Exercise practice 2: Power Play

4.1 Exercise practice 2: Figure it Out, pages 26-27

This page covers the second Figure it Out section. Every question and sub-part is worked below.

4.1.1 E2.1 Units digit of a quotient

Question. Find the units digit of 2^224 / 4^32. Hint: 4 = 2^2.

Answer. 6.

Key steps. Since 4 = 2^2, we have 4^32 = (2^2)^32 = 2^64. Therefore 2^224 / 2^64 = 2^160. The units digits of powers of 2 repeat in the cycle 2, 4, 8, 6 with period 4. Since 160 is divisible by 4, the units digit is the last member of the cycle, which is 6.

4.1.2 E2.2 Bottles after 40 days

Question. A container holds 5 bottles, and one extra container arrives each day. How many bottles are present after 40 days?

Answer. 200 = 2 x 10^2 bottles.

Key steps. One container arrives each day, so after 40 days there are 40 containers. Each container has 5 bottles, giving 40 x 5 = 200 bottles. In exponential form this is 2 x 10^2.

4.1.3 E2.3 Three ways as products of powers

Question. Express each number in three separate ways as a product of two or more powers; the exponents may be any integers. (i) 64^3; (ii) 192^8; (iii) 32^{-5}.

Answer. (i) 2^{10} x 2^8, 4^5 x 4^4, 8^3 x 8^3. (ii) 2^{48} x 3^8, 2^{40} x 2^8 x 3^8, 2^{40} x 6^8. (iii) 2^{-10} x 2^{-15}, 2^{-5} x 2^{-20}, 4^{-12} x 2^{-1}.

Key steps. Convert everything to a convenient base. (i) 64^3 = (2^6)^3 = 2^{18}; any split of 18 into two exponents works, and 4 = 2^2 and 8 = 2^3 give the other forms. (ii) 192 = 2^6 x 3, so 192^8 = 2^{48} x 3^8; splitting the 2-power and grouping 2^40 x 6^8 gives further ways. (iii) 32^{-5} = (2^5)^{-5} = 2^{-25}; split the exponent into sums that add to -25. Check one way: 2^{-10} x 2^{-15} = 2^{-25}.

4.1.4 E2.4 Always, sometimes or never true

Question. Decide whether each claim is always true, sometimes true or never true, and give a reason. (i) A cube can always be written as a square. (ii) Any fourth power is a square as well. (iii) A number's cube always divides its fifth power. (iv) Multiplying two cubes produces a cube. (v) q^{46} is both a fourth power and a sixth power, where q is prime.

Answer. (i) sometimes true; (ii) always true; (iii) always true; (iv) always true; (v) never true.

Key steps. (i) A cube is a square only when the exponent is a multiple of 6, as in 64 = 2^6 = 8^2 = 4^3. (ii) n^4 = (n^2)^2, so it is always a square. (iii) n^5 / n^3 = n^2, so the division is exact. (iv) (n1)^3 x (n2)^3 = (n1 n2)^3. (v) A number is a fourth power when its exponent is a multiple of 4 and a sixth power when it is a multiple of 6; 46 is neither, so the claim is never true.

4.1.5 E2.5 Simplifying powers

Question. Simplify: (i) 10^{-2} x 10^{-5}; (ii) 5^7 / 5^4; (iii) 9^{-7} / 9^4; (iv) (13^{-2})^{-3}; (v) m^5 n^{12} (mn)^9.

Answer. (i) 10^{-7}; (ii) 5^3; (iii) 9^{-11}; (iv) 13^6; (v) m^{14} n^{21}.

Key steps. For (i) add the exponents: -2 + -5 = -7. For (ii) subtract: 7 - 4 = 3. For (iii) subtract: -7 - 4 = -11. For (iv) multiply: -2 x -3 = 6. For (v) expand (mn)^9 = m^9 n^9, then add exponents for each base: m has 5 + 9 = 14 and n has 12 + 9 = 21.

4.1.6 E2.6 Squares of decimals

Question. If 12 squared is 144, work out the squares of 1.2, of 0.12 and of 0.012, and also of 120.

Answer. The first square is 1.44, the second 0.0144, the third 0.000144, and 120^2 is 14400.

Key steps. Move the decimal point in the base one place to the left; the square moves two places to the left. Starting from 12^2 = 144, (1.2)^2 has two decimal places, (0.12)^2 has four, and (0.012)^2 has six. For 120^2, move the point one place right and the square moves two places right.

4.1.7 E2.7 Same numbers, different forms

Question. Which of these values are equal: 2^4 x 3^6, 6^4 x 3^2, 6^{10}, 18^2 x 6^2, 6^{24}?

Answer. 2^4 x 3^6, 6^4 x 3^2 and 18^2 x 6^2 are equal.

Key steps. Convert each to prime bases. 2^4 x 3^6 has value 16 x 729 = 11664. 6^4 x 3^2 = 1296 x 9 = 11664. 18^2 x 6^2 = 324 x 36 = 11664. The remaining two are much larger: 6^{10} and 6^{24} involve ten and twenty-four factors of 6.

4.1.8 E2.8 Which is greater

Question. Which is greater: (i) 4^3 or 3^4; (ii) 2^8 or 8^2; (iii) 100^2 or 2^{100}?

Answer. (i) 3^4; (ii) 2^8; (iii) 2^{100}.

Key steps. (i) 4^3 = 64 and 3^4 = 81. (ii) 2^8 = 256 and 8^2 = 64. (iii) 100^2 = 10000 while 2^{100} is about 1.27 x 10^{30}; a repeated doubling passes a fixed square very quickly. For large exponents, compare the exponent of the growing power rather than computing fully.

4.1.9 E2.9 Length of a milk packet code

Question. A dairy plans 8.5 billion milk packets in a year and wants a unique code for each using digits 0 to 9. How many digits should the code have?

Answer. 10 digits.

Key steps. With n digits, there are 10^n possible codes. We need 10^n at least 8.5 x 10^9. Now 10^9 = 1,000,000,000 is too small, while 10^{10} = 10,000,000,000 is larger than 8.5 billion. So n = 10; codes of length 10 suffice, and 9 would not.

4.1.10 E2.10 Numbers that are squares and cubes

Question. 64 is both a square (8^2) and a cube (4^3). Find other such numbers and describe them in general.

Answer. Every sixth power works: 1^6 = 1, 2^6 = 64, 3^6 = 729, 4^6 = 4096, and so on.

Key steps. A number is a square when all exponents in its prime factorisation are even, and a cube when all are multiples of 3. Both conditions hold exactly when every exponent is a multiple of 6, so the numbers are the sixth powers n^6. This also explains why the first example is 64: 64 = 2^6.

4.1.11 E2.11 Alphanumeric passcodes

Question. A digital locker uses a 5-character passcode that may contain digits and letters. How many passcodes are possible?

Answer. 36^5 = 60,466,176.

Key steps. There are 10 digits and 26 letters, so each position has 36 choices. Five positions give 36 x 36 x 36 x 36 x 36 = 36^5. No repetition rule is stated, so each position is independent.

4.1.12 E2.12 Sheep and goats

Question. The world population of sheep is about 10^9, and goats are about equally numerous. What is their combined population? The choices are 20^9, 10^{11}, 10^{10}, 10^{18}, 2 x 10^9 and 10^9 + 10^9.

Answer. Total = 10^9 + 10^9 = 2 x 10^9. Both the option 2 x 10^9 and the option 10^9 + 10^9 represent this value; the printed answer key marks the sum form.

Key steps. Add equal powers of ten by adding the coefficients: one times 10^9 plus one times 10^9 gives two times 10^9. The option 20^9 is (2 x 10)^9 = 2^9 x 10^9, much larger; 10^{10} and 10^{11} are ten and a hundred times larger; 10^{18} is far larger. Note that two options are mathematically equal, so the question tests the idea of adding like terms rather than a single unique form.

4.1.13 E2.13 Scientific notation calculations

Question. Calculate and write in scientific notation: (i) clothing pieces if each of 8.2 x 10^9 people has 30; (ii) honeybees if there are 10^8 colonies with 50,000 bees each; (iii) bacterial cells if each of 8.2 x 10^9 people has about 38 trillion; (iv) total seconds spent eating in a lifetime, assuming 70 years and 1 hour a day.

Answer. (i) 2.46 x 10^{11}; (ii) 5 x 10^{12}; (iii) about 3.116 x 10^{23}; (iv) 9.198 x 10^7 seconds.

Key steps. (i) 8.2 x 10^9 x 30 = 8.2 x 3 x 10^{10} = 24.6 x 10^{10} = 2.46 x 10^{11}. (ii) 10^8 x 5 x 10^4 = 5 x 10^{12}. (iii) 38 trillion = 3.8 x 10^{13}; multiply by 8.2 x 10^9 to get 3.116 x 10^{23}. (iv) 3600 seconds x 365 days x 70 years = 91,980,000 seconds = 9.198 x 10^7 seconds.

4.1.14 E2.14 One billion seconds ago

Question. What was the date 1 billion seconds ago?

Answer. About 31.7 years ago, or roughly 11,574 days before today. For 31 October 2026 this is around February 1995.

Key steps. Divide 10^9 by the number of seconds in a day: 1,000,000,000 / 86,400 is about 11,574 days. Dividing 11,574 by 365.25 gives about 31.7 years. Since the exact date depends on the present day, the method matters more than a single printed date; subtract the number of days from today's date.