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Question 126

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

Solution

The First 3 digit number, which leaves a remainder of 1 when divided by 4 is 101 and the last is 997.

Total number of terms is 225. Hence applying the formula of AP and finding the sum

$$\frac{225}{2}\left(997+101\right)=\frac{1098}{2}\times\ 225=123525$$

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