Let N = 80pq2pq (7digit number). If N is exactly divisible by 120 then the sum…

Solution (step-by-step):

• N must be divisible by 120, and since 120 is divisible by 10, N must end in 0. Therefore q = 0.

• With q = 0 the number becomes 8 0 p 0 2 p 0 and the sum of digits is 8 + 0 + p + 0 + 2 + p + 0 = 10 + 2p.

• Divisibility by 3 requires 10 + 2p to be a multiple of 3. Testing p = 0..9 gives possible p values 1, 4, 7.

• Divisibility by 8 requires the last three digits 2p0 to be divisible by 8. Check the candidates:

• p = 1 gives 210, not divisible by 8; p = 4 gives 240, which is divisible by 8; p = 7 gives 270, not divisible by 8. So p = 4.

• Thus the digits are 8,0,4,0,2,4,0 and their sum is 8 + 0 + 4 + 0 + 2 + 4 + 0 = 18.

Answer: 18