37. How many possible values of (p + q + r) are there satisfying 1/p + 1/q + 1/r = 1, where p, q, and r are natural numbers (not necessarily distinct)?

1. A. None
2. B. One
3. C. Three
4. D. More than three

Answer: C

Explanation

We need to find combinations of natural numbers p, q, r (not necessarily distinct) such that 1/p + 1/q + 1/r = 1.
Without loss of generality, assume p ≤ q ≤ r.
* If p = 1, then 1/q + 1/r = 0, which is impossible for natural numbers q, r. So p cannot be 1.
* If p = 2:
1/2 + 1/q + 1/r = 1 => 1/q + 1/r = 1/2.
If q = 2, then 1/r = 0, impossible.
If q = 3, then 1/r = 1/2 – 1/3 = 1/6 => r = 6. So (2, 3, 6) is a solution. Sum = 11.
If q = 4, then 1/r = 1/2 – 1/4 = 1/4 => r = 4. So (2, 4, 4) is a solution. Sum = 10.
If q > 4, then 1/q 1/2 – 1/4 = 1/4. This means r < 4. But we assumed q <= r, so q must be 1/q + 1/r = 2/3.
If q = 3, then 1/r = 2/3 – 1/3 = 1/3 => r = 3. So (3, 3, 3) is a solution. Sum = 9.
If q > 3, then 1/q 2/3 – 1/3 = 1/3. This means r < 3. But we assumed q <= r, so q must be = 4, then 1/p <= 1/4. Then 1/p + 1/q + 1/r <= 1/4 + 1/4 + 1/4 = 3/4 = 4.
The distinct sets of (p, q, r) (up to permutation) are (2, 3, 6), (2, 4, 4), and (3, 3, 3).
The corresponding sums (p + q + r) are 11, 10, and 9. There are 3 distinct possible values for (p + q + r). This is a classic number theory problem often seen in competitive exams.