What three consecutive integers have a sum of 4887?

Here we will use algebra to find three consecutive integers whose sum is 4887. 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 4887. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4887

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 = 4887
3X + 3 = 4887

3X + 3 - 3 = 4887 - 3
3X = 4884

3X/3 = 4884/3
X = 1628

Which means that the first number is 1628, the second number is 1628 + 1 and the third number is 1628 + 2. Therefore, three consecutive integers that add up to 4887 are 1628, 1629, and 1630.

1628 + 1629 + 1630 = 4887

We know our answer is correct because 1628 + 1629 + 1630 equals 4887 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4888?
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