The Mayan numeral system was a vigesimal (base-20) positional numeral system:
The numerals are made up of three symbols: zero (shell shape), one (a dot), and five (a bar). For example, is written as three dots in a horizontal row above two horizontal bars (for a total of 5 symbols). With these three symbols, each of the twenty vigesimal digits could be written as in the diagram.
Numbers after were written vertically in powers of twenty. For example, would be written as one dot above three dots atop two bars (for a total of 6 symbols).
How many symbols would a Mayan scribe have to carve in order to write all the numbers from to
Inspired by this problem
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From the diagram given, the first group of 2 0 numbers ( 0 to 1 9 ) has 4 0 dots, 3 0 bars, and 1 shell for a total of 4 0 + 3 0 + 1 = 7 1 symbols.
The next group of 2 0 numbers ( 2 1 to 3 9 ) would have the same 7 1 symbols but 1 ⋅ 2 0 extra dot symbols in the row above, the next group of 2 0 numbers ( 4 0 to 5 9 ) would have the same 7 1 symbols but 2 ⋅ 2 0 extra dot symbols in the row above, and so on, so that the first 2 0 groups of 2 0 numbers or first 4 0 0 numbers ( 0 to 3 9 9 ) would have ( total groups of 2 0 in 0 to 3 9 9 ) ⋅ 7 1 + ( total number of symbols for the numbers 1 to 1 9 ) ⋅ 2 0 = 2 0 ⋅ 7 1 + 7 0 ⋅ 2 0 = 2 8 2 0 symbols.
The next group of 4 0 0 numbers ( 4 0 0 to 7 9 9 ) would have the same 2 8 2 0 symbols as the first 4 0 0 numbers but 2 0 extra shell symbols for the zero place holders for the numbers from 4 0 0 to 4 2 0 and 1 ⋅ 4 0 0 extra dot symbols for the top row, for a total of 2 8 2 0 + 2 0 + 1 ⋅ 4 0 0 = 3 2 4 0 symbols. Similary, the numbers 8 0 0 to 1 1 9 9 would have 2 8 2 0 + 2 0 + 2 ⋅ 4 0 0 = 3 6 4 0 symbols, the numbers 1 2 0 0 to 1 5 9 9 would have 2 8 2 0 + 2 0 + 3 ⋅ 4 0 0 = 4 0 4 0 symbols, and the numbers 1 6 0 0 to 1 9 9 9 would have 2 8 2 0 + 2 0 + 4 ⋅ 4 0 0 = 4 4 4 0 symbols.
Finally, the numbers 2 0 0 0 to 2 0 1 9 would have the same 7 1 symbols as the numbers 0 to 1 9 , but 2 0 extra shell symbols for the zero place holders and 2 0 extra bar symbols on the top row for a total of 7 1 + 2 0 + 2 0 = 1 1 1 symbols.
Therefore, the numbers 0 to 2 0 1 9 would have a total of 2 8 2 0 + 3 2 4 0 + 3 6 4 0 + 4 0 4 0 + 4 4 4 0 + 1 1 1 = 1 8 2 9 1 symbols. Since 0 has 1 symbol and 2 0 1 9 has 9 symbols, the numbers 1 to 2 0 1 8 would have a total of 1 8 2 9 1 − 1 − 9 = 1 8 2 8 1 symbols.