Difference between revisions of "Mock AIME I 2015 Problems/Problem 9"
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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=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\ | + | 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\ | + | 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\ | + | 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. | This is a total of <math>\fbox{115}</math> solutions. | ||
Latest revision as of 20:27, 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.
