What is the sum of all three-digit numbers that are divisible by | Math Question and Answer

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 = 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,
S_N = (
N
2
)[T₁ + (N-1)d]
⇒ S_N = (
69
2
)[2 × 104 + (69-1)(13)]
⇒ S_N = (
69
2
)[208 + 884]
⇒ S_N = (
69
2
)[1092]
⇒ S_N = 37674

You can reuse this answer
Creative Commons License