What three consecutive integers have a sum of 408?

Here we will use algebra to find three consecutive integers whose sum is 408. We start by assigning X to the first integer. Since they are consecutive, it means that the 2nd number will be X + 1 and the 3rd number will be X + 2 and they should all add up to 408. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 408

To solve for X, you first add the integers together and the X variables together. Then you subtract three from each side, followed by dividing by 3 on each side. Here is the work to show our math:

X + X + 1 + X + 2 = 408
3X + 3 = 408

3X + 3 - 3 = 408 - 3
3X = 405

3X/3 = 405/3
X = 135

Which means that the first number is 135, the second number is 135 + 1 and the third number is 135 + 2. Therefore, three consecutive integers that add up to 408 are 135, 136, and 137.

135 + 136 + 137 = 408

We know our answer is correct because 135 + 136 + 137 equals 408 as displayed above.

Three Consecutive Integers
Enter another number below to find what three consecutive integers add up to its sum.

What three consecutive integers have a sum of 409?
Here is the next algebra problem we solved.