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Thread: hee hee need help in Pri 4 Maths (again)

  1. #41
    Moderator ziploc's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Another possible answer where K=3.

    3397
    3641
    7038
    Last edited by ziploc; 15th July 2010 at 10:29 PM.

  2. #42
    Deregistered allenleonhart's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by night86mare View Post
    yes, seems correct.

    whoever thought this question up needs to be shot.

    if must give tedious question, at least make sure that you only have one answer, tsk.
    i like this. imagine u start shooting off 1-4 and u find out all will work.

    tats amazing

  3. #43

    Default Re: hee hee need help in Pri 4 Maths (again)

    when your K = 3

    you get

    3396
    3142 +
    _______
    6538
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  4. #44
    Moderator ziploc's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by nixontkl View Post
    My answer K = 3

    using logic K + K = H, H <=9 thus K can only be 1-4

    so if K = 1, then H = 2 11A2 + 2JED = 2F1G
    since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
    using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
    if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
    now left with 4,3,8,0
    3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out

    there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah


    when u get to K = 3 33A6 + 6JED = 6F3G
    Numbers left = 1,2,4,5,6,7,8,9,0

    same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
    set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
    so far numbers left are ( 2,7,8,0 )
    last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
    numbers left ( 7,0 )
    Eh, we already shown in previous posts that K=1,2,3 are all possible.
    Last edited by ziploc; 15th July 2010 at 10:26 PM.

  5. #45
    Member leowyien's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by nixontkl View Post
    My answer K = 3

    using logic K + K = H, H <=9 thus K can only be 1-4

    so if K = 1, then H = 2 11A2 + 2JED = 2F1G
    since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
    using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
    if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
    now left with 4,3,8,0
    3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out

    there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah


    when u get to K = 3 33A6 + 6JED = 6F3G
    Numbers left = 1,2,4,5,6,7,8,9,0

    same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
    set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
    so far numbers left are ( 2,7,8,0 )
    last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
    numbers left ( 7,0 )
    IS this Pri 4?
    indie

  6. #46

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Eh, we already shown in previous posts that K=1,2,3 are all possible.
    LOL by the time i type finish I already lap by 1 page of post liao LOL
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  7. #47
    Deregistered allenleonhart's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Eh, we already shown in previous posts that K=1,2,3 are all possible.
    now left to proof 4
    once u do that, we proved every possible variation for K already. can tell the sch go home sleep

  8. #48

    Smile Re: hee hee need help in Pri 4 Maths (again)

    hey hey.... another solution: K=4!

    4469
    4572 +
    -----
    9041
    -----



    what a joke question!!!!

  9. #49

    Default Re: hee hee need help in Pri 4 Maths (again)

    Its not primary 4... K=1-4... They havent learn inequality in the first place... There's prob missing info? Primary school i did a lot of guess&check. Bloody time consuming. Lol
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  10. #50
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by zaren View Post
    hey hey.... another solution: K=4!

    4469
    4572 +
    -----
    9041
    -----



    what a joke question!!!!
    win.

    GRATS GUYS. WE ALL PROVED 1-4 ALL CAN WORK. MEANING THIS QN CAN **********

  11. #51

    Default Re: hee hee need help in Pri 4 Maths (again)

    LOL seems like this math question

    all the student need is to deduce that K = 1 or 2 or 3 or 4

    and simply by using the 4th digits K + K + (any carry over) = H and H can only be <= 9 so K is 9/2 (round down) or less
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  12. #52

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by qwerty628 View Post
    Its not primary 4... K=1-4... They havent learn inequality in the first place... There's prob missing info? Primary school i did a lot of guess&check. Bloody time consuming. Lol
    Nah I don't think you really need knowledge of inequalities...just have to think in that way, which isn't too hard without knowing about inequalities. It's just some logic.

    But since there are multiple answers and inequalities are not in the pri 4 syllabus, this qn is really dumb.
    Last edited by brapodam; 15th July 2010 at 10:40 PM.

  13. #53
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    Nah I don't think you really need knowledge of inequalities...just have to think in that way, which isn't too hard without knowing about inequalities. It's just some logic.
    must prove all can work is not easy. especially when all the remaining variables have no links

  14. #54

    Default Re: hee hee need help in Pri 4 Maths (again)

    good thinking question, a trick question that filter out the best student from the good student

    its like most student who complete the test can only get 90++ but never 100 unless he/she figure out the trick. proving that the student is a thinking out of the box.
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  15. #55

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by allenleonhart View Post
    must prove all can work is not easy. especially when all the remaining variables have no links
    Yeah, true
    Quote Originally Posted by nixontkl View Post
    good thinking question, a trick question that filter out the best student from the good student

    its like most student who complete the test can only get 90++ but never 100 unless he/she figure out the trick. proving that the student is a thinking out of the box.
    I'm not sure how many pri 4 students can think like that under exam conditions though. As for thinking out of the box...well, in my opinion, it's just a regression in the thought process where you resort to experimenting with random numbers (with a bit of logic) when you are unable to think of a suitable mathematical method to solve the problem. It's logic, it's not much math. But maybe that's just me.

  16. #56
    Moderator ziploc's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by zaren View Post
    hey hey.... another solution: K=4!

    4469
    4572 +
    -----
    9041
    -----



    what a joke question!!!!
    Haha here is another possible answer where K=4:

    4479
    4562
    9041

    Similar to what you have.

  17. #57
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ed9119 View Post
    should have joined the Maths Club and not Photo Club while in school

    whats the answer?

    K K A H
    K J E D +
    ----------
    H F K G
    ----------

    What is K ? (0 to 9)

    Googled, youtubed and personal workings without success
    how to derive a number from ur problem (where K= [0-9]) when ur problem do not have a numerical figure inside? At most the numbers present in the problem would be like 2K but this is only representing that its 2 times of K rather than K itself meaning any numbers...

    i need to do some soul searching on what i learnt in primary school...
    Last edited by Fotophilic; 15th July 2010 at 10:47 PM.
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  18. #58
    Deregistered allenleonhart's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by brapodam View Post
    Yeah, true

    I'm not sure how many pri 4 students can think like that under exam conditions though. As for thinking out of the box...well, in my opinion, it's just a regression in the thought process where you resort to experimenting with random numbers (with a bit of logic) when you are unable to think of a suitable mathematical method to solve the problem. It's logic, it's not much math. But maybe that's just me.
    yea its logic. but u will need more when u move on too.

    now we are supposed to see f'x out of no where one. not easy. and must always know wad things cannot eb zero. a lot of guesswork too also

  19. #59
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by nixontkl View Post
    My answer K = 3

    using logic K + K = H, H <=9 thus K can only be 1-4

    so if K = 1, then H = 2 11A2 + 2JED = 2F1G
    since 1 and 2 is taken, the rest of the letters = 3,4,5,6,7,8,9,0
    using the additional of the second digit 'A' and 'E' A+E = 1 or 11 there for A,E = 5,6/4,7,3/8 (combination to form 11)
    if we choose A, E as 5, 6 then the 1st digit 2 + D must be less than 10 else the 2nd digit is not true, thus if we sub D as 7 then 2 + 7 = G = 9
    now left with 4,3,8,0
    3rd digit K + J = F, since 2nd digit A+E we set as 11 therefore the 3rd digit will have a carry over, become 1+1+J = F === 2+J = F (F = 3/4/8/0) F cannot = 3 cause J is not 1, F cannot = 4 cause J is not 2 (K = 1 and H = 2) leaving 8 and 0 but 2 + J = F cannot have a carry over thus F cannot be 8 as A, E is 5, 6 and 0 is out

    there for K = 1 is not possible, continue with K = 2 and H = 4, bah bah bah


    when u get to K = 3 33A6 + 6JED = 6F3G
    Numbers left = 1,2,4,5,6,7,8,9,0

    same method start with A, E A + E = 13 (combination for A, E = 4,9/5,8/6,7)
    set A, E as 4, 9 then your 3rd digit must be 1+3+J = F (F must be <9 so that there is no carry over) let J = 1 your 3d digit 1+3+1 = F (F=5)
    so far numbers left are ( 2,7,8,0 )
    last set of digits 6 + D = G (G < 10) thus 6 + D(D=2) = G(G=8)
    numbers left ( 7,0 )
    how do u arrive at the initial part where H <=9?
    cameras are not made of tofu

  20. #60

    Default Re: hee hee need help in Pri 4 Maths (again)

    Quote Originally Posted by ziploc View Post
    Haha here is another possible answer where K=4:

    4479
    4562
    9041

    Similar to what you have.
    yep, variables A and E are interchangeable. that would also apply to all the previous solutions mentioned in this thread.

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