What three consecutive integers have a sum of 9318?

Here we will use algebra to find three consecutive integers whose sum is 9318. 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 9318. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 9318

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 = 9318
3X + 3 = 9318

3X + 3 - 3 = 9318 - 3
3X = 9315

3X/3 = 9315/3
X = 3105

Which means that the first number is 3105, the second number is 3105 + 1 and the third number is 3105 + 2. Therefore, three consecutive integers that add up to 9318 are 3105, 3106, and 3107.

3105 + 3106 + 3107 = 9318

We know our answer is correct because 3105 + 3106 + 3107 equals 9318 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 9319?
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