Find three consecutive even integers such that the sum of the smallest to the largest 36

If they were all the same integer, then each one would be

1/3 of 36, and you would have

12 + 12 + 12 = 36.

But you don't want them to all be the same number. You

want them to be three consecutive even numbers.

Fine! Take 2 away from the first 12, and glue it onto the last 12.

Their sum doesn't change, but now you have

10 + 12 + 14 = 36.

First, I would set up an equation (assuming 36 is the sum of all 3 numbers).

x + (x+2) + (x+4) = 36

3x+6=36

-6 - 6

3x=30

- - - - - -

3 3

x=10

Now, we just plug in the numbers to find the consecutive integers

10 + (10+2) + (10+4)

10+12+14

So your 3 consecutive even integers are 10,12 and 14