What is the largest number of nuggets that you cannot order using any combination of the above?

Question

At McDonald's you can order Chicken McNuggets in boxes of 6,9, and 20. What is the largest number of nuggets that you cannot order using any combination of the above?

Answer 1

Consider that if n is any multiple of 3 other than 3 itself, you can order n nuggets by ordering:

n/9 boxes of 9 if n≡0 mod 9
⌊n/9⌋ boxes of 9 and one box of 6, if n≡6 mod 9
⌊n/9⌋-1 boxes of 9 and 2 boxes of 6, if n≡3 mod 9

Now, if n≡2 mod 3, then n-20≡0 mod 3, so assuming if n≡2 mod 3 and n-20>3 (i.e. that n>23), then you can order n nuggets by ordering one box of 20 and n-20 nuggets using the scheme above.

Finally, if n≡1 mod 3, then n-40≡0 mod 3, so if n>43, then either:

n≡0 mod 3, so you can order n nuggets using the scheme in the first paragraph, or
n≡1 mod 3 and n-40>3 (since n>43), so you can order two boxes of 20 and n-40 nuggets using the scheme in the first paragraph, or
n≡2 mod 3, so you can order one box of 20 and n-20 nuggets using the scheme in the first paragraph.

Thus you can order any number of nuggets greater than 43. You cannot, however, order 43 nuggets exactly. Why? Suppose you order no boxes of 20 -- then the number of nuggets you order will be congruent to 0 mod 3, and thus cannot be 43. Similarly, if you order exactly one box of 20, then the number of nuggets ordered will be congruent to 2 mod 3, and thus cannot be 43. So you must order at least 2 boxes of 20, and then would have to find some way to order the last 3 nuggets using some combination of boxes of 6, 9, and 20 nuggets -- clearly impossible, since all of them have more than 3 nuggets.

Therefore, 43 is the largest number of nuggets that cannot be ordered using any combination of the above.

Answer 2

All multiples of any of the 3 numbers are attainable as well as all multiples of 3 above 3 itself. This pattern of excluding multiples or 3 is reapeated after each multiple of 20 (for example 23 is unattainable while all other multiples of 3 added to 20 are not, or in another way of saying, all multiples of 3 added to 2) This is key. After the second multiple of 20 (40), with the exception of 43, all multiples of 3, multiples of 3 added to 2, and multiples of 3 added to 1 (in other words ALL integers whatsoever) are now excluded. So your answer is 43.

Answer 3

Tough question, because if you really wanted to, you could get into really high numbers. For example, you can't order 1,001 nuggets. Neither can you order 200,978 nuggets. You could keep going higher and higher as long as the number can't be divided by 6, 9, or 20.

Answer 4

an interesting question... sarai_kristi and remix are wrong. 1001 can be ordered by 9 times 9 and 46 times 20... i'll try to figure it out and edit this answer

Answer 5

any number larger than 20 that isn't divisible by 6, 9, or 20.

Answer 6

Haha. That number is infinite.

Answer 7

69,69 and 2 more fingers make it a 71

Answer 8

that question doesn't make any sense!!!

Answer 9

1,000,000.32659428465769874549876546354321346435

Answer 10

1,000,000,437,695,484,657,698,745,49......