# Three consecutive multiples of 3 have a sum of 36. What is the greatest number?

Nov 28, 2016

The greatest of the three numbers is 15.

The other two numbers are 9 and 12.

#### Explanation:

The three consecutive multiples of 3 can be written as;

$x$, $x + 3$ and $x + 6$ with $x + 6$ being the greatest.

We know from the problem the sum of these three numbers equal 36 so we can write and solve for $x$ through the following:

$x + x + 3 + x + 6 = 36$

$3 x + 9 = 36$

$3 x + 9 - 9 = 36 - 9$

$3 x = 27$

$\frac{3 x}{3} = \frac{27}{3}$

$x = 9$

Because we are looking for the largest we must add $6$ to $x$ to obtain the largest number:

$6 + 19 = 15$

Nov 28, 2016

15

#### Explanation:

A multiple of 3 can be written $3 n$ where $n$ is a positive integer.
So 3 consecutive multiples of 3 can be written $3 n , 3 n + 3 , 3 n + 6$
The sum of these is 36
$3 n + 3 n + 3 + 3 n + 6 = 36$
$9 n + 9 = 36$
Divide through by 9
$n + 1 = 4$
$n = 3$
If $n = 3$ then $3 n = 9$ and the three consecutive multiples of three are 9, 12 and 15 which do indeed total 36