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 = 22 x 3 x 5
126 = 2 x 32 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 126k is a perfect square.
- the smallest integer value of m for which 60m is a multiple of 126.
P.S. If you’re still stuck remember to look for Miss Loi’s primary school teacher friends Miss Hau Chu Fen and Mdm Lau Chio Min.
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13 Comments
曜
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Err...
120 = 2 x 3^2 x 7?
should be 126.. i think.
曜
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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!
曜
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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
曜
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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 🙂
曜
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2 x 7 = 17 ??
Minus mark for wrong answer, but still get working marks. (=
曜
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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!!!!
曜
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Miss Loi, care to explain 3 and 4 to me? thanks.
曜
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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 k such that 126 x k is 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 k must 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. 60m is a multiple of 126. But obviously 60m is also a multiple of 60!
ii. By this definition, do you agree that when m is at its 'smallest integer value for which 60m is a multiple of 126', this implies that 60m IS 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:
60m = 2 x 2 x 3 x 3 x 5 x 7
m = (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...
曜
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Hi Jala, LCM = Lowest Common Multiple. Care to elaborate on which part do you not understand?
曜
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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