What three consecutive integers have a sum of 9303?

Here we will use algebra to find three consecutive integers whose sum is 9303. 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 9303. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 9303

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 = 9303
3X + 3 = 9303

3X + 3 - 3 = 9303 - 3
3X = 9300

3X/3 = 9300/3
X = 3100

Which means that the first number is 3100, the second number is 3100 + 1 and the third number is 3100 + 2. Therefore, three consecutive integers that add up to 9303 are 3100, 3101, and 3102.

3100 + 3101 + 3102 = 9303

We know our answer is correct because 3100 + 3101 + 3102 equals 9303 as displayed above.

Three Consecutive Integers
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What three consecutive integers have a sum of 9304?
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