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. 6
• B. 4
• C. 7
• D. 5
• E. 3

Hint:

When solving for integer solutions in fraction equations, rewrite in divisibility form and check constraints on sign and size.

Answer 1

The Correct Option is B

We are given: \[ \frac{1}{m} + \frac{4}{n} = \frac{1}{12}. \] Rewriting: \[ \frac{1}{m} = \frac{1}{12} - \frac{4}{n} = \frac{n - 48}{12n}. \] So: \[ m = \frac{12n}{n - 48}. \] Since \( m \) and \( n \) are positive integers, \( n - 48 \) must divide \( 12n \). Also, \( n \) is an odd integer less than 60. So possible \( n \) values are odd numbers from 1 to 59. Check each odd \( n>48 \) (since denominator positive): \( n = 49, 51, 53, 55, 57, 59 \). - \( n = 49 \): \( m = \frac{12 \times 49}{1} = 588 \) integer.
- \( n = 51 \): \( m = \frac{612}{3} = 204 \) integer.
- \( n = 53 \): \( m = \frac{636}{5} \) not integer.
- \( n = 55 \): \( m = \frac{660}{7} \) not integer.
- \( n = 57 \): \( m = \frac{684}{9} = 76 \) integer.
- \( n = 59 \): \( m = \frac{708}{11} = 64.36\ldots \) not integer.
Thus valid pairs: \((m, n) = (588, 49), (204, 51), (76, 57)\).
Also check \( n<48 \) odd values that make \( m \) positive integer:
For \( n<48 \), \( n - 48<0 \), which makes \( m \) negative — not allowed.
Hence total = \( 3 \) pairs.
Wait — given options show 4 possible answer choices; let's recheck:
If \( n = 59 \) fails, only 3 valid pairs exist. So answer is (5) 3.