</math>
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We can rewrite the first and second conditions as:
(a) <math>(bk)bc</math> is a perfect square, or <math>ck</math> is a perfect square.<math>\newline</math>
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(a) <math>(bk)bc</math> is a perfect square, or <math>ck</math> is a perfect square.
(b) <math>b(k+7)c</math> is a power of <math>2</math>, so it follows that <math>b</math>, <math>c</math>, and <math>k+7</math> are all powers of <math>2</math>.<math>\newline</math>
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(b) <math>b(k+7)c</math> is a power of <math>2</math>, so it follows that <math>b</math>, <math>c</math>, and <math>k+7</math> are all powers of <math>2</math>.
 
Now we use casework on <math>k</math>. Since <math>k+7</math> is a power of <math>2</math>, <math>k</math> is <math>1, 9, 25, 57, 121,</math> or <math>249</math> or <math>k>500</math>.
 
Now we use casework on <math>k</math>. Since <math>k+7</math> is a power of <math>2</math>, <math>k</math> is <math>1, 9, 25, 57, 121,</math> or <math>249</math> or <math>k>500</math>.
If <math>k>500</math>, then no value of b makes <math>1<=a, b<=500</math>.
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If <math>k>500</math>, then no value of <math>b</math> makes <math>1\leq a, b\leq 500</math>.
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If <math>k=57</math> or <math>k=249</math>, then no value of <math>c</math> that is a power of <math>2</math> makes <math>ck</math> a perfect square.
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If <math>k=1</math>, then <math>c=1, 4, 16, 256</math> and <math>b=1, 2, 4, 8, 16, 32, 64, 128, 256</math> for <math>4\dot 9=36</math> solutions.
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If <math>k=9</math>, then <math>c=1, 4, 16, 256</math> and <math>b=1, 2, 4, 8, 16, 32</math> for <math>4\dot 6=24</math> solutions.
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If <math>k=25</math>, then <math>c=1, 4, 16, 256</math> and <math>b=1, 2, 4, 8, 16</math> for <math>4\dot 5=20</math> solutions.
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If <math>k=121</math>, then <math>c=1, 4, 16, 256</math> and <math>b=1, 2, 4</math> for <math>4\dot 3=12</math> solutions.
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This is a total of <math>\fbox{092}</math> solutions.

Revision as of 22:25, 10 January 2019

Since $a$ is a multiple of $b$, let $a=kb$. We can rewrite the first and second conditions as: (a) $(bk)bc$ is a perfect square, or $ck$ is a perfect square. (b) $b(k+7)c$ is a power of $2$, so it follows that $b$, $c$, and $k+7$ are all powers of $2$. Now we use casework on $k$. Since $k+7$ is a power of $2$, $k$ is $1, 9, 25, 57, 121,$ or $249$ or $k>500$. If $k>500$, then no value of $b$ makes $1\leq a, b\leq 500$. If $k=57$ or $k=249$, then no value of $c$ that is a power of $2$ makes $ck$ a perfect square. If $k=1$, then $c=1, 4, 16, 256$ and $b=1, 2, 4, 8, 16, 32, 64, 128, 256$ for $4\dot 9=36$ solutions. If $k=9$, then $c=1, 4, 16, 256$ and $b=1, 2, 4, 8, 16, 32$ for $4\dot 6=24$ solutions. If $k=25$, then $c=1, 4, 16, 256$ and $b=1, 2, 4, 8, 16$ for $4\dot 5=20$ solutions. If $k=121$, then $c=1, 4, 16, 256$ and $b=1, 2, 4$ for $4\dot 3=12$ solutions. This is a total of $\fbox{092}$ solutions.

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