Question

The sum of three consecutive integers is 123. What would be the three integers?

Answer 1

The 3 integers are 40, 41, and 42.

Explanation:

Let n = the first integer, then n + 1, is second, and n + 2, is the third:

`n + n + 1 + n + 2 = 123`

`3n + 3 = 123`

`3n = 120`

`n = 40`

`n + 1 = 41`

`n + 2 = 42`

Answer 2

The three integers are `40`, `41`, and `42`.

Explanation:

There are two ways to solve this problem.

A] Since we are looking for three consecutive integers that add up to `123`, if we divide the total by `3`, we get the average, which is also the middle integer.

It follows that the first and third integers will be that integer minus and plus one respectively.

`123/3=41`

`41` is the middle integer.

`41-1=40`

`40` is the first integer.

`41+1=42`

`42` is the third integer.

Hence, the three integers are `40`, `41`, and `42`.

---

B] The other way to solve this problem involves very basic algebra (even though this question is in Prealgebra), in which we represent the first integer as `x`, in which case the second and third integers will be `x+1` and `x+2`.

Since we know the sum of these three integers is `123`, we can write:

`x+x+1+x+2=123`

Adding the `x` separately and the numbers separately, we get:

`3x+3=123`

Subtract `3` from each side.

`3x+3-3=123-3`

`3x+cancel3-cancel3=120`

`3x=120`

Divide each side by `3`.

`(3x)/3=120/3`

`(cancel3x)/(cancel3)=40`

`x=40`

`:.x+1=40+1=41` and `x+2=40+2=42`

Hence the three consecutive integers are `40`, `41`, and `42`.