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

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    Moderator ed9119's Avatar
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    Default hee hee need help in Pri 4 Maths (again)

    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
    Last edited by ed9119; 15th July 2010 at 08:27 PM.
    shaddap and just shoot .... up close
    Walkeast

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    Deregistered allenleonhart's Avatar
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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
    wa i like. let the geek club handle. give me 20 mins i go shower first and cool down before thinking
    i'll just brute force it. btw there are 9 letters. meaning its most likely 1-9 or something
    Last edited by allenleonhart; 15th July 2010 at 08:32 PM.

  3. #3

    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
    I am blown!! I did PSLE 5 years ago and then 3 years ago... I am dreading my next one's !!

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

    ok. so far i figured out this far.

    k ranges from 1-4 and cannot be bigger.
    Last edited by allenleonhart; 15th July 2010 at 08:40 PM.

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

    ok i got 1 combination out already

    1132
    1784+
    2916

    in this case, k=1

    my logic

    k+k cannot be greater than 10. or else u get an extra number after H

    k+k will always be an even number.

    so hence i deduced from here, k+k must be between 0-4.

    in this case, h will range from 1-9. H cannot be 0. reason being, either K=0, or K+J must end up less than 10.

    but since K exists, k will never equal to H, so the K=0 and H=0 case is impossible.
    so leaves the next case: k+j less than 10. but if that happens, K is non zero, H will always give a value. reject 0 for H


    now ur left with a few conditions

    now i move on.
    K cannot be 0. why?

    if k=0, k not equal to H,
    H must equal to 1.
    yet for the condition to fufil, K+J and the remainder from A and E must be > 10. that means a+e must give a 10 at least.
    so if u take 1+9, thats the only case to make H non zero, so that u have a value. if that happens, k=f, which is wrong.

    so now i made it till k=1 to 4
    h>0
    now, u see another thing liao

    K+K is non zero, meaning, sub smallest value in.
    1+1.

    H must be equal or larger than 1+1, meaning H must be >=2
    now since i minimize my scope, i can try out and slowly test the cases 1 by 1.
    Last edited by allenleonhart; 15th July 2010 at 09:01 PM.

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    Moderator ed9119's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    can show working ?
    shaddap and just shoot .... up close
    Walkeast

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

    Quote Originally Posted by ed9119 View Post
    can show working ?
    no working. i just brute force with knowing that k not equal to H, K ranges from 1-4 and H is larger than 2

    but if u really need...

    place a value for k first.

    let k=1

    11??
    1???
    ??1?

    now can guess value for H. H must be larger or equal to 2, since k is non zero. so u have now H rangles from 2-9

    so i let it be 2, or 3

    11a2
    1jed
    2f1g

    now u can guess, if all numbers are different
    how to get 1 from A+e?
    a+e can either give an 11, or give a 10, meaning 2+d will result in a value greater than 10.
    i sub liao test for 2+d greater than 10, i get repeats. so its wrong.

    so i move on to test a+e=11
    so that means
    3+8
    4+7
    5+6

    i rest 5+6 i cannot, cause left 0,9,8,3,7. just wont work

    so i test and test...

    cause i know the value that K could be, it helps limit my scope. i know H range also. so i can guess liao.
    Last edited by allenleonhart; 15th July 2010 at 09:00 PM.

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

    btw ed, i think this is under the trial and error section. in psle, trial and error is given marks if u can do the table and show the tests u did.

    i just minimize the scope with logic only. so i dun have to test so many

    teach ur child logic base. let him understand how come u can limit the range. u dun have to test all combinations like this

    oh yea. my logic is always

    if add together, is it more than 10 or less? and i keep testing from there.
    Last edited by allenleonhart; 15th July 2010 at 09:11 PM.

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

    Wah why Pri 4 match so difficult? Are they teaching geniuses in Primary schools these days? Even a university student will have a hard time solving this.

  10. #10

    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
    is there more information?

    as allen has mentioned, should be 1-4, because K + K gives single digit number, it cannot be 5, since that would give H = 10, and i assume that each letter represents 0-9.

    K cannot be 0, because K + K gives max H = 1 (spilling over from K + J = 1F, if this is the case, and if K is 0, then 1 + 0 (K) + J will give max 10, and that would mean that F = 0, which means that K and F are 0, unlikely).


    after which you will have to brute-force it.

    1 1 8 2 + 1 5 3 4 = 2 7 1 6

    satisfies the requirement that each letter is a differnet number, but that is an assumed condition and is not something that we are told.

    K = 1 in this case.
    Last edited by night86mare; 15th July 2010 at 09:21 PM.

  11. #11

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

    Yeah since when were Pri 4 students taught such advanced algebra?

  12. #12

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

    It's crazy, I've no idea what kind of math is that.

  13. #13

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

    anyways, this is with all the assumptions i have made, this question is so skimpy...

    there has to be more conditions, or you can just free-ride it anyway you want, especially if different letters can represent same number.

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

    Quote Originally Posted by night86mare View Post
    is there more information?

    as allen has mentioned, should be 1-4, because K + K gives single digit number, it cannot be 5, since that would give H = 10, and i assume that each letter represents 0-9.

    K cannot be 0, because K + K gives max H = 1 (spilling over from K + J = 1F, if this is the case, and if K is 0, then 1 + 0 (K) + J will give max 10, and that would mean that F = 0, which means that K and F are 0, unlikely).


    after which you will have to brute-force it.

    1 1 8 2 + 1 5 3 4 = 2 7 1 6

    satisfies the requirement that each letter is a differnet number, but that is an assumed condition and is not something that we are told.

    K = 1 in this case.
    yep. u phrased my extremely messy logic into quite simple parts thanks!

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

    Quote Originally Posted by ziploc View Post
    Wah why Pri 4 match so difficult? Are they teaching geniuses in Primary schools these days? Even a university student will have a hard time solving this.
    in plimary 4 ur supposed to know how to brute force already
    i remember last time they used to ask for this kind of combinations one. usually they range it within 20-30 tries and u can get ans so its still quite ok.

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

    and ed, not too late. u can always join us here
    http://www.clubsnap.com/forums/group.php?groupid=37

  17. #17
    Moderator ziploc's Avatar
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    Default Re: hee hee need help in Pri 4 Maths (again)

    Putting it into equations:

    G = (H + D) % 10, X = INT((H + D) / 10)
    K = (A + E + X) % 10, Y = INT((A + E + X) / 10)
    F = (K + J + Y) % 10, Z = INT((K + J + Y) / 10)
    H = K + K + Z

    As you can see above, I don't think you can't solve it via algebra as there are simply not enough info - nothing that defines D, A, E, J. It can only be done by brute force substitution.

  18. #18

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

    Quote Originally Posted by allenleonhart View Post
    in plimary 4 ur supposed to know how to brute force already
    i remember last time they used to ask for this kind of combinations one. usually they range it within 20-30 tries and u can get ans so its still quite ok.
    No they're putting too much stress on pri sch students. What kind of rubbish is this? We learn simple algebra like 3x + 4y = 15, 2x + y = 10, calculate x and y this kind of qn in sec 2, what kind of question is this?

    Next thing you know they'll start teaching complex numbers and hypothesis testing in kindergarten...seriously

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

    Quote Originally Posted by ziploc View Post
    Putting it into equations:

    G = (H + D) % 10, X = INT((H + D) / 10)
    K = (A + E + X) % 10, Y = INT((A + E + X) / 10)
    F = (K + J + Y) % 10, Z = INT((K + J + Y) / 10)
    H = K + K + Z

    As you can see above, I don't think you can't solve it via algebra as there are simply not enough info - nothing that defines D, A, E, J. It can only be done by brute force substitution.
    yep. brute force is only way. if they had any more info i'll had thrown into GC and solved

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

    Quote Originally Posted by brapodam View Post
    No they're putting too much stress on pri sch students. What kind of rubbish is this? We learn simple algebra like 3x + 4y = 15, 2x + y = 10, calculate x and y this kind of qn in sec 2, what kind of question is this?

    Next thing you know they'll start teaching complex numbers and hypothesis testing in kindergarten...seriously
    LOL wont lar. brute force is just test numbers. pluck a few and try ur luck. not much to pluck in actually. K is the first one wanna identify, then guess H, (K+K) or (K+K+1)

    which leaves u with 7 more numbers all DIFFERENT to guess with.

    thats why i can do in 5 mins

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