Q37. 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)?

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 < 1/4, so 1/r = 1/2 - 1/q > 1/2 - 1/4 = 1/4. This means r < 4.

But we assumed q <= r, so q must be <= r. This means q can't be greater than 4. * If p = 3: 1/3 + 1/q + 1/r = 1 => 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 < 1/3. So 1/r = 2/3 - 1/q > 2/3 - 1/3 = 1/3. This means r < 3. But we assumed q <= r, so q must be <= r. This means q can't be greater than 3.

* If p >= 4, then 1/p <= 1/4. Then 1/p + 1/q + 1/r <= 1/4 + 1/4 + 1/4 = 3/4 < 1. So no solutions for p >= 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.