Question 643275
There is probably a much more streamlined, elegant solution, but I could not think of it yet.
Besides, I cannot speak the modular arithmetic language, and I do not know how what you are expected to know about it.
If you need the streamlined, elegant solution, ask the folks at the artofproblemsolving.com forum.
 
MY CLUMSY SOLUTION ({{{m}}} and {{{n}}} represent positive integers):
 
Combining only boxes of 6 and 9 nuggets, you can get only numbers of nuggets that are multiples of 3.
You cannot get 3 nuggets, but you can get amounts including all the multiples of 3 from 6 on.
For even multiples of 3, of the form {{{(2n)*3}}},
since {{{(2n)*3=n*(2*3)=n*6}}},
you can just use {{{n}}} boxes of 6.
For odd multiples of 3, of the form {{{(2n+1)*3}}},
since {{{(2n+1)*3=(2n-2+3)*3=(2n-2)*3+3*3=(2n-2)*3+9=2(n-1)*3+9=(n-1)*6+9}}},
you can just use 1 box of 9 and {{{n-1}}} boxes of 6.
 
Numbers that are not multiples of 3, require using at least 1 box of 20 nuggets.
Numbers that are not multiples of 3, are either 1 less than a multiple of 3, or 1 more than a multiple of 3. We can represent them as {{{3m-1}}} and {{{3m+1}}} respectively.
 
The number 20 is  of the form {{{3m-1}}}, because {{{20=21-1=3*7-1}}}.
Using 1 box of 20 nuggets, you can get 20 nuggets, 20+6=26, or  20+9=29, or 20 plus any other larger multiple of 3.
In other words, you cannot get 23, but you can get all the numbers of the form {{{3m-1}}} starting from 26.
 
Numbers of the form {{{3m+1}}} will require 2 boxes of 20 nuggets.
Since {{{40=39+1=3*13+1}}}, with 2 boxes of 20 nuggets, you can get all the numbers of the form {{{3m+1}}} too, but only starting from {{{40+6=46}}}.
The previous number of the form {{{3m+1}}} is 43, which cannot be formed by adding multiples of 20, 6, and 9.
The number {{{highlight(43)}}} is the largest number that cannot be formed by adding multiples of 20, 6, and 9.