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>. | + | 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: | + | |
| − | (a) <math>(bk)bc</math> is a perfect square, or <math>ck</math> is a perfect square. | + | We can rewrite the first and second conditions as: |
| − | (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>. | + | |
| + | (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>. | ||
| + | |||
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<= | + | |
| + | If <math>k>500</math>, then no value of <math>b</math> makes <math>1\leq a, b\leq 500</math>. | ||
| + | |||
| + | 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. | ||
| + | |||
| + | If <math>k=1</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32, 64, 128, 256</math> for <math>5\cdot 9=45</math> solutions. | ||
| + | |||
| + | If <math>k=9</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16, 32</math> for <math>5\cdot 6=30</math> solutions. | ||
| + | |||
| + | If <math>k=25</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4, 8, 16</math> for <math>5\cdot 5=25</math> solutions. | ||
| + | |||
| + | If <math>k=121</math>, then <math>c=1, 4, 16, 64, 256</math> and <math>b=1, 2, 4</math> for <math>5\cdot 3=15</math> solutions. | ||
| + | |||
| + | This is a total of <math>\fbox{115}</math> solutions. | ||
Latest revision as of 21:27, 1 March 2020
Since 
is a multiple of 
, let 
.
We can rewrite the first and second conditions as:
(a) 
is a perfect square, or 
is a perfect square.
(b) 
is a power of 
, so it follows that , 
, and 
are all powers of .
Now we use casework on 
. Since is a power of , is 
or 
or 
.
If , then no value of makes 
.
If 
or 
, then no value of that is a power of makes a perfect square.
If 
, then 
and 
for 
solutions.
If 
, then and 
for 
solutions.
If 
, then and 
for 
solutions.
If 
, then and 
for 
solutions.
This is a total of 
solutions.
