A Story of Numbers Class 8 Solutions

Early humans counted food, livestock, trade goods, ritual offerings, and passing days. The modern place-value system using the digits 0-9 developed in India, spread to the Arab world by about 800 CE, and later reached Europe and the rest of the world.

Pair one stick from the first herd with one stick from the second. If one collection has sticks left over, that herd is larger. The unpaired sticks give exactly how many cows the smaller herd needs to become equal.

Exactly as many as there are letters in the chosen ordered alphabet. For English, the one-letter method represents 26 numbers.

Introduce symbols for further landmark numbers and combine them by fixed rules. Roman numerals do this with L, C, D, and M for 50, 100, 500, and 1000, though ever-larger values still require more conventions.

For addition, join the collections. For subtraction, pair and remove sticks. For multiplication, make equal copies of one collection, one for each stick in the other. For division, repeatedly form equal groups of the divisor size; the number of groups is the quotient and any sticks left are the remainder.

Use strings as place-value digits, as spreadsheet columns do: a through z, then aa, ab, ..., az, ba, and so on. Every positive integer then receives a finite string.

One valid design is a base-3 place-value system with digits A, B, C for 0, 1, 2. Read positions from right to left as powers of 3; for example, BCA means 1\times3^2+2\times3+0=15.

Urapon means 1 and ukasar means 2. Larger names add these units: 3 is 2+1, 4 is 2+2, 5 is 2+2+1, and 6 is 2+2+2.

People commonly recognise groups up to about 4 immediately. At 5 or more, most people begin counting or mentally grouping the objects.

A single landmark must be repeated too many times as numbers grow. Since 1345=269\times5, a system with only a symbol for a group of 5 would need that symbol 269 times, unless it introduced larger landmarks.

(i) MCCXXII; (ii) MMCMXCIX; (iii) CCCII; (iv) DCCXV.

Combine like symbols and regroup five Xs as L and two Ls as C. The result is CLXV.

The products are CCL, 25 copies of M, MMD, and LXIII respectively. The second result illustrates why multiplication and very large numbers are awkward in this Roman system.

The product is 427812. With only the listed additive Roman landmarks, this would require 427 copies of M followed by DCCCXII, showing why an abacus or an extended overbar convention was needed for such calculations.

Different objects may be counted in customary bundles or with specialised classifiers. Separate sequences can reflect how a community trades, stores, or speaks about those objects.

(i) eight ukasar; (ii) ukasar-urapon; (iii) eighteen ukasar; (iv) four ukasar. These correspond to regrouping entirely in units of 2, with one urapon where needed.

It is a place-value system, uses only ten reusable digits including 0, represents arbitrary sizes compactly, and supports standard algorithms for addition, subtraction, multiplication, and division.

Give it a fixed base, a digit for zero, one digit for each value from 0 to base-1, and place values that are powers of the base. These rules make representation unambiguous and arithmetic systematic.

Use repeated landmark symbols for these decompositions: 10458=10000+4(100)+5(10)+8; 1023=1000+2(10)+3; 2660=2(1000)+6(100)+6(10); 784=7(100)+8(10)+4; 1111=1000+100+10+1; 70707=7(10000)+7(100)+7.

Counting each landmark symbol gives (i) 276 and (ii) 4322.

Since 143=1(125)+0(25)+3(5)+3(1), use one 5^3 symbol, no 5^2 symbol, three 5^1 symbols, and three unit symbols.

Their landmark counts are: 15=3(5); 50=2(25); 137=125+2(5)+2; 293=2(125)+25+3(5)+3; 651=625+25+1. Replace each term by the matching symbol shown in the textbook table.

Zero cannot be written because the system has no symbol for it. Every positive integer can be represented by grouping it into powers of 5.

Base 7 has landmarks 7^0=1,7^1=7,7^2=49,7^3=343,\ldots. In general, a base-n system has n^0,n^1,n^2,n^3,\ldots.

For (i), the values are 9608 and 507, so the sum is 10115: one 10000 symbol, one 10 symbol, and five unit strokes. For (ii), the values are 1110 and 46, so the sum is 1156: one 1000 symbol, one 100 symbol, five 10 symbols, and six unit strokes.

The first is 125+2(25)+5+2=182, and the second is 3(125)+25+2(5)+2=412. Their sum is 594, represented by four 5^3 symbols, three 5^2 symbols, three 5^1 symbols, and four unit symbols.

Multiplication by 10 advances each landmark one place: 10\times10=100, 100\times10=1000, 1000\times10=10000, and 10000\times10=100000.

Multiplication by 10^2 advances two landmarks: 10\times100=1000, 100\times100=10000, 1000\times100=100000, and 10000\times100=1000000.

The products are 10^5,10^5,10^6, and 10^{10}. Yes: in any base b, b^m b^n=b^{m+n}, so the product of two landmarks is another landmark. Multiplying by the base moves every landmark to the next one.

The first number is 522, so its product is 5220; the second is 1010, so its product is 10100. In the additive Egyptian system, replace every symbol by the next higher landmark symbol.

Bring counters of equal place value together. Ten unit counters exchange for one ten counter; then ten tens exchange for one hundred. The final abacus shows 2950.

No in a regrouped standard representation. Ten copies of any landmark must be exchanged for one copy of the next landmark.

Choose symbols U, F, and S for 1, 4, and 16. Then 1-3 are U, UU, UUU; 4 is F; 5-7 are FU, FUU, FUUU; 8-11 are FF, FFU, FFUU, FFUUU; 12-15 are FFF, FFFU, FFFUU, FFFUUU; and 16 is S.

Replace every landmark symbol by the next higher landmark symbol, because 5\times5^k=5^{k+1}. Regroup if five identical symbols arise.

Using place-value groups: 63=(1,3)_{60}, 132=(2,12)_{60}, 200=(3,20)_{60}, 60=(1,0)_{60}, and 3605=(1,0,5)_{60}. Draw each group with the textbook's symbols for 1 and 10.

Both may appear as a single unit symbol when trailing empty positions are not marked. Later Mesopotamians introduced a placeholder inside numerals, but because it was not consistently written at the end, some ambiguity remained.

From bottom upward, use the 1, 20, and 360 places: 77=(3,17); 100=(5,0); 361=(1,0,1); and 721=(2,0,1). In each place, dots mean 1, bars mean 5, and the shell is the zero placeholder.

The groups are 2 thousands, 6 hundreds, 3 tens, and 4 ones, so the numeral is 2\times10^3+6\times10^2+3\times10+4=2634.

Its landmarks are powers of 10: 1,10,100,1000,\ldots. Yes. A digit's position determines which power of 10 it multiplies, and 0 marks an empty position.

Alternating vertical and horizontal forms separates adjacent place values. With only Zong strokes, 41 could merge visually and be misread as another grouping, such as 23, 32, or 122, if spacing is unclear.

Let urapon be 0 and ukasar be 1. Then 1, 2, 3, 4, ... are ukasar; ukasar-urapon; ukasar-ukasar; ukasar-urapon-urapon; and so on. Unlike the additive Gumulgal names, position now gives powers of 2, making the system compact and unlimited.

They are essential in money, time, measurement, addresses, phones, computing, science, engineering, medicine, accounting, and trade. Without compact place value and zero, calculation, record keeping, digital technology, and scientific work would be far slower and more error-prone.

25=3\times8+1, so it is 31_8. Also 25=1\times5^2, so it is 100_5. Finally, 25=16+8+1, so it is 11001_2.