# 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.