find the sum of all three digit numbers which leave the remainder 3 when divided by 7

Dear Student!

Here is the answer to your query.

The three digits number which leave the remainder 3 when divided by 7 are

101, 108, 115, ..., 997

Clearly the sequence forms an A.P. with

First term (*a*) = 101

Common difference (*d*) = 108 – 101 = 7

and *n*^{th} term = 997

⇒ *a* + (*n* – 1)*d* = 997

⇒ 101 + (*n* – 1) 7 = 997

⇒ (*n* – 1) 7 = 997 – 101 = 896

⇒ *n* = 128 + 1 = 129

Now

Hence sum of all three digits number which leaves the remainder 3 when divided by 7 is 70821.

Cheers!

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