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Re: Probability - Integer [#permalink]
24 Jul 2011, 20:07
1
This post received KUDOS
Expert's post
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. _________________
Re: Probability - Integer [#permalink]
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.
Re: Probability - Integer [#permalink]
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. _________________
Re: Probability - Integer [#permalink]
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 _________________
Re: Probability - Integer [#permalink]
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. _________________
ARISE AWAKE AND REST NOT UNTIL THE GOAL IS ACHIEVED
Re: Probability - Integer [#permalink]
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
This issue would have occurred if the question had asked 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
Re: Probability - Integer [#permalink]
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
This issue would have occurred if the question had asked 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..
Re: Probability - Integer [#permalink]
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. _________________
Re: Probability - Integer [#permalink]
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
Re: Probability - Integer [#permalink]
04 Aug 2011, 20:27
1
This post received KUDOS
Expert's post
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. _________________
Re: Probability - Integer [#permalink]
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
Re: Probability - Integer [#permalink]
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....
Re: Probability - Integer [#permalink]
10 Jul 2013, 14:30
1
This post received KUDOS
Expert's post
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.
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