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.