What three consecutive integers have a sum of 3912?

Here we will use algebra to find three consecutive integers whose sum is 3912. 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 3912. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3912

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 = 3912
3X + 3 = 3912

3X + 3 - 3 = 3912 - 3
3X = 3909

3X/3 = 3909/3
X = 1303

Which means that the first number is 1303, the second number is 1303 + 1 and the third number is 1303 + 2. Therefore, three consecutive integers that add up to 3912 are 1303, 1304, and 1305.

1303 + 1304 + 1305 = 3912

We know our answer is correct because 1303 + 1304 + 1305 equals 3912 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 3913?
Here is the next algebra problem we solved.