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# If number N is randomly drawn from a set of all non-negative

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Intern
Joined: 09 Jan 2016
Posts: 3
Concentration: Technology, Finance
GMAT 1: 700 Q49 V37
If number N is randomly drawn from  [#permalink]

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20 Jan 2016, 12:27
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the
probability that 5N3/8 is an integer?

I am not aware of the options for the question
CEO
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
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Re: If number N is randomly drawn from  [#permalink]

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20 Jan 2016, 12:38
Nak07 wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the
probability that 5N3/8 is an integer?

I am not aware of the options for the question

If 5N3 is to be treated as a 3 digit number, then any number ending in 3 will never be divisible by 8. So the probability should be 0. What is the source of the question?
CEO
Joined: 20 Mar 2014
Posts: 2624
Concentration: Finance, Strategy
Schools: Kellogg '18 (M)
GMAT 1: 750 Q49 V44
GPA: 3.7
WE: Engineering (Aerospace and Defense)
Re: If number N is randomly drawn from a set of all non-negative  [#permalink]

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20 Jan 2016, 12:39
Nak07 wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the
probability that 5N3/8 is an integer?

I am not aware of the options for the question

Make sure to search for a question before you post.

Topics merged. Refer to the solution in this thread.
Intern
Joined: 04 Nov 2015
Posts: 37
Re: If number N is randomly drawn from a set of all non-negative  [#permalink]

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01 Aug 2016, 21:59
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For $$5N^3/8$$ to be an integer, $$5N^3$$ should be completely divisible by 8. This will happen only if $$N^3$$ is divisible by 8 i.e. if N has 2 as a factor. So for $$5N^3/8$$ to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that $$5N^3/8$$ is an integer is 1/2.

How can n be 0 for 5n^3/8 to be an integer .please explain
Veritas Prep GMAT Instructor
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Re: If number N is randomly drawn from a set of all non-negative  [#permalink]

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02 Aug 2016, 22:42
sandeep211986 wrote:
VeritasPrepKarishma wrote:
DeeptiM wrote:
If number N is randomly drawn from a set of all non-negative single-digit integers, what is the probability that 5N^3/8 is an Integer?

Set of all non-negative single digit integers - 0 to 9 i.e. 10 numbers

For $$5N^3/8$$ to be an integer, $$5N^3$$ should be completely divisible by 8. This will happen only if $$N^3$$ is divisible by 8 i.e. if N has 2 as a factor. So for $$5N^3/8$$ to be an integer, N should be even.

Out of the 10 numbers 0 - 9, 5 are even and 5 are odd.
Probability that $$5N^3/8$$ is an integer is 1/2.

How can n be 0 for 5n^3/8 to be an integer .please explain

If n = 0,

5*n^3 / 8 = 5 * 0^3 / 8 = 0/8 = 0 (an integer)
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Karishma
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Re: If number N is randomly drawn from a set of all non-negative  [#permalink]

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13 Apr 2019, 15:27
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Re: If number N is randomly drawn from a set of all non-negative   [#permalink] 13 Apr 2019, 15:27

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# If number N is randomly drawn from a set of all non-negative

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