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When converting from a base \(10\) number to a Mayan number using the greedy division algorithm, place values are filled from left to right, that is from largest place value down. As noted at the beginning, this process leads to a unique expression in a “pure'' base system such as ours. But the Mayan system is not pure; the \(18\) makes for some interesting choices, depending on whether you think of building a number from the top or from the bottom.

Sometimes it is easier to think of a base \(b\) place-value number as made up of bins. Each bin can have zero to \(b-1\) sets (or bags), each containing a full complement of the next lower place value. For example, in the base \(10\) number \(253,\) the left bin has \(2\) sacks, each with \(10\) bags of \(10\) counters; the next bin to the right contains \(5\) bags, each with \(10\) counters; and the last bin on the right contains \(3\) individual counters. So if a number is thought of as being filled from the top, as with the greedy algorithm, as many sets of the largest power of the base as possible are put in the left bin. Then from the remainder, as many complete sets of the next lower power as possible are put in the next bin, and so on.

On the other hand, if a number is thought of not in terms of sets of powers of the base, but as piles of individual counters, a natural method would be to start putting counters in a bin starting on the right with the units place. As soon as a set of size \(b\) is obtained, a single counter to represent a set of size \(b\) is put in the next bin to the left (the second bin from the right), the contents of the right bin are discarded, and one resumes putting counters in the right bin.

This is the practical procedure suggested by the second (bottom-up) algorithm we presented. This is also, in essence, how calculations are done on an abacus. When one column is full, a single counter on the next wire to the left is moved up and the previous column is cleared. In addition and multiplication we call this carrying (to the next highest power). For a pure base system, either method (filling from the top or the bottom) will give you the same encoded number.

But in the Mayan system, the third and higher place values are not pure. The third place is \(18\times20.\) If, in a certain expression, the number in the second place (the \(20\)'s place) is \(18\) or \(19,\) all or part of it could be carried over to the third (or \(18\times20\)’s) place. Or if the second place is only a \(0\) or a \(1,\) then an \(18\times20\) can be carried *backward* from the third place down to the second, making the second digit now \(18\) or \(19.\) Thus the same natural number can be encoded in two different ways in the Mayan system. We will make this explicit in a moment, but first some examples are in order. Please note the following. In a pure number system, when a place value is filled, \(1\) is carried to the next highest place and the lower bin is emptied. This is not the case in the second place of the Mayan numbers. When the second place is full, it has \(20\) sets of \(20,\) but the third place is \(18\times20.\) So the whole \(20\times20\) cannot be carried over to the third place; only \(18\times20\) can be carried, leaving \(2\times20\) in the \(20\)'s place. If the second place is \(18,\) then \(18\times20\) is carried and the place is emptied; if the second place has \(19\) or is full with \(20,\) then when \(18\times20\) is carried, a \(1\) or a \(2\) must be left behind in the second place.

Consider the Mayan number \(4.8.18.9\). In expanded form, this would be \[4(18\times20^2)+8(18\times20)+18(20)+9.\] Notice that the second and third places are both multiples of \(18\times20.\) So the whole \(18\times20\) in the second place can be carried over as one unit in the third place, resulting in the new, but equivalent Mayan number \(4.9.0.9\). Both of these representations equal \(32049\) in base \(10.\) See Figure 4 for two more (yet again amazing) serpent numbers depicting these two equivalent numbers. Likewise the Mayan number \(4.8.19.9\) could also be written as \(4.9.1.9\) by carrying from the second to the third place.

**Figure 4.** \(4.8.18.9\) on the left, \(4.9.0.9\) on the right, both equivalent to \(32049\) base \(10.\)

Working the same two examples the other way, if you have the Mayan numbers \(4.9.0.9\) or \(4.9.1.9,\) one unit of \(18\times20\) from the third place can be carried down to add \(18\) to the second place, giving \(4.8.18.9\) and \(4.8.19.9,\) respectively.

To summarize, given a Mayan number of the form \(a_k .\ldots.a_2.a_1.a_0,\) where \(a_1=18\) or \(19,\) an equivalent Mayan number would be \[a_k.\cdots.(a_2+1).(a_1-18).a_0,\] with a \(1\) carried to the fourth place as usual if \(a_2 + 1 = 20.\) Conversely, if a Mayan number is of the form \(a_k.\cdots.a_2.a_1.a_0,\) where \(a_2>0\) and \(a_1=0\) or \(1,\) then the equivalent Mayan number would be \[a_k.\cdots.(a_2-1).(a_1+18).a_0.\]

Outside of these two cases, any other combination of numerals in a Mayan number is unique. In other words, if we can find an expression for \(n\) in which \(a_1=18\) or \(19,\) we have two expressions for \(n.\) If this does not happen, the expression is unique. (Of course, any other non-pure place-value system would violate uniqueness. However, the Mayan system is the only non-pure system we are aware of.)

To check this, suppose \(n\) is a number that has no expression (in the Mayan system) in which the second place is \(18\) or \(19.\) Take two possible expressions:

\(n=\) | \(a_0+a_1\,(20)+a_2\,(18\times20)+a_3\,(18\times20^2)+\cdots+a_k\,(18\times20^{k-1})\) |

\(=\) | \(a^{\prime}_0+a^{\prime}_1\,(20)+a^{\prime}_2\,(18\times20)+a^{\prime}_3\,(18\times20^2)+\cdots+a^{\prime}_t\,(18\times20^{t-1}).\) |

According to our assumption, \(0\leq a_1,a^{\prime}_1\leq 17,\) and for \(i\neq 1,\) \(0\leq a_i\leq 19.\)

To start with, we have \(a_0=a^{\prime}_0,\) by the same argument used in the general proof. When we now take \(n_1=(n-a_0)/20,\) we get, using the very same argument, \(a_1=a^{\prime}_1,\) since \(a_1-a^{\prime}_1\) is both a multiple of \(18\) and a number in absolute value smaller than \(18.\)

We now take \(n_2=(n_1-a_1)/18\) and get equality of all other \(a_i\)'s, and \(k=t,\) since now we are dealing with a number in pure base \(20.\)

Thus for all cases when the second place number cannot be chosen to be \(18\) or \(19,\) the Mayan notation is unique.

So, for which base \(10\) natural numbers is the Mayan number not unique? As we said above, this happens when the second numeral can be chosen to be \(18\) or \(19.\) Then we can carry one \(18\times20\) to the next higher place. The smallest such numbers are found by assuming all higher place values are zero. Then we have the range of natural numbers \([360,399]\) corresponding to the Mayan numbers from \(18.0=1.0.0\) to \(19.19=1.1.19,\) for which the Mayan number is not unique. Other intervals of non-unique Mayan numbers would be found by filling in non-zero numerals in any and all of the higher places while retaining the restriction on the second place. The above ideas make for great open questions in many different classes, from elementary school through college courses, as long as the Mayan calendric number system has been introduced.

Pedro J. Freitas (Universidade de Lisboa) and Amy Shell-Gellasch (Beloit College), "When a Number System Loses Uniqueness: The Case of the Maya - Non-uniqueness," *Loci* (June 2012), DOI:10.4169/loci003883