What three consecutive integers have a sum of 117?

Here we will use algebra to find three consecutive integers whose sum is 117. 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 117. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 117

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 = 117
3X + 3 = 117

3X + 3 - 3 = 117 - 3
3X = 114

3X/3 = 114/3
X = 38

Which means that the first number is 38, the second number is 38 + 1 and the third number is 38 + 2. Therefore, three consecutive integers that add up to 117 are 38, 39, and 40.

38 + 39 + 40 = 117

We know our answer is correct because 38 + 39 + 40 equals 117 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 118?
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