What three consecutive integers have a sum of 3252?

Here we will use algebra to find three consecutive integers whose sum is 3252. 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 3252. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3252

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 = 3252
3X + 3 = 3252

3X + 3 - 3 = 3252 - 3
3X = 3249

3X/3 = 3249/3
X = 1083

Which means that the first number is 1083, the second number is 1083 + 1 and the third number is 1083 + 2. Therefore, three consecutive integers that add up to 3252 are 1083, 1084, and 1085.

1083 + 1084 + 1085 = 3252

We know our answer is correct because 1083 + 1084 + 1085 equals 3252 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 3253?
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