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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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]
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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

Final Answer:

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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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. :x

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

TIA
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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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. :x

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: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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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. :x

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. :-D

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! :)

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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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Here is my take:

We are given that n is a positive integer and need to determine whether (1/10)^n < 0.01. We can convert 0.01 to a fraction and display the question as:

Is (1/10)^n < 1/100 ?

Is (1/10)^n < (1/10)^2 ?

Using the negative exponent rule, we can take the reciprocal of our bases and switch the signs of the exponents.

Is 10^-n < 10^-2 ?

Because the bases are now the same, we equate the exponents.

Is -n < -2 ?

Is n > 2 ?

Statement One Alone:

n > 2

We see that statement one directly answers the question. We can eliminate answer choices B, C, and E.

Statement Two Alone:

(1/10)^(n-1) < 0.1

We can simplify the inequality in statement two.

(1/10)^(n-1) < 0.1

(1/10)^(n-1) < (1/10)^1

Using the negative exponent rule, we can take the reciprocal of our bases and switch the signs of the exponents.

10^-(n-1) < 10^-1

The bases are now equal, so we can equate the exponents.

-(n – 1) < -1

n – 1 > 1

n > 2

We see that n is greater than 2. Statement two alone is sufficient to answer the question.

Answer: D
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
I had a somewhat similar approach for statement 2, but I landed up with an answer contradicting 1.

I considered 0.1 as 1/10

so then the equation is: n-1 < 1 (because the base is the same i.e. 1/10)

therefore n < 2...where am I going wrong?


Bunuel wrote:
pavanpuneet wrote:

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]
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Hi ameyaprabhu,

If you choose to 'convert' the numbers into decimals, then that's fine, but you still have to remember how the rules of math 'work'

Which number is bigger: (1/10)^1 or (1/10)^2? WHY is that the case?

Since we're working with an inequality, we're bound by the rule that one value MUST be bigger than the other. If you're ever unsure about an 'algebraic' approach that you're using, you can easily verify whether you're correct or not by TESTing VALUES.

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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
If the base is same we can say Exponents are equal.
If we rephrase the prompt question, we get:
(1/10)^n < 0.01
= (1/10)^n < (1/100)
= (1/10)^n < (1/10)^2
= n < 2 [as the bases are same, so exponents are equal)

what is wrong here? Bunuel

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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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Hi RashedVai,

When dealing with exponents, you have to be careful about your 'assumptions' - and ultimately you still have to remember how the rules of math 'work.' The general pattern that you describe is SOMETIMES true (and it's certainly true when raising positive integers that are GREATER THAN 1), but it's not always true...

What if the base is equal to "1"....? What's bigger: 1^1 or 1^2? Since the base is 1, those 2 values are the SAME.

When raising a POSITIVE FRACTION to different exponents, what's the relationship in the resulting numbers....?

Which number is bigger: (1/10)^1 or (1/10)^2? WHY is that the case (and which number is bigger?)?

Since we're working with an inequality, we're bound by the rule that one value MUST be bigger than the other. If you're ever unsure about an 'algebraic' approach that you're using, you can easily verify whether you're correct or not by TESTing VALUES.

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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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I am no longer active on GMAT Club.

Originally posted by dabral on 04 Feb 2021, 02:20.
Last edited by dabral on 18 Aug 2023, 18:00, edited 1 time in total.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
is (1/10)^n < 0.01
1/10^n < 1/10^2
10^n > 10^2
Is n > 2

Stmt 1. n > 2. Sufficient

Stmt 2 (1/10)^n-1 < 0.1
1/10 ^ n-1 < 1/10^1
10^n-1 > 10^1
n-1>1
n>2. Sufficient

Answer is D.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
So, first we simply the equation given in the question.
(1/10)^n <0.01?

which is (0.1)^n < 0.01?

This situation will only be possible when n >2.

So, let's find this out.

(a) n>2

We want n>2 and the option also says the same. So sufficient.

(b) (1/10)^n−1 < 0.1

FOr this statement to be true, n >2 coz only than this will be possible.

n=3

(0.1)^3-1 < 0.1
(0.1)^2 < 0.1
(0.01) < 0.1

So, sufficient.

D is the answer.
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If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
If the base is same we can say Exponents are equal.
If we rephrase the prompt question, we get:
(1/10)^n < 0.01
= (1/10)^n < (1/100)
= (1/10)^n < (1/10)^2
= n < 2 [as the bases are same, so exponents are equal)


What is the real deal and what are the conditions for this rule to work and not work? Could you please type out the rule with conditions avigutman
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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Elite097 wrote:
If the base is same we can say Exponents are equal.
If we rephrase the prompt question, we get:
(1/10)^n < 0.01
= (1/10)^n < (1/100)
= (1/10)^n < (1/10)^2
= n < 2 [as the bases are same, so exponents are equal)


What is the real deal and what are the conditions for this rule to work and not work? Could you please type out the rule with conditions avigutman


You have to think about what's happening to the value of the base when you start multiplying it by itself over and over again, Elite097. There are no "rules" here, just logic. If your base is (1/10), and you start multiplying it by (1/10) over and over again, you're going to lost 90% of your distance from zero each time you multiply (this is always true: any value that you multiply by (1/10) will lose 90% of its distance from zero).
Please let me know what follow-up questions you may have.
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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Re: If n is a positive integer, is (1/10)^n < 0.01? [#permalink]
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