How to determine if $n (n \geq 2)$ given numbers can be summed up to a certain number? I'm interested in knowing if there's a theorem for checking if combinations of $n$ given positive numbers can be summed up to a certain number. 
For example, if I was given $n = 2$ numbers, which are $k_1$ and $k_2$, and I was asked if I could make a combination like $n_1 * k_1 + n_2 *k_2 = C$, all $\in \mathbb{N}^+$, I could check remainders to see if it is possible. 
However, if $n = 3$ or higher, how could I check if combinations of $k_1, k_2, ..., k_n$ can be summed up to a $C$?
Thank you for any inputs!
 A: What you're asking is closely related to the Frobenius Coin Problem (FCP). The idea is that your $k_i$'s are denominations of coins, and we want to know which amounts we can make out of them. The difference is that the FCP requires that $n_i\ge 0$ for each $i$, while you are requiring $n_i>0$.
This difference can be reconciled as follows (using $n=3$ for illustration): The number $C$ can be formed as a positive-integer combination of $k_1, k_2, k_3$ if and only if the number $C-k_1-k_2-k_3$ can be formed as a non-negative integer combination of the same set.
For $n=2$, there is a standard result: $k_1k_2-k_1-k_2$ is the largest value that cannot be reached in the FCP. Therefore $k_1k_2$ is the largest number than cannot be reached using your requirement that each coin appear at least once in the sum.
For example, if $k_1=6$ and $k_2=11$, then we cannot form $C=66$ unless we allow one of the $n_i$ to be $0$, but we can get $67=2\cdot 6+5\cdot 11$, and every value over $67$.
Finding the right coefficients is straightforward enough, inductively:
Assume $C\ge 67$, and suppose $C=n_1\cdot 6+n_2\cdot 11$. If $n_2>1$, then we can form $C+1$ by subtracting $1$ from $n_2$ and adding $2$ to $n_1$. If $n_2=1$, then $n_1\ge10$, and we can then form $C+1$ by reducing $n_1$ by $9$, and increasing $n_2$ by $5$.

As for the values less than $66$ that can be formed, sticking with this example... there are only a few, and simply listing them is probably the easiest way to find them all. We can start with $17+11j$ for a few values of $j\ge 0$, and then $23+11j$, $29+11j$, etc.

For $n\ge3$, there is no similar nice formula, but there are algorithms that work. You can learn more here: https://en.wikipedia.org/wiki/Coin_problem
