Exercise practice 2: A Story of Numbers

4.1 Exercise practice 2: Egyptian, base-5, place value and Mayan systems

This page covers the remaining Figure it Out sections. Some questions use symbol pictures; the numerical values and the regrouping rules are given below.

4.1.1 E2.1 Egyptian numerals

Question. Represent 10458, 1023, 2660, 784, 1111 and 70707 in the Egyptian system.

Answer. For 10458, take one 10000-symbol, four 100-symbols, five 10-symbols and eight unit symbols; for 1023, one 1000, two 10s and three units; for 2660, two 1000s, six 100s and six 10s; for 784, seven 100s, eight 10s and four units; for 1111, one each of 1000, 100, 10 and 1; and for 70707, seven 10000-symbols, seven 100-symbols and seven units.

Key steps. Egyptian symbols are base-10 landmarks, so write each number as a sum of 1, 10, 100, 1000, 10000 and so on. 784 is 7 hundreds + 8 tens + 4 units. 70707 needs care: it is 70000 + 700 + 7, not 70707 written as a single symbol; there is no symbol for 70707. Each count of a landmark becomes that many copies of its symbol.

4.1.2 E2.2 Reading Egyptian numerals

Question. What numbers do the two Egyptian numerals stand for?

Answer. The first is 276; the second is 4322.

Key steps. Add the values of the symbols: a 100-symbol is 100, a 10-symbol is 10 and a 1-symbol is 1. For 276, read as 2 hundreds + 7 tens + 6 units. For 4322, read as 4 thousands + 3 hundreds + 2 tens + 2 units. This is the same grouping rule used in E2.1, but in the reading direction.

4.1.3 E2.3 Base-5 numerals

Question. Write 15, 50, 137, 293 and 651 in the base-5 system, using the symbols of Table 2.

Answer. 15 = 3 x 5; 50 = 2 x 25; 137 = 125 + 5 + 5 + 1 + 1; 293 = 2 x 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1; 651 = 625 + 25 + 1.

Key steps. The landmarks are 1, 5, 25, 125, 625. Take as many of the largest landmark as possible, then move to the next. For 293: two 125s make 250, leaving 43; then one 25 leaves 18; then three 5s leave 3; then three 1s. Check the total each time.

4.1.4 E2.4 A number missing from base-5

Question. Can every number be written in this base-5 system? Explain.

Answer. Zero cannot be represented, because the system has no symbol for zero and no empty position. Every count starts from one symbol.

Key steps. A representation system needs a way to show "none of this landmark". If the system has no zero-like placeholder, an empty place is not distinguishable from an unwritten one. The Hindu system solves this with the digit 0, which is why it can write every number unambiguously with ten symbols.

4.1.5 E2.5 Landmark numbers of a base

Question. Work out the landmark numbers for base 7. What is the general pattern of landmarks in a base-n system?

Answer. Base-7 landmarks: 1, 7, 49, 343, 2401, ... In general, the base-n landmarks are the powers n^0, n^1, n^2, n^3, ...

Key steps. Each landmark is n times the previous one: 7^0 = 1, 7^1 = 7, 7^2 = 49, 7^3 = 343. The same pattern gives base-5 landmarks 1, 5, 25, 125 and base-10 landmarks 1, 10, 100, 1000.

4.1.6 E2.6 Egyptian addition

Question. Add the pairs of Egyptian numerals shown in the textbook.

Answer. (i) 10111, written with ten 1000-symbols (lotus), one 100-symbol, one 10-symbol and one unit symbol, after regrouping. (ii) 11206, written with one 10000-symbol (finger), one 1000-symbol, two 100-symbols and six unit symbols.

Key steps. For (i), count the symbols first: the first numeral has 9 lotus (9000), 6 coils (600) and 8 units; the second has 5 coils (500) and 3 units. The totals are 9 thousands, 11 hundreds and 11 units. Regroup ten hundreds as one thousand (making 10 thousands) and ten units as one ten, leaving 1 hundred, 1 ten and 1 unit: 10111. For (ii), the first numeral has one finger (10000) and 8 coils (800), while the second has 4 coils (400) and 6 units. The coil total is 12 hundreds, which regroups as 1 thousand and 2 hundreds, giving 11206.

4.1.7 E2.7 Addition in base-5

Question. Add the numerals shown in the base-5 system created in the chapter.

Answer. The sum is 595, written in the base-5 symbols as four triangle symbols (125s), three square symbols (25s), four hexagon symbols (5s) and no circle symbol (1s).

Key steps. In the base-5 symbols, circle = 1, hexagon = 5, square = 25 and triangle = 125. Count each numeral's symbols: the first has 1 circle, 2 hexagons, 1 square and 2 triangles; the second has 4 circles, 1 hexagon, 2 squares and 2 triangles. The totals are 5 circles, 3 hexagons, 3 squares and 4 triangles. Five circles regroup into one hexagon, so the final count is 4 triangles, 3 squares, 4 hexagons and 0 circles. As a check, 4 x 125 + 3 x 25 + 4 x 5 = 500 + 75 + 20 = 595.

4.1.8 E2.8 Landmark products in Egyptian numbers

Question. What is a landmark number multiplied by 10 and by 100? Find the products shown in the textbook.

Answer. Multiplying a landmark by 10 gives the next landmark; by 100 gives the landmark two places higher. The products are the corresponding landmark symbols.

Key steps. Each landmark is a power of 10, so 10^k x 10 = 10^{k+1} and 10^k x 100 = 10^{k+2}. The digit of 1 stays the same and only the landmark changes, which is why the Egyptian answer is a single symbol, not a longer numeral.

4.1.9 E2.9 General products and the ×5 rule

Question. Find the two products shown, and give a rule for multiplying a number by 5 in base-5.

Answer. Multiplying by a landmark produces another landmark. In base-5, multiplying by 5 moves every symbol one place to the left; a number written in base-5 is multiplied by 5 by shifting the places.

Key steps. In base-5, 5 = 10 (in base-5 notation), so multiplying by 5 is the same as multiplying by the base. That shifts the place values by one position, exactly as multiplying by 10 shifts decimal places in base 10. The distributive property then applies to every term of the number.

4.1.10 E2.10 Egyptian symbol repetition

Question. Can one symbol occur 10 or more times in an Egyptian numeral? Why not?

Answer. No. Ten copies of a landmark always regroup into one symbol of the next landmark, so a correctly written numeral never has ten or more copies of the same symbol.

Key steps. Suppose there are ten 10-symbols. Their total is 100, so they must be replaced by one 100-symbol. The same argument works for units, hundreds and higher places. This is the base-10 carrying rule in symbol form.

4.1.11 E2.11 A base-4 system

Question. Create a base-4 system and represent 1 to 16.

Answer. Use symbols for 1, 4 and 16, with 1 as a small unit mark, 4 as a triangle and 16 as a square. Then 1 = one unit mark, 2 = two unit marks, 3 = three unit marks, 4 = one triangle, 5 = one triangle and one unit mark, 6 = one triangle and two unit marks, 7 = one triangle and three unit marks, 8 = two triangles, 9 = two triangles and one unit mark, 10 = two triangles and two unit marks, 11 = two triangles and three unit marks, 12 = three triangles, 13 = three triangles and one unit mark, 14 = three triangles and two unit marks, 15 = three triangles and three unit marks, and 16 = one square.

Key steps. The landmarks are 4^0 = 1, 4^1 = 4, 4^2 = 16. Write each number as a sum of these landmarks: for example, 13 = 3 x 4 + 1, so three triangle symbols and one unit mark; 15 = 3 x 4 + 3, so three triangles and three unit marks. The representation is unique for numbers below the next landmark. In base-4 notation the same numbers would be written 1, 2, 3, 10, 11, 12, 13, 20, 21, 22, 23, 30, 31, 32, 33 and 100.

4.1.12 E2.12 Mesopotamian numerals

Question. Represent 63, 132, 200, 60 and 3605 in the Mesopotamian system.

Answer. 63 = 1 x 60 + 3; 132 = 2 x 60 + 12; 200 = 3 x 60 + 20; 60 = 1 x 60; 3605 = 1 x 3600 + 5.

Key steps. Mesopotamian landmarks are 1, 60, 3600, ... Write each number as a sum with coefficients below 60. For 132: two 60s and 12 units. For 3605: one 3600 and five units, with a blank or placeholder in the 60 place. The coefficients are then written in the Mesopotamian unit symbols.

4.1.13 E2.13 Mayan numerals

Question. Represent 77, 100, 361 and 721 in the Mayan system.

Answer. 77 = 3 x 20 + 17; 100 = 5 x 20 + 0; 361 = 1 x 360 + 0 x 20 + 1; 721 = 2 x 360 + 0 x 20 + 1.

Key steps. The Mayan places are 1, 20, 360, 7200, ... Write the number as a sum of these places, then draw each coefficient with dots (1) and bars (5). For 77 the lower place is 17 (three bars and two dots) and the upper place is 3 (three dots). For 100 the lower place is zero, shown by the shell placeholder.

4.1.14 E2.14 Chinese rod numerals

Question. Why did the Chinese alternate the Zong and Heng symbols? How is 41 written with only Zong symbols, and could that numeral be misread without spaces? Also form a base-2 system with ukasar and urapon, compare it with Gumulgal; discuss Hindu numerals in daily life; and work out base-8, base-5 and base-2 forms of 25.

Answer. Alternating forms distinguish neighbouring places. With only Zong symbols, 41 would use the symbols for 4 and 1; without spacing, the numeral could be read as 23, 32 or 122, because the same symbols could fall in different landmark positions. Base-2 with urapon = 0 and ukasar = 1 writes 1 as uk, 2 as uk-ur, 3 as uk-uk, 4 as uk-ur-ur, and so on; Gumulgal differs because it counts by twos with no zero and no place value. Hindu numerals and zero underpin commerce, science, computing and daily measurement; without them, large-scale calculation and technology would be far harder. In base-8, 25 = 31 (three 8s and one 1); in base-5, 25 = 100; in base-2, 25 = 11001.

Key steps. For base-8: 25 = 3 x 8 + 1, so the digits are 31. For base-5: 25 = 1 x 25 + 0 x 5 + 0, so 100. For base-2: 25 = 16 + 8 + 1 = 2^4 + 2^3 + 2^0, so 11001. Each conversion uses the same grouping rule with a different base.