What three consecutive integers have a sum of 3567?

Here we will use algebra to find three consecutive integers whose sum is 3567. 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 3567. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 3567

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 = 3567
3X + 3 = 3567

3X + 3 - 3 = 3567 - 3
3X = 3564

3X/3 = 3564/3
X = 1188

Which means that the first number is 1188, the second number is 1188 + 1 and the third number is 1188 + 2. Therefore, three consecutive integers that add up to 3567 are 1188, 1189, and 1190.

1188 + 1189 + 1190 = 3567

We know our answer is correct because 1188 + 1189 + 1190 equals 3567 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 3568?
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