What three consecutive integers have a sum of 7008?

Here we will use algebra to find three consecutive integers whose sum is 7008. 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 7008. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 7008

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 = 7008
3X + 3 = 7008

3X + 3 - 3 = 7008 - 3
3X = 7005

3X/3 = 7005/3
X = 2335

Which means that the first number is 2335, the second number is 2335 + 1 and the third number is 2335 + 2. Therefore, three consecutive integers that add up to 7008 are 2335, 2336, and 2337.

2335 + 2336 + 2337 = 7008

We know our answer is correct because 2335 + 2336 + 2337 equals 7008 as displayed above.

Three Consecutive Integers
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