Correct Answer: 35

Solution : Given number = 55p1067q9
If a number is divisible by 99, it should also be divisible by 9 and 11.
For the divisibility rule of 9, the sum of the digits of the number should be divisible by 9.
⇒ 5 + 5 + p + 1 + 0 + 6 + 7 + q + 9 = m (let m be a number divisible by 9)
⇒ 27 + 6 + p + q = m
⇒ 6 + p + q = m (since 27 is divisible by 9)
⇒ p + q = m – 6
So the possible values of (p + q) are 3 and 12 (for m = 9 or 18).
For the divisibility rule of 11, the difference between the sum of odd and even place digits should be divisible by 11.
Sum of odd places = 5 + p + 0 + 7 + 9 = 21 + p
Sum of even places = 5 + 1 + 6 + q = 12 + q
The difference = 21 + p – 12 – q = n (let n be a number divisible by 11)
⇒ 9 + p – q = n
⇒ p – q = n – 9
So, the possible values of (p – q) are – 9 or 2 (for n = 0 or 11).
Only p + q = 12 and p – q = 2 gives p = 7 and q = 5
All the other combinations of values give improper values of p and q.
$\therefore$ pq = 7 × 5 = 35
Hence, the correct answer is 35.