What three consecutive integers have a sum of 2302?

Here we will use algebra to find three consecutive integers whose sum is 2302. 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 2302. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 2302

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 = 2302
3X + 3 = 2302

3X + 3 - 3 = 2302 - 3
3X = 2299

3X/3 = 2299/3
X = 766 1/3

Since 766 1/3 is not an integer, there is no true answer to this problem.

However, there are three numbers that add up to 2302. The first number is (766 1/3), the second number is (766 1/3) + 1, and the third number is (766 1/3) + 2. Therefore, we could make this the answer to "Three consecutive numbers that add up to 2302 are?":

766 1/3 + 767 1/3 + 768 1/3 = 2302

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 2303?
Here is the next algebra problem we solved.