Question

Find the sum of all the three -digit numbers which leaves remainder 3 when divided by 7

Answer 1

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Hey Mate !!

Here's your answer !!

The first 3 digit number that leaves a remainder 3 when divided by 7 = 101

The last 3 digit number that leaves a remainder 3 when divided by 7 = 997

The common difference is 7 

So number of terms = ?

An = a + (n - 1) d

997 = 101 + (n-1) 7

997 - 101 = (n-1) 7

896 = (n-1) 7

896 / 7 = (n-1)

128 = (n-1)

n = 128 + 1 = 129 terms.

So sum of all the three -digit numbers which leaves remainder 3 when

divided by 7 are :-

Sn = n/2 (A + An )

Sn = 129 / 2 ( 101 + 997 )

Sn = 129 / 2 * 1098

Sn = 129 * 549

Sn = 70821 

So the sum of all the 3 digit numbers which leaves a remainder 3 when

divided by 7 is 70821.

Hope this helps !!

Cheers !!

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