hee hee need help in Pri 4 Maths (again)

[ziploc]

Another possible answer where K=3.

3397
3641
7038

 

[allenleonhart]

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

 

[nixontkl]

when your K = 3

you get

3396
3142 +
_______
6538

 

[ziploc]

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.

 

[leowyien]

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?

 

[nixontkl]

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

 

[allenleonhart]

Eh, we already shown in previous posts that K=1,2,3 are all possible.

now left to proof 4:think:
once u do that, we proved every possible variation for K already. can tell the sch go home sleep

 

[zaren]

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

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

:bsmilie:

what a joke question!!!!

 

[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

 

[allenleonhart]

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

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

:bsmilie:

what a joke question!!!!

win.:thumbsup:

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

 

[nixontkl]

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

 

[brapodam]

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.

 

[allenleonhart]

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

 

[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.

 

[brapodam]

must prove all can work is not easy. especially when all the remaining variables have no links

Yeah, true

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.

 

[ziploc]

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

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

:bsmilie:

what a joke question!!!!

Haha here is another possible answer where K=4:

4479
4562
9041

Similar to what you have.

 

[Fotophilic]

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 ;(

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...

 

[allenleonhart]

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:sweat:

 

[Fotophilic]

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?

 

[zaren]

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.