What three consecutive integers have a sum of 2001?

Here we will use algebra to find three consecutive integers whose sum is 2001. 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 2001. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 2001

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 = 2001
3X + 3 = 2001

3X + 3 - 3 = 2001 - 3
3X = 1998

3X/3 = 1998/3
X = 666

Which means that the first number is 666, the second number is 666 + 1 and the third number is 666 + 2. Therefore, three consecutive integers that add up to 2001 are 666, 667, and 668.

666 + 667 + 668 = 2001

We know our answer is correct because 666 + 667 + 668 equals 2001 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 2002?
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