IF (z)^n = 1, what is the value of z?

(1) n is a non zero integer
(2) z > 0

z^n = 1
if n=0 , then any value of z except 0 , will satisfy this
if n!=0
i). then z = 1, for all value of n, z^n = 1
ii). then z = -1, for all even value of n, z^n = 1


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.

Hence C.

Thanks

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.

Thanks for the quick response. Elaborating on your response...

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

anuptvm wrote:

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.

Thanks for the quick response. Elaborating on your response...

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

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:

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.

Thanks for the quick response. Elaborating on your response...

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 on

answer should take to choice E

even if we cancel anuptvm question about n<0, to match the 1st statement, we still miss condition: z is integer

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.

Like it? We LOVE it!

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.

tomada wrote:Like it? We LOVE it!

Hi tomada!

Thank you very much for your highly positive feedback!

See you in other BTG posts!

Cheers,
Fabio.

IF (z)^n = 1, what is the value of z?

If n = 0, then z could be any number, obviously.

now.

(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). Which one is it, +1 or -1?

(2) z > 0

Ah, this tells is that it's +1. Therefore, (C)

Woozler wrote:If n = 0, then z could be any number, obviously.

Except zero.

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).

You are right, Woozler (anshumishra had already "said" that in his first post, by the way).

(My work was done to guarantee the expression in red is a FACT, not just "faith by trials", although I agree that we can "convince ourselves" pretty easily...)