NMTC 2015 Final Q5
a + b + c + d + e + f − a + b − c + d − e + f 5 a + 4 b + 3 c + 2 d + e 5 a − 4 b + 3 c − 2 d + e 2 0 a + 1 2 b + 6 c + 2 d − 2 0 a + 1 2 b − 6 c + 2 d = = = = 0 = = 1 − 1 0 0 0 .
Find a b c d e f
The answer is 0.
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2 solutions
Nice but please give answers of other variables.
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Absolutely not required, for claiming the product abcdef zero, it is sufficient to show that least one of them is zero. Anyways a,b,c,d,e,f∈{0}.
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How did you know that "a,b,c,d,e,f∈R"?
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@Pi Han Goh – Made some edits.....Now is it correct??
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@Rishabh Jain – You still need to show that the there the values of a,c,e are finite values, else you wouldn't know whether the system of equations is satisfied.
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@Pi Han Goh – But from the equations it is obvious that for undefined values of a,c,e these 6 equations cannot turn into something finite. I am a bit confused.....
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@Rishabh Jain – No no, you misunderstood. When you're writing a solution, you need to make sure the problem statement is correct as well, instead of just assuming it to be correct.
If you only know that b=d=0, how do you know that two of the equations given can't be simplified to a + b = 4 and a − b = 5 (which is absurd)?
The issue here is that you ALREADY assumed a,b,c,d,e are finite values, when in fact you shouldn't make such an assumption. That's why after knowing that b=d=0, you still have to prove that a,c,e are finite values and all these values satisfy all the given equations.
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@Pi Han Goh – Ok ...I understood your point. I will edit my solution. Thanks for the valuable time.... :)
I have changed the problem. Now try it and post solutions.
Eqn3 - Eqn4 gives 2b+d=0 ------------(7)
Eqn5+ Eqn6 gives 6b+d=0 ------------(8)
From (7) and (8) , b=d=0
Adding (1) and (2) and using b=d=0,we get f=0.
Now we are left with 3 equations in 3 variables that are a,c and e.
a+c+e=1
5a+3c+e=0
20a+6c=0
These can be solved easily to get a=3/8,c=-5/4,e=15/8 and hence abcdef=0