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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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I loved it ! But want that explanation in a short way because we can attempt in a less period of time