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Donut Dilemma: Newsletter Challenge (January 2021)

Posted December 31, 2020 5:01 PM
Pathfinder Tags: challenge question donuts

This month's GlobalSpec Newsletter Challenge is:

A local restaurant sells donuts in boxes of 6, 9 and 20.
What is the greatest number of donuts that cannot be purchased (in that exact amount), no matter which boxes are purchased?

Check back later this month for the answer, right here on CR4.

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#1

Re: Donut Dilemma: Newsletter Challenge (January 2021)

12/31/2020 5:52 PM

43...

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#2

Re: Donut Dilemma: Newsletter Challenge (January 2021)

12/31/2020 9:26 PM

Donuts? It's actually the infamous McNugget number. Above 43, any integer can be expressed as a multiple of 6, 9, and 20, i.e. a*6 + b*9 + c*20.

https://mathworld.wolfram.com/McNuggetNumber.html

It sounds like something Paul Erdos would be involved in.

(Actually, Frobenius)

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#3
In reply to #2

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/01/2021 11:22 AM

The proof is not that difficult. I wrote a Matlab script to break down numbers 1 - 50.

1 Not Purchasable

2 Not Purchasable

3 Not Purchasable

4 Not Purchasable

5 Not Purchasable

6 = 0x20 + 0x9 + 1x6

7 Not Purchasable

8 Not Purchasable

9 = 0x20 + 1x9 + 0x6

10 Not Purchasable

11 Not Purchasable

12 = 0x20 + 0x9 + 2x6

13 Not Purchasable

14 Not Purchasable

15 = 0x20 + 1x9 + 1x6

16 Not Purchasable

17 Not Purchasable

18 = 0x20 + 2x9 + 0x6

19 Not Purchasable

20 = 1x20 + 0x9 + 0x6

21 = 0x20 + 1x9 + 2x6

22 Not Purchasable

23 Not Purchasable

24 = 0x20 + 2x9 + 1x6

25 Not Purchasable

26 = 1x20 + 0x9 + 1x6

27 = 0x20 + 3x9 + 0x6

28 Not Purchasable

29 = 1x20 + 1x9 + 0x6

30 = 0x20 + 2x9 + 2x6

31 Not Purchasable

32 = 1x20 + 0x9 + 2x6

33 = 0x20 + 3x9 + 1x6

34 Not Purchasable

35 = 1x20 + 1x9 + 1x6

36 = 0x20 + 4x9 + 0x6

37 Not Purchasable

38 = 1x20 + 2x9 + 0x6

39 = 0x20 + 3x9 + 2x6

40 = 2x20 + 0x9 + 0x6

41 = 1x20 + 1x9 + 2x6

42 = 0x20 + 4x9 + 1x6

43 Not Purchasable

44 = 1x20 + 2x9 + 1x6

45 = 0x20 + 5x9 + 0x6

46 = 2x20 + 0x9 + 1x6

47 = 1x20 + 3x9 + 0x6

48 = 0x20 + 4x9 + 2x6

49 = 2x20 + 1x9 + 0x6

50 = 1x20 + 2x9 + 2x6

Each number breaks down into n = 20a + 9b + 6c. There are 6 numbers in a row after 43 that can be "purchased". If number n is "purchaseable", n+6 can be purchased just be increasing 'c' by one. So once you have 6 in a row, you're home free for the rest of the integers.

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#4

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/01/2021 12:24 PM

six is 2 x 3 and 9 is 3 x 3

Using 2 and 3 you can easily get any number above 2

If the number is odd subtract 3 then divide by 2 to get the number of 2s

So all multiples of three are possible above 6.

Because 20 is 2 mod 3, above and including 26 all numbers equal to 0 mod 3 or 2 mod 3 are possible.

And because 40 is 1 mod 3, above and including 46 all numbers are possible.

45 is 0 mod 3

44 is 2 mod 3

43 is 1 mod 3 and so is the greatest number not possible.

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#5
In reply to #4

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/01/2021 2:39 PM

Boxes come with 6 donuts, 9 donuts, and 20 donuts inside. You can't buy part of a box. The question is "what number of donuts can you buy made up of a combination of full boxes?"

In #3, a is the number of 20 donut boxes, b the number of 9 donut boxes, and c the number of 6 donut boxes. The total number of donuts purchased is n = 20a + 9b + 6c. Not every number n can be broken down with a, b, and c as integers.

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#7
In reply to #5

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/02/2021 6:17 AM

Pardon?

I'm not splitting any boxes.

Perhaps the first couple of lines are not clear. I'm basically just saying that with 6 and 9, you can make any number ≥ 6 which is a multiple of 3 (0 modulo 3).

From there because 20 is 2 modulo 3: above 26 inclusive any number which is 0 or 2 modulo 3 is possible. And, because 40 is 1 modulo 3: above 46 inclusive any number which is 0, 1 or 2 modulo 3 is possible. So we only need to examine 45 (0 mod3 ) 44 (2 mod 3) and 43 which is the answer.

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#8
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Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/02/2021 11:38 AM

I see where you're coming from now. Every even 0 modulo 3 is taken care of by "6 boxes" (2 x 3) and every odd 0 modulo 3 is taken care of by a "9 box" (3 x 3). So you can get the "even" 0 modulo 3s with the "six boxes" and substitute in a "nine box" to get the odd 0 modulo 3s.

When you get up where you can use the "20 box", you have 2 extra donuts, so the same pattern from the "six" and "nine" boxes is shifted up by 2 donuts, and when you can use 2 "20 boxes", the same pattern is shifted up by 4 (or up by 1).

Very clever! I like it!

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#6

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/01/2021 9:10 PM

You've got to know your doughnut limit....!

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#9

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/04/2021 10:26 AM

Someone is making a lot of dough out of selling all these do[ugh]nuts...

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#10

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/05/2021 10:19 AM

It's possible to cut one (toroidal) donut into two interlocked rings. With some latitude on definition of shape, an answer becomes impossible.

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#11
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Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/05/2021 11:58 AM

Clearly it's possible. Just try to cut a möbius strip into thirds, you end up with two interlocked rings one the same length as the original, and one twice as long.

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#13
In reply to #11

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/05/2021 5:23 PM

Yet again, you defeat my efforts to find a loophole. I'm just being a bit 'hissy' waiting for the official answer to last Challenge Question (though fair enough if there hasn't been time to post it). It may not be pragmatic, but half the fun is in considering various interpretations.

Just a minor aside, the donut cutting situation was one I first saw on QI. It's perhaps a little added fun for those who have solved this question.

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#14
In reply to #11

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/07/2021 2:32 AM

"A smart mathematician confided

that a Möbius strip is one-sided.

You get quite a laugh

When you cut it in half

As it stays in one piece when divided." - Anon

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#12

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/05/2021 5:03 PM

"What is the greatest number" Infinity of course!

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#15
In reply to #12

Re: Donut Dilemma: Newsletter Challenge (January 2021)

01/07/2021 2:35 AM

<...Infinity...>

It is going to prove difficult to purchase that number, indeed...

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