Difference between revisions of "1962 AHSME Problems/Problem 36"
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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>. | ||
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+ | THE ABOVE IS WRONG |
Revision as of 02:16, 14 June 2018
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
.
THE ABOVE IS WRONG