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# If two distinct positive divisor of 64 are randomly selected

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Senior Manager
Joined: 21 Oct 2013
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If two distinct positive divisor of 64 are randomly selected  [#permalink]

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18 Mar 2014, 15:04
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65% (hard)

Question Stats:

61% (02:22) correct 39% (02:08) wrong based on 550 sessions

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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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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18 Mar 2014, 20:32
14
14
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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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05 Mar 2015, 08:27
1
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.
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If two distinct positive divisor of 64 are randomly selected  [#permalink]

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16 Sep 2015, 03:27
2
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 non-conflicting pool

(2) 5/7 * 2/6 - same as above but the non-conflicting 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 non-conflicting divisors and see the probability
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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01 Nov 2015, 01:29

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
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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02 Nov 2015, 21:11
2
bhatiavai wrote:

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.

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).
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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11 Jan 2016, 05:45
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?
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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11 Jan 2016, 06:33
1
1
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
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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16 Jan 2016, 13:13
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?
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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16 Jan 2016, 16:12
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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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09 Apr 2016, 13:58
1
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
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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25 Jul 2016, 04:13
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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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17 Apr 2017, 20:48
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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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17 Apr 2017, 20:59
1
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.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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19 Aug 2019, 18:18
Great approach.

@veritasprepkarishma your approach should be the official solution instead of what's contained as the official solution.
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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28 Dec 2019, 06:20
1
Top Contributor
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.

Here are the positive divisors of 64: {1, 2, 4, 8, 16, 32, 64}

We can see that, as long as we DON'T select the 32 or 64, then the sum of the two numbers is guaranteed to be less than 32
So, P(sum is less than 32) = P(neither is a 64 nor the 32 is selected)
= P(no 64 or 32 is selected on the 1st try AND no 64 or 32 is selected on the 2nd try)
= P(no 64 or 32 is selected on the 1st try) x P(no 64 or 32 is selected on the 2nd try)
= 5/7 x 4/6
= 20/42
= 10/21

Cheers,
Brent
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Re: If two distinct positive divisor of 64 are randomly selected  [#permalink]

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13 Jan 2020, 13:52
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.

The positive divisors of 64 are:

1, 2, 4, 8, 16, 32, and 64

If the sum of two distinct divisors of 64 is less than 32, each has to be less than 32. Therefore, they are 1, 2, 4, 8, and 16. The number of ways to select 2 divisors from these 5 divisors is 5C2 = (5 x 4)/2 = 10.

Since the total number of ways to select 2 divisors from all 7 divisors is 7C2 = (7 x 6)/2 = 21, the probability is 10/21.

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Re: If two distinct positive divisor of 64 are randomly selected   [#permalink] 13 Jan 2020, 13:52