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AP ICET 2016 Paper

Question : 133 of 200

Marks: +1, -0

The sum of all the 3 digit numbers which leave the remainder 1 when divided by 4 is

124575
123475
123525
124475

Solution:

The first and the last 3-digit numbers divisible by 4 are 100 and 996. So, the 3-digit numbers that would leave a remainder 1 in each case are 101, 105, .............., 997

Total number of such 3-digits numbers = 225

It forms an A.P with n= 225, a = 101, l = 997

∴ Required sum

= \frac{n}{2}(a+1) = \frac{225}{2}(101+997)

=123525

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