99 = 9 × 11, where 9 and 11 are co-primes. Clearly, a number is divisible by 99 if it is divisible by both 9 and 11. Since the number is divisible by 9, we have: (3 + 4 + 2 + 2 + 2 + 1 + 3 + p + q) = a multiple of 9 ⇒ 17 + (p + q) = 18 or 27 ⇒p+q=1 ...(i) or p + q = 10 ...(ii) Since the number is divisible by 11, we have: (q + 3 + 2 + 2 + 3) – (p + 1 + 2 + 4) = 0 or a multiple of 11 ⇒ (10 + q) – (7 + p) = 0 or 11 ⇒ 3 + (q – p) = 0 or 11 ⇒ q – p = – 3 or q – p = 8 ⇒p–q=3 ...(iii) or q – p = 8 ...(iv) Clearly, if (i) holds, then neither (iii) nor (iv) holds. So, (i) does not hold. Also, solving (ii) and (iii) together, we get: p = 6.5, which is not possible. Solving (ii) and (iv) together, we get: p = 1, q = 9.
Explanation as extracted from the printed page; notation may be imperfect.