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


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Answer:

37674

Step by Step Explanation:
  1. All numbers which are divisible by 13, forms. an arithmetic progression with differences between consecutive terms to be 13.
  2. 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.
  3. 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.
  4. 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
  5. Now, the sum of arithmetic progression can be found using standard formula,
    SN = ( 
    N
    2
     )[T1 + (N-1)d]
    ⇒ SN = ( 
    69
    2
     )[2 × 104 + (69-1)(13)]
    ⇒ SN = ( 
    69
    2
     )[208 + 884]
    ⇒ SN = ( 
    69
    2
     )[1092]
    ⇒ SN = 37674

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