Find the sum of three positive consecutive integers a, b, c such that a^2 + b^2 + c^2 = 110.

Guest Jan 5, 2020

#2**0 **

*Find the sum of three positive consecutive integers a, b, c such that a^2 + b^2 + c^2 = 110.*

a, b, & c are consecutive, and a^{2} + b^{2} + c^{2} = 110

A third of 110 is about 33 so start from there.

The closest square is 36, so the next two going up would be 49 and 64.

You can easily see that that total is way too large, try the other direction.

The closest square is 36, so the next two going down would be 25 and 16.

That total is way too small, so try the one below 36 and the one above it.

36 + 25 + 49 = 110 ... Like the Baby Bear's porridge, it's just right.

So a, b, & c are 5, 6, & 7, respectively, and so 5 + 6 + 7 = **18** and there's your final answer.

_{.}

Guest Jan 5, 2020

#3**+1 **

Middle number = x the other two numbers are x-1 and x+1

(x-1)^2 + x^2 + (x+1)^2 = 110

x^2 - 2x +1 + x^2 + x^2 + 2x +1 = 110

3x^2 +2 = 110

3x^2 = 108

x^2 = 36

x = 6 x-1 = 5 x+1 = 7 5+6+7 =18

ElectricPavlov Jan 5, 2020