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2007
Mon
16
Apr
Miss Loi avatar

Numbers – Remember What They Taught You In Primary School?

(13)
Tuition given in the topic of E-Maths Tuition Questions from the tuition centre desk of Miss Loi at 6:36 pm (Singapore time)

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

  1. the largest integer which is a factor of both 60 and 126.
  2. the smallest integer which is an exact multiple of both 60 and 126.
  3. the smallest integer k such that 126k is a perfect square.
  4. 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.

頑張って!!!

Revision Exercise

To show that you have understood what Miss Loi just taught you from the centre, you must:

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Comments & Reactions

13 Comments

  1. NTT's Avatar
    NTT commented in tuition class


    2007
    Apr
    16
    Mon
    8:30pm
     
    1

    Err...

    120 = 2 x 3^2 x 7?

    should be 126.. i think.

  2. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Apr
    16
    Mon
    8:45pm
     
    2

    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!

  3. NTT's Avatar
    NTT commented in tuition class


    2007
    Apr
    17
    Tue
    11:15pm
     
    3

    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

  4. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Apr
    18
    Wed
    10:25am
     
    4

    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 🙂

  5. yk's Avatar
    yk commented in tuition class


    2007
    Apr
    29
    Sun
    1:18am
     
    5

    2 x 7 = 17 ??

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

  6. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Apr
    29
    Sun
    8:25pm
     
    6

    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!!!!

  7. Prodigy's Avatar
    Prodigy commented in tuition class


    2007
    Oct
    14
    Sun
    9:32pm
     
    7

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

  8. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Oct
    15
    Mon
    12:00am
     
    8

    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.

  9. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Oct
    15
    Mon
    12:09am
     
    9

    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

  10. Jala's Avatar
    Jala commented in tuition class


    2007
    Dec
    20
    Thu
    1:40am
  11. Miss Loi's Avatar
    Miss Loi Friend Miss Loi on Facebook @MissLoi commented in tuition class


    2007
    Dec
    20
    Thu
    1:55pm
     
    11

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

  12. MathsRock's Avatar
    MathsRock commented in tuition class


    2009
    Oct
    11
    Sun
    4:19pm
     
    12

    *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? (:

  13. michelle's Avatar
    michelle commented in tuition class


    2009
    Nov
    9
    Mon
    10:06pm

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