Question

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

• A. 5
• B. 3
• C. 6
• D. 4

Answer 1

The correct option is B 3

$\frac{1}{m}=\frac{1}{12}-\frac{4}{n}\Rightarrow m=\frac{12n}{n-48}$
Since m is positive, n must be greater than 48. Possible odd values of n such that 48 < n < 60 are 49, 51, 53, 55, 57 and 59, and 49, 51, and 57 give integral values of m. Hence option (b) is the answer.