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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
4
2
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
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
If n is an integer, is (0.1)^n greater than (10)^n? [#permalink]
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26 Oct 2015, 06:24
9
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.
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
Answer: Option E
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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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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
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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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?
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23 Jan 2018, 19:31