What is the sum of all three-digit numbers that are divisible by 13?
Answer:
37674
- 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, Nth term is given by,
TN = T1 + (N-1)d
⇒ 988 = 104 + (N-1)(13)
⇒ 13(N - 1) = 988 - 104
⇒ 13(N - 1) = 884
⇒ N - 1 =884 13
⇒ N - 1 = 68
⇒ N = 68 + 1
⇒ N = 69 - Now, the sum of arithmetic progression can be found using standard formula,
SN = (
)[T1 + (N-1)d]N 2
⇒ SN = (
)[2 × 104 + (69-1)(13)]69 2
⇒ SN = (
)[208 + 884]69 2
⇒ SN = (
)[1092]69 2
⇒ SN = 37674