Author 
Message 
TAGS:

Hide Tags

Senior Manager
Joined: 21 Oct 2013
Posts: 429

If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
18 Mar 2014, 16:04
Question Stats:
59% (01:29) correct 41% (01:21) wrong based on 517 sessions
HideShow timer Statistics
If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32? A. 1/3 B. 1/2 C. 5/7 D. 10/21 E. 15/28 This is good question.
Official Answer and Stats are available only to registered users. Register/ Login.




Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
18 Mar 2014, 21:32
goodyear2013 wrote: If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32? 1/3 1/2 5/7 10/21 15/28
This is good question. 64 = 2^6 So there are 7 factors of 64: 1, 2, 4, 8, 16, 32, 64 For the sum of any two factors selected randomly to be more than 32, we need to select at least one of 32 and 64. If two of the first 5 factors are selected, the sum will always be less than 32. So probability of sum being less than 32 = 5C2/7C2 = 10/21 Also note something peculiar about the powers of 2: Sum of all factors less than 32 is 31. Sum of all factors less than 64 is 63 and so on... Figure out a similar pattern for powers of 3.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!




Senior Manager
Status: Math is psychological
Joined: 07 Apr 2014
Posts: 421
Location: Netherlands
GMAT Date: 02112015
WE: Psychology and Counseling (Other)

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
05 Mar 2015, 09:27
We can find the factors using factor pairs:
1....64 2....32 3..... 4....16 5.... 6.... 7.... 8....8
We can see that there are 7 factors of 64 (which we can also test after doing the prime factorization of 64, resulting in 2^6, which leads to 6+1=7 factors).
Then, as Karishma pointed out, we select 2 numbers out of 7 and leave out 5: 7!/2!5! = 21.
Alternatively, if this doesn't come up as an idea, we can rank the factors and count the possibilities ourselves, since there are only 7 factors:
1,2,4,8,16,32,64
32 and 64 cannot be included, as adding any number to those would result in a number more than 32. 1 can be added to 2,4,8,16 2 can be added to 4,8,16 4 can be added to 8,16 8 can be added to 16.
Counting the possibilities, we find that they are 10. Using the same way we can find the total possible sums of 2 numbers as 21.



Current Student
Joined: 12 Aug 2015
Posts: 287
Concentration: General Management, Operations
GMAT 1: 640 Q40 V37 GMAT 2: 650 Q43 V36 GMAT 3: 600 Q47 V27
GPA: 3.3
WE: Management Consulting (Consulting)

If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
16 Sep 2015, 04:27
counting approach: A. 5 step approach so there are 7 distinct divisors out of which only 2 (32 and 64) contradict the requirements (>32), i.e. when they get picked out either both or just one of them + some other divisor, their sum exceeds 32 (1) 2/7 * 5/6  we pick either 32 or 64 first and some other remaining divisor out of the nonconflicting pool (2) 5/7 * 2/6  same as above but the nonconflicting one comes first (3) 2/7 * 1/6  we pick either of the conflicting divisors and then pick the other one out of the remaining 6 divisors. (4) sum up the probabilities above because the events are independent = 11/21 is the probability we exceed 32 (5) 1  11/21 = 10/21 what we were asked to find. B. or 1 step approach (1) 5/7 * 4/6 = 10/21  we pick only the 5 nonconflicting divisors and see the probability
_________________
KUDO me plenty



Manager
Joined: 07 Dec 2009
Posts: 97
GMAT Date: 12032014

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
01 Nov 2015, 02:29
Can some one Please help me on this one...
Probability of sum < 32 = 1 (Probability of sum > 32)
Probability of sum > 32 = 2/7
My logic is that out of (1,2,4,8,16,32,64) if I were to pick 32 or 64 then my sum would be greater than 32 irrespective of what the other number is. what is the probability of picking (32,another number ) = P(picking 32) * P(any other number) = (1/7 * 1) what is the probability of picking (64,another number ) = P(picking 64) * P(any other number) = (1/7 * 1) Total probability = 2/7 Hence my answer is 5/7 , which is obviously wrong. I just cant figure where I'm going wrong.
Anyone ?
Cheers



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
02 Nov 2015, 22:11
bhatiavai wrote: Can some one Please help me on this one...
Probability of sum < 32 = 1 (Probability of sum > 32)
Probability of sum > 32 = 2/7
My logic is that out of (1,2,4,8,16,32,64) if I were to pick 32 or 64 then my sum would be greater than 32 irrespective of what the other number is. what is the probability of picking (32,another number ) = P(picking 32) * P(any other number) = (1/7 * 1) what is the probability of picking (64,another number ) = P(picking 64) * P(any other number) = (1/7 * 1) Total probability = 2/7 Hence my answer is 5/7 , which is obviously wrong. I just cant figure where I'm going wrong.
Anyone ?
Cheers Here are the issues with this method: 1. You have accounted for (32, 4) but not for (4, 32) 2. You have double counted (32, 64). You have counted it in both cases. Instead do this: Choose 32 with probability 1/7 and then any one of the 6 numbers with probability 1 > 1/7 Multiply this by 2 to select the other number first and then 32 > (1/7) * 2 Do the exact same thing for 64 so you get (1/7) * 2 * 2 Now you have counted (32, 64) and (64, 32) in both cases so subtract it out. Probability of choosing 32 and then 64 > (1/7)*(1/6) Multiply by 2 to account for 64 and then 32 > (1/7)*(1/6)*2 Finally you get > (1/7)*2*2  (1/7)*(1/6)*2 = 11/21 Probability of sum < 32 = 1 (Probability of sum > 32) Probability of sum < 32 = 1 11/21 = 10/21 Note that when you have to pick two distinct numbers out of n numbers, the combinations method is better (as shown in a post above).
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



Manager
Joined: 22 Sep 2015
Posts: 102

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
11 Jan 2016, 06:45
I got totally confused on this question and didn't even realize it was a combinations question. I still don't really understand why. Could someone pls explain this to me like I'm 5?



Math Expert
Joined: 02 Aug 2009
Posts: 6795

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
11 Jan 2016, 07:33
nycgirl212 wrote: I got totally confused on this question and didn't even realize it was a combinations question. I still don't really understand why. Could someone pls explain this to me like I'm 5? Quote: If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32? 1/3 1/2 5/7 10/21 15/28 ... Hi, the Q is a combinations problem because you rae required to pick 2 out of 6.. whereever you pick up some qty out of a given qty, it talks of combinations of those items and requires combinations. here we require to find divisors of 64.. 1,2,4,8,16,32,64... if we find the pairs whose sum is less than 32, 32 and 64 can be straight way eliminated... we are left with 5 numbers, of which 16 and 8 are the two largest... so 16 + 8,24, is the max sum we can have if we pick any two out of these 5... this means any pair out of these 5 will have a sum <32.. so our prob is ways to pick 2 out of these 5/ prob to pick any two out of total 7.. 5C2/7C2.. 5*4/7*6=10/21.. D .. Hope it helped
_________________
1) Absolute modulus : http://gmatclub.com/forum/absolutemodulusabetterunderstanding210849.html#p1622372 2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html 3) effects of arithmetic operations : https://gmatclub.com/forum/effectsofarithmeticoperationsonfractions269413.html
GMAT online Tutor



Manager
Joined: 22 Sep 2015
Posts: 102

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
16 Jan 2016, 14:13
chetan2u wrote: nycgirl212 wrote: I got totally confused on this question and didn't even realize it was a combinations question. I still don't really understand why. Could someone pls explain this to me like I'm 5? Quote: If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32? 1/3 1/2 5/7 10/21 15/28 ... Hi, the Q is a combinations problem because you rae required to pick 2 out of 6.. whereever you pick up some qty out of a given qty, it talks of combinations of those items and requires combinations. here we require to find divisors of 64.. 1,2,4,8,16,32,64... if we find the pairs whose sum is less than 32, 32 and 64 can be straight way eliminated... we are left with 5 numbers, of which 16 and 8 are the two largest... so 16 + 8,24, is the max sum we can have if we pick any two out of these 5... this means any pair out of these 5 will have a sum <32.. so our prob is ways to pick 2 out of these 5/ prob to pick any two out of total 7.. 5C2/7C2.. 5*4/7*6=10/21.. D .. Hope it helped yes thanks.. but why do we do combinations and not permutations here?



Current Student
Joined: 20 Mar 2014
Posts: 2639
Concentration: Finance, Strategy
GPA: 3.7
WE: Engineering (Aerospace and Defense)

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
16 Jan 2016, 17:12
nycgirl212 wrote: chetan2u wrote: nycgirl212 wrote: I got totally confused on this question and didn't even realize it was a combinations question. I still don't really understand why. Could someone pls explain this to me like I'm 5? Quote: If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32? 1/3 1/2 5/7 10/21 15/28 ... Hi, the Q is a combinations problem because you rae required to pick 2 out of 6.. whereever you pick up some qty out of a given qty, it talks of combinations of those items and requires combinations. here we require to find divisors of 64.. 1,2,4,8,16,32,64... if we find the pairs whose sum is less than 32, 32 and 64 can be straight way eliminated... we are left with 5 numbers, of which 16 and 8 are the two largest... so 16 + 8,24, is the max sum we can have if we pick any two out of these 5... this means any pair out of these 5 will have a sum <32.. so our prob is ways to pick 2 out of these 5/ prob to pick any two out of total 7.. 5C2/7C2.. 5*4/7*6=10/21.. D .. Hope it helped yes thanks.. but why do we do combinations and not permutations here? BY definition, permutations are combinations wherein the order of selection or distribution matters. In this case as we are adding 2 divisors of 6 it does not matter whether we choose 1+16 or we choose 16+1. If you end up applying permutations here, you will double the possible/allowed number of combinations.



Director
Joined: 24 Nov 2015
Posts: 544
Location: United States (LA)

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
09 Apr 2016, 14:58
list the distinct positive factors: 1,2,4,8,16,32,64 selecting 2 from this list = 7C2 = 21 as per the given condition there sum has to be less than 32 picking any 2 numbers from our list of first five numbers will render the sum less than 32 in any case selecting 2 number from 5 numbers = 5C2 = 10 proability = 10/21 correct answer  D



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
25 Jul 2016, 05:13
goodyear2013 wrote: If two distinct positive divisors of 64 are randomly selected, what is the probability that their sum will be less than 32?
A. 1/3 B. 1/2 C. 5/7 D. 10/21 E. 15/28
This is good question. Responding to a pm: Quote: For the question below, is it wrong to approach it is as follows:
There are 7 factors (1, 2, 4, 8, 16, 32, 64)  (5/7)*(4/6) = (20/42) = (10/21)
It is correct. As long as you observe that any two factors out of the first 5 will never add up to 32, you are good.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



SVP
Joined: 12 Dec 2016
Posts: 1794
Location: United States
GPA: 3.64

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
17 Apr 2017, 21:48
so, once again, I want a confirm that any question relating factor, or divisor should use only positive numbers. In this question, there are only 7 factors.



Veritas Prep GMAT Instructor
Joined: 16 Oct 2010
Posts: 8282
Location: Pune, India

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
17 Apr 2017, 21:59
chesstitans wrote: so, once again, I want a confirm that any question relating factor, or divisor should use only positive numbers. In this question, there are only 7 factors. Yes, factors/divisors are always positive.
_________________
Karishma Veritas Prep GMAT Instructor
Learn more about how Veritas Prep can help you achieve a great GMAT score by checking out their GMAT Prep Options >
GMAT selfstudy has never been more personalized or more fun. Try ORION Free!



NonHuman User
Joined: 09 Sep 2013
Posts: 8104

Re: If two distinct positive divisor of 64 are randomly selected
[#permalink]
Show Tags
18 Jul 2018, 08:01
Hello from the GMAT Club BumpBot! Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up  doing my job. I think you may find it valuable (esp those replies with Kudos). Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
_________________
GMAT Books  GMAT Club Tests  Best Prices on GMAT Courses  GMAT Mobile App  Math Resources  Verbal Resources




Re: If two distinct positive divisor of 64 are randomly selected &nbs
[#permalink]
18 Jul 2018, 08:01






