What three consecutive integers have a sum of 4908?

Here we will use algebra to find three consecutive integers whose sum is 4908. 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 4908. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4908

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 = 4908
3X + 3 = 4908

3X + 3 - 3 = 4908 - 3
3X = 4905

3X/3 = 4905/3
X = 1635

Which means that the first number is 1635, the second number is 1635 + 1 and the third number is 1635 + 2. Therefore, three consecutive integers that add up to 4908 are 1635, 1636, and 1637.

1635 + 1636 + 1637 = 4908

We know our answer is correct because 1635 + 1636 + 1637 equals 4908 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4909?
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