Difference between revisions of "1962 AHSME Problems/Problem 36"
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It is obvious that x>10 , thus x=12 , y=3. | It is obvious that x>10 , thus x=12 , y=3. | ||
Then, let <math>x-8=2n</math> , <math>x-10=2n-2</math> , it can be written as <math>2n*(2n-2) = 2^y</math> , | Then, let <math>x-8=2n</math> , <math>x-10=2n-2</math> , it can be written as <math>2n*(2n-2) = 2^y</math> , | ||
− | Also, <math>n*(n-1) = 2^y-2</math> , so, n only can be 2 , y=3, and the answer is <math>\boxed{B}</math>. | + | Also, <math>n*(n-1) = 2^(y-2)</math> , so, n only can be 2 , y=3, and the answer is <math>\boxed{B}</math>. |
Revision as of 05:26, 1 December 2017
Problem
If both and
are both integers, how many pairs of solutions are there of the equation
?
Solution
It is obvious that x>10 , thus x=12 , y=3.
Then, let ,
, it can be written as
,
Also,
, so, n only can be 2 , y=3, and the answer is
.