What three consecutive integers have a sum of 6507?

Here we will use algebra to find three consecutive integers whose sum is 6507. 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 6507. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 6507

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 = 6507
3X + 3 = 6507

3X + 3 - 3 = 6507 - 3
3X = 6504

3X/3 = 6504/3
X = 2168

Which means that the first number is 2168, the second number is 2168 + 1 and the third number is 2168 + 2. Therefore, three consecutive integers that add up to 6507 are 2168, 2169, and 2170.

2168 + 2169 + 2170 = 6507

We know our answer is correct because 2168 + 2169 + 2170 equals 6507 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 6508?
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