z^n = 1IF (z)^n = 1, what is the value of z?
(1) n is a non zero integer
(2) z > 0
Thanks for the quick response. Elaborating on your response...anshumishra wrote:
Statement 1 :
if n is a non-zero integer
z = 1 or -1 --- Insufficient
Statement 2 :
z > 0
Any value of n (except 0) would satisfy the equation --- Insufficient
Both combined, ensures Z = 1 --- Sufficient.
The question only asks about the value of "z" , so we shouldn't be bothered about possible values of "n", as long as that doesn't change value of "z".anuptvm wrote:Thanks for the quick response. Elaborating on your response...anshumishra wrote:
Statement 1 :
if n is a non-zero integer
z = 1 or -1 --- Insufficient
Statement 2 :
z > 0
Any value of n (except 0) would satisfy the equation --- Insufficient
Both combined, ensures Z = 1 --- Sufficient.
For Statement 2: z > 0
For any value of Z if n =0 then Z^n =1
And for n = 1 or -1 Z^n =1 when Z=1
So essentially Statement 2 gives us a set of values for N {-1,1,2,3,4,5...}
Combining Stmt 1 and Stmt 2 there would still be more than one answer for N
absolutely agree, adding another point would question z>0 so z belongs {0; +ve infitity} it can be 1/2, 1/3 so onanuptvm wrote:Thanks for the quick response. Elaborating on your response...anshumishra wrote:
Statement 1 :
if n is a non-zero integer
z = 1 or -1 --- Insufficient
Statement 2 :
z > 0
Any value of n (except 0) would satisfy the equation --- Insufficient
Both combined, ensures Z = 1 --- Sufficient.
For Statement 2: z > 0
For any value of Z if n =0 then Z^n =1
And for n = 1 or -1 Z^n =1 when Z=1
So essentially Statement 2 gives us a set of values for N {-1,1,2,3,4,5...}
Combining Stmt 1 and Stmt 2 there would still be more than one answer for N
fskilnik wrote:Hi there,
This is all a bit messed-up, isn´t it?
IMHO you are "loosing focus" in 2 ways:
First: you are doing "analysis" AFTER you all agree that something is not sufficient to answer the question. This is loosing time and energy ;
Second: some of you are trying to convince other that (1+2) is sufficient, but without technical arguments. Till now, nobody FOUND a different z, but nobody guaranteed that it cannot be found... so that the question (1+2) is still open... (I will solve below.)
My suggestion: do analysis if you have "hope" that a certain sttm is sufficient, but try to "KO" the problem with CONCRETE (numerical) examples if you think the sttm is insufficient, as quick as possible.
Have a look at all these below:
Being explicit:
(1) BIFURCATES (therefore insufficient):
TAKE z =1 and n = 1 then the answer would be (z=) 1.
TAKE z = -1 and n = 2 then the answer would be (z=) -1.
You see? It´s aggressive. You are looking for ways of getting different answers satisfying the question stem and sttm1, just that!
(2) Try to REUSE examples used before, therefore:
TAKE z =1 and n = 1 then the answer would be (z=) 1. (REUSED)
TAKE z = 2 and n=0 then the answer would be (z=) 2.
You see? Now I used the fact that 2^0 =1 , but I didn´t "analyse" that, I simply used that. It´s quicker, it´s focused, it´s stronger!
(1+2) Try to REUSE again...
TAKE z =1 and n = 1 then the answer would be (z=) 1. (REUSED)
Now... let´s analyse it JUST BECAUSE I cannot "see" an easy example of z>0, z <> 1 such that z^n =1 with integer n different from 0...
Important: the fact that I cannot see does NOT mean there isn´t one... even if the OA is "C" (I suppose it is)... the book/CD could be wrong!
TECHNICAL ANALYSIS:
Please note that if n > 0 then:
(i) 0 < z^n < 1 if z is between 0 and 1
(ii) z^n > 1 if z is greater than 1
Therefore for n>0, there is neither 0<z<1 nor z>1 such that z^n = 1, therefore till now z=1 necessarily.
What if n<0 ? In this case -n > 0 and applying the analysis done before:
there is neither 0<z<1 nor z>1 such that z^(-n) = 1, for sure.
That means 1/(z^n) = 1 does not happen, that means z^n = 1 does not happen (for n<0 and z> 0 different from 1).
Conclusion: the answer is (C), with full rigour and pretty quick.
I hope you like it.
Best Regards,
Fabio.
Hi tomada!tomada wrote:Like it? We LOVE it!
If n = 0, then z could be any number, obviously.IF (z)^n = 1, what is the value of z?
(1) n is a non-zero integer
Ah, this tells is that it's +1. Therefore, (C)(2) z > 0
Except zero.Woozler wrote:If n = 0, then z could be any number, obviously.
You are right, Woozler (anshumishra had already "said" that in his first post, by the way).Woozler wrote: (1) n is a non-zero integer
- so now our choice for z is limited only to +1 (for any n) and -1 (for even n).
