What three consecutive integers have a sum of 1968?

Here we will use algebra to find three consecutive integers whose sum is 1968. 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 1968. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 1968

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 = 1968
3X + 3 = 1968

3X + 3 - 3 = 1968 - 3
3X = 1965

3X/3 = 1965/3
X = 655

Which means that the first number is 655, the second number is 655 + 1 and the third number is 655 + 2. Therefore, three consecutive integers that add up to 1968 are 655, 656, and 657.

655 + 656 + 657 = 1968

We know our answer is correct because 655 + 656 + 657 equals 1968 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 1969?
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