GCSE number skills appear in TMUA as fast mental arithmetic under pressure: prime factorisation for HCF/LCM and divisor counts, standard form, bounds, recurring decimals and systematic counting. The traps are careless ones - keep the working tidy.
Contents: Primes, factors and multiples (M2.3)Decimals, standard form, bounds (M2.8, M2.9, M2.13, M2.14)Units and counting (M1, M2.5)Must-knowTrapsReal questions
Primes, factors and multiples (M2.3)
- Prime factorise: \(360 = 2^3\cdot3^2\cdot5\). HCF = lowest powers of common primes; LCM = highest powers of all primes.
- Number of divisors of \(p^a q^b r^c\) is \((a + 1)(b + 1)(c + 1)\).
- \(\text{lcm}(1, \ldots, 10) = 2^3\cdot3^2\cdot5\cdot7 = 2520\).
- Among any three numbers \(n, n + 2, n + 4\) one is a multiple of \(3\) - so \(3, 5, 7\) is the only prime triple of that form.
Decimals, standard form, bounds (M2.8, M2.9, M2.13, M2.14)
- Recurring decimal: \(x = 0.\dot2\dot7\), \(100x = 27.\dot2\dot7\), \(99x = 27\), \(x = \frac{3}{11}\). A fraction terminates iff its lowest-terms denominator has only the prime factors \(2\) and \(5\).
- Standard form \(a \times 10^n\) with \(1 \leq a < 10\): multiply the \(a\)'s, add the powers, then renormalise.
- Bounds: a value \(4.6\) to 1 d.p. lies in \([4.55, 4.65)\). Upper bound of a quotient = upper/lower; lower bound = lower/upper.
- Truncated (not rounded) to \(3.14\): \(3.14 \leq x < 3.15\).
- Estimation: round each number to 1 s.f. before calculating.
Units and counting (M1, M2.5)
- \(90\) km/h \(= 90000/3600 = 25\) m/s. \(1\) m\(^3 = 1000\) litres; \(1\) m\(^2 = 10^4\) cm\(^2\).
- Product rule for counting: \(m\) ways then \(n\) ways gives \(mn\). Without repetition the choices shrink: \(5 \times 4 \times 3\).
Must-know (no formula booklet)
- divisors of \(p^aq^b = (a+1)(b+1)\)
- terminating \(\Leftrightarrow\) denominator \(2^a5^b\)
- \(x\) to 1 d.p. \(= 4.6 \Rightarrow 4.55 \leq x < 4.65\)
- km/h \(\div 3.6 =\) m/s
Classic traps
- The upper bound of an interval is not attainable: write \(< 4.65\), not \(\leq\).
- Adding in standard form needs equal powers of \(10\) first.
- Codes with 'different digits' use \(9 \times 8\), not \(9 \times 9\).
Worked example
Q. How many positive divisors has \(360\)?
A. \(360 = 2^3\cdot3^2\cdot5\), so \((3+1)(2+1)(1+1) = 24\).