CBSE Class 8 Mathematics A Story of Numbers Long Answer and Case Study Questions with Answers

Long Answer Questions

Question 1. Before numerals were invented, people still had to count their animals, their grain and the passing days.
(i) Describe the three methods of counting discussed in the chapter — using objects, using names or sounds, and using written symbols — with one example of each.
(ii) Explain what a one-to-one mapping is and why it is the basic idea behind all three methods.
(iii) State one drawback of the stick method and one drawback of the method that uses the letters ‘a’ to ‘z’, and write what a good number system must therefore be.
Answer:
(i) Using objects: one stick (or pebble) is kept for each cow, so the heap of sticks shows the number of cows.
Using names or sounds: a fixed sequence of sounds, such as the letters ‘a’, ‘b’, ‘c’, …, ‘z’, is said in order while pointing at the objects.
Using written symbols: a fixed sequence of written marks, such as I, II, III, IV, V, …, X, …, XX of the Roman system.
(ii) In a one-to-one mapping, each object counted is paired with exactly one member of the standard sequence, and no two objects share a member. This matching, done in the fixed order, is what actually does the counting; the last member reached gives the size of the collection.
(iii) Sticks never run out, but one needs as many sticks as objects, so large collections become unmanageable. Letters are easy to say, but they stop at 26.
So a good number system needs a standard sequence that is both unending and convenient — exactly what the Hindu number system provides.
In simple words: All counting pairs objects one-to-one with a sequence; the best sequence never ends and is easy to use.

Teacher's Note:
1. Give one clear example for each method.
2. End with the two qualities: unending and convenient.

Question 2. The numerals we write today began their journey in India.
(i) What do ancient Indian texts such as the Yajurveda Samhita tell us about number names based on powers of 10?
(ii) Name the manuscript in which the digit 0 (written as a dot) first appears, and name the mathematician of about 499 CE who made elaborate scientific computations with the ten Indian symbols.
(iii) Describe briefly how these numerals reached the Arab world and then Europe, naming the scholars involved, and explain why they came to be wrongly called ‘Arabic numerals’.
Answer:
(i) The Yajurveda Samhita already lists number names based on powers of 10, much as we say them today — eka (1), dasha (10), shata (100), sahasra (1000), ayuta (10,000) — going up to 10¹² and beyond. So the spoken structure of our numbers is thousands of years old.
(ii) The Bakhshali manuscript (about the 3rd century CE) has the first known use of ten digits with 0 as a dot. Aryabhata (about 499 CE) first fully explained the system and computed with it.
(iii) The system reached the Arab world by about 800 CE. Al-Khwarizmi (about 825) and Al-Kindi (about 830) wrote about calculating with Hindu numerals. It reached Europe by about 1100 CE, and Fibonacci argued strongly for it around 1200.
Europeans learnt the numerals from Arab scholars, so they called them ‘Arabic numerals’, although the Arab scholars themselves called them ‘Hindu numerals’.
In simple words: India named and wrote numbers in base 10; Arab scholars carried them to Europe, which wrongly called them Arabic.

Teacher's Note:
1. Keep the dates in order: 3rd century, 499, 825, 830, 1100, 1200.
2. Correct names: Hindu, Indian or Hindu-Arabic numerals.

Question 3. The Roman number system was used in Europe for centuries.
(i) Write the seven Roman landmark numbers and the value of each.
(ii) Write 1368 and 2742 in Roman numerals, showing the grouping into landmark numbers in each case.
(iii) Give two reasons why this system was finally replaced by the Hindu number system.
Answer:
(i) I = 1, V = 5, X = 10, L = 50, C = 100, D = 500, M = 1000.
(ii) 1368 = 1000 + 100 + 100 + 100 + 50 + 10 + 5 + 1 + 1 + 1, so 1368 = MCCCLXVIII. Check: 1000 + 300 + 50 + 10 + 5 + 3 = 1368.
2742 = 1000 + 1000 + 500 + 100 + 100 + 40 + 1 + 1, so 2742 = MMDCCXLII (40 is written XL, ten less than fifty). Check: 2000 + 700 + 40 + 2 = 2742.
(iii) First, very large numbers cannot be written without inventing more and more new symbols.
Second, arithmetic, especially multiplication and division, is very difficult, because the landmark numbers are not powers of one number; only trained people could compute, and they needed an abacus.
In simple words: Roman numerals use seven symbols, but they cannot write all numbers easily and make calculation hard.

Teacher's Note:
1. Group from the largest landmark number down.
2. Remember the 'one less' forms: IV, IX, XL, XC, CD, CM.

Question 4. The Egyptians developed a written number system around 3000 BCE.
(i) Explain, using the idea of making collections of pebbles, how each Egyptian landmark number is obtained from the one before it, and write the first five landmark numbers.
(ii) Show how 3254 is grouped in this system and state how many of each symbol are used.
(iii) State the main shortcoming of the Egyptian system when very large numbers have to be written.
Answer:
(i) Start with 1 pebble: the first landmark number is 1. Put 10 such collections together to get 10, the second landmark number. Put 10 collections of 10 together: 10 × 10 = 100, then 1000, and so on.
Each landmark number is 10 times the one before: 1, 10, 100, 1000, 10000.
(ii) 3254 = 1000 + 1000 + 1000 + 100 + 100 + 10 + 10 + 10 + 10 + 10 + 1 + 1 + 1 + 1.
So it uses 3 symbols for 1000, 2 for 100, 5 for 10 and 4 for 1 — 14 symbols in all. Check: 3000 + 200 + 50 + 4 = 3254.
(iii) Each new power of 10 needs a brand new symbol. The Egyptians had symbols up to about 10⁷, so beyond that an unending list of new symbols would be needed. The problem was solved only by place value.
In simple words: Egyptian landmark numbers go up by ×10, but every new power needs a new symbol.

Teacher's Note:
1. Count the symbols by adding the digits: 3 + 2 + 5 + 4 = 14.
2. The shortcoming leads to the idea of place value.

Question 5. The chapter builds a brand new number system by changing just one thing — the group size.
(i) State when a number system is called a base-n system, and write the landmark numbers of a base-4 system up to the fourth one.
(ii) Using the symbols □ for 16, △ for 4 and L for 1, write the numbers 7, 11 and 14 in this base-4 system.
(iii) Convert 150 into a base-4 place value numeral, showing the grouping, and check your answer.
Answer:
(i) A number system is base-n when its first landmark number is 1 and each next landmark number is the current one multiplied by the same fixed number n. For base 4: 1, 4, 16, 64.
(ii) 7 = 4 + 1 + 1 + 1, so 7 is △ L L L.
11 = 4 + 4 + 1 + 1 + 1, so 11 is △ △ L L L.
14 = 4 + 4 + 4 + 1 + 1, so 14 is △ △ △ L L.
No symbol is ever needed four times, because four of a symbol make the next one.
(iii) 150 = 2 × 64 + 22; 22 = 1 × 16 + 6; 6 = 1 × 4 + 2.
So 150 = 2 × 64 + 1 × 16 + 1 × 4 + 2 × 1, and the base-4 numeral is 2112. Check: 128 + 16 + 4 + 2 = 150.
In simple words: Base 4 uses 1, 4, 16, 64; 150 becomes 2112.

Teacher's Note:
1. None of 7, 11 and 14 needs the 16 symbol.
2. In base 4 each digit is 0, 1, 2 or 3.

Question 6. A base-60 system may look strange today, yet we still live with it.
(i) Give two reasons that scholars have suggested for the Mesopotamians choosing 60 as their base, and state where base 60 still survives in our daily life.
(ii) Express 4000 in the Mesopotamian system, writing it as a sum of multiples of 3600, 60 and 1.
(iii) Explain the ambiguity that blank spaces caused in their numerals, and how the later Mesopotamians tried to remove it.
Answer:
(i) Any two: 60 links with the periods of important events, such as their lunar month of 30 days or the yearly journey of the Sun; fractions are easy to write in base 60; their older chain of landmark numbers 1, 10, 60, 600, 3600, … settled into powers of 60 alone.
Base 60 survives in time: 1 hour = 60 minutes and 1 minute = 60 seconds.
(ii) 4000 = 1 × 3600 + 400 = 1 × 3600 + 6 × 60 + 40 × 1. So the numeral has three groups: 1, 6, 40. Check: 3600 + 360 + 40 = 4000.
(iii) They showed each power of 60 by position alone and left a blank where a power was missing. The spacing differed from scribe to scribe, so a reader could not tell which cluster went with which power or how many blanks lay between clusters; the same numeral could be read as different numbers.
The later Mesopotamians introduced a placeholder symbol for an empty position — the idea behind our 0 — though they still did not use it at the end of a numeral.
In simple words: 60 was handy for time and fractions; blanks caused misreading until a placeholder was used.

Teacher's Note:
1. 60 has many factors, so fractions are simple.
2. A placeholder is the first step towards zero.

Teacher's Note:
1. Count Roman symbols one by one: M M D C C C X L V I I.
2. Powers of 10 multiply into powers of 10.

Question 8. “Position decides value.” Explain this idea.
(i) Define a positional (place value) number system.
(ii) Name the four civilisations mentioned in the chapter that used place value representations, and state the base used by each where the chapter gives it.
(iii) Explain why the Mesopotamian and Mayan systems are still not counted as fully developed place value systems, and what the Hindu system did differently.
Answer:
(i) A number system that has a base and uses the position of each symbol to decide which landmark number it counts is a positional or place value system. Symbols for the landmark numbers are then not needed, so writing becomes very compact.
(ii) Mesopotamian (Babylonian): base 60. Mayan: landmark numbers 1, 20, 360. Chinese rod numerals: base 10. Hindu (Indian): base 10.
(iii) The Mesopotamians first left blanks for missing powers of 60; uneven spacing let one numeral be read in more than one way, and even their later placeholder was not used at the end of a numeral, so a number like 3600 stayed ambiguous.
The Mayan system had a shell-like placeholder, but its third landmark number is 360 instead of 20 × 20 = 400, so it is not a true base-20 system and loses the advantages of a base.
The Hindu system uses exactly one digit in every position and a genuine digit 0, so every number has only one way of being written, and 0 is also a number in its own right.
In simple words: Place value means position sets value; only the Hindu system made it complete with a true base and a real zero.

Teacher's Note:
1. The Mayan system does not give a base number in the chapter's sense.
2. Zero at the end of a numeral is what removes the last ambiguity.

Question 9. The chapter ends with five great ideas that together produced the way we write numbers today — counting in groups of a single number, grouping using landmark numbers, choosing powers of a number as landmark numbers, using positions to show the landmark numbers, and the idea of 0. Take these five ideas in order. For each one, name a number system of the chapter that shows it, give one short example of a number written using that idea, and state the particular difficulty of the earlier stage that the idea removed. End by stating which single idea you think was the most powerful, with a reason.
Answer:
Stage 1 — Counting in groups of a single number (Gumulgal). 5 = 2 + 2 + 1 is ukasar-ukasar-urapon. It improved on plain tally marks, where one mark was needed for every object and the eye cannot take in five or more at a glance. Its own difficulty: a number like 1345 would need hundreds of repetitions of one word.
Stage 2 — Grouping with landmark numbers (Roman). Symbols for 1, 5, 10, 50, 100, 500, 1000, so 27 = 10 + 10 + 5 + 1 + 1 is XXVII, far shorter than 27 marks. It removed endless repetition of one group size, but since the landmark numbers are not powers of one number, multiplication stayed very difficult.
Stage 3 — Powers of a number as landmark numbers: a base (Egyptian). 1, 10, 100, 1000, …, so 324 is three hundred-symbols, two ten-symbols and four unit-symbols. Now the product of two landmark numbers is another landmark number and regrouping is always by the same size, so computation becomes easy. Remaining difficulty: every new power needs a new symbol.
Stage 4 — Position shows the landmark number: place value (Mesopotamian). 640 = 10 × 60 + 40 needs only two clusters. Finitely many symbols now write every number. Remaining difficulty: the blank space for a missing power.
Stage 5 — Zero as a digit and as a number (Hindu). A placeholder fills the empty position, and in India 0 became a full number, as in 1729 or 2026.
In my view the fifth idea, zero, was the most powerful: it alone made every numeral unambiguous and gave arithmetic, algebra and modern science their foundation. (Any well-reasoned choice is acceptable.)
In simple words: Each idea fixed the weakness of the one before, and zero completed the system.

Teacher's Note:
1. Keep the five stages in order with one system and one example each.
2. The final choice is open; the reason is what earns the mark.

Question 10. Suppose human beings had been born with only eight fingers.
(i) Explain how many digit symbols the Hindu number system would then have had, what its landmark numbers would be, and how a number would be written as a sum of multiples of those landmark numbers.
(ii) Write 1000 as a base-8 numeral and as a base-5 numeral, showing the grouping in both, and verify each answer.
(iii) Using your two answers, explain the trade-off between the size of the base and the length of the numeral, and say why a base of 10 is a comfortable middle choice.
Answer:
(i) We would most likely have used base 8, with eight digits, 0 to 7, and landmark numbers the powers of 8: 1, 8, 64, 512, 4096, …
Every number N would be written as \( N = a_k \times 8^k + a_{k-1} \times 8^{k-1} + \dots + a_1 \times 8 + a_0 \), with each digit \( a_i \) one of 0 to 7 and \( a_k \ne 0 \); the numeral is \( a_k a_{k-1} \dots a_1 a_0 \).
(ii) Base 8: 1000 = 1 × 512 + 488; 488 = 7 × 64 + 40; 40 = 5 × 8 + 0. So 1000 = 1 × 512 + 7 × 64 + 5 × 8 + 0 × 1, numeral 1750. Check: 512 + 448 + 40 + 0 = 1000.
Base 5 (landmark numbers 1, 5, 25, 125, 625): 1000 = 1 × 625 + 375 and 375 = 3 × 125. So 1000 = 1 × 625 + 3 × 125 + 0 × 25 + 0 × 5 + 0 × 1, numeral 13000. Check: 625 + 375 = 1000.
(iii) The same number needs 5 digits in base 5 and 4 digits in base 8 (and 4 in base 10 as well, while base 2 would need 10 digits: 1111101000).
A larger base gives shorter numerals (each group holds more) but needs more digit symbols to remember and bigger addition and multiplication tables. A smaller base needs few symbols but long numerals.
Base 10 sits comfortably between these extremes, and it matches our ten fingers.
In simple words: Bigger bases mean shorter numerals but more symbols; base 10 balances the two.

Teacher's Note:
1. Count the digits carefully: 1000 itself has four digits in base 10.
2. Always check a conversion by expanding it back.

Case Study Based Questions

Question 11. The Gallery of Early Counting
A city museum has opened a gallery called “Before Numerals”. The first showcase holds casts of two famous bones. A label reads: “The Lebombo bone, found in South Africa, carries 29 notches and is about 44,000 years old. The Ishango bone, found in the Democratic Republic of Congo, is 20,000 to 35,000 years old and has notches arranged in columns, which may record a calendar.” A child in the gallery cuts 29 small lines on a strip of clay to copy the Lebombo bone, and her teacher tells her that this is exactly how the earliest people recorded how many goats they owned or how many days had passed.
(i) What are the marks cut on these bones called?
(ii) Which of the three counting methods of the chapter — sticks, sounds or written symbols — do these notches resemble most?
(iii) How many objects would the maker of the Lebombo bone have been counting?
(iv) State one difficulty a person would face in using this method to record 2,000 objects.
Answer:
(i) They are called tally marks (notches).
(ii) The stick method: one mark per object, just as one stick per cow; a mark is cut instead of a stick being added.
(iii) 29 objects, since the bone has 29 notches (perhaps 29 days of a lunar month).
(iv) 2,000 separate notches would be needed, taking a huge surface and a long time, and the total could only be read by counting every notch one by one.
In simple words: Tally marks record one notch per object, which is simple but impractical for large numbers.

Teacher's Note:
1. Tally marks are one-to-one mapping cut into a surface.
2. Large counts need grouping, not single marks.

Question 12. The Station Clock
The clock above the platform of an old railway station has a brass face on which the hours are marked in Roman numerals instead of the usual digits. A traveller looks up and sees the numeral IX on the lower left of the dial and the numeral IV on the right. Her little brother asks why 4 is written as IV and not as IIII, and why some hour marks take so many more symbols than others. Their grandfather explains that the Romans grouped a number into as many 10s as possible, then as many 5s as possible, then 1s, and that they also wrote one less than a landmark number by placing a smaller symbol before it.
(i) Which hour is marked by the numeral IX?
(ii) Write the numerals for the hours 7 and 12 as they appear on this clock face.
(iii) Explain why 4 can be written as IV.
(iv) Which Roman landmark numbers appear anywhere on a 12-hour clock face?
Answer:
(i) IX stands for 9, one less than ten.
(ii) 7 = 5 + 1 + 1 is VII; 12 = 10 + 1 + 1 is XII.
(iii) 5 is a landmark number, V. Writing the smaller I before V means one less than five, so IV = 4. In the same way XL = 40.
(iv) Only I (1), V (5) and X (10). L, C, D and M are not needed below 13.
In simple words: Clock numerals use only I, V and X, with IV and IX meaning one less than 5 and 10.

Teacher's Note:
1. A smaller symbol before a larger one means subtract.
2. A smaller symbol after means add.

Question 13. Reema’s Sheet of Symbols
A sheet of paper slipped out of an old book and floated to the floor. Reema picked it up and found it covered with wedge-like marks arranged in small clusters. Her father smiled and said that about 4000 years ago a civilisation flourished in Mesopotamia, in the western part of Asia, in a region that today lies mostly in Iraq, and that these clusters were their way of writing numbers. He added that their system had one symbol for 1 and another for 10, that it later grew into a base-60 system, and that although the civilisation has long disappeared, we still use their base 60 every single day.
(i) Which civilisation’s numerals were on Reema’s sheet?
(ii) In which present-day country does most of that region lie?
(iii) What is the base-60 system also called?
(iv) Give one place in everyday life where base 60 is still used.
Answer:
(i) The Mesopotamian civilisation, about 4000 years ago in western Asia.
(ii) Iraq.
(iii) The sexagesimal system (or the Babylonian number system).
(iv) Measuring time: 1 hour = 60 minutes and 1 minute = 60 seconds (also degrees and minutes of angle).
In simple words: The marks were Mesopotamian base-60 numerals, which survive in our clocks.

Teacher's Note:
1. Sexagesimal means base 60.
2. Clock time is the everyday link.

Question 14. The Mango Vendor
Shyam sells mangoes at a wholesale market. He never counts mangoes one by one. He first fills small trays that hold 10 mangoes each; when he has 10 full trays, he empties them into a crate that holds 100 mangoes; when 10 crates are full, they go into a truck-load of 1000. At the end of the morning his record book shows 3 truck-loads, 4 crates, 7 trays and 5 loose mangoes. His helper laughs and says, “You are counting exactly the way the Egyptians wrote their numbers.”
(i) Write the landmark numbers that Shyam is using.
(ii) How many mangoes did Shyam sell that morning?
(iii) Name the base of the number system Shyam is using.
(iv) Why does Shyam never leave 10 filled trays in his final record?
Answer:
(i) 1 (loose mango), 10 (tray), 100 (crate), 1000 (truck-load); each is 10 times the one before.
(ii) 3 × 1000 + 4 × 100 + 7 × 10 + 5 × 1 = 3000 + 400 + 70 + 5 = 3475 mangoes.
(iii) Base 10 (decimal), the same grouping the Egyptians used.
(iv) 10 full trays are at once exchanged for one crate. In base 10 no landmark number may occur 10 or more times, since 10 of any landmark number make exactly one of the next.
In simple words: Trays, crates and trucks are 10, 100 and 1000, so the record 3, 4, 7, 5 means 3475.

Teacher's Note:
1. The counts 3, 4, 7, 5 read straight off as the digits of 3475.
2. Exchanging 10 for 1 is 'carrying'.

Question 15. The Archaeologist’s Tablet
An archaeologist cleans a small clay tablet from Mesopotamia. On it she can make out three clusters of wedge marks separated by gaps. Working from the left, the first cluster shows 2, the second shows 11 and the third shows 35. She knows that in this system the rightmost cluster counts the 1s, the cluster to its left counts the 60s and the next one counts the 3600s. She also notes in her diary that the gaps on the tablet are uneven, so another reader might group the marks differently and read out a completely different number.
(i) Which number is written on the tablet?
(ii) Why can no cluster on such a tablet ever stand for 60 or more?
(iii) Explain the ambiguity the archaeologist is worried about.
(iv) What idea did the later Mesopotamians use to reduce this ambiguity?
Answer:
(i) 2 × 3600 + 11 × 60 + 35 × 1 = 7200 + 660 + 35 = 7895.
(ii) 60 of any power of 60 make exactly one of the next power, so they would be regrouped into the position to the left. Each cluster can show only 1 to 59.
(iii) Because the spacing is uneven, another reader may split the marks into clusters differently or misjudge how many empty positions lie between clusters. The marks then go with the wrong powers of 60 and the tablet is read as a different number.
(iv) A placeholder symbol written in an empty position instead of a blank — the idea behind our 0. It was still not used at the end of a numeral, so some ambiguity remained.
In simple words: The tablet shows 7895, but uneven gaps can mislead a reader until a placeholder is used.

Teacher's Note:
1. Multiply each cluster by its power of 60.
2. A placeholder fixes blanks only where it is actually used.

Question 16. The Granary Scribe
An Egyptian scribe keeps the grain accounts of a royal granary, counting in sacks. For the first store he writes 2 of the symbol for 1000, 4 of the symbol for 100, no symbol for 10 and 7 of the symbol for 1. For the second store he writes 1 of the symbol for 1000, 8 of the symbol for 100, 6 of the symbol for 10 and 5 of the symbol for 1. He must now record the two stores together on a single tablet, and he knows the rule that 10 copies of any symbol are replaced by one copy of the next symbol.
(i) How many sacks of grain are in the first store?
(ii) How many separate symbols does the scribe write for the first store?
(iii) Find the total for both stores and state how many of each symbol the combined numeral has after regrouping.
(iv) Why would this scribe be in difficulty if he had to record a number as large as 10¹²?
Answer:
(i) 2 × 1000 + 4 × 100 + 0 × 10 + 7 × 1 = 2407 sacks.
(ii) 2 + 4 + 0 + 7 = 13 symbols.
(iii) Second store: 1000 + 800 + 60 + 5 = 1865. Total = 2407 + 1865 = 4272 sacks.
Putting the symbols together gives 3 thousand-, 12 hundred-, 6 ten- and 12 unit-symbols. 12 unit-symbols become 1 ten + 2 units, making 7 ten-symbols; 12 hundred-symbols become 1 thousand + 2 hundreds, making 4 thousand-symbols.
Combined numeral: 4 thousand-symbols, 2 hundred-symbols, 7 ten-symbols and 2 unit-symbols, which reads 4272.
(iv) Every higher power of 10 needs a new Egyptian symbol, and symbols existed only up to about 10⁷. For 10¹² he would first have to invent several new symbols, and the list of new symbols would never end.
In simple words: The stores hold 2407 and 1865 sacks; together 4272 after regrouping.

Teacher's Note:
1. Regroup from the smallest symbol upwards.
2. No symbol for 10 means nothing is written there, not a zero.

Question 17. The Calendar Stone
A carved stone from a Mayan city shows a column of symbols. Reading from the bottom upwards, the lowest group has two bars and three dots, the group above it is a single shell-like sign, and the topmost group has two dots. The guide explains that the lowest position counts 1s, the next counts 20s and the one above counts 360s, that a dot stands for 1 and a bar for 5, and that the shell-like sign is their placeholder, used in the same way as our 0.
(i) Write the three landmark numbers used on this stone.
(ii) Which number does the carving stand for?
(iii) How is the number 13 drawn in Mayan symbols?
(iv) Why is the Mayan system not a true base-20 system?
Answer:
(i) 1, 20 and 360, from the bottom position upwards.
(ii) Bottom: two bars and three dots = 5 + 5 + 1 + 1 + 1 = 13. Middle: shell = 0. Top: two dots = 2.
Number = 2 × 360 + 0 × 20 + 13 × 1 = 720 + 0 + 13 = 733.
(iii) Two bars (5 + 5 = 10) together with three dots (1 + 1 + 1 = 3).
(iv) In a true base-20 system the landmark numbers would be 1, 20, 400, 8000, … Here the third is 360, not 400, so they are not all powers of one number and the system loses the advantages of a real base.
In simple words: The stone shows 2 × 360 + 0 + 13 = 733.

Teacher's Note:
1. Read from the bottom up.
2. The third Mayan place is 360, a common trap.

Question 18. The Base-5 Counter at the Maths Mela
At a school maths mela, one counter is run entirely in base 5. Visitors are given loose beads; five loose beads are exchanged for one stick, five sticks for one ring, and five rings for one star. A notice on the table says: “No one may leave this counter holding five of anything.” A visitor arrives at the counter with 237 beads and the volunteers begin exchanging them.
(i) Write the landmark numbers of this counter.
(ii) After all the exchanges, how many stars, rings, sticks and loose beads will the visitor have?
(iii) Write the visitor’s total as a base-5 place value numeral.
(iv) Explain the reason behind the notice on the table.
Answer:
(i) 1 (bead), 5 (stick), 25 (ring), 125 (star).
(ii) 237 = 1 × 125 + 112; 112 = 4 × 25 + 12; 12 = 2 × 5 + 2.
So the visitor gets 1 star, 4 rings, 2 sticks and 2 loose beads. Check: 125 + 100 + 10 + 2 = 237.
(iii) Writing the counts from the largest landmark number: 1422.
(iv) Five of anything is always exchanged for one of the next kind, so no place can hold 5 or more. That is why base-5 numerals use only the digits 0 to 4 and every number has just one correct form.
In simple words: 237 beads become 1 star, 4 rings, 2 sticks and 2 beads, that is 1422 in base 5.

Teacher's Note:
1. One ring is 5 sticks = 25 beads; one star is 125 beads.
2. The notice is the rule 'no digit equal to the base'.

Question 19. Two Symbols Are Enough
In the school computer lab, Aarav is told that the machine in front of him stores every number using only two symbols, 0 and 1, because each tiny switch inside it can only be off or on. His teacher says that this is simply a base-2 number system, built on exactly the same idea as the Egyptian base-10 and Mesopotamian base-60 systems they had read about: the landmark numbers are the powers of the base, and the position of a digit tells which landmark number it counts. Aarav decides to test the idea by writing a few numbers for himself.
(i) Write the first eight landmark numbers of the base-2 system.
(ii) Write 83 as a base-2 numeral, showing the grouping.
(iii) Which number does the base-2 numeral 110110 stand for?
(iv) How many digits are needed to write 500 in base 2, and what does this tell you about choosing a very small base?
Answer:
(i) 1, 2, 4, 8, 16, 32, 64, 128.
(ii) 83 = 64 + 19; 19 = 16 + 3; 3 = 2 + 1.
83 = 1 × 64 + 0 × 32 + 1 × 16 + 0 × 8 + 0 × 4 + 1 × 2 + 1 × 1, so the numeral is 1010011. Check: 64 + 16 + 2 + 1 = 83.
(iii) Place values 32, 16, 8, 4, 2, 1: 32 + 16 + 0 + 4 + 2 + 0 = 54.
(iv) 500 = 256 + 128 + 64 + 32 + 16 + 4, so it is 111110100, which needs 9 digits. A very small base needs only a couple of symbols (good for on-off switches), but numerals become very long.
In simple words: Base 2 uses only 0 and 1, so numbers like 83 and 500 become long strings of digits.

Teacher's Note:
1. Write a 0 for every power of 2 that is skipped.
2. Fewer symbols means longer numerals.

Question 20. The Tablet of a Lost People
A team digging near a dry river bed finds a tablet of a people whose number system is unknown. The tablet records a quantity of grain written with three symbols that the team is confident mean the digits 2, 1 and 3, read from left to right in a place value system, exactly as we read 213. A second, undamaged tablet from the same site gives the same quantity in a known script as 58. The team therefore has one numeral, 213, written in an unknown base n, and its value, 58, in our own system.
(i) Write an equation in n for the value of the numeral 213 in base n.
(ii) Solve your equation and find the base used by these people.
(iii) Verify your answer by converting the numeral back to base ten.
(iv) Elsewhere the same tablet records 100 units of oil. Write 100 as a numeral in the base you have found.
Answer:
(i) The places have values n², n and 1, so 213 means 2n² + 1 × n + 3. Hence 2n² + n + 3 = 58.
(ii) 2n² + n − 55 = 0, which factorises as (2n + 11)(n − 5) = 0, giving n = \( -\frac{11}{2} \) or n = 5.
A base must be a positive whole number greater than every digit used (here greater than 3), so n = 5. These people counted in base 5.
(iii) 2 × 5² + 1 × 5 + 3 = 50 + 5 + 3 = 58, which matches the undamaged tablet.
(iv) Base-5 landmark numbers: 1, 5, 25, 125. 100 = 4 × 25 + 0 × 5 + 0 × 1, so the numeral is 400. Check: 4 × 25 = 100.
In simple words: Solving 2n² + n + 3 = 58 gives base 5, and 100 is written 400 in that base.

Teacher's Note:
1. Reject the fractional root because a base is a whole number.
2. In base 5, 100 = 4 × 25 gives 400.