What three consecutive integers have a sum of 1812?

Here we will use algebra to find three consecutive integers whose sum is 1812. 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 1812. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 1812

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 = 1812
3X + 3 = 1812

3X + 3 - 3 = 1812 - 3
3X = 1809

3X/3 = 1809/3
X = 603

Which means that the first number is 603, the second number is 603 + 1 and the third number is 603 + 2. Therefore, three consecutive integers that add up to 1812 are 603, 604, and 605.

603 + 604 + 605 = 1812

We know our answer is correct because 603 + 604 + 605 equals 1812 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 1813?
Here is the next algebra problem we solved.