What three consecutive integers have a sum of 6408?

Here we will use algebra to find three consecutive integers whose sum is 6408. 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 6408. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 6408

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 = 6408
3X + 3 = 6408

3X + 3 - 3 = 6408 - 3
3X = 6405

3X/3 = 6405/3
X = 2135

Which means that the first number is 2135, the second number is 2135 + 1 and the third number is 2135 + 2. Therefore, three consecutive integers that add up to 6408 are 2135, 2136, and 2137.

2135 + 2136 + 2137 = 6408

We know our answer is correct because 2135 + 2136 + 2137 equals 6408 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 6409?
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