Wednesday, June 23, 2010

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?

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.