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Given N = 80pq2pq is exactly divisible by 120.

Since the last digit of the dividend is 0, the last digit of N also should be 0.

Then obviously, q = 0.

i.e., N = 80p02p0

Now we have to find p.

We know 120 = 3 x 5 x 8 and 3, 5 and 8 are co-primes.

If N is divisible by 120 then it is divisible by 3, 5 and 8.

We know that, \"If a number is divisible by 3 then the sum of its digits

also divisible by 3\"

Then, 8 + 0 + p + 0 + 2 + p + 0 = 10 + 2p

ie., 10 + 2p must be divisible by 3

Then the possible values of p are 1,4,7,10,13 and so on

Since p is a digit, the possible values are 1, 4, 7

Now, N must be divisible by 8 also.

We know that, \"If a number is divisible by 8 then its last 3 digits

also divisible by 8\"

Here 2p0 is a multiple of 8.

Now put all the above possible values of p then we have 2p0 = 210 or 240 or 270

From these, 240 is a multiple of 8.

Then the value of p = 4.

and N = 8040240

Hence the required sum = 8 + 0 + 4 + 0 + 2 + 4 + 0 = 18

**Video solution:**

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