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

Ahh yes. I missed 0 in the first case

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.

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'

thanks karishma....

Never thought of that .....

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

The probability is not 1/2. You have assumed that the result would be an integer with n=0.

0 is an integer.
When N=0, then 5(N^3)/8 = 0
So the answer will be 5/10 or 1/2

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

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..

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.

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

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.