Chapter 3 exercise answers: A Story of Numbers

Open NCERT p. 54One way to think about it

1. Keep every number as a pile of sticks. To compare two piles, pair them off one stick from each pile at a time; the pile with sticks left over is the bigger one.
2. Add: push the two piles together. The combined pile is the sum.
3. Subtract: pair each stick of the smaller pile with a stick of the bigger pile and remove both. The sticks of the bigger pile left without a partner are the difference.
4. Multiply: for every stick in the first pile, lay out one copy of the second pile (make a copy by placing one new stick beside each stick of the second pile). Push all the copies together. That pile is the product.
5. Divide: take out of the first pile a group that matches the divisor pile stick for stick, and put one stick into a separate answer pile. Repeat until a full matching group can no longer be taken out. The answer pile is the quotient and the sticks still left are the remainder.
6. This is one method; other methods that only move and match sticks are just as correct.

In shortEvery operation becomes an action on sticks: join the piles to add, pair off and remove to subtract, lay one copy of the second pile for each stick of the first to multiply, and take out divisor-sized groups (one answer stick per group) to divide.

Watch the lesson Why any number system needs a fixed, ordered sequence of symbols · हिंदी में देखें

Open NCERT p. 54One way to think about it

1. Keep a to z for the numbers 1 to 26.
2. After z, use two-letter strings in dictionary order: aa = 27, ab = 28, …, az = 52, ba = 53, …, zz = 702 (26 + 26 × 26).
3. Then use three-letter strings: aaa = 703, aab = 704, and so on; after those come four-letter strings, and so on without end.
4. A string can be as long as we like, so the names never run out, and the order is fixed in advance, so counting by one-to-one mapping still works.
5. This is one way; there are many others. For example, z could stand for each full 26 with a letter for what is left, so 27 = za and 53 = zza.

In shortOne way: a to z for 1 to 26, then aa to zz for 27 to 702 in dictionary order, then aaa = 703 and so on with longer and longer strings; the strings never run out, so every number gets a name. Other ways work too.

Watch the lesson Why any number system needs a fixed, ordered sequence of symbols · हिंदी में देखें

Open NCERT p. 54One way to think about it

1. Choose symbols: a dot • for one, a square □ for a group of 3 dots, and a triangle △ for a group of 3 squares (9 dots).
2. Fix the order: •, ••, □, □•, □••, □□, □□•, □□••, △, △•, … Whenever 3 of a symbol would appear, they are replaced by 1 of the next symbol.
3. Check it against Section 3.1: the order is fixed and agreed before counting starts, numbers stay short enough to read at a glance, and it never ends, because a new symbol can always be given to a group of 3 of the largest symbol.
4. Any system whose sequence is fixed and never ends is a correct answer; this is only one example.

In shortExample: • = 1, □ = 3 dots, △ = 3 squares, so 1 to 9 are •, ••, □, □•, □••, □□, □□•, □□••, △. Many other systems are equally valid.

Watch the lesson Why any number system needs a fixed, ordered sequence of symbols · हिंदी में देखें

Open NCERT p. 59Matches NCERT’s answer

(i) 1222

1. Group 1222 into landmark numbers from the largest down: 1000 + 200 + 20 + 2.
2. Write each landmark's numeral as many times as needed: M for 1000, CC for 200, XX for 20, II for 2.

AnswerMCCXXII

(ii) 2999

1. Group 2999 as 2000 + 900 + 90 + 9.
2. 900 is 100 less than 1000, so it is CM; 90 is 10 less than 100, so it is XC; 9 is 1 less than 10, so it is IX.
3. Put the landmarks together: MM (2000) + CM (900) + XC (90) + IX (9).

AnswerMMCMXCIX

(iii) 302

1. Group 302 as 300 + 2.
2. 300 is CCC, and 2 is II.

AnswerCCCII

(iv) 715

1. Group 715 as 700 + 10 + 5.
2. 700 is D + CC (500 + 100 + 100), 10 is X, and 5 is V.

AnswerDCCXV

Watch this explained “A four-figure number in ten signs”, 3:33 into Roman numerals: grouping a number into tens, fives and ones · हिंदी में देखें

Open NCERT p. 60One way to think about it

1. This is an open question; the book does not give one answer. Here are two sensible reasons.
2. Each sequence is still a proper number system: its order is fixed and agreed, and objects are counted by one-to-one mapping. So counting works whichever sequence is used.
3. Reason 1: the counting words can also tell the listener what kind of thing is being counted, so 'three' said about canoes sounds different from 'three' said about fish.
4. Reason 2: different things are handled in different groups (some in pairs, some in bundles), so each sequence can be built around the group size that suits those objects, the way the Gumulgal built their names from twos.

In shortMost likely because the number words also carry information about the objects, such as their kind or the groups they are usually counted in. Each sequence is still a fixed, agreed order, so each works as a number system. Other reasons are possible too.

Open NCERT p. 60Matches NCERT’s answer

(i) (ukasar-ukasar-ukasar-ukasar-urapon) + (ukasar-ukasar-ukasar-urapon)

1. To add, put the two names together: ukasar-ukasar-ukasar-ukasar-urapon and ukasar-ukasar-ukasar-urapon give seven ukasar and two urapon.
2. Two urapon make one ukasar, so this becomes eight ukasar and no urapon.
3. Check with our own numerals, only as a check: 9 + 7 = 16, which is eight 2s.

Answerukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(ii) (ukasar-ukasar-ukasar-ukasar-urapon) – (ukasar-ukasar-ukasar)

1. To subtract, strike out of the first name every word of the second name: remove three ukasar from ukasar-ukasar-ukasar-ukasar-urapon.
2. One ukasar and the urapon are left.
3. Check: 9 − 6 = 3.

Answerukasar-urapon

(iii) (ukasar-ukasar-ukasar-ukasar-urapon) × (ukasar-ukasar)

1. Multiplying by ukasar means doubling: write every word twice, then turn each pair of urapon into one ukasar.
2. ukasar-ukasar is ukasar twice, so double the first number twice.
3. First doubling: four ukasar and one urapon become eight ukasar and two urapon, which is nine ukasar.
4. Second doubling: nine ukasar become eighteen ukasar.
5. Check: 9 × 4 = 36, which is eighteen 2s.

Answerukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar

(iv) (ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar-ukasar) ÷ (ukasar-ukasar)

1. To divide, keep taking the divisor ukasar-ukasar out of the first number, and write one urapon each time it comes out.
2. Eight ukasar hold four groups of ukasar-ukasar with nothing left over, so we write four urapon.
3. Two urapon make one ukasar, so four urapon are ukasar-ukasar.
4. Check: 16 ÷ 4 = 4.

Answerukasar-ukasar

Watch this explained “A name is an addition”, 0:40 into Counting in twos, and what number-names reveal about a culture's base · हिंदी में देखें

Open NCERT p. 61One way to think about it

1. This is an open question; these are the main features.
2. Place value: a digit's position tells which landmark it counts, so the 3 in 300 means three hundreds. Every landmark is 10 times the one before, so bigger landmarks need no new symbols. The Roman system has no place value and repeats letters instead: 2888 needs 4 digits but MMDCCCLXXXVIII needs 14 letters.
3. 0 as a digit: 0 marks an empty place, so 205 and 25 can never be confused. The Roman system has no zero.
4. Only ten symbols: 0 to 9 are enough for every number, however large. The Roman system stops at M (1000), so 25,000 would need M written 25 times.
5. Ease of calculation: because of these, addition, subtraction, multiplication and division can be done digit by digit in columns. With Roman numerals the landmarks go up by 5, then 2, then 5, then 2, so regrouping changes from step to step and multiplying is hard.

In shortPlace value, 0 used as a digit, and only ten symbols for every number. Together they keep numerals short and make calculation easy, which the Roman system does not.

Watch this explained “The one missing property”, 8:23 into Roman numerals: grouping a number into tens, fives and ones · हिंदी में देखें

Open NCERT p. 61One way to think about it

1. This is an open question; the refinement depends on the system you made. Here is how to refine one, shown on a plain tally (one mark for each thing).
2. Count in groups: choose a group size, say 5, and give a group of 5 marks a new symbol, as the Gumulgal named numbers by twos and the Romans used V for five.
3. Use landmark numbers: give new symbols to bigger groups too, for example one for 5 fives (25) and one for 5 twenty-fives (125). Write a number by taking as many of the largest landmark as fit, then the next, and so on, as in 2367 = 1000 + 1000 + 100 + 100 + 100 + 50 + 10 + 5 + 1 + 1.
4. Check the result: 38 was 38 tally marks; now it is one 25-symbol, two 5-symbols and three marks, six symbols in all.

In shortRefine your system by counting in groups and giving new symbols to landmark numbers, then write each number from the largest landmark down; numbers become much shorter to write and read. Many different refinements are correct.

Watch this explained “From twos to groups of any size”, 5:19 into Counting in twos, and what number-names reveal about a culture's base · हिंदी में देखें

Open NCERT p. 62Matches NCERT’s answer

1. Symbols: stroke = 1, arch = 10, coil = 100, lotus = 1,000, bent finger = 10,000.
2. Group each number into landmarks, taking as many of the largest as fit and then moving down, as the book writes 324 = 100 + 100 + 100 + 10 + 10 + 4. Draw each symbol that many times; a landmark that is not needed is simply left out.

3. Number	Fingers (10,000)	Lotus (1,000)	Coils (100)	Arches (10)	Strokes (1)
   10458	1	–	4	5	8
   1023	–	1	–	2	3
   2660	–	2	6	6	–
   784	–	–	7	8	4
   1111	–	1	1	1	1
   70707	7	–	7	–	7

Answer10458: 1 finger, 4 coils, 5 arches, 8 strokes. 1023: 1 lotus, 2 arches, 3 strokes. 2660: 2 lotus, 6 coils, 6 arches. 784: 7 coils, 8 arches, 4 strokes. 1111: 1 lotus, 1 coil, 1 arch, 1 stroke. 70707: 7 fingers, 7 coils, 7 strokes.

Watch this explained “Writing three hundred and twenty-four”, 3:27 into The Egyptian system, and what a landmark number is for · हिंदी में देखें

Open NCERT p. 62Matches NCERT’s answer

(i) What numbers do these numerals stand for?

1. Count the symbols of each kind; the rows they sit in do not matter. There are 2 coils, 7 arches (3 + 3 + 1) and 6 strokes (3 + 3).
2. 2 × 100 + 7 × 10 + 6 × 1 = 200 + 70 + 6 = 276.

Answer276

(ii) What numbers do these numerals stand for?

1. There are 4 lotus symbols, 3 coils, 2 arches and 2 strokes.
2. 4 × 1000 + 3 × 100 + 2 × 10 + 2 × 1 = 4000 + 300 + 20 + 2 = 4322.

Answer4322

Watch this explained “Position carries nothing”, 4:52 into The Egyptian system, and what a landmark number is for · हिंदी में देखें

Open NCERT p. 63Matches NCERT’s answer

1. Symbols (Table 2): triangle = 1, square = 5, hexagon = 25, circle = 125, wave = 625.
2. Group each number, taking as many of the largest landmark as fit, as the book does for 143:
3. 15 = 5 + 5 + 5
4. 50 = 25 + 25
5. 137 = 125 + 5 + 5 + 1 + 1
6. 293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1
7. 651 = 625 + 25 + 1
8. The count of each symbol, largest first, gives the base-5 digits (a 0 where a symbol is not used).

Answer15: 3 squares (base-5 digits 30). 50: 2 hexagons (200). 137: 1 circle, 2 squares, 2 triangles (1022). 293: 2 circles, 1 hexagon, 3 squares, 3 triangles (2133). 651: 1 wave, 1 hexagon, 1 triangle (10101).

Watch this explained “Writing a hundred and forty-three”, 1:35 into What "base n" means, and why ten is a choice not a law · हिंदी में देखें

Open NCERT p. 63Matches NCERT’s answer

1. The landmark numbers are 1, 5, 25, 125, 625, …; each is 5 times the one before, and each has its own symbol (triangle, square, hexagon, circle, wave, …).
2. Every counting number 1, 2, 3, … can be written: take as many of the largest landmark as fit, then move down, as the book does for 143 = 125 + 5 + 5 + 5 + 1 + 1 + 1. The landmarks never run out, so there is always one big enough to start from.
3. Zero is different. It contains no landmark at all, so grouping draws no symbol, and none of the symbols stands for 0. An empty space cannot be read as a number.
4. So 0 is the number that cannot be represented, and the chapter treats 0 as a number like any other.

AnswerYes: 0. Every counting number can be written using the symbols for 1, 5, 25, 125, …, but there is no symbol for zero, so 0 cannot be represented.

Watch this explained “Why there, and not somewhere else”, 2:20 into Why a base alone still runs out of symbols · हिंदी में देखें

Open NCERT p. 63Matches NCERT’s answer

1. For base 7, start at 1 and keep multiplying by 7: 7⁰=1, 7¹=7, 7²=49, 7³=343.
2. The same rule for any base n: start at n⁰=1 and multiply by n each time to get n¹, n², n³, and so on.

AnswerThe first four landmark numbers of a base-7 system are 1, 7, 49, 343; in general, the landmark numbers of a base-n system are the powers nᵏ for k = 0, 1, 2, 3, …

Watch this explained “The definition, in two clauses”, 2:11 into What "base n" means, and why ten is a choice not a law · हिंदी में देखें

Open NCERT p. 65Matches NCERT’s answer

(i) Add the following Egyptian numerals:

1. First numeral: 9 lotus (1,000s), 6 coils (100s) and 8 strokes (2 + 3 + 3). Second numeral: 5 coils and 7 strokes (3 + 4).
2. Put them together and count each kind: 9 lotus, 11 coils, 15 strokes.
3. 10 strokes make 1 arch: 5 strokes and 1 arch are left.
4. 10 coils make 1 lotus: 1 coil is left, and there are now 10 lotus.
5. 10 lotus make 1 bent finger (10,000).
6. Check: 9608 + 507 = 10115.

Answer1 bent finger, 1 coil, 1 arch and 5 strokes (10115)

(ii) Add the following Egyptian numerals:

1. First numeral: 1 lotus and 8 arches (3 + 3 + 2). Second numeral: 4 arches (3 + 1) and 6 strokes (3 + 3).
2. Together: 1 lotus, 12 arches, 6 strokes.
3. 10 arches make 1 coil: 1 lotus, 1 coil, 2 arches, 6 strokes.
4. Check: 1080 + 46 = 1126.

Answer1 lotus, 1 coil, 2 arches and 6 strokes (1126)

Watch this explained “Adding in a base”, 3:03 into What "base n" means, and why ten is a choice not a law · हिंदी में देखें

Open NCERT p. 65Matches NCERT’s answer

1. First numeral: 1 circle (125), 2 hexagons (25 each), 1 square (5), 2 triangles (1 each).
2. Second numeral: 3 circles, 1 hexagon, 2 squares, 2 triangles.
3. Put them together and count each kind: 4 circles, 3 hexagons, 3 squares, 4 triangles.
4. No kind reaches 5, so nothing needs regrouping.
5. Check: 182 + 412 = 594 = 4 × 125 + 3 × 25 + 3 × 5 + 4 × 1.

Answer4 circles, 3 hexagons, 3 squares and 4 triangles (4334 in base-5 digits, which is 594)

Watch this explained “One sum, three systems”, 4:30 into What "base n" means, and why ten is a choice not a law · हिंदी में देखें

Open NCERT p. 69Matches NCERT’s answer

1. Each Egyptian landmark is 10 times the one before, so 10 copies of any symbol are worth exactly 1 of the next symbol.
2. A number is written by taking as many of the largest landmark as fit, so at most 9 of any symbol are left. If 10 appear, for example after adding, they are regrouped into 1 of the next symbol, as the book's addition turns 15 strokes into 1 arch and 5 strokes.
3. The one exception is at the top: the Egyptians had symbols only up to 10,000,000. The number 100,000,000 would need that largest symbol 10 times unless a new symbol were invented, which is the shortcoming this section describes.

AnswerNo. 10 of any symbol always regroup into 1 of the next symbol, so no symbol appears 10 or more times, as long as a next symbol exists; beyond the largest symbol (10,000,000) a new symbol would be needed.

Watch this explained “What it got right”, 0:00 into Why a base alone still runs out of symbols · हिंदी में देखें

Open NCERT p. 70Checked by computerAnswers can differ: one example

1. This is an open question; any symbols can be used. Here is one base-4 system.
2. The landmarks are powers of 4: 1, 4 and 16. Symbols: ● = 1, ■ = 4 (four dots), ★ = 16 (four squares).
3. Write a number by taking as many of the largest landmark as fit, then moving down. No symbol is ever used 4 times, because 4 of a symbol make 1 of the next.

4. Number	Numeral
   1	●
   2	●●
   3	●●●
   4	■
   5	■●
   6	■●●
   7	■●●●
   8	■■
   9	■■●
   10	■■●●
   11	■■●●●
   12	■■■
   13	■■■●
   14	■■■●●
   15	■■■●●●
   16	★

Answer● = 1, ■ = 4, ★ = 16: 1 ●, 2 ●●, 3 ●●●, 4 ■, 5 ■●, 6 ■●●, 7 ■●●●, 8 ■■, 9 ■■●, 10 ■■●●, 11 ■■●●●, 12 ■■■, 13 ■■■●, 14 ■■■●●, 15 ■■■●●●, 16 ★. Other symbols work just as well.

Watch this explained “A system small enough to watch”, 5:45 into Why a base alone still runs out of symbols · हिंदी में देखें

Open NCERT p. 70One way to think about it

1. Each landmark times 5 is the next landmark: 1 × 5 = 5, 5 × 5 = 25, 25 × 5 = 125, 125 × 5 = 625.
2. So to multiply by 5, replace every symbol by the next one up (triangle → square, square → hexagon, hexagon → circle, circle → wave, wave → arrow) and keep the number of each symbol the same.
3. Example: 143 is 1 circle, 3 squares, 3 triangles. Times 5 it becomes 1 wave, 3 hexagons, 3 squares, which is 625 + 75 + 15 = 715 = 143 × 5.
4. No regrouping is needed, because no count changes.

In shortTo multiply by 5, move every symbol one step up the landmark list (triangle → square → hexagon → circle → wave → arrow), keeping the same number of each. It is like writing a 0 at the end of a Hindu numeral to multiply by 10.

Watch this explained “One rule, any base”, 8:07 into Why a base alone still runs out of symbols · हिंदी में देखें

Open NCERT p. 73Matches NCERT’s answer

(i) 63

1. 63 = 1×60 + 3.
2. Write 1 wedge in the 60s place, and 3 wedges in the 1s place.

Answerone wedge, then three wedges (1, 3 in base 60).

(ii) 132

1. 132 = 2×60 + 12.
2. 12 = 1×10 + 2, so the 1s place uses one angle-wedge and two wedges.

Answertwo wedges, then one angle-wedge and two wedges (2, 12 in base 60).

(iii) 200

1. 200 = 3×60 + 20.
2. 20 = 2×10, so the 1s place uses two angle-wedges.

Answerthree wedges, then two angle-wedges (3, 20 in base 60).

(iv) 60

1. 60 = 1×60 + 0.
2. There is nothing in the 1s place, so that position is simply left blank.

Answerone wedge, then a blank space (1, 0 in base 60).

(v) 3605

1. 3605 = 1×3600 + 0×60 + 5.
2. The 60s place is empty (a blank), and the 1s place uses five wedges.

Answerone wedge, a blank space, then five wedges (1, 0, 5 in the 3600s, 60s, 1s places).

Watch this explained “Writing bigger numbers”, 2:23 into Mesopotamian base-60: place value in a sexagesimal system · हिंदी में देखें

Open NCERT p. 80Matches NCERT’s answer

1. Why alternate: neighbouring places are drawn differently (Zong upright, Heng lying flat), so the eye can see where one digit ends and the next begins, even without spaces.
2. Zong only: 41 has 4 in the tens place and 1 in the ones place, both drawn upright: |||| |.
3. Without a clear space this becomes |||||, five upright rods, which is also the Zong symbol for 5. It could equally be read as 14, 23 or 32, or even 113.

AnswerThe Chinese alternated Zong and Heng so that neighbouring places look different. With Zong only, 41 is |||| |; without a space it becomes |||||, which could also be read as 5, 14, 23 or 32, so yes, it can be misread.

Watch this explained “Try it without the turn”, 3:39 into Chinese rod numerals: base-10 place value, one symbol short · हिंदी में देखें

Open NCERT p. 80Checked by computerAnswers can differ: one example

1. This is an open question; here is one way. Let urapon be the digit 0 and ukasar the digit 1 (the other way round works just as well).
2. The places, from the right, are worth 1, 2, 4, 8, 16, …; each is 2 times the one before, as Hindu places are worth 1, 10, 100, …

3. Number	Numeral
   0	urapon
   1	ukasar
   2	ukasar-urapon
   3	ukasar-ukasar
   4	ukasar-urapon-urapon
   5	ukasar-urapon-ukasar
   6	ukasar-ukasar-urapon
   7	ukasar-ukasar-ukasar
   8	ukasar-urapon-urapon-urapon

4. Compare with the Gumulgal system: there ukasar means 2 and urapon means 1, and a name is simply added up (ukasar-ukasar-urapon = 2 + 2 + 1 = 5), so the place of a word does not matter. Here a word's value depends on its place: ukasar-urapon-ukasar = 4 + 0 + 1 = 5.
5. The Gumulgal names grow long quickly (16 is eight ukasar), while here 16 is ukasar-urapon-urapon-urapon-urapon, five words. This system also needs a word for zero, which the Gumulgal system does not.

AnswerWith urapon = 0 and ukasar = 1 in places worth 1, 2, 4, 8, …: 0 urapon, 1 ukasar, 2 ukasar-urapon, 3 ukasar-ukasar, 4 ukasar-urapon-urapon, 5 ukasar-urapon-ukasar, 6 ukasar-ukasar-urapon, 7 ukasar-ukasar-ukasar, 8 ukasar-urapon-urapon-urapon. Unlike Gumulgal names, which are added up, each word's place fixes its value, so names stay much shorter.

Watch this explained “Reading one, from the right”, 4:53 into Mesopotamian base-60: place value in a sexagesimal system · हिंदी में देखें

Open NCERT p. 80One way to think about it

1. This is an open question; these are examples, and many other answers are right.
2. Daily life: prices, bills and bank balances, time and dates, phone numbers, measurements of length, weight and temperature, marks and scores.
3. Professions: shopkeepers and accountants, bankers, engineers and architects, doctors and pharmacists (doses), scientists, and computer programmers (computers store numbers in base 2, using 0 and 1).
4. Zero matters in three ways: it marks an empty place, so 205 is not confused with 25; it means 'nothing', like a zero balance; and it is a number we can calculate with, as Brahmagupta described.
5. Without place value and 0, numbers would be written the Roman or Egyptian way: long numerals, new symbols for bigger numbers, and slow, hard calculation, so banking, science and computers would have been much harder to build.

In shortHindu numerals and 0 are used wherever numbers are written or calculated: money, time, measurement, science, engineering, medicine and computing. Without them, numbers would be long and clumsy to write and slow to calculate with, and much of modern science and technology would have been far harder. Other examples are equally valid.

Watch this explained “A digit among digits”, 4:38 into The Hindu number system, and why treating 0 as a digit changed everything · हिंदी में देखें

Open NCERT p. 80Checked by computer

1. Base 8: the powers of 8 are 1, 8, 64, ...; 25 = 3×8 + 1, so 25 is written '31' in base 8.
2. Base 5: the powers of 5 are 1, 5, 25, ...; 25 = 1×25 + 0×5 + 0, so 25 is written '100' in base 5.
3. Base 2: the powers of 2 are 1, 2, 4, 8, 16, ...; 25 = 16 + 8 + 1, so 25 is written '11001' in base 2.

Answer25 in base 8 is 31; in base 5 is 100; in base 2 is 11001.

Watch this explained “Two promotions”, 9:10 into The Hindu number system, and why treating 0 as a digit changed everything · हिंदी में देखें