Question

find the sum of all 3 digit numbers which leave the remainder one when divided by 4

Answer 1

The first 3-digit number that leaves a remainder of 2 when divided by 3 = 101.

The next numbers being 104, 107, 110, 113,……………

The last 3-digit number that leaves a remainder of 2 when divided by 3 = 998.

So, practically, these numbers form an AP series 101, 104, 107, 110, 113,…….., 998.

First term of this AP series = 101.

Common difference = 3.

Final (n-th) term of the series = 998.

Therefore, 101 + (n - 1)*3 = 998; which leads to n = 300.

So, sum of the numbers in this AP series = 300*(101 + 998) / 2 = 164850.


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Answer 2

Step-by-step explanation:

997 divided by 4 leaves remainder 1. So 997 is the final term.