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secondary 4 | A Maths
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secondary 4 chevron_right A Maths chevron_right Singapore

need hrlp w part (iii) only, pls explain too

Date Posted: 3 years ago
Views: 541

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Ong Jing Long
Ong Jing Long's answer
116 answers (Tutor Details)
1) Larger than 405: cannot use 405
2) Highest common factor is 405: you need to be able to factorise such that you get 405, so you can start with the prime factorisation of 405 = 3^4 × 5
3) Lowest common multiple is 4860: number should be smaller than or equals to 4860

let a be the first number, b be the second number

we explore a case:
a = 4860, b = 2 × 3^5 × 5, then this violates rule 2 since b would be the highest common factor (this would be the case for any other b, e.g. 2^2 × 3^4 × 5)

you can try other cases but they will be the same where 1 of the rules will be violated

all in all, a tip for this would be to multiply by the two numbers with maximum powers of different prime numbers. This ensures HCF and LCM
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3 years ago
is this question something like guess and check?
Ong Jing Long
Ong Jing Long
3 years ago
you can do that but ultimately there is a fixed way to solve the question.

edit: i realised the description i included in the answer is in a mess so I'm repeating it here.

Usually the way to solve the question would be to multiply by the maximum powers of the different prime numbers