What three consecutive integers have a sum of 4980?

Here we will use algebra to find three consecutive integers whose sum is 4980. 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 4980. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4980

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 = 4980
3X + 3 = 4980

3X + 3 - 3 = 4980 - 3
3X = 4977

3X/3 = 4977/3
X = 1659

Which means that the first number is 1659, the second number is 1659 + 1 and the third number is 1659 + 2. Therefore, three consecutive integers that add up to 4980 are 1659, 1660, and 1661.

1659 + 1660 + 1661 = 4980

We know our answer is correct because 1659 + 1660 + 1661 equals 4980 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4981?
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