Three consecutive multiples of 3 have a sum of 36.

Let 3n be the first number, 3n + 3  be the second number and 3n + 6 be the third number....then....

3n + [3n+ 3] + [ 3n + 6]  = 36    simplify

9n +  9  = 36     subtract 9 from both sides 

9n =  27    divide both sides by 9

n = 3

So.....the first number 3(3)   = 9     tjhe second = 3(3) + 3  =  12     and the last = 3(3) + 6  = 15

And   9 + 12 + 15 =  36

So...the greatest number is 15

Set up an equation.  Your equation should look like this: n+(n+3)+(n+6)=36

n is the smallest number,  therefore n+6 is the largest.

To simplify your equation,  it looks like:  3n+9=36

Subtract 9 from both sides.

3n+9-9=36-9

3n=27

Now divide both sides by 3.

3n/3=27/3

n=9

The smallest number is 9.

To find the largest number,  add 6 to n.

9 + 6 = 15

The greatest number is 15.

The middle number is 9 + 3 = 12.  The middle number is 12.

We can go back and check our work.

9 + 12 + 15 = 36

9,  12,  and 15 are consecutive multiples of 3.

Therefore,  your answer is 15.

@-Happy7

OK, H7....quit showing off...you're making me look bad....LOL!!!!

{ Actually....you're getting pretty good at this stuff ]