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

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==Solution==
 
==Solution==
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It is obvious that x>10 , thus x=12 , y=3.
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The equality implies <math>x-8</math> and <math>x-10</math> are both powers of two; since they differ by two, it must be the case that <math>(x-8,x-10) = (4,2)</math> or <math>(x-8,x-10) = (-2,-4)</math>.  (Note that <math>(1,-1)</math> is not allowed because then the product is negative.) These yield <math>(x,y) = (12,3)</math> or <math>(x,y) = (6,3)</math>, for a total of <math>\boxed{2\textbf{ (C)}}</math> solutions.
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Then, let x-8=2n , x-10=2n-2 , it can be written as 2n*(2n-2) = 2^y ,
 
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Also, n (n-1) = 2^y-2 , so, n only can be 2 , y=3, and the answer is <math>\boxed{B}</math>.
 

Latest revision as of 14:01, 15 June 2018

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

The equality implies $x-8$ and $x-10$ are both powers of two; since they differ by two, it must be the case that $(x-8,x-10) = (4,2)$ or $(x-8,x-10) = (-2,-4)$. (Note that $(1,-1)$ is not allowed because then the product is negative.) These yield $(x,y) = (12,3)$ or $(x,y) = (6,3)$, for a total of $\boxed{2\textbf{ (C)}}$ solutions.

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