Forty three, or McNugget Numbers

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I am 43 years old, as of today.  

I don’t normally think much about such things (turning 40 was an exception to that one), so I’ve been surprised by the way “43” has impressed me, looming on the horizon.  I was trying to figure out what was going on with that, and eventually realized it was the fact that 43 seemed like such a dull number.

So I set out to discover cool facts about the number 43.

There were a few obvious things:  43 is prime, for example [but there are LOTS of those].  It can be written as a sum of consecutive integers in two ways (21+22, or -42+-41+…+42+43), but again, no big deal there.

Inspired by a recent observation that no matter what base your number system is, you’ll always describe it as being base 10 [akin to the old joke: “There are 10 kinds of people, those who understand binary and those who don’t.”], I was playing with alternate bases — maybe 43 has cooler properties in some other base.  The best thing I’ve found along that path is 43 = 21_{21}.  How cool is that?   Well, maybe not the coolest thing ever, but I now have a new favorite piecewise linear function: f(n) = n_n (n base n).

And then I finally found some cool math involving 43!  The Wikipedia page on 43 pointed out that 43 is the largest natural number that is not a McNugget Number; that is, you cannot buy 43 McNuggets at McDonald’s by buying the usual packages of 6, 9, and 20 McNuggets each.  But you can get 44, 45, 46, … and all larger numbers of McNuggets.  For example, 44 = 20 + 24 = 20 + 6×4, so you can order four packages of 6 McNuggets, and 1 pack of 20, and have your 44.  45 is five nine-packs, 46 is a sixer and two 20s.  But if you need to purchase exactly 43 McNuggets, you’re out of luck.

One way to verify that 43 cannot be so gotten:  It is easy to tell if a small number is a McNugget number (e.g. it is obvious that 8 is not).  And note that if a larger number is a McNugget number, then subtracting 6, 9, or 20 should give another McNugget number.  So we start by letting A = {43}, and at each stage we replace each member x of A with x-6, x-9, and x-20.  If at any point we get a member of A that is a McNugget number, reversing that path would show that 43 is also a McNugget number. (And conversely, if 43 were a McNugget number, such a path would exist.)

We have: 

Initial step:  {43}

Second step: {37, 34, 23}  [so e.g. if 34 could be built out of 6, 9, and 20, then adding 9 gives 43]

Third step: {31, 28, 17, 28, 25, 14, 17, 14, 3} = {31, 28, 25, 17, 14, 3}  [we can ignore 3, obviously not a McNugget number]

Fourth step: {25, 22, 11, 22, 19, 8, 19, 16, 5, 11, 8, 8, 5} = {25, 22, 19, 16, 11, 8, 5} 

Fifth step: {19, 16, 5, 16, 13, 2, 13, 10, 10, 7, 5, 2, 2} = {19, 16, 13, 10, 7, 5, 2}

Sixth step: {13, 10, 10, 7, 7, 4, 4, 1, 1} = {13, 10, 7, 4, 1}

Seventh step: {7, 4, 4, 1, 1} = {7, 4, 1}

At this point we can probably stop, as we clearly cannot build any of these three numbers as sums of sixes, nines and twenties, and hence we cannot build 43 as such a sum either.

Proving that every number greater than 43 is a McNugget number is a nice application of generalized induction:  one shows that 44, 45, 46, 47, 48, and 49 are McNugget numbers as the base case (by direct calculation for each).  The inductive step is to assume that n > 49 and that every number between 44 and n-1 is a McNugget number.  Then n-1 > n-6 > 43, so n-6 is a McNugget number, and hence n = (n-6) + 6 is another McNugget number.  Thus every number greater than 43 is a McNugget number.

Okay, I’ve decided 43 is cool after all. 

Perhaps I’ll celebrate with a birthday lunch…

6 Responses to “Forty three, or McNugget Numbers”

  1. jd2718 Says:

    You are younger than me. And prime. But back when I was almost 43 (and prime), I posted the puzzle and got some nice discussion in the comments, from basic “How do I do this?” to some extensions.

    Jonathan

  2. Ξ Says:

    Happy Birthday TwoPi! You should have brought McNuggets to your class today as a treat. 🙂

    I think the 21 base 21 is neat. I was thinking about that, and from 10 [which is 10 base 10] to 990 [which is 99 base 99] the ab mod ab numbers seem to go up by 2’s [11 base 11 is 12 (base 10), 12 base 12 is 14 (base 11) Then the gap increases at 20, 30, etc.– it’s interesting to think about.

    Thanks Batman for pointing out the pattern!

  3. dpaste Says:

    Found you via Puntabulous. This post made my head hurt. Bad memories of high school math class also resurfaced. But Happy Birthday. I will be McNugget-ing in October.

  4. Maria H. Andersen Says:

    Happy Birthday! Prime birthdays are always nice, all though I did quite like 2^5 myself. : )

  5. mbork Says:

    Happy birthday (although a bit late;))!

    And for another nice (birthday) puzzle; I used to tell my students back in 2001 that my age equals the sum of the digits of the year I was born. It’s a pity it doesn’t work anymore…

    PS. I’m looking forward to being 2^32 years old!

  6. TwoPi Says:

    Naturally that sum of digits puzzle is only usable for one year. It is tempting to try to reuse it by using the sum of the digits of the calendar date (e.g. 14/8/1977), but sadly you lose uniqueness once you go that route.

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