Today I was given the following problem.
Imagine walking into a chocolate store where they only sell boxes of $6$, $9$ or $20$ chocolates. What is the largest amount of chocolates you can't buy without eating some of it? For example, one couldn't buy $17$ chocolates from the store, as $17$ isn't a linear combination of $6$, $9$ and $20$.
Although this originally was a programming exercise, I (correctly) expected it to be more challenging when approached with only mathematics.
It all boils down to this. Given set $B = \{b_1,\, b_2,\, \dots,\, b_n \}$ of which all linear combinations with scalars in $\mathbb{N}$ define the set $V$. Let there be some $x \in V$ for which all natural numbers greater than $x$ are in $V$. So for all $m \in \mathbb{N}$ there is the implication $m \geq x \implies (m + 1) \in V$.
Because $m \in V$, and therefore $m$ is a lineair combination of elements of $B$, we can say $m = \lambda_1 b_1 + \lambda_2 b_2 + \dots + \lambda_n b_n$ (btw, might be important to note I'm using $\mathbb{N} = \{0,\, 1,\, 2,\, \dots\}$). Because we know $m + 1$ is an element of $V$ too we can do the same, thus $m + 1 = \mu_1 b_1 + \mu_2 b_2 + \dots + \mu_n b_n$. By subtracting $m$ from this last equation, we get $b_1 (\mu_1 - \lambda_1) + b_2 (\mu_2 - \lambda_2) + \dots + b_n(\mu_n - \lambda_n) = 1$. If we say $k_i = \mu_i - \lambda_i$ we get the lineair equation $b_1 k_2 + b_2 k_2 + \dots + b_n k_n = 1$, and therefore it can be written as $\vec{b} \cdot \vec{k} = 1$ with $\vec{b},\, \vec{k} \in \mathbb{N}^n$. However, what to do next?
I know this is a linear Diophantine equation, which means it has answers if at least two of the elements from $B$ are co-primes and $0$ if not, but even then, I don't know how to solve such an equation.
If I were able to, however, I suppose it would result in some set of linearly independent vectors $W$. However, if I were able to find it, I still can't see how I could extract the right value for $x$ from this.