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Difference between revisions of "1962 AHSME Problems/Problem 36"

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==Solution==
 
==Solution==
 
It is obvious that x>10 , thus x=12 , y=3.
 
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 ,
+
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, n (n-1) = 2^y-2 , 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 04:25, 1 December 2017

Problem

If both $x$ and $y$ are both integers, how many pairs of solutions are there of the equation $(x-8)(x-10) = 2^y$?

$\textbf{(A)}\ 0\qquad\textbf{(B)}\ 1\qquad\textbf{(C)}\ 2\qquad\textbf{(D)}\ 3\qquad\textbf{(E)}\ \text{more than 3}$

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 $\boxed{B}$.

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