What is the sum of all 3 digit numbers that leave a remainde

The use of arithmetic formula will help you find the sum of such integers quickly.

We want all three digit integers that gives remainder of 2 upon dividing by 3.

some important points regarding this problem.

1) there are only 3 possible remainders for any integer upon dividing by 3 which are 0, 1, 2. You will get the remainder of 2 when the integer being divided is 1 less than the integer that is evenly divisible by 3. First of such kind of integer in 101 and last is 998 (since 102 and 999 are evenly divisible by 3)

2) the next integer that gives remainder of 2 when divided by 3 after 101 will be 101+3 i.e. 104. We can form here the sequence of all the integers that gives remainder of 2 when divided by 3 in which every integer will be 3 greater than the previous integer. 101, 104, 107, 110, ...........,998.

3) So we have arithmetic progression with first term 101, last term 998, and common difference 3.

4) Formula to calculate number of terms in the sequence with common difference 3 is
(i) Including both ends [(last term - first term)/3] + 1
(ii) Including only one end [(last term - first term)/3]
(iii) Excluding both ends [(last term - first term)/3] - 1

5) Since we want both ends to be included in the counting, we will use the first case. So number of terms is [(998 - 101)/3] +1 ---------> 299 + 1 -------> 300

6) Sum of arithmetic progression with first term a, last term l and number of terms n will be n(a+l)/2 -------> 300(1099)/2 ---------> 164850

Hope that's clear.