# Sum of range of numbers divisble by 6 but not by 9

1. Jan 22, 2016

### cmkluza

1. The problem statement, all variables and given/known data
Find the sum of the numbers between 200 and 800 inclusive, which are multiples of 6, but not multiples of 9.

2. Relevant equations

3. The attempt at a solution
Numbers that are multiples of 6 should be: a = 6n, n ∈ ℤ and a is any multiple of six.
200 = 6nn1 = $\frac{200}{6}$ = 33.3, so the first multiple of 6 would be at n1 = 34.
800 = 6nn2 = $\frac{800}{6}$ = 133.3, so the last multiple of 6 would be at n2 = 133.
I'm then taking the sum of an arithmetic series starting at n1 = 34 and ending at n2 = 133 with a common difference d = 6. If 34 is the starting point then the number of terms I'm looking for, n, is 133 - 34 = 99.

$a_1 = 6(34) = 204$
$a_{99} = 204 + (99-1)6 = 204 + 588 = 792$
$S_{99} = \frac{99(204 + 792)}{2} = 49,302$.

How would I go about finding the sum for all numbers divisible by both 6 and 9, between 200 and 800, so as to take it out of my current sum, finding the answer?

2. Jan 22, 2016

### Samy_A

What is the least common multiple of 6 and 9?

By the way, 49302 is not the sum of all multiples of 6 between 200 and 800. (Doesn't that 792 look suspicious?)

Last edited: Jan 22, 2016
3. Jan 22, 2016

### haruspex

How many numbers in the range 1 to 2?

4. Jan 22, 2016

### 2nafish117

Σ(multiples of six)-Σ(multiples of LCM(6,9))

5. Jan 22, 2016

### SammyS

Staff Emeritus
What must be true about n, if 6n is to be divisible by 9 ?