The answer
(a)(i) \(2^4 \times 3^2\)
(ii) 36 and 48
(b) \(p = 2\), \(q = 7\)
O-Level E-Math 2021 Paper 1 Question 20 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a)(i) Write 144 as prime factors. [1] (ii) HCF 12, LCM 144, both \(> 20\); find the two numbers. [2] (b) \(784 = 2^4 \times 7^2\); \(784 \div \frac{p}{q}\) is a perfect cube (\(p\), \(q\) prime); find \(p\), \(q\). [2]
(a)(i) \(144 = 12^2 = (2^2 \times 3)^2 = 2^4 \times 3^2\).
(ii) Each number is \(2^a \times 3^b\) with \(\{a\} \subseteq \{2, 3, 4\}\), \(\{b\} \subseteq \{1, 2\}\). HCF \(= 2^2 \times 3^1\) and LCM \(= 2^4 \times 3^2\) force the index pairs to be \(\{2, 4\}\) for 2 and \(\{1, 2\}\) for 3. The pair \(36 = 2^2 \times 3^2\) and \(48 = 2^4 \times 3\) are both \(> 20\) (the alternative \(12\) and \(144\) fails the "\(> 20\)" condition). So the numbers are 36 and 48.
(b) \(784 \div \dfrac{p}{q} = 784 \times \dfrac{q}{p} = 2^4 \times 7^2 \times \dfrac{q}{p}\). For a perfect cube every index must be a multiple of 3: divide by \(2\) (so \(2^4 \to 2^3\)) and multiply by \(7\) (so \(7^2 \to 7^3\)). Hence \(p = 2\), \(q = 7\), giving \(2^3 \times 7^3 = 14^3 = 2744\).
Answer: (a)(i) \(2^4 \times 3^2\)
(ii) 36 and 48
(b) \(p = 2\), \(q = 7\)
Same structure, different numbers
Swap the constants, dress a quadratic as a length, hide a derivative inside an integral, and a student sees a brand new problem. The structure underneath is the same, and so is the method. Once a student can name the structure, a whole row of questions that look different start to open the same way.
That is where marks really leak: in choosing the method, not in the algebra that follows. We call it Lock and Key, name the lock, then the key follows.
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It is a prime factorisation question from Numbers & operations, worth 5 marks: (a) 1 + 2, (b) 2.
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