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Number, units and estimation — solutions

21 full solutions · spec M1.1, M1.2, M2.1-M2.14

Section 1 Part 2: Higher GCSE mathematics21 questions
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Every question on this topic with its answer, the working, and a full method: what the question is testing, the reason for each step, the mistakes that are easy to make and what they would give you, and the principle to take away.

1. NUM-01Number, units and estimation · spec M2.3 · Paper 1 style · ★☆☆

Find the highest common factor of \(2^{3}\times 3^{2}\times 5\) and \(2^{2}\times 3^{3}\times 7\).

  1. A\(18\)
  2. B\(6\)
  3. C\(12\)
  4. D\(36\)
  5. E\(72\)
  6. F\(108\)
Answer: D. Take the lower power of each common prime: \(2^2 \times 3^2 = 36\).
Full method — how and why
HCF from prime factorisations: take each common prime to its lower power.
  1. List the primes in each: \(2^3 \times 3^2 \times 5\) and \(2^2 \times 3^3 \times 7\).
    Both are already factorised, which is the form the rule needs.
  2. Take the common primes only: \(2\) and \(3\) appear in both; \(5\) and \(7\) do not.
    A prime missing from one number cannot divide both.
  3. Take the lower power of each: \(2^{\min(3,2)} = 2^2\) and \(3^{\min(2,3)} = 3^2\).
    The HCF must divide both, so it cannot use more copies of a prime than the smaller number has.
  4. Multiply: \(4 \times 9 = 36\).
    Check: \(36\) divides \(360\) and \(756\) ✓.
Where it goes wrong
  • Taking the higher powers (that gives the LCM contribution, not the HCF).
  • Including \(5\) or \(7\), which gives a number dividing only one of them.
Takeaway HCF: common primes, lowest powers. LCM: all primes, highest powers.
2. NUM-03Number, units and estimation · spec M2.9 · Paper 1 style · ★☆☆

Express the recurring decimal \(0.\dot{2}\dot{7}\) as a fraction in its lowest terms.

  1. A\(\dfrac{9}{33}\)
  2. B\(\dfrac{5}{18}\)
  3. C\(\dfrac{3}{11}\)
  4. D\(\dfrac{27}{99}\)
  5. E\(\dfrac{3}{10}\)
  6. F\(\dfrac{27}{100}\)
Answer: C. \(x = 0.2727\ldots\), \(100x = 27.2727\ldots\), so \(99x = 27\) and \(x = \frac{27}{99} = \frac{3}{11}\).
Full method — how and why
The standard algebraic trick for recurring decimals: multiply by a power of \(10\) that shifts one full period, then subtract.
  1. Let \(x = 0.272727\ldots\) The repeating block has two digits, so multiply by \(10^2 = 100\): \(100x = 27.272727\ldots\)
    Shifting by exactly one period makes the decimal tails identical, so they cancel.
  2. Subtract: \(100x - x = 27.2727\ldots - 0.2727\ldots = 27\), so \(99x = 27\).
    The infinite tails cancel exactly - this is the whole method.
  3. Solve: \(x = \dfrac{27}{99}\).
    The denominator is always \(10^k - 1\), i.e. a string of nines matching the period length.
  4. Simplify by \(9\): \(\dfrac{3}{11}\).
    The question demands lowest terms, and \(\frac{27}{99}\) is offered as a trap.
Where it goes wrong
  • Choosing \(\frac{27}{99}\) - correct in value but not in lowest terms.
  • Multiplying by \(10\) instead of \(100\), which does not align the tails.
Takeaway Multiply by \(10^{(\text{period length})}\), subtract, then simplify fully.
3. NUM-04Number, units and estimation · spec M2.13 · Paper 1 style · ★☆☆

A length is given as \(4.6\) cm correct to \(1\) decimal place, and a width as \(3\) cm correct to the nearest cm. Find the upper bound of the area.

  1. A\(16.1\) cm\(^2\)
  2. B\(11.375\) cm\(^2\)
  3. C\(16.45\) cm\(^2\)
  4. D\(16.275\) cm\(^2\)
  5. E\(14.375\) cm\(^2\)
  6. F\(13.8\) cm\(^2\)
Answer: D. Upper bounds \(4.65\) and \(3.5\): \(4.65 \times 3.5 = 16.275\).
Full method — how and why
Bounds: each measurement is an interval. For a maximum product, take the top of each interval.
  1. Length \(4.6\) to 1 d.p. means \(4.55 \leq \ell < 4.65\), so the upper bound is \(4.65\).
    Rounding to 1 d.p. means the true value is within \(0.05\) either side.
  2. Width \(3\) to the nearest cm means \(2.5 \leq w < 3.5\), so the upper bound is \(3.5\).
    Nearest whole number means within \(0.5\) either side - a different half-width from the length.
  3. Both factors are positive, so the area is largest when both are largest: \(4.65 \times 3.5\).
    Products of positive numbers increase with each factor.
  4. Compute: \(4.65 \times 3.5 = 16.275\) cm\(^2\).
    \(4.65 \times 3 = 13.95\), plus \(4.65 \times 0.5 = 2.325\).
Where it goes wrong
  • Using \(4.65 \times 3.0 = 13.95\) by forgetting the width also has a bound.
  • Using half-widths of \(0.05\) for both, giving \(3.05\) for the width.
Takeaway Each measurement has its own accuracy: 1 d.p. gives \(\pm0.05\), nearest whole gives \(\pm0.5\).
4. NUM-05Number, units and estimation · spec M2.8 · Paper 1 style · ★☆☆

Evaluate \(\dfrac{(3 \times 10^{4}) \times (8 \times 10^{-2})}{6 \times 10^{-3}}\), giving your answer in standard form.

  1. A\(4 \times 10^{4}\)
  2. B\(4 \times 10^{6}\)
  3. C\(4 \times 10^{-1}\)
  4. D\(2.4 \times 10^{5}\)
  5. E\(1.5 \times 10^{5}\)
  6. F\(4 \times 10^{5}\)
Answer: F. \(\frac{24 \times 10^{2}}{6 \times 10^{-3}} = 4 \times 10^{5}\).
Full method — how and why
Standard-form arithmetic: handle the numbers and the powers of ten separately, then renormalise.
  1. Multiply the numerator: \((3 \times 8) \times 10^{4 + (-2)} = 24 \times 10^{2}\).
    Multiply the leading numbers, add the indices.
  2. Divide by \(6 \times 10^{-3}\): \(\dfrac{24}{6} \times 10^{2 - (-3)} = 4 \times 10^{5}\).
    Divide the numbers, subtract the indices - and subtracting \(-3\) adds \(3\).
  3. Check the form: \(1 \leq 4 < 10\) ✓, so no renormalising is needed.
    Standard form requires the leading number to be between \(1\) and \(10\).
Where it goes wrong
  • Getting \(10^{-1}\) by adding instead of subtracting the \(-3\).
  • Leaving \(24 \times 10^{2}\), which is not in standard form.
Takeaway Numbers multiply/divide; indices add/subtract. Then check \(1 \leq a < 10\).
5. NUM-06Number, units and estimation · spec M2.5 · Paper 1 style · ★☆☆

A code consists of two letters from A, B, C, D, E followed by two digits from \(1\) to \(9\). Letters may be repeated but the two digits must be different. How many codes are possible?

  1. A\(3240\)
  2. B\(2025\)
  3. C\(1440\)
  4. D\(900\)
  5. E\(1620\)
  6. F\(1800\)
Answer: F. \(5 \times 5 \times 9 \times 8 = 1800\).
Full method — how and why
The product rule for counting, with one part allowing repeats and the other not.
  1. Letters: \(5\) choices for the first and, since repeats are allowed, \(5\) again for the second: \(5 \times 5\).
    'May be repeated' means the second choice is unaffected by the first.
  2. Digits: \(9\) choices for the first, but only \(8\) for the second because they must differ: \(9 \times 8\).
    'Must be different' removes exactly one option from the second slot.
  3. Multiply all the stages: \(5 \times 5 \times 9 \times 8 = 1800\).
    The product rule: independent stages multiply.
Where it goes wrong
  • Using \(9 \times 9 = 81\) for the digits, giving \(2025\) - an offered option.
  • Using \(5 \times 4\) for the letters, giving \(1440\).
Takeaway Multiply the number of choices at each stage, reducing by one only where repetition is forbidden.
6. NUM-07Number, units and estimation · spec M1.2 · Paper 1 style · ★☆☆

A car travels at \(90\) km/h. What is this speed in metres per second?

  1. A\(30\) m/s
  2. B\(54\) m/s
  3. C\(15\) m/s
  4. D\(25\) m/s
  5. E\(324\) m/s
  6. F\(1.5\) m/s
Answer: D. \(90\,000\) m in \(3600\) s: \(90000/3600 = 25\) m/s.
Full method — how and why
A compound-unit conversion: convert distance and time separately.
  1. \(90\) km \(= 90 \times 1000 = 90\,000\) m.
    Convert the distance unit first.
  2. \(1\) hour \(= 60 \times 60 = 3600\) s.
    Convert the time unit.
  3. Divide: \(\dfrac{90\,000}{3600} = 25\) m/s.
    Equivalently, divide km/h by \(3.6\) - worth memorising.
Where it goes wrong
  • Multiplying by \(3.6\) instead of dividing, giving \(324\) - an offered option.
  • Converting only one of the two units.
Takeaway km/h \(\div 3.6 =\) m/s; convert numerator and denominator units separately.
7. NUM-08Number, units and estimation · spec M2.14 · Paper 1 style · ★☆☆

Estimate the value of \(\dfrac{4.92 \times \sqrt{99.3}}{0.198}\) by rounding each number to \(1\) significant figure.

  1. A\(2500\)
  2. B\(250\)
  3. C\(500\)
  4. D\(100\)
  5. E\(50\)
  6. F\(25\)
Answer: B. \(\frac{5 \times 10}{0.2} = 250\).
Full method — how and why
Estimation by rounding to one significant figure, choosing roundings that make the arithmetic exact.
  1. Round each: \(4.92 \approx 5\), \(99.3 \approx 100\), \(0.198 \approx 0.2\).
    One significant figure; \(99.3\) rounds to \(100\), which is also a perfect square - convenient.
  2. Take the root: \(\sqrt{100} = 10\).
    Choosing \(100\) rather than \(99\) is what makes the root exact.
  3. Compute: \(\dfrac{5 \times 10}{0.2} = \dfrac{50}{0.2} = 250\).
    Dividing by \(0.2\) is multiplying by \(5\).
Where it goes wrong
  • Dividing by \(0.2\) as if it were \(2\), giving \(25\).
  • Forgetting the square root and using \(100\) directly, giving \(2500\).
Takeaway Round to 1 s.f., but choose the rounding that makes roots and divisions exact; dividing by \(0.2\) multiplies by \(5\).
8. NUM-10Number, units and estimation · spec M2.13 · Paper 1 style · ★☆☆

A number \(x\) is truncated (not rounded) to \(2\) decimal places and the result is \(3.14\). Which inequality describes the possible values of \(x\)?

  1. A\(3.14 < x < 3.15\)
  2. B\(3.135 \leq x < 3.145\)
  3. C\(3.13 < x \leq 3.14\)
  4. D\(3.135 < x < 3.145\)
  5. E\(3.14 \leq x < 3.15\)
  6. F\(3.14 \leq x \leq 3.15\)
Answer: E. Truncation removes the later digits, so \(x\) is at least \(3.14\) and less than \(3.15\).
Full method — how and why
Truncation, not rounding. Truncating throws away everything after the cut, so the value can only be above the stated one.
  1. Truncating to \(2\) d.p. means simply deleting the third and later decimals.
    No rounding up ever happens, which makes the interval one-sided.
  2. The smallest such \(x\) is \(3.14\) itself (with nothing after it).
    \(3.14\) truncates to \(3.14\), so equality is allowed at the bottom.
  3. The largest is anything below \(3.15\): at \(3.15\) the truncation would read \(3.15\).
    Strict inequality at the top.
  4. So \(3.14 \leq x < 3.15\).
    Contrast with rounding, which would give \(3.135 \leq x < 3.145\) - an offered distractor.
Where it goes wrong
  • Using the rounding interval \(3.135 \leq x < 3.145\).
  • Making the lower end strict, excluding \(3.14\) itself.
Takeaway Truncation: \([a, a + 10^{-n})\). Rounding: \([a - \frac12 10^{-n}, a + \frac12 10^{-n})\).
9. NUM-11Number, units and estimation · spec M2.2 · Paper 1 style · ★☆☆

Evaluate \(\left(2\dfrac{1}{3} - 1\dfrac{3}{4}\right) \div \dfrac{7}{6}\).

  1. A\(\dfrac{2}{3}\)
  2. B\(\dfrac{7}{12}\)
  3. C\(\dfrac{49}{72}\)
  4. D\(\dfrac{1}{2}\)
  5. E\(\dfrac{7}{24}\)
  6. F\(1\)
Answer: D. \(\frac73 - \frac74 = \frac{28 - 21}{12} = \frac{7}{12}\); \(\frac{7}{12} \times \frac67 = \frac12\).
Full method — how and why
Mixed numbers and division of fractions. Convert to improper fractions, subtract, then multiply by the reciprocal.
  1. Convert: \(2\frac13 = \frac73\) and \(1\frac34 = \frac74\).
    Whole part times denominator plus numerator.
  2. Subtract with a common denominator \(12\): \(\dfrac{28}{12} - \dfrac{21}{12} = \dfrac{7}{12}\).
    \(\frac73 = \frac{28}{12}\) and \(\frac74 = \frac{21}{12}\).
  3. Divide by \(\frac76\) means multiply by \(\frac67\): \(\dfrac{7}{12} \times \dfrac{6}{7}\).
    Dividing by a fraction is multiplying by its reciprocal.
  4. Cancel and multiply: the \(7\)s cancel and \(\frac{6}{12} = \frac12\).
    Cancel before multiplying to keep numbers small.
Where it goes wrong
  • Multiplying by \(\frac76\) instead of its reciprocal, giving \(\frac{49}{72}\) - an offered option.
  • Subtracting the whole parts and fractions separately and mishandling the borrow.
Takeaway Improper fractions, common denominator, then multiply by the reciprocal - and cancel before multiplying.
10. NUM-13Number, units and estimation · spec M2.7 · Paper 1 style · ★☆☆

Which of the following is equal to \(\dfrac{8^{4} \times 2^{-5}}{4^{3}}\)?

  1. A\(1\)
  2. B\(2\)
  3. C\(16\)
  4. D\(8\)
  5. E\(4\)
  6. F\(\dfrac12\)
Answer: B. \(\frac{2^{12}\cdot2^{-5}}{2^{6}} = 2^{1} = 2\).
Full method — how and why
Everything is a power of \(2\). Convert and use the index laws.
  1. Convert: \(8 = 2^3\) so \(8^4 = 2^{12}\); \(4 = 2^2\) so \(4^3 = 2^6\).
    A common base turns the whole expression into index arithmetic.
  2. Numerator: \(2^{12} \times 2^{-5} = 2^{7}\).
    Multiplying adds indices.
  3. Divide: \(2^{7-6} = 2^1 = 2\).
    Dividing subtracts indices.
Where it goes wrong
  • Using \(8^4 = 2^{7}\) (adding instead of multiplying the indices).
  • Sign slip on the \(-5\), giving \(2^{11-6} = 32\).
Takeaway Rewrite every number as a power of the same base, then just add and subtract indices.
11. NUM-14Number, units and estimation · spec M2.12 · Paper 1 style · ★☆☆

Simplify \(\sqrt{48} \times \sqrt{27} \div \sqrt{12}\).

  1. A\(18\)
  2. B\(12\)
  3. C\(9\sqrt{3}\)
  4. D\(6\)
  5. E\(36\)
  6. F\(6\sqrt3\)
Answer: F. \(\sqrt{48} = 4\sqrt3\), \(\sqrt{27} = 3\sqrt3\), \(\sqrt{12} = 2\sqrt3\). So \(\dfrac{4\sqrt3 \times 3\sqrt3}{2\sqrt3} = \dfrac{36}{2\sqrt3} = \dfrac{18}{\sqrt3} = 6\sqrt3\).
Full method — how and why
Simplify each surd first; they all reduce to multiples of \(\sqrt3\), so the expression collapses.
  1. Simplify: \(\sqrt{48} = \sqrt{16 \times 3} = 4\sqrt3\), \(\sqrt{27} = 3\sqrt3\), \(\sqrt{12} = 2\sqrt3\).
    Take out the largest square factor each time.
  2. Multiply the first two: \(4\sqrt3 \times 3\sqrt3 = 12 \times 3 = 36\).
    \(\sqrt3 \times \sqrt3 = 3\) - the surd disappears.
  3. Divide: \(\dfrac{36}{2\sqrt3} = \dfrac{18}{\sqrt3}\).
    Only one surd remains, in the denominator.
  4. Rationalise: \(\dfrac{18}{\sqrt3} \times \dfrac{\sqrt3}{\sqrt3} = \dfrac{18\sqrt3}{3} = 6\sqrt3\).
    Multiply top and bottom by \(\sqrt3\).
Where it goes wrong
  • Answering \(18\) by forgetting to rationalise (or by cancelling the \(\sqrt3\) incorrectly).
  • Combining under one root: \(\sqrt{48 \times 27 \div 12} = \sqrt{108}\) - correct but then needing simplification to \(6\sqrt3\) anyway.
Takeaway Simplify each surd, do the arithmetic, then rationalise any surd left in a denominator.
12. NUM-15Number, units and estimation · spec M2.5 · Paper 1 style · ★☆☆

How many three-digit positive integers have digits that are all different and all odd?

  1. A\(120\)
  2. B\(24\)
  3. C\(125\)
  4. D\(100\)
  5. E\(30\)
  6. F\(60\)
Answer: F. Odd digits \(1, 3, 5, 7, 9\): \(5 \times 4 \times 3 = 60\).
Full method — how and why
Counting with two restrictions: digits from a fixed set, and all different.
  1. List the allowed digits: \(1, 3, 5, 7, 9\) - five odd digits.
    'All odd' restricts the pool; note \(0\) is even so no leading-zero problem arises.
  2. First digit: \(5\) choices. Second: \(4\) (one used). Third: \(3\).
    'All different' removes one option at each stage.
  3. Multiply: \(5 \times 4 \times 3 = 60\).
    The product rule.
Where it goes wrong
  • Using \(5^3 = 125\) by ignoring 'all different' - an offered option.
  • Worrying about leading zeros, which cannot occur here.
Takeaway Count position by position, reducing the pool by one each time when repetition is banned.
13. NUM-16Number, units and estimation · spec M1.1 · Paper 1 style · ★☆☆

Water flows through a pipe at \(1.2\) litres per second. How long does it take to fill a tank of volume \(3.6\) m\(^3\)? (1 m\(^3\) = 1000 litres.)

  1. A\(30\) minutes
  2. B\(50\) minutes
  3. C\(5\) minutes
  4. D\(1\) hour
  5. E\(3\) hours
  6. F\(500\) seconds
Answer: B. \(3600\) litres at \(1.2\) litres/s takes \(3000\) s \(= 50\) minutes.
Full method — how and why
A rate problem with a unit conversion. Convert the volume to litres, divide by the rate, then convert the time.
  1. Convert the volume: \(3.6\) m\(^3 = 3.6 \times 1000 = 3600\) litres.
    The conversion is given in the question.
  2. Divide by the rate: \(\dfrac{3600}{1.2} = 3000\) seconds.
    Volume divided by volume-per-second gives seconds.
  3. Convert to minutes: \(\dfrac{3000}{60} = 50\) minutes.
    The options are in minutes, so finish the conversion.
Where it goes wrong
  • Leaving the answer as \(3000\) seconds and picking \(500\) seconds by a slip.
  • Dividing by \(1.2\) as if it were \(12\), giving \(5\) minutes.
Takeaway Make the units match before dividing, and convert the answer into the units the options use.
14. NUM-17Number, units and estimation · spec M2.13 · Paper 1 style · ★☆☆

\(a = 5.0\) and \(b = 2.0\), each correct to \(2\) significant figures. Find the lower bound of \(\dfrac{a}{b}\).

  1. A\(\dfrac{4.5}{2.5}\)
  2. B\(\dfrac{4.95}{1.95}\)
  3. C\(\dfrac{5.05}{1.95}\)
  4. D\(\dfrac{4.95}{2.05}\)
  5. E\(2.4\)
  6. F\(\dfrac{5.05}{2.05}\)
Answer: D. Smallest numerator over largest denominator: \(\frac{4.95}{2.05}\).
Full method — how and why
Bounds for a quotient: smallest over largest gives the smallest result.
  1. \(a = 5.0\) to 2 s.f. means \(4.95 \leq a < 5.05\).
    The last significant figure is in the first decimal place, so the half-width is \(0.05\).
  2. \(b = 2.0\) to 2 s.f. means \(1.95 \leq b < 2.05\).
    Same reasoning.
  3. For the smallest quotient, make the numerator as small as possible and the denominator as large as possible.
    Increasing the denominator decreases a positive fraction - this is the key insight.
  4. So the lower bound is \(\dfrac{4.95}{2.05}\).
    Left in fraction form, which is what the options offer.
Where it goes wrong
  • Pairing both lower bounds as \(\frac{4.95}{1.95}\): a smaller denominator makes the fraction bigger, so that is not the lower bound.
  • Using \(\pm 0.5\) instead of \(\pm 0.05\) - for these numbers, 2 significant figures means one decimal place.
Takeaway Quotients: lower bound \(= \frac{\text{lower}}{\text{upper}}\), upper bound \(= \frac{\text{upper}}{\text{lower}}\).
15. NUM-18Number, units and estimation · spec M2.3 · Paper 1 style · ★☆☆

What is the smallest positive integer that is divisible by every integer from \(1\) to \(10\)?

  1. A\(3628800\)
  2. B\(1260\)
  3. C\(840\)
  4. D\(720\)
  5. E\(5040\)
  6. F\(2520\)
Answer: F. \(\text{lcm}(1, \ldots, 10) = 2^3\cdot3^2\cdot5\cdot7 = 2520\).
Full method — how and why
The smallest number divisible by all of \(1\) to \(10\) is their LCM: take the highest power of each prime up to \(10\).
  1. List the primes up to \(10\): \(2, 3, 5, 7\).
    Only primes matter; composite numbers are built from them.
  2. Take the highest power of each that is at most \(10\): \(2^3 = 8\), \(3^2 = 9\), \(5\), \(7\).
    \(8\) needs three \(2\)s and \(9\) needs two \(3\)s - the largest demands from any single number.
  3. Multiply: \(8 \times 9 \times 5 \times 7 = 2520\).
    \(72 \times 35 = 2520\).
  4. Check: \(2520\) is divisible by \(4\), \(6\), \(10\) automatically, since those are built from the primes included.
    Covering the highest prime powers covers every composite too.
Where it goes wrong
  • Multiplying all the numbers \(1\) to \(10\) (\(3628800\)) - divisible, but far from smallest.
  • Using \(2^2\) instead of \(2^3\), giving \(1260\), which is not divisible by \(8\).
Takeaway LCM of a range: each prime to the highest power that appears within the range.
16. NUM-19Number, units and estimation · spec M2.9 · Paper 1 style · ★☆☆

Which of the following fractions has a terminating decimal expansion?

  1. A\(\dfrac{7}{40}\)
  2. B\(\dfrac{3}{22}\)
  3. C\(\dfrac{4}{15}\)
  4. D\(\dfrac{11}{14}\)
  5. E\(\dfrac{2}{9}\)
  6. F\(\dfrac{5}{12}\)
Answer: A. A fraction in lowest terms terminates iff its denominator has no prime factors other than \(2\) and \(5\): \(40 = 2^3 \cdot 5\).
Full method — how and why
A fraction terminates exactly when its lowest-terms denominator is built only from \(2\)s and \(5\)s.
  1. Check each denominator is in lowest terms with the numerator.
    A shared factor could change the denominator - here none do.
  2. Factorise: \(40 = 2^3 \times 5\) ✓; \(12 = 2^2\times3\) ✗; \(15 = 3\times5\) ✗; \(14 = 2\times7\) ✗; \(9 = 3^2\) ✗; \(22 = 2\times11\) ✗.
    Any prime other than \(2\) or \(5\) forces a recurring expansion.
  3. So only \(\dfrac{7}{40}\) terminates (it is \(0.175\)).
    Because \(10 = 2\times5\), only these primes can divide a power of ten.
Where it goes wrong
  • Assuming a 'nice-looking' denominator like \(12\) terminates.
  • Forgetting to reduce a fraction first in questions where the numerator shares a factor.
Takeaway Terminating \(\Leftrightarrow\) denominator (in lowest terms) is \(2^a 5^b\).
17. NUM-20Number, units and estimation · spec M2.11 · Paper 1 style · ★☆☆

Find the exact value of \(\left(\sqrt{5} - 1\right)^2 + \left(\sqrt{5} + 1\right)^2\).

  1. A\(12\)
  2. B\(8\)
  3. C\(10\)
  4. D\(6\)
  5. E\(12 + 4\sqrt5\)
  6. F\(4\sqrt5\)
Answer: A. \((6 - 2\sqrt5) + (6 + 2\sqrt5) = 12\).
Full method — how and why
A sum of conjugate squares. The surd terms cancel, so the answer is rational.
  1. Expand the first: \((\sqrt5 - 1)^2 = 5 - 2\sqrt5 + 1 = 6 - 2\sqrt5\).
    \((a-b)^2 = a^2 - 2ab + b^2\) with \((\sqrt5)^2 = 5\).
  2. Expand the second: \((\sqrt5 + 1)^2 = 6 + 2\sqrt5\).
    Only the middle sign differs.
  3. Add: the \(\mp2\sqrt5\) terms cancel, leaving \(12\).
    Conjugates always cancel their surd parts when added.
Where it goes wrong
  • Dropping the cross terms: here it happens to give \(12\) anyway, but the same slip fails whenever the two brackets are not conjugates.
  • Adding the surd terms instead of cancelling, giving \(12 + 4\sqrt5\) - an offered option.
Takeaway \((a+b)^2 + (a-b)^2 = 2(a^2+b^2)\): here \(2(5 + 1) = 12\), with no surds surviving.
18. NUM-02Number, units and estimation · spec M2.3 · Paper 1 style · ★★☆

The lowest common multiple of \(12\), \(18\) and \(n\) is \(180\). Which of the following could be \(n\)?

  1. A\(20\)
  2. B\(24\)
  3. C\(8\)
  4. D\(40\)
  5. E\(50\)
  6. F\(27\)
Answer: A. \(\text{lcm}(12, 18) = 36 = 2^2\cdot3^2\) and \(180 = 2^2\cdot3^2\cdot5\), so \(n\) must be a divisor of \(180\) that is a multiple of \(5\). Only \(20\) qualifies: \(24\), \(8\) and \(27\) have no factor \(5\), and \(50\) and \(40\) do not divide \(180\).
Full method — how and why
An LCM condition. Work out what the other two numbers already supply, then see what \(n\) must add - and must not exceed.
  1. Factorise: \(12 = 2^2 \times 3\), \(18 = 2 \times 3^2\), so \(\text{lcm}(12,18) = 2^2 \times 3^2 = 36\).
    LCM takes the highest power of each prime.
  2. The target is \(180 = 2^2 \times 3^2 \times 5\), so \(n\) must supply the factor \(5\).
    \(36\) has no factor \(5\), so \(n\) is the only possible source.
  3. \(n\) must also divide \(180\): it cannot contain \(2^3\), \(3^3\) or \(5^2\), or the LCM would be larger than \(180\).
    The LCM is at least \(n\), so \(n\) must divide the target exactly.
  4. Check the options: \(20 = 2^2 \times 5\) divides \(180\) and supplies the \(5\) ✓. The others either lack a factor \(5\) (\(24\), \(8\), \(27\)) or do not divide \(180\) (\(50 = 2\times5^2\), \(40 = 2^3\times5\)).
    Test both conditions on each candidate.
Where it goes wrong
  • Choosing \(50\) or \(40\) because they contain a \(5\), forgetting they must divide \(180\).
  • Assuming \(n\) must equal \(180 \div 36 = 5\).
Takeaway For \(\text{lcm}(a,b,n) = L\): \(n\) must divide \(L\) and supply any prime powers \(a\) and \(b\) lack.
19. NUM-09Number, units and estimation · spec M2.3 · Paper 1 style · ★★☆

How many positive divisors does \(360\) have?

  1. A\(24\)
  2. B\(20\)
  3. C\(18\)
  4. D\(36\)
  5. E\(12\)
  6. F\(16\)
Answer: A. \(360 = 2^3\cdot3^2\cdot5\), so the number of divisors is \((3+1)(2+1)(1+1) = 24\).
Full method — how and why
Counting divisors from the prime factorisation: each prime's exponent can be chosen independently.
  1. Factorise: \(360 = 8 \times 45 = 2^3 \times 3^2 \times 5\).
    Split into convenient factors and factorise each.
  2. A divisor has the form \(2^a 3^b 5^c\) with \(0 \leq a \leq 3\), \(0 \leq b \leq 2\), \(0 \leq c \leq 1\).
    Any such combination divides \(360\), and every divisor has this form uniquely.
  3. Count the choices: \(4 \times 3 \times 2 = 24\).
    Each exponent has (power \(+ 1\)) possible values, because \(0\) is allowed.
Where it goes wrong
  • Multiplying the exponents (\(3 \times 2 \times 1 = 6\)) instead of (exponent \(+1\)).
  • Forgetting that \(1\) and \(360\) are themselves divisors - the '\(+1\)'s account for them.
Takeaway Number of divisors of \(p^aq^br^c\) is \((a+1)(b+1)(c+1)\).
20. NUM-12Number, units and estimation · spec M2.3 · Paper 2 style · ★★☆

\(n\) is a positive integer such that \(n\), \(n + 2\) and \(n + 4\) are all prime. Which of the following statements is/are true?

  1. IThere is exactly one such value of \(n\).
  2. II\(n\) must be odd.
  3. IIIOne of \(n\), \(n + 2\), \(n + 4\) must be divisible by \(3\).
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: H. Among any three numbers of the form \(n, n+2, n+4\) one is a multiple of \(3\) (they cover all residues mod \(3\)) - III. So one of them must equal \(3\) itself for all to be prime, giving \(n = 3\) only (\(3, 5, 7\)) - I. And \(3\) is odd - II.
Full method — how and why
A prime-triple question. One observation about remainders on division by \(3\) settles all three statements.
  1. III: among \(n\), \(n+2\), \(n+4\), the remainders on division by \(3\) are all different (they differ by \(2\) and \(4\), i.e. by \(2\) and \(1\) modulo \(3\)), so one of them is a multiple of \(3\). True.
    Three numbers hitting all three remainder classes must include one divisible by \(3\).
  2. I: that multiple of \(3\) can only be prime if it equals \(3\). Checking the possibilities gives \(n = 3\) (\(3,5,7\) are all prime); \(n = 1\) fails (\(1\) is not prime) and larger cases contain a composite multiple of \(3\). True.
    So the triple is unique - this is why \(3,5,7\) is famous as the only 'prime triplet' of this form.
  3. II: \(n = 3\) is odd. (Also, any even \(n > 2\) would not be prime.) True.
    Both routes give the same verdict.
  4. All three.
    Match to the option list.
Where it goes wrong
  • Testing a few values, finding only \(3,5,7\), and still doubting uniqueness - the mod-3 argument proves it.
  • Thinking \(n\) could be \(2\): then \(4\) is not prime.
Takeaway Any three numbers spaced \(2\) apart cover all remainders mod \(3\), so one is divisible by \(3\).
21. X2-18Number, units and estimation · spec M2.3 · Paper 2 style · ★★☆

\(n\) is a positive integer greater than \(1\). Which of the following statements is/are true?

  1. IIf \(n\) is not prime, then \(n\) has a prime factor at most \(\sqrt n\).
  2. IIIf \(n\) has no prime factor at most \(\sqrt n\), then \(n\) is prime.
  3. IIIIf \(n\) is prime, then \(n\) has no factor between \(1\) and \(n\) exclusive.
  1. Anone of them
  2. BI only
  3. CII only
  4. DIII only
  5. EI and II only
  6. FI and III only
  7. GII and III only
  8. HI, II and III
Answer: H. If \(n = ab\) with \(1 < a \leq b\) then \(a \leq \sqrt n\) and \(a\) has a prime factor \(\leq \sqrt n\) (I). II is the contrapositive of I. III is the definition of prime.
Full method — how and why
Three number-theory statements about primality testing, two of which are contrapositives of each other.
  1. I: if \(n\) is composite write \(n = ab\) with \(1
    The smaller factor of a pair cannot exceed the square root, or the product would be too big.
  2. \(a>1\) has a prime factor, which is at most \(a \leq \sqrt n\). So \(n\) has a prime factor \(\leq \sqrt n\). True.
    Every integer greater than \(1\) has a prime factor, and it cannot exceed the number itself.
  3. II: 'no prime factor \(\leq\sqrt n\) \(\Rightarrow\) \(n\) prime' is exactly the contrapositive of I, so it is True automatically.
    Contrapositives share truth values - no separate work needed.
  4. III: 'if \(n\) is prime it has no factor strictly between \(1\) and \(n\)' is the definition of prime. True.
    A prime's only positive factors are \(1\) and itself.
  5. So all three.
    Match to the list.
Where it goes wrong
  • Treating II as the converse of I and rejecting it; it is the contrapositive (negate and swap).
  • Wording III as 'no factors at all', which is wrong: \(1\) and \(n\) are factors.
Takeaway To test \(n\) for primality, trial-divide by primes up to \(\sqrt n\) - and I/II are the two readings of why that works.