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Nak07
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If number N is randomly drawn from a set of all non-negative 20 Jan 2016, 12:27
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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
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If number N is randomly drawn from a set of all non-negative 20 Jan 2016, 12:38
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Nak07
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
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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?
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If number N is randomly drawn from a set of all non-negative 20 Jan 2016, 12:39
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Nak07
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
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Make sure to search for a question before you post.

Topics merged. Refer to the solution in this thread.
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If number N is randomly drawn from a set of all non-negative 01 Aug 2016, 21:59
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VeritasPrepKarishma
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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.
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How can n be 0 for 5n^3/8 to be an integer .please explain
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If number N is randomly drawn from a set of all non-negative 02 Aug 2016, 22:42
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sandeep211986
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DeeptiM
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
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If n = 0,

5*n^3 / 8 = 5 * 0^3 / 8 = 0/8 = 0 (an integer)
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If number N is randomly drawn from a set of all non-negative 27 May 2020, 21:36
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If 0 cubed =0 and 5(0)=0, then how does 0/8 = an integer?

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If number N is randomly drawn from a set of all non-negative 28 May 2020, 00:42
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If 0 cubed =0 and 5(0)=0, then how does 0/8 = an integer?

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Integer a is divisible by integer b means that a/b=integer. Since 0/8 = 0 = integer, then 0 is divisible by 8. In fact, Zero is divisible by every integer except zero itself: 0/integer=0.
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If number N is randomly drawn from a set of all non-negative 27 Apr 2025, 06:53
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