it's not that you can't.
can't suggests a lack of ability to do so.
you can, but brute force does it much more efficiently.. ironically.
hee hee need help in Pri 4 Maths (again)
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can't suggests a lack of ability to do so.
you can, but brute force does it much more efficiently.. ironically.
well, no one ever said that examinations should be just about regurgitating methods taught to you in class, with just a few modifications here and there to make them "more challenging" (when they are not).
they are about differentiating academic ability, is that right?
if that is the case, if academic ability is about brute force memorisation, then it can encompass brute force trial and error methods too, right?
i fail to see how setting a goal for primary school students where they could get full marks in mathematics examinations by studying enough will aid them in the long run. it's high time that kids get a taste of real life earlier on, so that the singaporean education system doesn't just produce a bunch of kids waltzing into secondary school, falling flat on their face when they can't get 100 marks for maths, and then start crying in class.
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btw, i realised ur using programming logic:thumbsup:
cause suddenly i realise the mod and int looks familiar to wad i did for my noi training:think:
hey don't forget, you're a jc1 student, and one who does decently in math. This qn is given to a pri 4 student...can they really solve it? maybe those prc scholars can, but what about locals?
no lar. my jc 1 didnt teach me this:embrass:
this one was familiar cause i encountered when i was p4-6 that time:embrass:
u can imagine. i used to be a good boy and do 10 pages per assessment book per topic per day one:embrass:
encounter this always use logic. what are the stuff i can link and limit first?
then brute force
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are you telling me that you need a few more years of education to dial in random numbers and do trial and error?
essentially, the skills needed here are:
1) carefulness
2) determination to get the answer
3) simple addition skills
i believe 1 and 2 are part and parcel of examinations, 3 is taught by primary 4. i hope.
Here is another possible answer:
1172
1643
2815
So as you can see, the answer is not exact, and there are many permutations. The question looks like something that was conjured up by a teacher too creative. And it looks more like IQ test than math.
1172
1643
2815
So as you can see, the answer is not exact, and there are many permutations. The question looks like something that was conjured up by a teacher too creative. And it looks more like IQ test than math.
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Yes but I dont think math should be about planting in random numbers and see if they fit. That should perhaps be part of a life skills workshop to develop people's perseverance and patience to solve the problem by testing a random range of numbers in the equation.
Sure, some logic is involved, but nothing much more than that. I thought they were already trying to discourage the "trial and error" method since the start of our education, why else would they teach us ways to work out "number pattern" style questions?
i am wondering if you can get K = 2.
i tried for a while, and i couldn't really get anything like that out.
perhaps, but that's your view.
i think as long as the entire examination doesn't centre around 50 questions of that sort, it should be fine.
such questions can be a way to differentiate better students from the poorer ones - or perhaps how exam-smart you are. obviously, if i were a pri 4 student and i got this for my final examinations, i would realise quickly that this would be a waste of my time relative to the other "safer" questions, and leave it for last. then if i check my answers and i still have time, then can play with numbers for fun.
ziploc, try for a permutation for k is not equal to 1.
so far the ones we did, K=1 only:think:
qn wants another value of K
if not K=1 means something else. got hidden logic that limits k=1
anyway to do it via Cprog? input value of K and guess value.
i think possible
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:thumbsup: at that age, i would have probably wasted the whole time trying to figure out how to do that, and will only give up after i realise the exam is about to end lolperhaps, but that's your view.
i think as long as the entire examination doesn't centre around 50 questions of that sort, it should be fine.
such questions can be a way to differentiate better students from the poorer ones - or perhaps how exam-smart you are. obviously, if i were a pri 4 student and i got this for my final examinations, i would realise quickly that this would be a waste of my time relative to the other "safer" questions, and leave it for last. then if i check my answers and i still have time, then can play with numbers for fun.
later they change the qn to "find all the possible combinations of values of A, D, E, F, G, H, and K".
Right, go for it, math geek! :sweatsm:
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.
It is actually quotient and remainder, but since there is no symbol for quotient, I use programming representation. Here is another way of putting it (in words):
G = remainder((H + D) / 10), X = quotient((H + D) / 10)
K = remainder((A + E + X) / 10), Y = quotient((A + E + X) / 10)
F = remainder((K + J + Y) / 10), Z = quotient((K + J + Y) / 10)
H = K + K + Z
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Well, nowadays they also want to teach creativity to students and not only confine to rules. This question does look like they are teaching trial-and-error.
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 )
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 )
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