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
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
Last edited by allenleonhart; 15th July 2010 at 08:32 PM.
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.
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.
can show working ?
shaddap and just shoot .... up close
Walkeast
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.
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.
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.
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.
Yeah since when were Pri 4 students taught such advanced algebra?
It's crazy, I've no idea what kind of math is that.
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.
and ed, not too late. u can always join us here
http://www.clubsnap.com/forums/group.php?groupid=37
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.
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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