Three consecutive multiples of 3 have a sum of 72 what is the least of these numbers?

Three consecutive multiples of 3 are:

3n, 3 (n+1), 3 (n+2)

We are told that the sum of these multiples is 72 so:

3n+3 (n+1) + 3 (n+2) = 72 divide both sides of the equation by 3

n+n+1+n+2=24 combine like terms on the left side

3n+3=24 subtract 3 from both sides

3n=21 divide both sides by 3

n=7

Since the smallest number was 3n, the smallest number is 3 (7) = 21

So the smallest number is 21.

(the numbers are 21, 24, 27)