Question

Find three consecutive integers whose sum is 201?

Answer 1

`66, 67, 68`

Explanation:

Let's start with one integer, and call it `x.`

We now want two integers each consecutive to `x`, or each `1` above x. These will each be `1` above `x`, because integers do not contain decimals.

So, for our first integer consecutive to `x`, we have `x+1`, as this is `1` above `x`.

For our second integer consecutive to `x,` we have `x+2,` as `x+2` is consecutive to `x+1`.

Let's set the sum our integers equal to `201` and solve for `x:`

`(x)+(x+1)+(x+2)=201`

Combine like terms. This means combining all regular integers and combining all `x.`

`3x+3=201`

Solve for `x:`

`3xcancel(+3-3)=201-3`

`3x=198`

`cancel3x/cancel3=198/3`

`x=66`

So, our first integer, `x`, is `66`. The integers consecutive to this are just `x+1` and `x+2,` or `66+1=67` and `66+2=68`.