Question

Four consecutive integers add up to 30. What is the greatest of these integers?

Answer 1

We can express the sum of these four consecutive integers in the following way:

`n+(n+1)+(n+2)+(n+3)=30`

Then, we need to sum the like terms, that is n's with n's and the integers. Then, we have:

`(n+n+n+n)+(1+2+3)=30`
`4n+6=30`

Then, solving for n, we can subtract 6 to both sides of the equation, and then divide by 4 to both sides of the equation too:

`4n+6-6=30-6\Rightarrow4n=24\Rightarrow(4)/(4)n=(24)/(4)\Rightarrow n=6`

Since n = 6, the consecutive integers are:

`6+(6+1)+(6+2)+(6+3)=6+7+8+9=30`

Therefore, the greatest of these integers is 9 (since we have 6, 7, 8, and 9).