Three consecutive multiples of 3 have a sum of 36.

What is the greatest number?

GAMEMASTERX40 Nov 10, 2016

#2**+5 **

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

CPhill Nov 10, 2016

#1**+5 **

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

happy7 Nov 10, 2016

#2**+5 **

Best Answer

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

CPhill Nov 10, 2016