3. A Story of Numbers

Q2: Keep one stick for each of our cows and ask the neighbour to do the same. Now pair one of our sticks with one of the neighbour’s sticks again and again. The one whose sticks are left over has more cows.

Q3: The sticks left over with the neighbour show how many more cows we need.

In English we can represent only 26 numbers (a to z).

In Marathi or Hindi (Devanagari) there are about 48 letters (vowels and consonants). So we can represent about 48 numbers.

Yes. We keep making groups and give new symbols to bigger groups. For example, L for 50, C for 100, D for 500 and M for 1000.

Then 21 = XXI, 30 = XXX, 40 = XL, 50 = L, and so on.

• Addition: Put both collections together. The new collection is the sum.
• Subtraction: Pair each stick of the smaller collection with a stick of the bigger one and remove the pairs. The sticks left show the difference.
• Multiplication: For every stick in the second collection, place one full copy of the first collection. Put all of them together.
• Division: Make groups from the first collection, each group having as many sticks as the second collection. The number of groups is the quotient. The sticks left over are the remainder.

After z (26), we can use two-letter strings in order: aa (27), ab (28), …, az (52), ba (53), …, zz.

After zz, we use three-letter strings: aaa, aab, and so on. In this way every number gets a name.

Sample: Let ● = 1 and ▲ = 3 (a group of three ●). Then 1 = ●, 2 = ●●, 3 = ▲, 4 = ▲●, 5 = ▲●●, 6 = ▲▲, 7 = ▲▲●, and so on.

Most of us can see up to 4 objects at a glance. For 5 or more objects we have to count.

For big numbers we get too many groups. Then we have to count the groups also, which is slow and confusing.

1345 = 269 groups of 5 (since 269 × 5 = 1345). So we must write the symbol for 5 as many as 269 times.

(i) 1222 = 1000 + 100 + 100 + 10 + 10 + 1 + 1 = MCCXXII

(ii) 2999 = 1000 + 1000 + 900 + 90 + 9 = MMCMXCIX

(iii) 302 = 100 + 100 + 100 + 1 + 1 = CCCII

(iv) 715 = 500 + 100 + 100 + 10 + 5 = DCCXV

(a) Together we have 6 Cs, 4 Xs and 5 Is. 5 Is make a V. 5 Cs make a D.

So the sum = D C XXXX V = DCXLV (that is 645).

(b) Together we have 2 Ls, 5 Xs, 2 Vs and 5 Is. 5 Is make a V, so we get 3 Vs. 2 Vs make an X, so we get 6 Xs and 1 V. 5 Xs make an L, so we get 3 Ls and 1 X. 2 Ls make a C.

So the sum = C L X V = CLXV (that is 165).

V × L = 5 groups of L = LLLLL = CCL

L × D = 50 groups of D = 25000 = MMMMMMMMMMMMMMMMMMMMMMMMM (25 Ms)

V × D = 5 groups of D = DDDDD = MMD

VII × IX = (V + I + I) × IX = XLV + IX + IX = LXIII (63)

CCXXXI × MDCCCLII = 231 × 1852 = 4,27,812. In Roman numerals this needs 427 Ms followed by DCCCXII.

They may count different things in different groups. For example, fish may be counted in pairs and coconuts in bunches.

They have not yet separated the idea of number from the objects being counted.

Here urapon = 1 and ukasar = 2.

• It uses only ten symbols (0 to 9) to write any number.
• It is a place value system with base 10.
• It has a symbol for zero.
• Addition, subtraction, multiplication and division are easy.

Sample: In the ●, ▲ system, add a new symbol ■ for 3 × 3 = 9, and ★ for 9 × 3 = 27. Then 40 = ★ ■ ▲ ● (27 + 9 + 3 + 1).

10458 = 1 finger, 4 coiled ropes, 5 heel bones, 8 strokes

1023 = 1 lotus, 2 heel bones, 3 strokes

2660 = 2 lotuses, 6 coiled ropes, 6 heel bones

784 = 7 coiled ropes, 8 heel bones, 4 strokes

1111 = 1 lotus, 1 coiled rope, 1 heel bone, 1 stroke

70707 = 7 fingers, 7 coiled ropes, 7 strokes

(i) 2 coiled ropes, 7 heel bones and 6 strokes = 200 + 70 + 6 = 276

(ii) 4 lotuses, 3 coiled ropes, 2 heel bones and 2 strokes = 4000 + 300 + 20 + 2 = 4322

143 = 125 + 5 + 5 + 5 + 1 + 1 + 1 = ○ □□□ △△△

15 = 5 + 5 + 5 = □□□

50 = 25 + 25 = ⬡⬡

137 = 125 + 5 + 5 + 1 + 1 = ○ □□ △△

293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1 = ○○ ⬡ □□□ △△△

651 = 625 + 25 + 1 = ∼ ⬡ △

Every counting number can be represented. We group it into powers of 5, and no symbol is needed more than 4 times.

But zero (0) cannot be represented because there is no symbol for it.

Base-7: 7⁰ = 1, 7¹ = 7, 7² = 49, 7³ = 343, 7⁴ = 2401, …

Base-n: 1, n, n², n³, n⁴, … (powers of n).

Circles: 1 + 3 = 4. Hexagons: 2 + 1 = 3. Squares: 1 + 2 = 3. Triangles: 2 + 2 = 4.

No symbol reaches 5, so no regrouping is needed.

Sum = ○○○○ ⬡⬡⬡ □□□ △△△△ (182 + 412 = 594)

Multiplying by 10 gives the next landmark number.

(i) coiled rope (100) (ii) lotus (1000) (iii) finger (10,000) (iv) tadpole (1,00,000)

Multiplying by 100 moves two landmark numbers ahead.

(i) lotus (10³) (ii) finger (10⁴) (iii) tadpole (10⁵) (iv) astonished man (10⁶)

(i) 10 × 10⁵ = 10⁶ = astonished man

(ii) 10² × 10³ = 10⁵ = tadpole

(iii) 10³ × 10³ = 10⁶ = astonished man

(iv) 10⁴ × 10⁶ = 10¹⁰. The Egyptians had no symbol for this.

Yes. In base 5, □ × □ = 5 × 5 = 25 = ⬡, and ⬡ × □ = 125 = ○.

Yes. In any base-n system, the landmark numbers are powers of n, and n^a × n^b = n^{a + b}, which is again a landmark number.

Multiply each symbol by 10, that is, change it to the next landmark symbol.

(i) 5 lotuses, 2 coiled ropes, 2 heel bones (522 × 10 = 5220)

(ii) 1 finger, 1 coiled rope (1010 × 10 = 10,100)

Replace every symbol by the symbol of the next landmark number. A stroke becomes a heel bone, a heel bone becomes a coiled rope, and so on.

Bring the counters on each line together. Ones: 7 + 3 = 10. Remove these 10 counters and put 1 counter on the tens line.

Tens: 4 + 1 = 5. Hundreds: 9. Thousands: 2.

So the sum = 2950. Whenever a line gets 10 counters, we replace them by 1 counter on the next line.

No. 10 copies of any symbol are always replaced by one symbol of the next landmark number. So a symbol appears at most 9 times.

Let ● = 1, ▲ = 4, ■ = 16.

Replace every symbol by the next landmark symbol: △ → □, □ → ⬡, ⬡ → ○, ○ → ∼, ∼ → ↑.

For example, ○ □□ △ (136) × 5 = ∼ ⬡⬡ □ (680).

Here ▼ stands for 1 and ◀ stands for 10. Places (from right) show 1s, 60s and 3600s.

(i) 63 = (1) × 60 + 3 → ▼   ▼▼▼

(ii) 132 = (2) × 60 + 12 → ▼▼   ◀▼▼

(iii) 200 = (3) × 60 + 20 → ▼▼▼   ◀◀

(iv) 60 = (1) × 60 + 0 → ▼   (blank)

(v) 3605 = (1) × 3600 + (0) × 60 + 5 → ▼   (blank)   ▼▼▼▼▼

3600 = (1) × 60². It is written as a single ▼ followed by two blank places.

So it looks just like 1 and 60. This creates confusion.

Dot = 1, bar = 5, shell = 0. Landmark numbers from bottom: 1, 20, 360.

It is the final and most developed stage. It is a base-10 system with the digit 0.

Its landmark numbers are 1, 10, 10², 10³, … (powers of 10).

Yes, it is a place value system. The position of a digit tells its landmark number.

They alternated so that two neighbouring places can be told apart easily.

With only Zongs, 41 would be |||| | (four rods and one rod).

If there is no clear space, these five rods can also be read as 5, 14, 23, 32, 113, 122, 311 and so on.

Let urapon (ur) = 0 and ukasar (uk) = 1. Places from right are 1, 2, 4, 8, …

The base-2 system can write every number. The Gumulgal system has names only up to 6. It needs longer names as numbers grow.

They are used in shopping, banking, phone numbers, dates, time, measurement, computers, science and engineering. Shopkeepers, accountants, engineers, doctors and scientists use them every day.

Without them, writing and calculating big numbers would be very slow and confusing. Computers and modern technology would be very hard to build.

In base 8 we would use only the digits 0 to 7. In base 5 we would use only 0 to 4.

Base 8: 25 = (3) × 8 + 1 → 31

Base 5: 25 = (1) × 25 + (0) × 5 + 0 → 100

Base 2: 25 = 16 + 8 + 1 = (1)×16 + (1)×8 + (0)×4 + (0)×2 + 1 → 11001