Question

find the sum of all three digit which leave the remainder 3 when divided by 5??
ANSWER IT QUICKLY
NO SPAM!!!

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