If you add a multiple of 6 to a multiple of 9 you always get a multiple of which number When I add a multiple of 5 with a multiple of 9, I always get a multiple of.......
I know that the multiples of 5 are 5, 10, 15, 20, 25...
and the multiples of 9 are 9, 18, 27, 36, 45....
If I add 5+9=14, I get a multiple of 2,7,14
If I add 10+18= 28, I get a multiple of 2,7,14, 28
If I add 15+27=42, I get a multiple of 2,7,6, 14, 42
If I add 18+15=33, I get a multiple of 3,11 and 33
I cannot find any common multiple of something that will always exist when I add a multiple of 5 with a multiple of 9.  
 A: $$9n+6m=3\cdot3n+2\cdot 3m=3(3n+2m)$$
Since $n$ and $m$ are whole, you will get a multiple of $3$. This works because both $6$ and $9$ are multiples of $3$.

Edit: 
If you want to solve the problem for $9$ and $5$ instead, we have $9n+5m$.
This will only be a multiple of some number for some pairs of $n$ and $m$. For instance, if $m=3k$ for any $k$, then $9n+15k$ will always be divisible by $3$ (since $3$ again is a common factor).
A: When you add a multiple of $n$ to a multiple of $k$, you always get a multiple of at least $\gcd(n,k)$.
In the case of $5$ and $9$, you always get a multiple of at least $\gcd(5,9)=1$.
A: The general answer is that the set of  multiples of $a$ added to a multiple of $b$ is the set of multiples of $\gcd(a,b)$.
More than that, it even is the abstract definition of a g.c.d. in PIDs.
A: If you add $10$ (a multiple of $5$) to $9$, then you get $19$, which is prime.
In particular, $19$ is only a multiple of $1$ and itself, $19$.
So these are the only potential candidates to answer your question: $1$ and $19$.
Yes, every sum will be a multiple of $1$, since they are all whole numbers.
Is every such sum divisible by $19$? Well, $5 + 9 = 14$ is not divisible by $19$; so, no.
Therefore, the only answer is $1$.
A: If two numbers both have a common divisor:  Say $14 = 2*7$ and $21 = 3*7$  Any sum of multiple.  $21*k + 14*m = 3*7k + 2*7m = 7(3k + 2m)$ will be a multiple of the common divisor.  
So any multiple of $6$ and $9$ will be a multiple of $3$ because $6k + 9m = 2*3k + 3*3m = 3(2k + 3m)$.
There is a result called Euclid's algorithm that show that you can find a sum or difference of multiples of  $M$ and $K$ so that $Mk + Kj = \gcd(M,K)$.
So to answer your question.  If you add a sum of any multiple of M to a multiple of K what must the result be a multiple of; the result will be a multiple of any common divisors but common divisors are the only numbers it has to be a multiple of.
If you add a sum of multiples of $28$ and $12$ every sum must be a multiple of $1,2$ and $4$ as those are the common divisors.  Nothing else has to happen.
If you add a sum of multiples of $5$ and $9$ every sum must be a multiple of ....$1$ as $1$ is the only common divisor.
Note:  $1$ is always a common divisor.
(Also note: I'm talking about whole numbers.)
