If z^n = 1, what is the value of z?

(1) n is a non-zero integer implies n can be positive or negative integer, which is NOT SUFFICIENT to find z.

(2) z > 0 implies that z is any positive value, which is again NOT SUFFICIENT.

Combining (1) and (2), we have n is a non-zero integer and z > 0. So, the only possible value of z = 1, and n can be any positive integer, say, 1, 2, 3, 4....
Hence, combining the statements the question can be answered.

I guess my question is , why cant it be B by itself ? Because for the equation to equal 1, what other value can z have knowing its positive, and in that case what would n equal to?

Can you give an example of b insufficient

Thanks

Since z > 0, so z can be 2, 3, 4, 5, 6...or even any fractional value say, \(\frac{1}{2}\), \(\frac{3}{4}\)...
Now if we take n = 0, then \(2^0 = 1\)
\(3^0 = 1\)
Similarly \((\frac{1}{2})^0 = 1\)
We don't have any unique value of z. So, statement 2 alone is NOT SUFFICIENT.

If z^n = 1, what is the value of z?

(1) n is a nonzero integer.
(2) z > 0

Zero raised to any power is zero and any number raised to the power of 0 equals one?
Is that the rule of it is reversed?

This is not a difficult question by itself, but clarity of approach matters a lot in getting it right. The low accuracy for this question suggests that most students could not think through this question clearly.

At eGMAT, we strongly advocate that in DS Questions, the student should first analyze the question statement thoroughly and only then move on to analyzing the two statements. You'll see how elegantly this question will simplify with this approach.

We are given that z^n = 1. So, what cases are possible for the value of z and n?

Case 1: z = 1; n has any integral value
Case 2: z = -1; n is an even integer
Case 3: z has any non-zero value; n = 0

Please note that only after this analysis are we going to the first Statement.

As per the first statement,
n is a non-zero integer
This rules out Case 3.
However, this still leaves out Case 1 and 2. So, z can either be equal to 1 or z can be equal to -1. So, Statement 1 alone is not sufficient.

As per the second statement,
z > 0
This rules out Case 2. However, Case 1 and 3 still remain. Again, we have not been able to determine a unique value of z. So, Statement 2 alone is not sufficient either.

Combining both the Statements,
From Statement 1, z could either be 1 or -1
From Statement 2, z > 0
Therefore, only possible value of z is 1.

Thus, by combining both the statements together, we have been able to determine a unique value of z. So, the correct answer is Choice C.

Takeaway: The correct answer is only a byproduct of a clear approach.

Hope this helps.

Japinder

Can we take \((\sqrt{1})^2\) as a plug in value to check these statements..

\(\sqrt{1}=1\), so are you asking whether we can plug 1 for n? Well, yes we can...

Oh.. ya.. I forgot that the square root of a perfect number is always +ve.

\(\sqrt{36}\) = 6 (not -6)

So \(\sqrt{1}\) = 1 (not -1)

I had +/- 1 in my head while combining statements I and II together.

Thanks for your clarification..

Any even root from any postie number is positive.

Hi All,

This exponent rule 'concept' is something of a classic in the realm of standardized testing - it serves as a relatively simple way to assess a Test Taker's 'thoroughness of understanding' on a specific concept:

Here, the concept is "using exponent rules, and one number raised to one exponent, how many different ways can you get to the number 1?"

The first ('obvious') answer is "1 raised to any power = 1"

eg. 1^2, 1^50, 1^(-3), etc.

There are OTHER possibilities though. If your base is -1, then any EVEN exponent will lead us to a total of 1...

eg. (-1)^2, (-1)^4, (-1)^(-2), etc.

Finally, raising any number to the '0 power' will also give us a total of 1...

eg. 1^0, 537^0, (-13)^0, etc.

When dealing with this specific situation, it's important to pay careful attention to the information that you're given. What do you really know about the 'base' and the 'power' involved? If you don't know anything, then you have to consider all of the above possibilities.

GMAT assassins aren't born, they're made,
Rich

(1) n is a non zero integer

We have: z^n = 1. Taking the n-th root of both sides, we get,

|z| = 1^(1/n)

|z| = 1

We need to insert the absolute value sign, because either of positive z or negative z can produce the value of one. Since we have two values for z, this statement is INSUFFICIENT

(2) z > 0

To satisfy the equation z^n = 1, we can simply let n be equal to zero. Then the quantity z^n will always have the value of one, regardless of the actual value of z. INSUFFICIENT

Statements (1) and (2) together:

|z| = 1 AND z > 0.

This statement implies that z can only equal +1 and we can eliminate the value of -1. SUFFICIENT

ANSWER: (C)

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

\(-1^{even exponent} = 1\)
\(1^{any non-zero integer} = 1\)

(1) n is a non zero integer

z can still be -1 or 1; insufficient.

(2) z > 0

\(z^0 = 1\)
Z can be any integer if n is 0; insufficient.

(1+2) If n is a non-zero integer AND z is greater than 0 then:

z = 1. Sufficient.

Answer is C.

Target question: What is the value of z?

Given: \(z^n = 1\)
There are 3 possible cases in which the above equation holds true. When a given piece of information yields a small handful of possible cases, I often find it useful to list the possible cases before dealing with the statements (which I've already scanned)
case i: z is any integer, and n = 0 (e.g., \(5^0 = 1\))
case ii: z = 1, and n is any number (e.g., \(1^3 = 1\))
case iii: z = -1, and n is any EVEN INTEGER (e.g., \((-1)^4 = 1\))

Statement 1: n is a non zero integer
This means we're dealing with EITHER case ii OR case iii. Since cases ii and iii yield different answers to the target question, statement 1 is NOT SUFFICIENT

Statement 2: z > 0
This means we're dealing with EITHER case i OR case ii. Since cases i and ii yield different answers to the target question, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
Statement 1 tells us we're dealing with case ii or case iii
Statement 2 tells us we're dealing with case i or case ii
Since only case ii satisfies both statements, it must be the case that z = 1, and n is any number
Since we can be certain that z = 1, the combined statements are SUFFICIENT

Answer: C

Cheers,
Brent

Stmt 1:
\(z^{n}\)=1
\(n\neq{0}\) means n = -1, -2, ..., 1, 2, 3 ....
Since power is not zero, base or z has to be 1 or -1
\(1^{2}\)=1 and \(-1^{2}\)=1
Not sufficient

Stmt 2:
z>0
z=1, 2, 3,
\(1^{0}\)=1
\(2^{0}\)=1
Many solutions for z
Not sufficient

Combining Stmt 1 and 2:
z is positive and power is not zero
hence only solution for z is 1
sufficient
Ans. C

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