# 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 nth 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!

• 38
as the number leaves remainder 3 when divided by 5 , and of three digits
therfore the first number will be 103
second number will be 108 (103+5) as 5 is the common difference
third number 113 (103+2*5)
......................
last number of the series will be 998 (995+3)
now total number = n
An = a+(n-1)d
998= 103+(n-1)5
895=(n-1)5
n=179+1 =180

now Sn = n/2 {2a+(n-1)d}
Sn =180/2{2*103+(180-1)5}
sn= 90{206+895}
Sn=1101*90 =99090 = answer
• 6
What are you looking for?