What three consecutive integers have a sum of 6810?

Here we will use algebra to find three consecutive integers whose sum is 6810. 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 6810. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 6810

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 = 6810
3X + 3 = 6810

3X + 3 - 3 = 6810 - 3
3X = 6807

3X/3 = 6807/3
X = 2269

Which means that the first number is 2269, the second number is 2269 + 1 and the third number is 2269 + 2. Therefore, three consecutive integers that add up to 6810 are 2269, 2270, and 2271.

2269 + 2270 + 2271 = 6810

We know our answer is correct because 2269 + 2270 + 2271 equals 6810 as displayed above.

Three Consecutive Integers
Enter another number below to find what three consecutive integers add up to its sum.

What three consecutive integers have a sum of 6811?
Here is the next algebra problem we solved.