Maths is all about foundation. Thus Miss Loi is sometimes a little disappointed that **many** students got themselves stuck in the question below. This question appears *almost without fail* and is meant to test your basics from primary school to Secondary One.

The numbers 60 and 126, written as products of their prime factors are:

60 = 2^{2} x 3 x 5

126 = 2 x 3^{2} x 7

Find

- the largest integer which is a factor of both 60 and 126.
- the smallest integer which is an exact multiple of both 60 and 126.
- the smallest integer
`k`such that 126`k`is a perfect square. - the smallest integer value of
`m`for which 60`m`is a multiple of 126.

P.S. If you’re still stuck remember to look for Miss Loi’s primary school teacher friends Miss *H*au *C*hu *F*en and Mdm *L*au *C*hio *M*in.

## 13 Comments

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Err...

120 = 2 x 3^2 x 7?

should be 126.. i think.

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Arigato gozaimasu NTT-san!

Everywhere it says 126 except the spot you highlighted ... sigh ... need to chase the optician for that new pair of glasses soon!

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Hehe.. NP.. Can I answer?? :p (Feels soooo secondary school... hehe)

1. 2 x 3 = 6

2. 2 x 2 x 3 x 3 x 5 x 7 = 1260

3. 2 x 7 = 17

4. 3 x 7 = 21

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Tupolev,

Of coz you will feel like sec sch - Miss Loi already mentioned this is a Sec One question!

No sweat right? Sigh ... but still many students got this wrong. But then again you're overaged! Bully kechil izit??!

BTW thanx for dropping by again 🙂

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2 x 7 = 17 ??

Minus mark for wrong answer, but still get working marks. (=

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yk, actually to Miss Loi the LHS is the more important portion which normally means that the RHS is just a formality *excuses herself for not checking properly* but having said that THIS IS AN UNFORGIVEABLE SLAP-YOUR-HAND WHAT-A-PITY WHAT-A-WASTE HOW-ON-EARTH-COULD THIS HAPPEN CARELESS MISTAKE!!!!

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Miss Loi, care to explain 3 and 4 to me? thanks.

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Hey Prodigy,

Thanks for rekindling this post from the early days, when Miss Loi was just a budding blogger making her first baby steps into the blogging world (and she still is) ...

Anyway to explain this better ...

First of all do understand that part 1 deals with HCF and part 2 deals with LCM.

For part 3, you need to find the least value of

ksuch that 126 xkis a perfect square (i.e. can be square-rooted, if there's such a word :P).For a number to be a perfect square,

there must be at least two of each of its factors.Hence for 126 = 2 x 3 x 3 x 7, we need at least another 2 and 7 to 'complete the square' (i.e. make it into 2 x 2 x 3 x 3 x 7 x 7). So

kmust be 2 x 7 = 14.曜

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Part 4 is a little tricky, so let's explain this one step at a time.

i. 60

mis a multiple of 126. But obviously 60mis also a multiple of 60!ii. By this definition, do you agree that when

mis at its 'smallest integer value for which 60mis a multiple of 126', this implies that 60mIS the LCM of both 60 and 126?iii. The LCM of both 60 and 126 is 2 x 2 x 3 x 3 x 5 x 7 (obtained in part 2). So we can equate:

60

m= 2 x 2 x 3 x 3 x 5 x 7m= (2 x 2 x 3 x 3 x 5 x 7)/(2 x 2 x 3 x 5)... cancel here ... cancel there ...

m= 3 x 7 = 21曜

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what is the LCM i do not understand...

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Hi Jala, LCM = Lowest Common Multiple. Care to elaborate on which part do you not understand?

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*For no.1, find HCF, which is 2x3 = 6.

*For no.2, find LCM, which is 2squarex3squarex5x7 = 1260.

*For no.3:

The prime factors of 126 is 2x3squarex7.

2 and 7 is not a perfect square, so make 2 and 7 become, 2square and 7square. So just add another 2 and another 7. So k is = 2x7 = 14.

*For no.4, if you how to do no.2, you should know how to do this.

Simply, 1260 divide by 60 = 21.

Correct? (:

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hi i dun understand 1234 can help