Archive for the 'Papers' Category

Mayan Number Paper…Revealed!

Ok, so the paper I co-wrote with my professor has finally been published! I didn’t even get any warning. About two weeks ago a packet shows up on my doorstep containing five copies of the magazine Mathematics Teacher with my article in it. The subject of our article was even on the front cover!


From what I can tell, Mathematics Teacher is basically like an industry magazine for math teachers, going over various teaching topics and ideas and the like, which is cool. I never heard about it before but, a publication is a publication.

Unfortunately, my contribution seems to be almost entire cut out of the article entirely. It’s not too surprising given that I only solved on specific problem in a textbook and the article is about teaching Mayan numerals in general, but is still a bit disappointing. At least my name is still on the byline:


So anyway, what was the problem I solved? Well to understand that, let’s delve a little bit into the Mayan number system. Mayan numbers are actually not too different from ours. They are positional, so each digit is a multiple of some base unit like ours. For example, in our system, each number is multiplied by a factor of ten. So 50 is 5 × 10, 3,000 is 3 × 103­, and so on. Each time you move a digit to the left, you’re multiplying it by 10. Same thing in the Mayan system, except that for them, you’re multiplying by twenty. Also, the Mayans wrote their numbers from top to bottom, instead of left to right, but that’s just a small detail.

The Mayans had unique symbols for each digit from 0 to 19 to stick in for each digit (well really, they only had three symbols, the dot (for 1), the line (for 5), and the shell (for 0). Kind of similar to tally marks). I’m not going to use the symbols for this post since it would be too confusing and too much work for me (you can see them here), so I’m going to use the typical notation of A.B.C.D… and so on, where each letter is a digit and the larger digit places are towards the left. So, for a quick example, 1024 in our system would be 2.15.4.

The only really weird thing about the Mayan numeral system is that there is a little “jog” in it. While almost every place is twenty times the value of the previous place, the third place (which would be the 202, or 400s place) is actually only 18 times the previous one. Why? Who knows? Although 18 × 20 is 360 which is very close to the number of days per year, so it’s possible that’s why. In the Mayan Long Count Calendar (responsible for the 12/21/2012 end of the world nonsense), 1.0.0 days is called a tun, about equal to a year, and is obviously a nice round number, rather than 18.0 (or really, 18.5 to be more accurate) if it were truly a pure base-20 system.

So what was the exact problem I solved? Well, given what I said about the numbering system with its strange twist, is there an easy and quick way to multiply any given Mayan number by 20? If it were purely base-20 the answer would be trivial: just stick a zero on the end (like multiplying any number by 10 in our system). But if you actually do that, you haven’t actually multiplied the entire number by 20. The second digit moves into the third digit which means it’s only been multiplied by 18, so the resulting number is a bit smaller than 20 times the number.

Nobody in any class the professor has taught had ever figured this out, nor had she, nor even the book publishers apparently had an answer (the professor had called and asked them). So here’s the answer:

1)      Stick a zero on the end.

2)      Take what is now the third digit and multiply it by two (don’t change it in the actual number, just see what it is and multiply it by two in your head)

3)      Add that number to the second digit.

4)      Regroup to account for possible roll-over in the second digit (i.e. carry the one…or two as the case may be)

5)      There is no step 5 because you’ve already finished!

It really is that easy. Basically, what you’re doing is this: When you add a zero on the end every digit is indeed multiplied by 20, except for the second digit. Therefore, you need to add that difference back into the number to get the correct answer. This can result in the second digit being to large for its place (i.e. being 18 or over), so if you need to, you can subtract 18 one or two times (you won’t need to do it more than twice, I promise), and add the number of times you had to subtract 18 to the third digit. It’s like if you were to add 9 + 7. Well, you can’t cram “16” into a single digit, so you subtract 10 once, and add the one to next place.

Let’s do two examples:

A) Multiply 8.2.13 by 20

1)      Add a zero:

2)      Take the third digit and multiply by two (so that would be 4)

3)      Add that to the second digit:

4)      Regroup if necessary (since the second digit is less than 18, it doesn’t roll over)

Therefore 8.2.13 × 20 =

B) Multiply by 20

1)  Add a zero:

2)  Take the third digit and multiply by two (so, 24)

3)  Add that to the second digit:

4)  Regroup if necessary (40 goes into 18 twice, with a remainder of 4, so we make the second digit 4 and add 2 to the third digit):

So anyway, that’s it. The only problem left is to see if the rolling-over ever stops. I mean, what if the number you’re dealing with keeps making higher digits continue to roll over all the way up? For example, if you have the number (in our system) 9999999999999 and add 1, the ones’ place becomes 0 and a 1 rolls over into the tens’ place, which also becomes zero and rolls over into the hundreds’ place…which also becomes zero and rolls over into the thousands’ place, and so on. It would be unfortunate if this were to happen in Mayan numerals, since all that writing would take time.

Fortunately, it doesn’t. Let’s take a Mayan number that would be like 999999etc: If you were to add 1 to this number, the places would keep rolling over until you were left with So, let’s see if that happens when you multiply by 20:

1)      Add a zero:

2)      Take the third place and multiply by two (17 × 2 = 34)

3)      Add that number into the second place digit (19 + 34 = 53):

4)      Regroup: 53 ÷ 18 = 2 remainder 17. So the third digit becomes 19 and the second becomes 17:

Phew! The rollover stops at the third place in all cases (I can say this since this is the case where the number are at their largest).

So that’s my solution. As my professor might say “it’s good clean fun!” So anyway…wait…hold on, what’s that at the end of the article?…


I have a program that can convert our number system into Mayan numbers? I don’t have a problem that can convert our number system into Mayan numbers. That’s means I need to make one, quick! My e-mail’s on the thing! And the deadline where everyone loses interest is in three days! ARRRRRGGGG!!!!!

See you on the 22nd!