What three consecutive integers have a sum of 9612?

Here we will use algebra to find three consecutive integers whose sum is 9612. 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 9612. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 9612

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 = 9612
3X + 3 = 9612

3X + 3 - 3 = 9612 - 3
3X = 9609

3X/3 = 9609/3
X = 3203

Which means that the first number is 3203, the second number is 3203 + 1 and the third number is 3203 + 2. Therefore, three consecutive integers that add up to 9612 are 3203, 3204, and 3205.

3203 + 3204 + 3205 = 9612

We know our answer is correct because 3203 + 3204 + 3205 equals 9612 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 9613?
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