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# If n is an integer, is (0.1)^n greater than (10)^n?

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If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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Updated on: 21 Jul 2018, 14:19
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If n is an integer, is $$(0.1)^n$$ greater than $$(10)^n$$?

(1) $$n > -10$$

(2) $$n < 10$$

Kudos for a correct solution.

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Originally posted by Bunuel on 25 Oct 2015, 09:01.
Last edited by carcass on 21 Jul 2018, 14:19, edited 2 times in total.
Edited by Carcass
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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26 Oct 2015, 06:24
15
3
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Question : is (0.1)^n > (10)^n?
Question : is (1/10)^n > (10)^n?
Question : is (10)^{-n} > (10)^n?
Question : is (-n) > n?
Question : is (2n) < 0?

Question : is n < 0?

Statement 1: n > −10
n may be Negative or Positive. Hence
NOT SUFFICIENT

Statement 2: n < 10
n may be Negative or Positive. Hence
NOT SUFFICIENT

Combining the two statements
even after combining the two
-10 < n < 10
i.e. n may be Negative or Positive. Hence
NOT SUFFICIENT

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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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25 Oct 2015, 09:33
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1
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Is (0.1)^n > (10)^n ---> $$10^{-n} > 10^n$$ ---> $$10^{2n} < 1$$ ---> $$10^{2n} < 10^0$$ ---> 2n < 0 ---> is n<0?

Per statement 1, n>-10, yes for n=-5 but no for n=5. Not sufficient.

Per statement 2, n<10, yes for n=-5 but no for n=5. Not sufficient.

Combining, the above 2 cases still apply giving you an ambiguous answer.

Thus, E is the correct answer.
##### General Discussion
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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25 Oct 2015, 16:29
6
4
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Target question: Is (0.1)^n greater than (10)^n?

REPHRASED target question: Is (1/10)^n greater than (10)^n?

Statement 1: n > −10
This statement doesn't FEEL sufficient, so I'll TEST some values.
There are several values of n that satisfy statement 1. Here are two:
Case a: n = 1, in which case (1/10)^n is NOT greater than (10)^n
Case b: n = -1, in which case (1/10)^-1 = 10 and 10^-1 = 1/10. Here, (1/10)^n IS greater than (10)^n
Since we cannot answer the target question with certainty, statement 1 is NOT SUFFICIENT

Aside: For more on this idea of plugging in values when a statement doesn't feel sufficient, you can read my article: http://www.gmatprepnow.com/articles/dat ... lug-values

Statement 2: n < 10
There are several values of n that satisfy statement 2. Here are two:
Case a: n = 1, in which case (1/10)^n is NOT greater than (10)^n
Case b: n = -1, in which case (1/10)^-1 = 10 and 10^-1 = 1/10. Here, (1/10)^n IS greater than (10)^n
Since we cannot answer the target question with certainty, statement 2 is NOT SUFFICIENT

IMPORTANT - Notice that I tested the SAME VALUES for both statements. This means that, the STATEMENTS COMBINED are also NOT SUFFICIENT

Cheers,
Brent
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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26 Oct 2015, 04:41
2
1
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Sol. (10)^-n > (10)^n or -n>n or n<0 ?
1) n > -10 Not sufficient
2) n < 10 Not sufficient

1) + 2) -10 < n <10 Not sufficient

Hence E
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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14 Jan 2017, 03:49
1
Please note that for n=0, both the expressions will be equal to 1.

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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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22 Jan 2017, 09:21
Hello All,

Got tricked by the question, finally got it

S-1) If n = -9 then Yes as (0.1)^-9 > 10^-9. But Not true of n = 9. Not Sufficient as we dont know value of n.
S-2) Same info as S-1 in different way. So If n = -9 Yes. n = 9 its not sufficient.
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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22 Jan 2017, 17:20
Would one way to solve this question to log expressions so you end up with n*log(0.1) and n*log(10) and then trying range of values?
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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30 Jul 2017, 01:04
Engr2012 wrote:
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Is (0.1)^n > (10)^n ---> $$10^{-n} > 10^n$$ ---> $$10^{2n} < 1$$ ---> $$10^{2n} < 10^0$$ ---> 2n < 0 ---> is n<0?

Per statement 1, n>-10, yes for n=-5 but no for n=5. Not sufficient.

Per statement 2, n<10, yes for n=-5 but no for n=5. Not sufficient.

Combining, the above 2 cases still apply giving you an ambiguous answer.

Thus, E is the correct answer.

how do you get 10^(2n) <1 from $$10^{-n} > 10^n$$
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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30 Jul 2017, 08:03
Engr2012 wrote:
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Is (0.1)^n > (10)^n ---> $$10^{-n} > 10^n$$ ---> $$10^{2n} < 1$$ ---> $$10^{2n} < 10^0$$ ---> 2n < 0 ---> is n<0?

Per statement 1, n>-10, yes for n=-5 but no for n=5. Not sufficient.

Per statement 2, n<10, yes for n=-5 but no for n=5. Not sufficient.

Combining, the above 2 cases still apply giving you an ambiguous answer.

Thus, E is the correct answer.

How do you get $$10^{2n} < 1$$ from $$10^{-n} > 10^n$$
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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30 Jul 2017, 08:13
1
pclawong wrote:
Engr2012 wrote:
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Is (0.1)^n > (10)^n ---> $$10^{-n} > 10^n$$ ---> $$10^{2n} < 1$$ ---> $$10^{2n} < 10^0$$ ---> 2n < 0 ---> is n<0?

Per statement 1, n>-10, yes for n=-5 but no for n=5. Not sufficient.

Per statement 2, n<10, yes for n=-5 but no for n=5. Not sufficient.

Combining, the above 2 cases still apply giving you an ambiguous answer.

Thus, E is the correct answer.

How do you get $$10^{2n} < 1$$ from $$10^{-n} > 10^n$$

$$10^{-n} > 10^n$$

Divide both sides by $$10^{-n}$$

$$1>10^n/10^{-n}$$ -->$$1>10^{2n}$$
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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30 Jul 2017, 08:41
ramyanag wrote:
pclawong wrote:
Engr2012 wrote:
[quote="Bunuel"]If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Is (0.1)^n > (10)^n ---> $$10^{-n} > 10^n$$ ---> $$10^{2n} < 1$$ ---> $$10^{2n} < 10^0$$ ---> 2n < 0 ---> is n<0?

Per statement 1, n>-10, yes for n=-5 but no for n=5. Not sufficient.

Per statement 2, n<10, yes for n=-5 but no for n=5. Not sufficient.

Combining, the above 2 cases still apply giving you an ambiguous answer.

Thus, E is the correct answer.

How do you get $$10^{2n} < 1$$ from $$10^{-n} > 10^n$$

$$10^{-n} > 10^n$$

Divide both sides by $$10^{-n}$$

$$1>10^n/10^{-n}$$ -->$$1>10^{2n}$$[/quote]

Thank you so much! That's clear

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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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02 Aug 2017, 13:08
1
Top Contributor
1
Bunuel wrote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Kudos for a correct solution.

Here's another approach:

Target question: Is (0.1)^n > (10)^n?
This is a good candidate for rephrasing the target question.

Since (0.1)^n is always POSITIVE, we can safely divide both sides of the inequality by (0.1)^n to get: 1 > [(10)^n]/[(0.1)^n]
There's a nice rule that says (a^n)/(b^n) = (a/b)^n
When we apply this rule to the right side of the inequality, we get: 1 > (10/0.1)^n
Simplify to get: Is 1 > 100^n?
Notice that, when n = 0, then 100^n = 100^0 = 1
So, when n > 0, then 100^n > 1, and when n < 0, then 100^n < 1
So, we can REPHRASE the target question as....
REPHRASED target question: Is n < 0?

Statement 1: n > -10
There are several values of n that satisfy statement 1. Here are two:
Case a: n = -9, in which case n < 0
Case b: n = 2, in which case n > 0
Since we cannot answer the REPHRASED target question with certainty, statement 1 is NOT SUFFICIENT

Statement 2: n < 10
There are several values of n that satisfy statement 1. Here are two:
Case a: n = -9, in which case n < 0
Case b: n = 2, in which case n > 0
Since we cannot answer the REPHRASED target question with certainty, statement 2 is NOT SUFFICIENT

Statements 1 and 2 combined
IMPORTANT: Notice that I was able to use the same counter-examples to show that each statement ALONE is not sufficient. So, the same counter-examples will satisfy the two statements COMBINED.
Since we cannot answer the REPHRASED target question with certainty, the combined statements are NOT SUFFICIENT

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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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08 Aug 2017, 03:58
Bunuel can we play around with the statement in question like this?
I mean to say:

Its given that is 10^-n>10^n

So can we divide/subtract/multiple/add on both sides of an expression that is to be proved?

When should we not do it?

Thanks
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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08 Aug 2017, 04:02
1
Shiv2016 wrote:
Bunuel can we play around with the statement in question like this?
I mean to say:

Its given that is 10^-n>10^n

So can we divide/subtract/multiple/add on both sides of an expression that is to be proved?

When should we not do it?

Thanks

How to manipulate inequalities (adding, subtracting, squaring etc.).
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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23 Jan 2018, 17:06
Bunuel chetan2u niks18 amanvermagmat

Quote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Why is it required for n to be an integer in Q stem?
How does one solve 1^0.5 or more specifically why does n
have to be an integer for 1^n to be 1?
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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23 Jan 2018, 19:31
Bunuel chetan2u niks18 amanvermagmat

Quote:
If n is an integer, is (0.1)^n greater than (10)^n?

(1) n > −10
(2) n < 10

Why is it required for n to be an integer in Q stem?
How does one solve 1^0.5 or more specifically why does n
have to be an integer for 1^n to be 1?

1^0.5=1^(1/2) which is square root of 1.
Mathematically it is possible to solve for non integer powers but then the question will become complicated and you might have to use calculator or log function.
Hence for simplicity it is mentioned that n is an integer. That’s my take on this question

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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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22 Aug 2018, 01:20
Can we divide both sides by 10^(-n)( to simplify to question if n<0) if we dont know whether n is positive or negative in inequality. As we dont know what is n and hence the sign may flip. Please clear this doubt.
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If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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22 Aug 2018, 01:52
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brains wrote:
Can we divide both sides by 10^(-n)( to simplify to question if n<0) if we dont know whether n is positive or negative in inequality. As we dont know what is n and hence the sign may flip. Please clear this doubt.

Hi brains

$$10^{-n}$$ will always be positive because a positive number that is 10 is being raised to some power. irrespective of the value of the power, the resulting number will always be positive

for e.g $$10^{-2}=\frac{1}{10^2}=0.01>0$$

similarly if $$10^2=100>0$$
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Re: If n is an integer, is (0.1)^n greater than (10)^n?  [#permalink]

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