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.

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