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

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.

@@ Line 5: / Line 5: @@
 ==Solution==
-It is obvious that x>10 , thus x=12 , y=3.
+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.
-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>.

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

Latest revision as of 14:01, 15 June 2018

Problem

Solution