Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Solution)
(→‎Solution)
Line 7: Line 7:
 
It is obvious that x>10 , thus x=12 , y=3.
 
It is obvious that x>10 , thus x=12 , y=3.
 
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>.

Revision as of 04:26, 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 $\boxed{B}$.

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1962_AHSME_Problems/Problem_36&oldid=88735"