What three consecutive integers have a sum of 4809?

Here we will use algebra to find three consecutive integers whose sum is 4809. 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 4809. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 4809

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 = 4809
3X + 3 = 4809

3X + 3 - 3 = 4809 - 3
3X = 4806

3X/3 = 4806/3
X = 1602

Which means that the first number is 1602, the second number is 1602 + 1 and the third number is 1602 + 2. Therefore, three consecutive integers that add up to 4809 are 1602, 1603, and 1604.

1602 + 1603 + 1604 = 4809

We know our answer is correct because 1602 + 1603 + 1604 equals 4809 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 4810?
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