The smallest possible value of p + q + r (where p > 6 ), if 1 + 3 + 5 + . . . + p + 1 + 3 + 5 + . . . + q = 1 + 3 + 5 + . . . + r where each set of parentheses contains the sum of consecutive odd integers as shown.

Step-by-step solution

We know that 1 + 3 + 5 + .... + (2k - 1) = k 2 Thus, the given equation can be written as p + 1 2 2 + q + 1 2 2 = r + 1 2 2 ⇒ p + 1 2 + q + 1 2 = r + 1 2 Therefore, p + 1 ,     q + 1 ,     r + 1 forms a Pythagoren triple. As     p > 6 ,     p + 1 > 7 . The first Pythagorean triple containing a number > 7 is (6, 8, 10). We may take p + 1 = 8, q + 1 = 6, r + 1 = 10 ⇒ p + q + r = 21