### What is the sum of all three-digit numbers that are divisible by 13?

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**Answer: **37674

**Step by Step Explanation: **- All numbers which are divisible by 13, forms. an arithmetic progression with differences between consecutive terms to be 13.
- We know that the smallest three-digit number is 100. On dividing 100 by 13, we get a remainder of 9.

Therefore, if we add remaining (13 - 9 = 4) to 100, resultant number (100 + 4 = 104) will be fully divisible by 13.

Therefore, the first number of the arithmetic progression is 104. - Similarly, the largest three-digit number is 999. On dividing 999 by 13 we get a remainder of 11.

Therefore, if we subtract 11 from 999, resultant number (999 - 11 = 988) will be fully divisible by 13.

Therefore, the last number of the arithmetic progression is 988. - If there are total N terms in series, N
^{th} term is given by,

T_{N} = T_{1} + (N-1)d

⇒ 988 = 104 + (N-1)(13)

⇒ 13(N - 1) = 988 - 104

⇒ 13(N - 1) = 884

⇒ N - 1 =

⇒ N - 1 = 68

⇒ N = 68 + 1

⇒ N = 69
- Now, the sum of arithmetic progression can be found using standard formula,

S_{N} = ( )[T_{1} + (N-1)d]

⇒ S_{N} = ( )[2 × 104 + (69-1)(13)]

⇒ S_{N} = ( )[208 + 884]

⇒ S_{N} = ( )[1092]

⇒ S_{N} = 37674