What three consecutive integers have a sum of 4704?

Here we will use algebra to find three consecutive integers whose sum is 4704. 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 4704. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4704

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 = 4704
3X + 3 = 4704

3X + 3 - 3 = 4704 - 3
3X = 4701

3X/3 = 4701/3
X = 1567

Which means that the first number is 1567, the second number is 1567 + 1 and the third number is 1567 + 2. Therefore, three consecutive integers that add up to 4704 are 1567, 1568, and 1569.

1567 + 1568 + 1569 = 4704

We know our answer is correct because 1567 + 1568 + 1569 equals 4704 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4705?
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