What three consecutive integers have a sum of 4125?

Here we will use algebra to find three consecutive integers whose sum is 4125. 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 4125. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4125

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 = 4125
3X + 3 = 4125

3X + 3 - 3 = 4125 - 3
3X = 4122

3X/3 = 4122/3
X = 1374

Which means that the first number is 1374, the second number is 1374 + 1 and the third number is 1374 + 2. Therefore, three consecutive integers that add up to 4125 are 1374, 1375, and 1376.

1374 + 1375 + 1376 = 4125

We know our answer is correct because 1374 + 1375 + 1376 equals 4125 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4126?
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