Question

find the sum of all 3 digit numbers which leave remainder 5 on the division by 8.
pls answer ​

Answer 1

The smallest 3 digit no. = 100 is divisble by 5

=> 103 is the 1st 3 digit no. which will give rem. 3

The largest 3 digit no. = 999

999/5 gives rem. 4 => 998 will give rem. 3

this series is obviouly in AP , ( 103,108,111,...998 )

998 = 103 + ( n - 1 ) 5

=> n = 895/5 +1 = 180

So sum

= n/2 ( first term + last term )

= 180/2 ( 103 + 998 ) = 90 ( 1101 ) = 99090