What three consecutive integers have a sum of 3909?

Here we will use algebra to find three consecutive integers whose sum is 3909. 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 3909. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3909

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 = 3909
3X + 3 = 3909

3X + 3 - 3 = 3909 - 3
3X = 3906

3X/3 = 3906/3
X = 1302

Which means that the first number is 1302, the second number is 1302 + 1 and the third number is 1302 + 2. Therefore, three consecutive integers that add up to 3909 are 1302, 1303, and 1304.

1302 + 1303 + 1304 = 3909

We know our answer is correct because 1302 + 1303 + 1304 equals 3909 as displayed above.

Three Consecutive Integers
Enter another number below to find what three consecutive integers add up to its sum.

What three consecutive integers have a sum of 3910?
Here is the next algebra problem we solved.