What three consecutive integers have a sum of 3708?

Here we will use algebra to find three consecutive integers whose sum is 3708. 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 3708. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3708

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 = 3708
3X + 3 = 3708

3X + 3 - 3 = 3708 - 3
3X = 3705

3X/3 = 3705/3
X = 1235

Which means that the first number is 1235, the second number is 1235 + 1 and the third number is 1235 + 2. Therefore, three consecutive integers that add up to 3708 are 1235, 1236, and 1237.

1235 + 1236 + 1237 = 3708

We know our answer is correct because 1235 + 1236 + 1237 equals 3708 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 3709?
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