Question

Find the sum of all three digits numbers each of which leave the remainder 3 when divided by 5.

Answer:

The sum of all three digits numbers each of which leave the remainder 3 when divided by 5 is 99090.

The smallest three digit number which leaves a remainder of 3 when divided by 5 is 100+3 = 103. The largest three digit number which leaves a remainder of 3 when divided by 5 is (1000-5)+3 = 998. Since all these numbers have a common difference, 5, they are in AP.
Let 998 be the nth term of the AP. Therefore,
103+(n-1)5 = 998 which implies that (n-1)5 = 895 which further implies that n-1 = 179 and this gives us n=180. Therefore, their sum is as follows:
180(103+998)/2 = 90(1101) = 99090.


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