2026 PHS P1 Q12

2026 PHS P1 Q12

Secondary 4
5 marks
  1. Use prime factors to explain why \(96\times45\) is not a perfect square.[2]
  2. The number \(96\times45\div\dfrac mn\) is a perfect cube where \(m\) and \(n\) are prime numbers.
    Find the value of \(m\) and the value of \(n\).[2]
  3. The number \(S=2^5\times3^3\times5\).
    The lowest common multiple (LCM) of \(45\), \(96\) and \(P\) is \(S\).
    \(P\) is between \(30\) and \(300\).
    Give a possible value of \(P\).[1]

Solution:

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Answer:(a) \(2^5\times3^3\times5\), with odd exponents. (b) \(m=5,n=2\) (c) Any of \(54,108,135,216,270\).

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