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?

Accepted Answer

E. $\frac{1}{m} = \frac{1}{12} - \frac{4}{n} = \frac{n - 48}{12n} \implies m = \frac{12n}{n - 48}$. Since $m > 0$, $n > 48$. Given $n < 60$ and $n$ is odd, possible values of $n$ are 49, 51, 53, 55, 57, 59. - For $n = 49$, $m = \frac{12 	imes 49}{1} = 588$ (integer). - For $n = 51$, $m = \frac{12 	imes 51}{3} = 204$ (integer). - For $n = 57$, $m = \frac{12 	imes 57}{9} = 76$ (integer). Other values of $n$ do not give integer $m$. So there are 3 pairs.

Number of integral pairs satisfying reciprocal equation

Explanation

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