What three consecutive integers have a sum of 1972?

Here we will use algebra to find three consecutive integers whose sum is 1972. 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 1972. Therefore, you can write the equation as follows:

(X) + (X + 1) + (X + 2) = 1972

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 = 1972
3X + 3 = 1972

3X + 3 - 3 = 1972 - 3
3X = 1969

3X/3 = 1969/3
X = 656 1/3

Since 656 1/3 is not an integer, there is no true answer to this problem.

However, there are three numbers that add up to 1972. The first number is (656 1/3), the second number is (656 1/3) + 1, and the third number is (656 1/3) + 2. Therefore, we could make this the answer to "Three consecutive numbers that add up to 1972 are?":

656 1/3 + 657 1/3 + 658 1/3 = 1972

Three Consecutive Integers
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What three consecutive integers have a sum of 1973?
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