Quick solution. Addition: join the two stick collections. Subtraction: match and remove sticks. Multiplication: make one equal group for each stick in the other collection. Division: form equal groups matching the divisor collection and note full groups plus leftover sticks.

1. Put the two stick collections together to add them.
2. Pair and remove sticks from one collection to subtract.
3. For multiplication, make several equal groups with the same number of sticks.
4. For division, share the sticks into equal groups and check how many groups are formed.

Addition: join the two stick collections. Subtraction: match and remove sticks. Multiplication: make one equal group for each stick in the other collection. Division: form equal groups matching the divisor collection and note full groups plus leftover sticks.

Quick solution. One valid extension is: a to z for 1 to 26, then aa, ab, ac, ..., zz, then aaa, aab, and longer strings in a fixed order. This gives an unending sequence.

1. Fix an order for the letters, a, b, c, ..., z.
2. Use these for the first 26 numbers.
3. After z, use two-letter strings such as aa, ab, ac, ..., zz.
4. After all two-letter strings, continue with three-letter strings.

One valid extension is: a to z for 1 to 26, then aa, ab, ac, ..., zz, then aaa, aab, and longer strings in a fixed order. This gives an unending sequence.

Quick solution. Example: dot = 1, bar = 5, star = 25. Then 38 is star + bar + bar + dot + dot + dot. Other answers are valid if symbols and order are fixed.

1. Let dot = 1, bar = 5 and star = 25.
2. To write 38, use one star for 25.
3. The remaining number is 13, so use two bars for 10.
4. The remaining number is 3, so use three dots.

Example: dot = 1, bar = 5, star = 25. Then 38 is star + bar + bar + dot + dot + dot. Other answers are valid if symbols and order are fixed.

Quick solution. 1222 = MCCXXII; 2999 = MMCMXCIX; 302 = CCCII; 715 = DCCXV.

1. 1222 = 1000 + 200 + 20 + 2 = MCCXXII.
2. 2999 = 2000 + 900 + 90 + 9 = MMCMXCIX.
3. 302 = 300 + 2 = CCCII.
4. 715 = 700 + 15 = DCCXV.

1222 = MCCXXII; 2999 = MMCMXCIX; 302 = CCCII; 715 = DCCXV.

Quick solution. V x L = CCL; L x D = M repeated 25 times; V x D = MMD; VII x IX = LXIII.

1. For V x L, repeated groups of V make many fives; regrouping ten fives as L and then five L as CCL gives CCL.
2. For L x D, repeated L groups regroup into thousands; the result is M repeated 25 times.
3. For V x D, repeated V groups regroup to two M symbols and one D, so the product is MMD.
4. For VII x IX, nine groups of VII regroup as L + X + III, so the product is LXIII.

V x L = CCL; L x D = M repeated 25 times; V x D = MMD; VII x IX = LXIII.

Quick solution. Different objects may have had different counting needs, so separate number-name sequences helped people count traded, stored or culturally important items clearly.

1. Counting systems grow from daily needs.
2. Objects that are traded, shared or stored often get special counting words.
3. A separate sequence can reduce confusion in a community.

Different objects may have had different counting needs, so separate number-name sequences helped people count traded, stored or culturally important items clearly.

Quick solution. The results are: sum = ukasar repeated 8 times; difference = ukasar-urapon; product = ukasar repeated 18 times; quotient = ukasar-ukasar.

1. ukasar-ukasar-ukasar-ukasar-urapon means 9.
2. ukasar-ukasar-ukasar-urapon means 7. Their sum is 16, written as ukasar repeated 8 times.
3. 9 minus 6 is 3, written as ukasar-urapon.
4. 9 x 4 is 36, written as ukasar repeated 18 times.
5. 16 divided by 4 is 4, written as ukasar-ukasar.

The results are: sum = ukasar repeated 8 times; difference = ukasar-urapon; product = ukasar repeated 18 times; quotient = ukasar-ukasar.

Quick solution. The Hindu number system is efficient because it has place value, a zero symbol, only ten digits, and easier algorithms for addition, subtraction, multiplication and division.

1. Place value changes the value of a digit according to its position.
2. Zero works as a number and as a placeholder.
3. The same ten digits can represent all whole numbers.
4. Addition, subtraction, multiplication and division follow compact algorithms.

The Hindu number system is efficient because it has place value, a zero symbol, only ten digits, and easier algorithms for addition, subtraction, multiplication and division.

Quick solution. 10458 = 10000 + 400 + 50 + 8; 1023 = 1000 + 20 + 3; 2660 = 2000 + 600 + 60; 784 = 700 + 80 + 4; 1111 = 1000 + 100 + 10 + 1; 70707 = 70000 + 700 + 7.

1. 10458 = 10000 + 400 + 50 + 8.
2. 1023 = 1000 + 20 + 3.
3. 2660 = 2000 + 600 + 60.
4. 784 = 700 + 80 + 4.
5. 1111 = 1000 + 100 + 10 + 1.
6. 70707 = 70000 + 700 + 7.

10458 = 10000 + 400 + 50 + 8; 1023 = 1000 + 20 + 3; 2660 = 2000 + 600 + 60; 784 = 700 + 80 + 4; 1111 = 1000 + 100 + 10 + 1; 70707 = 70000 + 700 + 7.

Quick solution. The first Egyptian numeral stands for 276. The second Egyptian numeral stands for 4322.

1. First numeral: 2 hundreds + 7 tens + 6 ones = 276.
2. Second numeral: 4 thousands + 3 hundreds + 2 tens + 2 ones = 4322.

The first Egyptian numeral stands for 276. The second Egyptian numeral stands for 4322.

Quick solution. 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.

1. 15 = 5 + 5 + 5.
2. 50 = 25 + 25.
3. 137 = 125 + 5 + 5 + 1 + 1.
4. 293 = 125 + 125 + 25 + 5 + 5 + 5 + 1 + 1 + 1.
5. 651 = 625 + 25 + 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.

Quick solution. Yes. Zero cannot be represented in this base-5 landmark system because every written symbol has positive value and there is no symbol for 0.

1. The system begins with the landmark number 1.
2. Every written numeral is made by adding one or more landmark symbols.
3. No collection of positive landmark symbols can add to 0.

Yes. Zero cannot be represented in this base-5 landmark system because every written symbol has positive value and there is no symbol for 0.

Quick solution. Base-7 landmarks are 1, 7, 49, 343, ... . In base n, the landmarks are n0, n1, n2, n3, ... .

1. 70 = 1.
2. 71 = 7.
3. 72 = 49.
4. 73 = 343.
5. In base n, the landmark numbers are n0, n1, n2, n3, and so on.

Base-7 landmarks are 1, 7, 49, 343, ... . In base n, the landmarks are n0, n1, n2, n3, ... .

Quick solution. Add by combining like symbols, then regrouping: 10 equal Egyptian symbols make the next landmark, and 5 equal base-5 symbols make the next landmark.

1. In Egyptian numerals, 10 copies of one landmark are replaced by one copy of the next landmark.
2. In the base-5 system, 5 copies of one landmark are replaced by one copy of the next landmark.
3. After regrouping, write the final symbols from largest landmark to smallest.

Add by combining like symbols, then regrouping: 10 equal Egyptian symbols make the next landmark, and 5 equal base-5 symbols make the next landmark.

Quick solution. Multiplying an Egyptian landmark by 10 gives the next landmark. Multiplying by 100 moves two landmark places ahead. The same power rule works in any base system.

1. Each Egyptian landmark is a power of 10.
2. 10 x 10k = 10(k+1), so multiplying by 10 gives the next landmark.
3. 100 x 10k = 10(k+2), so multiplying by 100 gives the landmark after the next.
4. The same property holds in any base system.

Multiplying an Egyptian landmark by 10 gives the next landmark. Multiplying by 100 moves two landmark places ahead. The same power rule works in any base system.

Quick solution. No. A correct Egyptian numeral cannot have 10 or more copies of one symbol, because 10 copies regroup as one next landmark symbol.

1. Ten ones become one ten.
2. Ten tens become one hundred.
3. Ten hundreds become one thousand.
4. So a properly grouped Egyptian numeral should not show 10 or more copies of the same symbol.

No. A correct Egyptian numeral cannot have 10 or more copies of one symbol, because 10 copies regroup as one next landmark symbol.

Quick solution. With A = 1, B = 4 and C = 16: 1=A, 2=AA, 3=AAA, 4=B, 5=BA, 6=BAA, 7=BAAA, 8=BB, 9=BBA, 10=BBAA, 11=BBAAA, 12=BBB, 13=BBBA, 14=BBBAA, 15=BBBAAA, 16=C.

1. 1 = A, 2 = AA, 3 = AAA, 4 = B.
2. 5 = BA, 6 = BAA, 7 = BAAA, 8 = BB.
3. 9 = BBA, 10 = BBAA, 11 = BBAAA, 12 = BBB.
4. 13 = BBBA, 14 = BBBAA, 15 = BBBAAA, 16 = C.

With A = 1, B = 4 and C = 16: 1=A, 2=AA, 3=AAA, 4=B, 5=BA, 6=BAA, 7=BAAA, 8=BB, 9=BBA, 10=BBAA, 11=BBAAA, 12=BBB, 13=BBBA, 14=BBBAA, 15=BBBAAA, 16=C.

Quick solution. Mesopotamian-style base-60 places are: 63 = [one sixty] | [three ones]; 132 = [two sixties] | [one ten + two ones]; 200 = [three sixties] | [two tens]; 60 = [one sixty] | [blank ones place]; 3605 = [one 3600] | [blank 60s place] | [five ones].

1. 63 = 1 x 60 + 3, so the places are [one sixty] | [three ones].
2. 132 = 2 x 60 + 12, so the places are [two sixties] | [one ten + two ones].
3. 200 = 3 x 60 + 20, so the places are [three sixties] | [two tens].
4. 60 = 1 x 60 with no ones, so the places are [one sixty] | [blank ones place].
5. 3605 = 1 x 3600 + 5, so the places are [one 3600] | [blank 60s place] | [five ones].

Mesopotamian-style base-60 places are: 63 = [one sixty] | [three ones]; 132 = [two sixties] | [one ten + two ones]; 200 = [three sixties] | [two tens]; 60 = [one sixty] | [blank ones place]; 3605 = [one 3600] | [blank 60s place] | [five ones].

Quick solution. Mayan place forms: 77 = 3 dots above 3 bars and 2 dots; 100 = one bar above zero; 361 = one dot above zero above one dot; 721 = two dots above zero above one dot.

1. 77 = 3 x 20 + 17.
2. 100 = 5 x 20 + 0.
3. 361 = 1 x 360 + 0 x 20 + 1.
4. 721 = 2 x 360 + 0 x 20 + 1.

Mayan place forms: 77 = 3 dots above 3 bars and 2 dots; 100 = one bar above zero; 361 = one dot above zero above one dot; 721 = two dots above zero above one dot.

Quick solution. The Chinese alternated Zong and Heng to separate neighbouring places. With only Zong symbols, 41 is four tens-rods followed by one ones-rod; without clear spacing the same five rods could be split as 14, 23, 32, 41, or simply 5.

1. Place value systems need clear separation between adjacent positions.
2. Zong and Heng symbols make neighbouring places visually different.
3. Using only Zong, 41 looks like four rods in the tens place followed by one rod in the ones place.
4. If the five rods are written too close together, they may be split as 14, 23, 32, 41, or even read as a simple group of 5.

The Chinese alternated Zong and Heng to separate neighbouring places. With only Zong symbols, 41 is four tens-rods followed by one ones-rod; without clear spacing the same five rods could be split as 14, 23, 32, 41, or simply 5.

Quick solution. One base-2 system is ukasar = 1 and urapon = 0: 1 = ukasar, 2 = ukasar urapon, 3 = ukasar ukasar, 4 = ukasar urapon urapon. Gumulgal is additive grouping; base-2 is positional.

1. 1 is ukasar.
2. 2 is ukasar urapon, meaning 10 in base 2.
3. 3 is ukasar ukasar, meaning 11 in base 2.
4. 4 is ukasar urapon urapon, meaning 100 in base 2.
5. Gumulgal repeats groups of 2, while base 2 uses place value.

One base-2 system is ukasar = 1 and urapon = 0: 1 = ukasar, 2 = ukasar urapon, 3 = ukasar ukasar, 4 = ukasar urapon urapon. Gumulgal is additive grouping; base-2 is positional.

Quick solution. Hindu numerals and zero are used in money, time, measurements, phones, computers, banking, engineering, medicine and science. Without them, large calculations and modern technology would be much slower.

1. Shops and banks need them for prices, bills and accounts.
2. Doctors and engineers need them for doses, measurements and designs.
3. Computers and phones depend on place value and zero in data systems.
4. Without zero and place value, large calculations would be much slower.

Hindu numerals and zero are used in money, time, measurements, phones, computers, banking, engineering, medicine and science. Without them, large calculations and modern technology would be much slower.

Quick solution. 25 in base 8 is 31_8; in base 5 it is 100_5; in base 2 it is 11001_2.

1. In base 8: 25 = 3 x 8 + 1, so 25 is 31_8.
2. In base 5: 25 = 1 x 25 + 0 x 5 + 0, so 25 is 100_5.
3. In base 2: 25 = 16 + 8 + 1, so 25 is 11001_2.

25 in base 8 is 31_8; in base 5 it is 100_5; in base 2 it is 11001_2.