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)

B12/23 is your answer