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# If n is a positive integer, is (1/10)^n < 0.01?

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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  16 Mar 2012, 09:04
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If $$n$$ is a positive integer, is $$\frac{1}{10}^n < 0.01?$$

(1) $$n > 2$$

(2) $$\frac{1}{10}^{n-1} < 0.1$$

First post here, so I hope I got the format right. I understand the OG explanation to this problem, but I tried taking a slightly alternate route and am coming up with the wrong answer. It's a rather simple one, but hope someone can shed some light on to where I've gone wrong. I included the OG explanation as well as my own in the spoiler.

[Reveal] Spoiler:
OG Explanation: Manipulate both sides to be expressed as powers of 10.

$$\frac{1}{10}^n < 0.01$$

$$(10^{-1})^n < 10^{-2}$$

$$10^{-n} < 10^{-2}$$

$$n > 2$$

1) $$n > 2$$. SUFFICIENT

2) $$\frac{1}{10}^{n-1} < 0.1$$

$$(10^{-1})^{n-1} < 10^{-1}$$

$$10^{-n+1} < 10^{-1}$$

$$-n+1 < -1$$

$$n > 2$$
SUFFICIENT

My slightly modified solution for statement 2 was to first manipulate the 0.1 on the right side of the inequality to become a fraction and to leave the left side as a fraction (my first instinct is to see that 0.01 is the same as 1/10). You would have:

$$\frac{1}{10}^{n-1} < \frac{1}{10}^1$$

$$n-1 < 1$$

$$n < 2$$

As you can see, I get an opposite answer. I know this is super simple, but where am I going wrong?
[Reveal] Spoiler: OA
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  16 Mar 2012, 09:32
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tkaelle wrote:
First post here, so I hope I got the format right. I understand the OG explanation to this problem, but I tried taking a slightly alternate route and am coming up with the wrong answer. It's a rather simple one, but hope someone can shed some light on to where I've gone wrong. I included the OG explanation as well as my own in the spoiler.

If $$n$$ is a positive integer, is $$\frac{1}{10}^n < 0.01?$$

1) $$n > 2$$

2) $$\frac{1}{10}^{n-1} < 0.1$$

OG Explanation: Manipulate both sides to be expressed as powers of 10.

$$\frac{1}{10}^n < 0.01$$

$$(10^{-1})^n < 10^{-2}$$

$$10^{-n} < 10^{-2}$$

$$n > 2$$

1) $$n > 2$$. SUFFICIENT

2) $$\frac{1}{10}^{n-1} < 0.1$$

$$(10^{-1})^{n-1} < 10^{-1}$$

$$10^{-n+1} < 10^{-1}$$

$$-n+1 < -1$$

$$n > 2$$
SUFFICIENT

My slightly modified solution for statement 2 was to first manipulate the 0.1 on the right side of the inequality to become a fraction and to leave the left side as a fraction (my first instinct is to see that 0.01 is the same as 1/10). You would have:

$$\frac{1}{10}^{n-1} < \frac{1}{10}^1$$

$$n-1 < 1$$

$$n < 2$$

As you can see, I get an opposite answer. I know this is super simple, but where am I going wrong?

From $$(\frac{1}{10})^{n-1} < (\frac{1}{10})^1$$ since the base, 1/10, is a fraction in the range (0,1) then it should be $$n-1>1$$. For example: $$(\frac{1}{10})^{2} < (\frac{1}{10})^1$$ --> $$2>1$$.

Hope it helps.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  16 Mar 2012, 16:07
Thanks for the quick response. I knew that you had to switch the inequality sign if you were multiplying or dividing by a negative value, but the same is also true when working with a value 0 < x < 1?

I'm a little confused because in this case, we're not doing any multiplying or dividing to the equation, but just ignoring the common base and comparing their exponents. Not quite sure why we change the sign
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  09 Aug 2012, 00:08
Hi Bunuel,

I actually landed up doing in the exact same way as tkaelle did. I understand what you mentioned .. however, is there any rule because I am sure I might up land up doing the same in the exam if I do not understand why we switch the signs/ or why cant we manipulate the fraction 1/10 and continue?

Thank you.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  09 Aug 2012, 00:34
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pavanpuneet wrote:
Hi Bunuel,

I actually landed up doing in the exact same way as tkaelle did. I understand what you mentioned .. however, is there any rule because I am sure I might up land up doing the same in the exam if I do not understand why we switch the signs/ or why cant we manipulate the fraction 1/10 and continue?

Thank you.

If you have a problem with fractions in powers, then manipulate to get rid of the them:
$$(\frac{1}{10})^{n-1} < \frac{1}{10}$$ --> $$\frac{1}{10^{n-1}}< \frac{1}{10}$$ --> cross-multiply: $$10<10^{n-1}$$ --> $$1<n-1$$ --> $$n>2$$.

OR:
$$(\frac{1}{10})^{n-1} < \frac{1}{10}$$ --> $$(10^{-1})^{n-1}<10^{-1}$$ --> $$10^{1-n}<10^{-1}$$ --> $$1-n<-1$$ --> $$n>2$$.

Hope it helps.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  15 Mar 2013, 08:59
given 10^(-n)<10^(-2)
n>2 ?

1) n>2 suff

2) 10^(1-n)<10^(-1)
1-n<-1
n>2 suff

ans is D
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  18 Mar 2013, 12:13
Great explanation Bunuel, thanks.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  21 Dec 2013, 23:18
Hi !

I have an issue with the 2nd equation :

(1/10)^n-1 < 0.1

What I would do is (1/10)^n-1 < 1/10)

n-1 < 1

My qn. is why is it necassary to change the sign to make n> 2??
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  22 Dec 2013, 03:48
Expert's post
mba1010 wrote:
Hi !

I have an issue with the 2nd equation :

(1/10)^n-1 < 0.1

What I would do is (1/10)^n-1 < 1/10)

n-1 < 1

My qn. is why is it necassary to change the sign to make n> 2??

Please check here: if-n-is-a-positive-integer-is-1-10-n-129176.html#p1059737 and here: if-n-is-a-positive-integer-is-1-10-n-129176.html#p1111563

Hope it helps.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  13 Jan 2015, 19:40
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  13 Jan 2015, 19:52
Expert's post
Hi All,

This DS question is essentially about arithmetic rules (decimals and exponents). It can be solved conceptually or you solve it by TESTing VALUES.

We're told that N is a POSITIVE INTEGER. We're asked (1/10)^N < 0.01 This is a YES/NO question

Fact 1: N > 2

IF....
N = 3
(1/10)^3 = 1/1000 = .001 and the answer to the question is YES

N = 4
(1/10)^4 = 1/10,000 = .0001 and the answer to the question is YES

As N gets bigger, the resulting decimal point gets smaller and the answer to the question is ALWAYS YES.
Fact 1 is SUFFICIENT

Fact 2: (1/10)^(N-1) < 0.1

Here, we have an interesting "restriction" - we can only use certain values for N....

IF....
N = 1
(1/10)^0 = 1 BUT this does NOT fit with the given information in Fact 2, so we CANNOT use this TEST CASE.

IF....
N = 2
(1/10)^1 = .1 BUT this also does NOT fit with the given information in Fact 2 EITHER.

This means that N CANNOT be 1 or 2. Since it has to be a POSITIVE INTEGER and we already have proof of what happens when N > 2 (the answer to the question is ALWAYS YES), we can stop working.
Fact 2 is SUFFICIENT

[Reveal] Spoiler:
D

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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]  27 Apr 2015, 23:38
The question says that n is any positive number.

So when I use n=2 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold,

Similarly when i use n=1, then 1<0.1 again a different answer
and when n=3, then it satisfies.

So how come answer is D??

I know i am doing something wrong.

Would hope to have a reasoning which can clarify where i am going wrong.

TIA
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here is an OG problem in which i have a real doubt in [#permalink]  27 Apr 2015, 23:51
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believer700 wrote:
The question says that n is any positive number.

So when I use n=1 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold.

So how come answer is D??

I know i am doing something wrong.

Would hope to have a reasoning which can clarify where i am going wrong.

TIA

Hello believer700

You should use information not only from task but from statement too.
When you use $$n = 1$$ you break condition of second statement $$(\frac{1}{10})^{(n-1)} <0.1$$ and this mean that you can't use such value for $$n$$
So you should combine conditions from task and statement.

This mean that you can take only those values for $$n$$ that will satisfy condition of task and statement.

So when $$n = 1$$ then it will be $$(\frac{1}{10})^{(1-1)} <0.1$$; $$\frac{1}{1}<0.1$$; not possible

when $$n = 2$$ then it will be $$(\frac{1}{10})^{(2-1)} <0.1$$; $$\frac{1}{10}<0.1$$; not possible

when $$n = 3$$ then it will be $$(\frac{1}{10})^{(3-1)} <0.1$$; $$\frac{1}{10}<0.1$$; possible
So $$n$$ should be at least $$3$$
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Re: here is an OG problem in which i have a real doubt in [#permalink]  28 Apr 2015, 00:04
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believer700 wrote:
The question says that n is any positive number.

So when I use n=2 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold,

Similarly when i use n=1, then 1<0.1 again a different answer
and when n=3, then it satisfies.

So how come answer is D??

I know i am doing something wrong.

Would hope to have a reasoning which can clarify where i am going wrong.

TIA

Dear believer700

It's good that you're being inquisitive about your mistakes. Analyzing the mistake you make once ensures that you don't ever make it again.

The part that you did wrong here that you considered only one piece of information about n: that n is a positive integer (this is given in the ques statement)

So, you thought that you could take n = 1 or 2. And then, got confused when these values of n did not satisfy the information given in Statements 1 and 2.

The correct way of talking about n is:

n is a positive integer such that (1/10)^(n-1) <0. (info in St. 2)

OR

n is a positive integer such that n > 2 (info in St. 1)

So, the possible values of n will be those that satisfy:

i) The information given in question statement AND
ii) The information given in Statements 1 and 2

I hope this clarifies your doubt!

- Japinder
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Re: here is an OG problem in which i have a real doubt in [#permalink]  28 Apr 2015, 01:57
Harley1980 wrote:
believer700 wrote:
The question says that n is any positive number.

So when I use n=1 in the second statement then both the equations become equal and the equality fails i.e 0.1<0.1 doesn't hold.

So how come answer is D??

I know i am doing something wrong.

Would hope to have a reasoning which can clarify where i am going wrong.

TIA

Hello believer700

You should use information not only from task but from statement too.
When you use $$n = 1$$ you break condition of second statement $$(\frac{1}{10})^{(n-1)} <0.1$$ and this mean that you can't use such value for $$n$$
So you should combine conditions from task and statement.

This mean that you can take only those values for $$n$$ that will satisfy condition of task and statement.

So when $$n = 1$$ then it will be $$(\frac{1}{10})^{(1-1)} <0.1$$; $$\frac{1}{1}<0.1$$; not possible

when $$n = 2$$ then it will be $$(\frac{1}{10})^{(2-1)} <0.1$$; $$\frac{1}{10}<0.1$$; not possible

when $$n = 3$$ then it will be $$(\frac{1}{10})^{(3-1)} <0.1$$; $$\frac{1}{10}<0.1$$; possible
So $$n$$ should be at least $$3$$

Thanks for the Explanation!!
Did not concentrate enough!
Re: here is an OG problem in which i have a real doubt in   [#permalink] 28 Apr 2015, 01:57
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