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?

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
