Find three consecutive whole numbers such that the sum of the squares of the numbers is equal to 869?

12 Answers

  • Numbers are: 16, 17. and 18

  • Numbers are: 16, 17. and 18

  • Numbers are: 16, 17. and 18

  • Let the 3 consecutive #s be n-1 , n , n+1. Sum of squares = 3n^2 +2. Then

    3n^2 = 867, ie., n^2 = 289 = (17)^2, ie., n = 17 & #s are 16,17,18.

  • You can estimate it very quickly.

    Divide the sum by 3:

    869/3 = 289 2/3

    That must be close to the square of the middle number. Square root of that is 17.0195… The numbers must be 16 17 18.

    Work out he square to verify that.

  • Let the consecutive whole number be ‘n’, ‘n+1’, ‘n+2’ .

    Their squares are

    n^2 , (n+1)^2 , (n+2)^2

    Their sum is

    n^2 + (n^2 + 2n + 1) + ( n^2 + 4n + 4) = 869

    3n^2 + 6n + 5 = 869

    3n^2 + 6n = 864

    n^2 + 2n = 288

    (n + 1)^2 – (1)^2 = 288

    (n + 1)^2 = 289

    n + 1 = sqrt(289) = 17

    Hence n = 16

    & n + 2 = 18

    Hence the three consecutive numbers are 16,17,& 18.

  • x² + (x + 1)² + (x + 2)² = 869

    x² + x² + 2x + 1 + x² + 4x + 4 = 869

    3x² + 6x – 864 = 0

    x² + 2x – 288 = 0

    [ x + 18 ] [ x – 16 ] = 0

    Accept x = 16

    Integers are 16 , 17 and 18

  • Let n – 1, n, and n + 1 be the consecutive numbers

    Then

    (n – 1)^2 + n^2 + (n + 1)^2 = 869

    n^2 – 2n + 1 + n^2 + n^2 + 2n + 1 = 869

    3n^2 + 2 = 869

    3n^2 = 867

    n^2 = 287

    n = √287

    = 17

    Numbers are: 16, 17. and 18

  • (x – 1)^2 + x^2 + (x + 1)^2 = 869

    x^2 – 2x + 1 + x^2 + x^2 + 2x + 1 = 869

    3x^2 + 2 = 869

    3x^2 = 867

    x^2 = 289

    x = 17

    x – 1 = 16

    x + 1 = 18

    hence the numbers are: 16, 17, 18

  • Did you try 16, 17, and 18?

    You would be adding three numbers to add up to 869. So the average is 289.67. The square root of 289.67 is 17.02. So I would try numbers around 17. The first set to try would be 16, 17, and 18. This happens to work, so it’s the solution.

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