What three consecutive integers have a sum of 3808?

Here we will use algebra to find three consecutive integers whose sum is 3808. 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 3808. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3808

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 = 3808
3X + 3 = 3808

3X + 3 - 3 = 3808 - 3
3X = 3805

3X/3 = 3805/3
X = 1268 1/3

Since 1268 1/3 is not an integer, there is no true answer to this problem.

However, there are three numbers that add up to 3808. The first number is (1268 1/3), the second number is (1268 1/3) + 1, and the third number is (1268 1/3) + 2. Therefore, we could make this the answer to "Three consecutive numbers that add up to 3808 are?":

1268 1/3 + 1269 1/3 + 1270 1/3 = 3808

Three Consecutive Integers
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What three consecutive integers have a sum of 3809?
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