Question

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, the smallest possible value of p + q + r, (where p > 6) is –

• A. 12
• B. 21
• C. 45
• D. 54

Answer 1

Answer: (B)

21

Answer 2

We know that

1 + 3 + 5 + … + (2k – 1) = k²

Thus, the given equation can be written as

$\left( \frac{p + 1}{2} \right)^{2} + \left( \frac{q + 1}{2} \right)^{2} = \left( \frac{r + 1}{2} \right)^{2}$

̃ (p + 1)² + (q + 1)² = (r + 1)²

Therefore, (p + 1, q + 1, r + 1) forms a Pythagorean triplet. As p > 6, p + 1 > 7.

The first Pythagorean triplet containing a number > 7

is (6, 8, 10).

\ We may take p + 1 = 8, q + 1 = 6, r + 1 = 10

̃ p + q + r = 21.