what's the biggest integer that can not be made up by 6, 9, and 20? what's the biggest integer that can not be made up by 6, 9, and 20?
for example, 35 can be expressed as 35 = 20 + 9 + 6
46 can be expressed as 46 = 20 * 2 + 6;
27 can be expressed as 27 = 9 * 3 or 27 = 6 * 3 + 9
so, what's the biggest integer that can not be made up by 6, 9, and 20?
 A: Finding the maximum "unreachable" number seems hard to do without a bunch of trial and error, but once found, it's relatively easy to prove, since the numbers involved are relatively small. In this case the answer is $43$.
Since it's not a multiple of $3$, the number $43$ cannot be written as a combination of $6$ and $9$ alone, but neither can $43-20=23$ or $43-2\cdot20=3$, so $43$ can't be written with one or two $20$'s either. Therefore $43$ is unreachable.
Now consider the next six numbers. Each is reachable:
$$\begin{align}
44&=4\cdot6+20\\
45&=5\cdot9\\
46&=6+2\cdot20\\
47&=3\cdot9+20\\
48&=8\cdot6\\
49&=9+2\cdot20
\end{align}$$
Consequently every number greater than these is reachable by simply adding an appropriate multiple of $6$. So $43$ is the largest unreachable number.
Remark: This problem is an instance of the three-number Frobenius coin problem, for which a good deal of research has been done. According the the Wikipedia entry, "Formulae and fast algorithms are known for three numbers though the calculations can be very tedious if done by hand." That's why I say that finding $43$ here "seems hard to do without a bunch of trial and error." If there is an easy calculation that produces it, I'd be keen to see it.
A: In simple words: we construct a set of rules for making $n+1$ from $n$ if we already have $n=6a+9b+20c$ (e.g. $n+1=6(a-5)+9(b-1)+20(c+2)$) and then we find maximum $N$ such that we can apply no rule of the set. Thus for every $n>N$ we can make $n+1$ from $n=6a+9b+20c$ (if we can find such $a,b,c$) thus we can have any number (say, by induction), starting with a reachable one.
Rigorously:
Consider $(a,b,c)\in$ $S=\{(2, 1, -1),$ $(-1, 3, -1),$ $(-2, -3, 2),$ $(5, -1, -1),$ $(1, -5, 2),$ $(-5, -1, 2),$ $(-4, 5, -1),$ $(-8, 1, 2)\}$.
For every $(a,b,c)\in S\ \ 6a+9b+20c=1$.
Suppose we have $n=6a+9b+20c$ for some $n> N$ and we want to construct $n+1$.
For which $(a,b,c)$ there does not exist $(d_a,d_b,d_c)\in S$ such, that $(a+d_a,b+d_b,c+d_c)\in \mathbb{Z}_+^3$?
We have $$\begin{cases}c<1\\b<5\\a<8\\
\left[\begin{array}{l}a<2\\b<3\end{array}\right.\\
\left[\begin{array}{l}a<5\\b<1\end{array}\right.
\end{cases}$$
and it can be shown with exhaustive search that maximum of $6a+9b+20c$ we can get with that conditions, is $42=7\cdot 6=6+4\cdot 9$, i.e. for every $(a,b,c)\in \mathbb{Z}_+^3$ if $6a+9b+20c>42$ we have such $(d_a,d_b,d_c)\in S$, that $(a+d_a,b+d_b,c+d_c)\in \mathbb{Z}_+^3$.
So starting from $44=4\cdot 6+20$ we can always add $1$ to a number we have.
$43$ is unreachable can be proven with exhaustive search for $c\le\frac{43}{20},b\le\frac{43}{9},a\le\frac{43}{6}$ that is $3\cdot 5\cdot 8=120$ cases.
Python behind the scene
