Question

How many pairs of positive integers m, n satisfy $\left(\frac{1}{m}\right)+\left(\frac{4}{n}\right)=\left(\frac{1}{12}\right)$, where n is an odd integer less than 60? (CAT 2007)

• A. 4
• B. 7
• C. 5
• D. 3
• E. 6

Answer 1

The correct option is D 3

Soln:d.
Given conditions :
$\left(\frac{1}{m}\right)+\left(\frac{4}{n}\right)=\left(\frac{1}{12}\right)$
M and n are positive integers
N is an odd integer less than 60.
On solving the equation, we get $m=\frac{12n}{(n-48)};$
Since m and n are positive integers, $n>48$, and from the question, $48<n<60$.
Hence possible values on $n=49,51,53,55,57,59.$
For $n=49,51$ and $57$ we get integral values of m. Hence the required no of integral pairs = 3