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

3397
3641
7038

2. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by night86mare
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

you get

3396
3142 +
_______
6538

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

Originally Posted by nixontkl

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.

5. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by nixontkl

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?

6. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by ziploc
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

7. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by ziploc
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. ## Re: hee hee need help in Pri 4 Maths (again)

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

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

what a joke question!!!!

9. ## 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

10. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by zaren
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. ## 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

12. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by qwerty628
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.

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

Originally Posted by brapodam
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. ## 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.

15. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by allenleonhart
must prove all can work is not easy. especially when all the remaining variables have no links
Yeah, true
Originally Posted by nixontkl
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. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by zaren
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. ## Re: hee hee need help in Pri 4 Maths (again)

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

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

What is K ? (0 to 9)

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...

18. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by brapodam
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. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by nixontkl

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?

20. ## Re: hee hee need help in Pri 4 Maths (again)

Originally Posted by ziploc
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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