Difference between revisions of "1962 AHSME Problems/Problem 36"
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| − | { | + | It is obvious that x>10 , thus x=12 , y=3. |
| + | Then, let x-8=2n , x-10=2n-2 , it can be written as 2n*(2n-2) = 2^y , | ||
| + | Also, n (n-1) = 2^y-2 , so, n only can be 2 , y=3, and the answer is <math>\boxed{B}</math>. | ||
Revision as of 05:24, 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 x-8=2n , x-10=2n-2 , it can be written as 2n*(2n-2) = 2^y ,
Also, n (n-1) = 2^y-2 , so, n only can be 2 , y=3, and the answer is 
.
