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

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

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24 Jul 2011, 00:40
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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?

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1/2

Last edited by Bunuel on 10 Jul 2013, 14:22, edited 2 times in total.
Renamed the topic, edited the question, added the OA and moved to PS forum.
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24 Jul 2011, 03:58
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?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5
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24 Jul 2011, 20:07
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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.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 31 May 2011 Posts: 88 Location: India Concentration: Finance, International Business GMAT Date: 12-07-2011 GPA: 3.22 WE: Information Technology (Computer Software) Followers: 1 Kudos [?]: 51 [0], given: 4 Re: Probability - Integer [#permalink] ### Show Tags 25 Jul 2011, 00:51 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. Ahh yes. I missed 0 in the first case Senior Manager Joined: 08 Nov 2010 Posts: 417 WE 1: Business Development Followers: 7 Kudos [?]: 105 [0], given: 161 Re: Probability - Integer [#permalink] ### Show Tags 25 Jul 2011, 22:54 If i can give my 2 cents tip. When you see the words non-negative you should immediately think about zero in your solution. Same way about non-positive of course. _________________ Manager Joined: 30 Sep 2009 Posts: 124 Followers: 0 Kudos [?]: 27 [0], given: 183 Re: Probability - Integer [#permalink] ### Show Tags 25 Jul 2011, 23:05 As per my understanding zero is neither postive nor negitive integer. why should zero be included in this question as the question is asking about the non-negative single-digit. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7125 Location: Pune, India Followers: 2134 Kudos [?]: 13650 [0], given: 222 Re: Probability - Integer [#permalink] ### Show Tags 26 Jul 2011, 00:16 abhi398 wrote: As per my understanding zero is neither postive nor negitive integer. why should zero be included in this question as the question is asking about the non-negative single-digit. Since 0 is neither positive nor negative, 'non negative' means 'positive and 0' _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199

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26 Jul 2011, 00:20
thanks karishma....

Never thought of that .....
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27 Jul 2011, 09:11
VeritasPrepKarishma wrote:
abhi398 wrote:
As per my understanding zero is neither postive nor negitive integer.

why should zero be included in this question as the question is asking about the non-negative single-digit.

Since 0 is neither positive nor negative, 'non negative' means 'positive and 0'

Absolutely! I've fell to this GMAT's one of the favorite traps so many times that I had this pinned on my wall (the real one. not FB's :D
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30 Jul 2011, 17:56
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.

The probability is not 1/2. You have assumed that the result would be an integer with n=0.
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30 Jul 2011, 19:20
0 is an integer.
When N=0, then 5(N^3)/8 = 0
So the answer will be 5/10 or 1/2
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30 Jul 2011, 19:26
Sudhanshuacharya 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?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5

Whenever the question says
5N^3/8 first evaluate the exponent and then the Multiply/Divide

So you only need to consider 5N^3/8 will always be (5N^3)/8 or 5(N^3/8) or (5/8)N^3...All are the same
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30 Jul 2011, 19:31
Sudhanshuacharya 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?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5

5N^3^3^3/8
Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right
So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3

Hope you got my point.
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31 Jul 2011, 21:10
krishp84 wrote:
Sudhanshuacharya 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?

is it (5N^3)/8 or 5N^(3/8)???

For the first case we need to check for the values of N for which 5(N/2)^3 is integer or N/2 is integer

therefore all even number satisfies this condition i.e. 2,4,6,8.
Number of terms = from 0 to 9. i.e. 10 numbers

Hence probability should be = 4/10 = 2/5

For the second case i guess the probability will be 1/5. as the only case when 5N^(3/8) is an integer when N is either 0 or 1. hence 2/10 = 1/5

5N^3^3^3/8
Now first calculate the exponents : 5N^3^3^3 from Right to Left the exponents and not Left to Right
So 5N^3^3^3/8 = 5N^3^27/8 and not 5N^27^3/8....In other words 3^27 <> 27^3

Hope you got my point.

Hi..I didnt get thi spart..can u help elaborate pls..
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01 Aug 2011, 08:33
ok - you have a question asking - what is the value of 2^3^4 ?
Is it 2^81 or 8^4 ?

It will be 2^81

This was easy...

Now let us say the question asked - what is the value of 2^3^4^2 ?
Will it be 2^3^16 or 8^4^2 ?

It will be 2^3^16

Takeaway - Always evaluate exponents from Right to Left if they are of the form - x^y^z^....
DeeptiM - Hope you got the point. Kudos are welcome if the post is useful.
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04 Aug 2011, 16:47
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'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8
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04 Aug 2011, 20:27
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pinchharmonic wrote:

how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8

Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).

So $$N^3/8$$ should be an integer. This means $$N^3$$ should be divisible by 8. But we have values for N, not $$N^3$$ so how do we figure what it means for N?
Let's say, N has 2 as a factor. Then $$N = 2a$$
$$N^3 = (2a)^3 = 8a^3$$
This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and $$N^3$$ is divisible by 8.

If we consider that $$N^3$$ has 2 as a factor and is greater than 8 (as you suggested above), is it essential that $$N^3$$ will be divisible by 8?
Consider $$N^3 = 10$$ (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.

Consider now, what would have happened if instead of 5, we had 4.
We would need $$4N^3/8$$ to be an integer i.e. $$N^3/2$$ to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it.
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05 Aug 2011, 09:59
VeritasPrepKarishma wrote:
pinchharmonic wrote:

how'd you determine in detail that as long as N has 2 as a factor it is divisible by 8? I'm assuming the 5 in front means nothing except that it is not divisible by 8, so that the next item in the product must be. I thought it must be that N^3 has 2 as a factor and is >= 8

Yes, 5 doesn't have any role to play except to tell you that it is not divisible by 8 and has no 2s in it. If instead of 5, we had 4, things would have been different (but let's not get into it right now).

So $$N^3/8$$ should be an integer. This means $$N^3$$ should be divisible by 8. But we have values for N, not $$N^3$$ so how do we figure what it means for N?
Let's say, N has 2 as a factor. Then $$N = 2a$$
$$N^3 = (2a)^3 = 8a^3$$
This number is divisible by 8 since it has 8 as a factor. Since N gets cubed, if it has a 2 in it, that becomes 8 and $$N^3$$ is divisible by 8.

If we consider that $$N^3$$ has 2 as a factor and is greater than 8 (as you suggested above), is it essential that $$N^3$$ will be divisible by 8?
Consider $$N^3 = 10$$ (has 2 as a factor and is >8) This is not divisible by 8 and for that matter, N is a not an integer in this case.

Consider now, what would have happened if instead of 5, we had 4.
We would need $$4N^3/8$$ to be an integer i.e. $$N^3/2$$ to be an integer. Yet again, we need N^3 to be even, so N should be even too i.e. it must have a 2 in it.

karishma, thanks.

So is there a way to make this factoring process generic? It seems like your thought process is to find the lowest factor of the denominator, which is 2, ie not 4 or 8. Then make sure you have that same factor in the numerator. Then the product of all factors in the numerator must be >= than the denominator. Meaning if in numerator there are other products that are not factors you just ignore them. like 5 below

(5x2^3)/8

and if you happeneted to choose 3

(5x3^3)/8 = (5x3x3x3)/8

since 5 and 3 are both not factors you still dont have the factor portion of the numerator >=8
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10 Jul 2013, 14:10
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.

what if N=8 or 4 rather than 2 ?
i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8
hence we have 3 fav conditions
..... now i got stuck here....

can not decide total cond.

m i going wrong way ? pls explain
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10 Jul 2013, 14:30
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jimmy0220 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.

what if N=8 or 4 rather than 2 ?
i think 8 should be a factor of n^3 means either N=8, or NxN=8 or NxNxN=8
hence we have 3 fav conditions
..... now i got stuck here....

can not decide total cond.

m i going wrong way ? pls explain

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 non-negative single-digit integers = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}, 10 elements.

Now, $$\frac{5N^3}{8}=\frac{5N^3}{2^3}$$ to be an integer N should be any even number. There are 5 even numbers in the set (0, 2, 4, 6, and 8), so the probability is 5/10=1/2.

Hope it's clear.
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Re: Probability - Integer   [#permalink] 10 Jul 2013, 14:30

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