What three consecutive integers have a sum of 710?

Here we will use algebra to find three consecutive integers whose sum is 710. 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 710. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 710

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 = 710
3X + 3 = 710

3X + 3 - 3 = 710 - 3
3X = 707

3X/3 = 707/3
X = 235 2/3

Since 235 2/3 is not an integer, there is no true answer to this problem.

However, there are three numbers that add up to 710. The first number is (235 2/3), the second number is (235 2/3) + 1, and the third number is (235 2/3) + 2. Therefore, we could make this the answer to "Three consecutive numbers that add up to 710 are?":

235 2/3 + 236 2/3 + 237 2/3 = 710

Three Consecutive Integers
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What three consecutive integers have a sum of 711?
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