Question
CSATMediumPrelims 2025Maths

How many possible values of (p+q+r)(p + q + r) are there satisfying 1p+1q+1r=1\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1, where p, q, and r are natural numbers (not necessarily distinct)?

Explanation

We need triples (p, q, r) of natural numbers with 1p+1q+1r=1\frac{1}{p} + \frac{1}{q} + \frac{1}{r} = 1.

All solutions (up to permutation) are:

  • (4, 4, 2): 14+14+12=1\frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1p+q+r=10p+q+r = 10
  • (2, 3, 6): 12+13+16=1\frac{1}{2} + \frac{1}{3} + \frac{1}{6} = 1p+q+r=11p+q+r = 11
  • (3, 3, 3): 13+13+13=1\frac{1}{3} + \frac{1}{3} + \frac{1}{3} = 1p+q+r=9p+q+r = 9

Distinct values of (p + q + r): {9, 10, 11} → 3 values.

Hence, the answer is Three.

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