Nuggets in packs of 6, 9 and 20.

Question:
At a fast food restaurant you can buy chicken nuggets in packs of 6, 9 or 20. Is there such a number N, that for all numbers greater than it, it is possible to buy any number of chicken nuggets?

Answer:
Keep in mind the key fact we need to prove that numbers are not possible. The equation we have is 6x + 9y + 20z = n. The minimum step we can take is 6 so lets work from there.
If we assume y and z are 0 then the equation becomes 6x = n. Hence the possible numbers become

 1  2  3  4  5  x
 7  8  9 10 11  x
13 14 15 16 17  x 
19 20 21 22 23  x 

If we assume y=1, z=0 then the equation becomes 9 + 6x = n. Hence the possible numbers become

 1  2  3  4  5  x
 7  8  x 10 11  x
13 14  x 16 17  x 
19 20  x 22 23  x 

Any further 9s will take us back to the same used columns so more 9s is pointless that leaves 20. If we assume y=0, z=1 then the equation becomes 20 + 6x = n

 1  2  3  4  5  x
 7  8  x 10 11  x
13 14  x 16 17  x
19  x  x 22 23  x
25  x  x 28 29  x
31  x  x 34 35  x
37  x  x 40 41  x
43  x  x 46 47  x

If we assume y=1, z=1 then the equation becomes 20 + 9 + 6x = 29 + 6n = n

 1  2  3  4  5  x
 7  8  x 10 11  x
13 14  x 16 17  x
19  x  x 22 23  x
25  x  x 28  x  x
31  x  x 34  x  x
37  x  x 40  x  x
43  x  x 46  x  x

If we go y=2,z=1 now then it becomes 38 + 6n which is already taken. So we try another 20 If we assume y=0, z=2 then the equation becomes 40 + 6x = n

 1  2  3  4  5  x
 7  8  x 10 11  x
13 14  x 16 17  x
19  x  x 22 23  x
25  x  x 28  x  x
31  x  x 34  x  x
37  x  x  x  x  x
43  x  x  x  x  x
 x  x  x  x  x  x

If we assume y=1, z=2 then the equation becomes 40 + 9 + 6x = 49 + 6x = n

Hence N = 43.