- Thread starter ed9119
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should have joined the Maths Club and not Photo Club while in school :bsmilie:

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 ;(

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'll just brute force it. btw there are 9 letters. meaning its most likely 1-9 or something

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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 ;(

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.

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.

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can show working ? :bsmilie: :bsmilie: :bsmilie:

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.

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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.

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.

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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 ;(

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.

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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.

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.

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. :sweat:

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.

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.

in plimary 4 ur supposed to know how to brute force already:sweat:

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.

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.

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

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

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

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

thats why i can do in 5 mins:sweat: