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

Jul 23, 2017

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:

$n + n + 1 + n + 2 = 123$

$3 n + 3 = 123$

$3 n = 120$

$n = 40$

$n + 1 = 41$

$n + 2 = 42$

Jul 24, 2017

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.

$\frac{123}{3} = 41$

$41$ 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:

$3 x + 3 = 123$

Subtract $3$ from each side.

$3 x + 3 - 3 = 123 - 3$

$3 x + \cancel{3} - \cancel{3} = 120$

$3 x = 120$

Divide each side by $3$.

$\frac{3 x}{3} = \frac{120}{3}$

$\frac{\cancel{3} x}{\cancel{3}} = 40$

$x = 40$

$\therefore x + 1 = 40 + 1 = 41$ and $x + 2 = 40 + 2 = 42$

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