At McDonalds you can order Chicken McNuggets in boxes of 6, 9, and 20. What is the largest number such that you can not order any combination of the above to achieve exactly the number you want?

Answer

# Problem 9 Answer

The answer is 43.Michael Shackleford, A.S.A.

Solution

For any desired number if it is divisible by 3 it can easily be made with 6 and 9 packs, except if the number is 3 itself. If you can't use all six packs then use one 9 pack and you can do the rest with six packs.If the number is not divisible by 3 then use one 20 pack. If the remaining number is divisible by 3 then use the above method for the rest.

If the number still isn't divisible by 3 use a second 20 pack. The remainder

The largest impossible number would be such that you would have to subtract 20 twice to get a remainder divisible by 3. However, you can't make 3 itself with 6 and 9 packs. So the largest impossible number is 2*20+3=43.

Michael Shackleford, A.S.A.

If the number still isn't divisible by 3 use a second 20 pack. The remainder

__must__be divisible by 3, in which case use the 6 and 9 packs as above.The largest impossible number would be such that you would have to subtract 20 twice to get a remainder divisible by 3. However, you can't make 3 itself with 6 and 9 packs. So the largest impossible number is 2*20+3=43.

Michael Shackleford, A.S.A.

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