What three consecutive integers have a sum of 5412?

Here we will use algebra to find three consecutive integers whose sum is 5412. 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 5412. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 5412

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 = 5412
3X + 3 = 5412

3X + 3 - 3 = 5412 - 3
3X = 5409

3X/3 = 5409/3
X = 1803

Which means that the first number is 1803, the second number is 1803 + 1 and the third number is 1803 + 2. Therefore, three consecutive integers that add up to 5412 are 1803, 1804, and 1805.

1803 + 1804 + 1805 = 5412

We know our answer is correct because 1803 + 1804 + 1805 equals 5412 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 5413?
Here is the next algebra problem we solved.