Take the two collections of sticks. Then:

• Add: push both collections together into one pile — the combined pile is the sum.
• Subtract: from the bigger collection, remove (pair off) one stick for each stick of the smaller collection. The sticks left over are the difference.
• Multiply: make several equal groups (one group per stick of the second number, each group as big as the first number) and combine them all.
• Divide: from the dividend, keep taking away groups of the divisor’s size. The number of complete groups is the quotient, and the sticks that cannot form a full group are the remainder.

With 26 letters we can name 1–26 as \(a,b,\dots,z\). One simple way to continue:

a, b, c, …, z  →  aa, bb, cc, …, zz  →  aaa, bbb, …, zzz  → …  (longer and longer strings)

• \(1222 = 1000+200+20+2 = M+CC+XX+II\)  →  MCCXXII
• \(2999 = 2000+900+90+9 = MM+CM+XC+IX\)  →  MMCMXCIX
• \(302 = 300+0+2 = CCC+II\)  →  CCCII
• \(715 = 500+200+10+5 = D+CC+X+V\)  →  DCCXV

First read each numeral as a value, compute, then write the answer back in Gumulgal style.

\((uk\,uk\,uk\,uk\,ur) + (uk\,uk\,uk\,ur) = (2{+}2{+}2{+}2{+}1) + (2{+}2{+}2{+}1) = 9 + 7 = 16 = 8\times 2.\)

= ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar  (eight ukasars)

\((uk\,uk\,uk\,uk\,ur) – (uk\,uk\,uk) = 9 – 6 = 3 = 2+1.\)

= ukasar-urapon

\((uk\,uk\,uk\,uk\,ur)\times(uk\,uk) = 9 \times 4 = 36 = 18\times 2.\)

= ukasar-ukasar-…-ukasar  (eighteen ukasars)

\((\text{eight ukasars}) \div (uk\,uk) = 16 \div 4 = 4 = 2+2.\)

= ukasar-ukasar

Because of place value and the digit 0, the Hindu system writes every number unambiguously with just ten symbols and makes calculation simple.

Break each number into powers of 10, then draw that many of each landmark symbol.

\(2\times100 + 7\times10 + 6\times1 = 200 + 70 + 6 = \mathbf{276}\)

\(4\times1000 + 3\times100 + 2\times10 + 2\times1 = 4000 + 300 + 20 + 2 = \mathbf{4322}\)

The base-5 landmark numbers and their symbols are:

Yes — the number 0. The system is purely additive and has no symbol for zero, so there is no way to write “nothing”. (Every positive whole number can be written, since the landmark numbers \(1,5,25,125,\dots\) grow without end.)

Base-7 landmarks are the powers of 7:

$$7^{0}=1,\; 7^{1}=7,\; 7^{2}=49,\; 7^{3}=343,\;\dots$$

In general, a base-\(n\) system has landmark numbers

$$n^{0}=1,\; n^{1}=n,\; n^{2},\; n^{3},\;\dots\quad(\text{the powers of } n).$$

Combine the like symbols, then regroup: 10 of any symbol = 1 of the next-higher symbol.

Combine and regroup:

• Ones: \(8+7 = 15 = 10+5\) → carry one 10, keep 5 ones.
• Tens: \(0+0+1\,(\text{carry}) = 1\) ten.
• Hundreds: \(6+5 = 11 = 10+1\) → carry one 1000, keep 1 hundred.
• Thousands: \(9+0+1\,(\text{carry}) = 10 = 1\times10^4\) → keep 0 thousands, carry one 10⁴.

$$9608 + 507 = 10115 = 1\times10^4 + 1\times100 + 1\times10 + 5\times1.$$

• Ones: \(0+6 = 6\) ones.
• Tens: \(8+3 = 11 = 10+1\) → carry one 100, keep 1 ten.
• Hundreds: \(0+1\,(\text{carry}) = 1\) hundred.
• Thousands: \(1+0 = 1\) thousand.

$$1080 + 36 = 1116 = 1\times1000 + 1\times100 + 1\times10 + 6\times1.$$

Combine like symbols:

$$382+512 = 7\times125 + 3\times5 + 4\times1 = 875+15+4 = 894.$$

No. Ten copies of any landmark number equal exactly the next landmark number, e.g. \(10\times10 = 100\). So whenever a symbol would appear 10 times, those ten are replaced by a single symbol of the next size. Each symbol can therefore appear at most 9 times.

Choose landmark numbers \(4^{0}=1,\; 4^{1}=4,\; 4^{2}=16\) and give each its own symbol:

∟ = 1  ·  △ = 4  ·  □ = 16

Now group each number into 16s, 4s and 1s:

Replace every symbol with the next-higher landmark symbol — exactly like “shifting up one place”:

$$1\to5,\quad 5\to25,\quad 25\to125,\;\dots$$

This works because the landmarks are powers of 5, so multiplying any landmark by 5 gives the next one. (It is the base-5 version of “add a 0 to multiply by 10” in base 10.)

The Mesopotamians used a vertical wedge for 1 and a corner wedge for 10. Numbers are grouped into powers of 60 and written by position (highest place on the left).

The Chinese alternated the Zong (vertical) and Heng (horizontal) rod numerals from one place to the next so that neighbouring digits look different and the boundary between places is clear.

If only Zong (vertical) rods were used, 41 would be four vertical strokes for the tens and one for the ones:

With no clear gap between the positions, the same five strokes could be misread as 5 (one place), or as 23, 32, 122, and so on. The alternation removes this confusion.

Let urapon = 0 and ukasar = 1. Then write each number in base 2 (using the place values \(1,2,4,8,\dots\)) and translate the binary digits.

Both systems use only two basic words, but they are very different:

• Base-2 is a place value system — position matters and it includes 0 (urapon), so it can write every number compactly.
• Gumulgal is purely additive (just repeat ukasar/urapon), it grows longer and longer, and originally had names only up to 6.

The Hindu numerals and 0 are everywhere: in money and shopping prices, phone numbers, dates and clocks, addresses, measurements, and exam marks. Professions that depend heavily on them include accounting and banking, engineering and science, computer programming, surveying, medicine, and trade/commerce.

Without place value and 0, calculations would be slow and clumsy (as with Roman numerals). Modern technology — computers, which rely on binary built on the idea of 0 and place value — would have been extremely hard to develop.

To convert, group 25 into powers of the new base.

\(25 = 3\times 8 + 1 = (3)\,8^{1} + (1)\,8^{0}\)  →  31 in base 8.

\(25 = 1\times 25 + 0\times 5 + 0\times 1 = (1)\,5^{2} + (0)\,5^{1} + (0)\,5^{0}\)  →  100 in base 5.

\(25 = 16 + 8 + 1 = (1)2^{4} + (1)2^{3} + (0)2^{2} + (0)2^{1} + (1)2^{0}\)  →  11001 in base 2.