find the sum of all 3 digit numbers which leave remainder 3 when divided by 5.
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Step-by-step explanation:
The smallest 3 digit no. = 100 is divisible 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 obviously 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
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