The sum of 3 consecutive integral numbers is 117. Find the numbers.?

7 Answers

  • I think you mean Consecutive integers (integrals are calculus material)

    So:

    x + (x+1) + (x+2) = 117

    3x + 3 = 117

    3x = 114

    x = 38

    so, 38, 39, 40

  • 38+39+40 = 117

    Source(s): 🙂 x
  • x+(x+1)(x+2)=117

    x+x+1+x+2=117

    3x + 3 = 117

    3x = 114

    x = 38

    38+1 = 39

    38+2 = 40

    Check 38+39+40 = 117

    so three consecutive integral numbers are 38 ,39 and 40

  • Let the three consecutive numbers be n-1, n, n+1 then their sum is 3n i.e. 3n=117 therefore n=39 and the three numbers are 38, 39,40

  • a+b+c=117

    b=a+1

    c=a+2

    a+a+1+a+2=117

    Therefore: 3a=117-3=114 . So a=38

    The numbers are: 38, 39 and 40.

  • 38, 39 and 40

  • n + (n+1) + (n + 2) = 117

    3n + 3 = 117

    3n = 114

    n = 38

    so 38, 39, 40

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