What three consecutive integers have a sum of 7608?

Here we will use algebra to find three consecutive integers whose sum is 7608. 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 7608. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 7608

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 = 7608
3X + 3 = 7608

3X + 3 - 3 = 7608 - 3
3X = 7605

3X/3 = 7605/3
X = 2535

Which means that the first number is 2535, the second number is 2535 + 1 and the third number is 2535 + 2. Therefore, three consecutive integers that add up to 7608 are 2535, 2536, and 2537.

2535 + 2536 + 2537 = 7608

We know our answer is correct because 2535 + 2536 + 2537 equals 7608 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 7609?
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