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# If x is positive and not equal to 1, then the product of x^(1/n) for a

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Math Expert
Joined: 02 Sep 2009
Posts: 58339
If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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24 Apr 2015, 02:55
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Difficulty:

85% (hard)

Question Stats:

54% (02:25) correct 46% (02:31) wrong based on 267 sessions

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If x is positive and not equal to 1, then the product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is between

A. 1 and x^(1/6)
B. x^(1/6) and x^(1/3)
C. x^(1/3) and x^(1/2)
D. x^(1/2) and x^(2/3)
E. x^(2/3) and x^(5/6)

Kudos for a correct solution.

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Posts: 58339
Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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27 Apr 2015, 02:05
5
1
6
Bunuel wrote:
If x is positive and not equal to 1, then the product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is between

A. 1 and x^(1/6)
B. x^(1/6) and x^(1/3)
C. x^(1/3) and x^(1/2)
D. x^(1/2) and x^(2/3)
E. x^(2/3) and x^(5/6)

Kudos for a correct solution.

MANHATTAN GMAT OFFICIAL SOLUTION:

Let’s list out the product, using “dot-dot-dot” for missing parts in the middle.

Product = (x^(1/21)) (x^(1/22)) (x^(1/23))…(x^(1/30))

Since the bases are all the same, we just add the exponents to get the final exponent.

Exponent of product = 1/21 + 1/22 + 1/23 + … + 1/30 (ten terms in the sum)

Now, we are not asked to compute the exact value of this number (a brutal task that would require us to find a common multiple of every integer from 21 to 30!). Rather, we want to know what the product lies between.

One way to find simple upper and lower bounds is to replace all the different fractions with the same fraction that is definitely larger or smaller. Then compute these “fake” sums and see what you get.

Start at the small end. The real sum is definitely greater than this fake sum:

1/30 + 1/30 + 1/30 + … + 1/30

We’ve replaced 1/21, 1/22, etc. with 1/30 in each position. 1/30 is definitely less than 1/21, 1/22, etc. through 1/29, so the sum is definitely less.

Lower fake sum = 1/30 + 1/30 + 1/30 + … + 1/30 (ten terms) = 10/30 = 1/3

We have a good lower bound. Let’s set an upper bound by replacing every fraction in the real sum with 1/20, which is definitely larger than every fraction in the real sum.

Upper fake sum = 1/20 + 1/20 + 1/20 + … + 1/20 (ten terms) = 10/20 = 1/2

We now have bounds on either end. The exponent must be between 1/3 and 1/2, so the product in question must be between x^(1/3) and x^(1/2).

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Joined: 22 Jan 2014
Posts: 170
WE: Project Management (Computer Hardware)
Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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24 Apr 2015, 03:20
2
Bunuel wrote:
If x is positive and not equal to 1, then the product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is between

A. 1 and x^(1/6)
B. x^(1/6) and x^(1/3)
C. x^(1/3) and x^(1/2)
D. x^(1/2) and x^(2/3)
E. x^(2/3) and x^(5/6)

Kudos for a correct solution.

x^(1/21) + x^(1/22) + ... + x^(1/30)

S = 1/21 + 1/22 + ... + 1/30

or S > 1/21 + 1/21 ... + 1/21
S > 10/21
and S < 10/30

10/21 < S < 1/3

so C
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Illegitimi non carborundum.
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Joined: 15 May 2014
Posts: 62
Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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25 Apr 2015, 08:09
2
1
x is positive and not equal to 1
Product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is

= $$x^{(\frac{1}{21}+\frac{1}{29})+(\frac{1}{22}+\frac{1}{28})+(\frac{1}{23}+\frac{1}{27})+(\frac{1}{24}+\frac{1}{26})+\frac{1}{25}+\frac{1}{30}}$$

= $$x^{(\frac{1}{20+1}+\frac{1}{20+9})+(\frac{1}{20+2}+\frac{1}{20+8})+(\frac{1}{20+3}+\frac{1}{20+7})+(\frac{1}{20+4}+\frac{1}{20+6})+\frac{1}{25}+\frac{1}{30}}$$

= $$x^{\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{12}+\frac{1}{25}+\frac{1}{30}}$$

= $$x^{\frac{61}{150}}$$

= $$x^{\frac{2}{5}}$$

$$x^{\frac{1}{3}} < x^{\frac{2}{5}} < x^{\frac{1}{2}}$$

Manager
Joined: 24 Jan 2015
Posts: 68
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WE: Consulting (Pharmaceuticals and Biotech)
Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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25 Apr 2015, 22:43
3
1
The product of x^(1/n) for all positive integers n such that 21 ≤ n ≤ 30 is

= $$x ^{1/21} * x ^{1/22} * x ^{1/23} * x ^{1/24} * x ^{1/25} * x ^{1/26} * x ^{1/27} * x ^{1/28} * x ^{1/29} * x ^{1/30}$$

=$$x^ {(\frac{1}{21} + \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} + \frac{1}{28} + \frac{1}{29} + \frac{1}{30})}$$

==> S = $$\frac{1}{21} + \frac{1}{22} + \frac{1}{23} + \frac{1}{24} + \frac{1}{25} + \frac{1}{26} + \frac{1}{27} + \frac{1}{28} + \frac{1}{29} + \frac{1}{30}$$

In order to find where the sum of exponents (S) lie, We can use the Max/Min concept.

To find the minimum possible value of the S, Lets assume all ten values to be $$\frac{1}{30}$$ [ Fact : 21 < 30 and $$\frac{1}{21} >\frac{1}{30}$$]

S = $$\frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30} + \frac{1}{30}$$

= $$\frac{10}{30}$$ = $$\frac{1}{3}$$ ==> $$x^S$$ > $$x^{\frac{1}{3}}$$

To find the maximum possible value of the S, Lets assume all ten values to be $$\frac{1}{21}$$ [ Fact : 21 < 30 and $$\frac{1}{21} >\frac{1}{30}$$]

S = $$\frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21} + \frac{1}{21}$$

= $$\frac{10}{21}$$ = $$\frac{1}{2}$$ (approximation) ==> $$x^S$$ < $$x^{\frac{1}{2}}$$

So $$x^{\frac{1}{3}}$$ < $$x^S$$ <$$x^{\frac{1}{2}}$$

Manager
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Location: United States
Schools: Duke '20 (D)
GMAT 1: 600 Q48 V27
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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27 Feb 2016, 21:39
thefibonacci wrote:

x^(1/21) + x^(1/22) + ... + x^(1/30)

S = 1/21 + 1/22 + ... + 1/30

or S > 1/21 + 1/21 ... + 1/21
S > 10/21
and S < 10/30

10/21 < S < 1/3

so C

Should not it be the other way around? 1/3 < S < 1/2
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a  [#permalink]

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14 Jun 2019, 02:48
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Re: If x is positive and not equal to 1, then the product of x^(1/n) for a   [#permalink] 14 Jun 2019, 02:48
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