Find three consecutive integers whose sum is 201?

1 Answer
Feb 27, 2018

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.