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# Is the nth root of n greater than the cube root of 3?

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Re: Is the nth root of n greater than the cube root of 3? [#permalink]
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

Is the nth root of n greater than the cube root of 3?

1) The nth root of n is equal to the 4th root of 4
2) The nth root of n is equal to the square root of 2

When you modify the original condition and the question, n^3>3^n? is calculated by multiplying 3n to the both equation, n^1/n>3^1/3. Then, there is 1 variable(n), which should match with the number of equations. So you need 1 more equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), when n=4, 4^3>3^4, which is no and sufficient.
For 2), when m=2, 4^2>2^4, which is also no and sufficient.
Therefore, the answer is D.

-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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Re: Is the nth root of n greater than the cube root of 3? [#permalink]
I completely agree this is an odd question and I haven't found anything like it in the OG questions. Having said so:

1. It clearly identifies what The nth root of n is equal to, therefore you could check the inequility of the steam SUFFICIENT
2. It clearly identifies what The nth root of n is equal to, therefore you could check the inequility of the steam SUFFICIENT

GMATinsight wrote:
zxcvbnmas wrote:
Is the nth root of n greater than the cube root of 3?

1) The nth root of n is equal to the 4th root of 4
2) The nth root of n is equal to the square root of 2

Question : Is $$n^{1/n} > n^{1/3}$$?

HINT: To answer the question we require some information about sign of n and range of values of n

Statement 1: $$n^{1/n} = 4^{1/4}$$
This is possible only for n = 4, Hence
SUFFICIENT

Statement 2: $$n^{1/n} = 2^{1/2}$$
This is possible only for n = 4, Hence
SUFFICIENT

IMPORTANT: I would like to mark this question as BAD and INVALID question for a test like GMAT.

GMAT never gives information in two statements that have even possibility of contradicting with each other i.e. value of n as per 2nd statement can't be 4 if the first statement has already brought it as 4. However 2nd statement may have more possible value of n in addition to 4 if the 1st statement has already brought us to get the value of n as 4
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Re: Is the nth root of n greater than the cube root of 3? [#permalink]
zxcvbnmas wrote:
Is the nth root of n greater than the cube root of 3?

1) The nth root of n is equal to the 4th root of 4
2) The nth root of n is equal to the square root of 2

We need to determine whether ^n√n > ^3√3.

Statement One Alone:

The nth root of n is equal to the 4th root of 4

Since ^n√n = ^4√4, the question becomes: is ^4√4 > ^3√3?

Let’s raise both sides to the 12th power:

(^4√4)^12 > (^3√3)^12 ?

4^3 > 3^4 ?

64 > 81 ?

We see that the answer is no. Statement one alone is sufficient to answer the question.

Statement Two Alone:

The nth root of n is equal to the square root of 2

Thus, we know that:

Since ^n√n = √2, the question becomes: is √2 > ^3√3?

Let’s raise both sides to the 6th power:

(√2)^6 > (^3√3)^6 ?

2^3 > 3^2 ?

8 > 9 ?

We see that the answer is no again. Statement two alone is also sufficient to answer the question.

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Re: Is the nth root of n greater than the cube root of 3? [#permalink]
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Re: Is the nth root of n greater than the cube root of 3? [#permalink]
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