Question

Using algebra, how do you find the smallest three consecutive integers whose sum is greater than 20?

Answer 1

Find that the three integers are: `6, 7, 8`

Explanation:

Suppose the middle consecutive integer is `n`.

Then we want:

`20 < (n-1) + n + (n+1) = 3n`

Dividing both ends by `3` we find:

`n > 20/3 = 6 2/3`

So the smallest integer value of `n` which satisfies this is `n = 7`, making the three integers: `6, 7, 8`