What three consecutive integers have a sum of 2142?

Here we will use algebra to find three consecutive integers whose sum is 2142. 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 2142. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 2142

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 = 2142
3X + 3 = 2142

3X + 3 - 3 = 2142 - 3
3X = 2139

3X/3 = 2139/3
X = 713

Which means that the first number is 713, the second number is 713 + 1 and the third number is 713 + 2. Therefore, three consecutive integers that add up to 2142 are 713, 714, and 715.

713 + 714 + 715 = 2142

We know our answer is correct because 713 + 714 + 715 equals 2142 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 2143?
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