How many 3 digit numbers are such that when divided by 7, leave a remainder 3 in each case.

The first and last numbers between 100 and 999 which on divisibility by 7 leaves the remainder 3 are 101 and 997 respectively.

Since, all the numbers between 100 and 999 which are divisible by 7 and leaves the remainder 3 are in AP, where, *a* = 101, *l *= 997 and *d* = 7.

*Now, l* = *a* + (*n* – 1)*d*

Hence, there are 129 numbers between 100 and 999 which are exactly divisible by 7 and leaves the remainder 3.

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