The answer
(a) all indices even
(b)(i) \(k = 2\)
(ii) \(\text{LCM} = 2^{20} \times 7^{15} \times 11^{15}\)
O-Level E-Math 2023 Paper 1 Question 4 · Verified worked solution by the Genius Plus Academy teaching team
The question
(a) \(M = p^{16} \times q^4 \times r^{12}\), where \(p\), \(q\) and \(r\) are prime numbers. Explain why \(M\) is a perfect square. [1]
(b) \(S = 2^8 \times 7^{15} \times 11^{15}\). (i) \(N = S \times k\) is a perfect cube. Find the smallest possible integer value of \(k\), as a product of its prime factors. [1] (ii) \(T = 2^{20} \times 7^8 \times 11^{15}\). Find the LCM of \(S\) and \(T\) as a product of its prime factors. [1]
(a) \(M = p^{16} q^4 r^{12} = \left(p^8 q^2 r^6\right)^2\). Every index is even, so \(M\) is the square of an integer, a perfect square.
(b)(i) For a perfect cube every index must be a multiple of 3. In \(S\), the indices are \(8, 15, 15\); only \(8\) is not a multiple of 3, and the next multiple is \(9\), so multiply by \(2^1\). Hence \(k = 2\).
(ii) The LCM takes the highest power of each prime: \(2^{\max(8,20)} \times 7^{\max(15,8)} \times 11^{\max(15,15)} = 2^{20} \times 7^{15} \times 11^{15}\).
Answer: (a) all indices even
(b)(i) \(k = 2\)
(ii) \(\text{LCM} = 2^{20} \times 7^{15} \times 11^{15}\)
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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Genius Plus Academy · O-Level & IP Mathematics
Our O-Level E-Math tuition trains the same recognise-the-structure method these worked solutions show, taught by a team that has marked these papers for years. It runs within our weekly Secondary Math programme, Sec 1 to 4 and IP.
It is a prime factorisation question from Numbers & operations, worth 3 marks: 1 + 1 + 1.
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