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) ?

• Assume without loss of generality that p ≤ q ≤ r to avoid counting permutations of the same set of values.

• Since 1/p + 1/q + 1/r = 1 and p ≤ q ≤ r, we must have 3/p ≥ 1, implying p ≤ 3. Also, 1/p < 1, so p > 1. Thus, p can be 2 or 3.

• If p = 2, the equation becomes 1/q + 1/r = 1/2. Since q ≥ 2, 2/q ≥ 1/2 implies q ≤ 4. Also q > 2 (otherwise 1/q + 1/r > 1/2). So q can be 3 or 4.

• If q = 3, 1/r = 1/2 - 1/3 = 1/6, so r = 6. The set is {2, 3, 6}, sum = 11.

• If q = 4, 1/r = 1/2 - 1/4 = 1/4, so r = 4. The set is {2, 4, 4}, sum = 10.

• If p = 3, the equation becomes 1/q + 1/r = 2/3. Since q ≥ 3, 2/q ≥ 2/3 implies q ≤ 3. Thus q = 3.

• If q = 3, 1/r = 2/3 - 1/3 = 1/3, so r = 3. The set is {3, 3, 3}, sum = 9.

• The possible values for (p + q + r) are 11, 10, and 9. There are exactly three distinct values.

• Therefore, the correct option is (c) Three.