Difference between revisions of "1962 AHSME Problems/Problem 36"
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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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Latest revision as of 15:01, 15 June 2018
Problem
If both and
are both integers, how many pairs of solutions are there of the equation
?
Solution
The equality implies and
are both powers of two; since they differ by two, it must be the case that
or
. (Note that
is not allowed because then the product is negative.) These yield
or
, for a total of
solutions.