If two distinct positive divisor of 64 are randomly selected

Question

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.

KarishmaB · Accepted Answer

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.

chetan2u · Answer

...
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

BrentGMATPrepNow · Answer

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

Answer: D

Cheers,
Brent

pacifist85 · Answer

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.

shasadou · Answer

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

bhatiavai · Answer

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

KarishmaB · Answer

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).

nycgirl212 · Answer

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?

nycgirl212 · Answer

yes thanks.. but why do we do combinations and not permutations here?

ENGRTOMBA2018 · Answer

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.

rhine29388 · Answer

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

KarishmaB · Answer

Responding to a pm:

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.

chesstitans · Answer

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.

KarishmaB · Answer

Yes, factors/divisors are always positive.

dcummins · Answer

Great approach.

@veritasprepkarishma your approach should be the official solution instead of what's contained as the official solution.

ScottTargetTestPrep · Answer

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.

Answer: D

Shamitjain · Answer

I did this in a slightly different way:

Total Factors of 64 are 64 , 1 , 32, 2 , 4 , 16 , 18

Sum of factors less than 32 - 1 , 2 , 4 , 16 , 18

therefore total no. of ways you can pick the first factor = 5/7 (as the same or either two numbers will be < 32) 
Second number = 4/6 (since one number is already selected in the first fraction.

5/7 * 4/6 = 20/42 = 10/21 And since in selection AB = BA , we do not need to multiply by 2

Fdambro294 · Answer

Distinct (+)Positive Factors of 64:

1 ; 2 ; 4 ; 8 ; 16 ; 32 ; 64

DEN = Total Possible Outcomes = All the Distinct Possible ways to Choose 2 Integers from the Above Set of 7 Divisors

"7 choose 2" = 7! / 2!5! = 21 Ways

NUM = # of Favorable Outcomes = the 2 Divisors chosen SUM to LESS THAN< 32

64 and 32 can not be chosen.

16? Can be chosen with 4 other Integers less than it ---- 4 ways

8? Can be chosen with 3 other Integers less than it ---- 3 ways

4? can be chosen with 2 other Integers less than it ---- 2 ways

and finally the combination of (1 + 2)

4 + 3 + 2 + 1 = 10 Favorable Outcomes

OR

since any combination of the Factors:

1 ; 2 ; 4 ; 8 ; 16

will give a SUM < 32, we can find the # of Distinct Ways to Choose a Combination of 2 out of these 5 Divisors.

Probability = "5 choose 2" / "7 choose 2" = (5!/2!3!) / (7!/2!5!) = 10/21

Probability = 10/21

Shamitjain · Answer

Was there a problem with the way I solved it?

If two distinct positive divisor of 64 are randomly selected

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If two distinct positive divisor of 64 are randomly selected

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