M40-44

Question

If one of the positive factors of  10!  is picked at random, what is the probability that it will be odd?

A.  \frac{1}{8} 
B.  \frac{1}{9} 
C.  \frac{1}{10} 
D.  \frac{1}{11} 
E.  \frac{1}{12}

Bunuel · Accepted Answer

Official Solution:

If one of the positive factors of  10!  is picked at random, what is the probability that it will be odd?

A.  \frac{1}{8} 
B.  \frac{1}{9} 
C.  \frac{1}{10} 
D.  \frac{1}{11} 
E.  \frac{1}{12}

To determine the probability, first calculate the highest power of 2 in  10!  by summing the quotients of 10 divided by powers of 2. This is computed as  \frac{10}{2} + \frac{10}{4} + \frac{10}{8} = 5 + 2 + 1 = 8 . Therefore,  10!  includes a factor of  2^8 .
 
Now, consider that EACH odd factor of 10! can be combined with powers of 2, ranging from  2^0 , which is 1, to  2^8 . This results in 9 possible combinations for each odd factor:
   
 	ext{(odd factor)}*2^0, \ 	ext{(odd factor)}*2^1, \ 	ext{(odd factor)}*2^2, \ ..., \ 	ext{(odd factor)}*2^8 .
   
Among these combinations, only the first one,  	ext{(odd factor)*1} , is odd. Thus, the probability of selecting an odd factor from the factors of  10!  is 1 out of these 9 options, or  \frac{1}{9} .

Answer: B

einstein801 · Answer

How does this solution include the odd factors like 1 or 3 or 5 or 7 or 9 or 1*3, 1*5, etc?

Bunuel · Answer

We can write each factor of 10! as represneted below:
 
 	ext{(odd factor_1)}*2^0, \ 	ext{(odd factor_1)}*2^1, \ 	ext{(odd factor_1)}*2^2, \ ..., \ 	ext{(odd factor_1)}*2^8 .

ext{(odd factor_2)}*2^0, \ 	ext{(odd factor_2)}*2^1, \ 	ext{(odd factor_2)}*2^2, \ ..., \ 	ext{(odd factor_2)}*2^8 .

ext{(odd factor_3)}*2^0, \ 	ext{(odd factor_3)}*2^1, \ 	ext{(odd factor_3)}*2^2, \ ..., \ 	ext{(odd factor_3)}*2^8 .

...
 	ext{(odd factor_x)}*2^0, \ 	ext{(odd factor_x)}*2^1, \ 	ext{(odd factor_x)}*2^2, \ ..., \ 	ext{(odd factor_x)}*2^8 .
 
 
As you can see that for each row, the only product that is odd occurs when we multiply by  2^0 , which is 1. Thus, every odd factor, whether it is 1, 3, 5, 7, 9, 13, 15, etc., will remain odd only in the first term of each sequence out of 9.

nihalharis · Answer

I prime factored 10! to get 2^8*3^4*5^2*7

Total number of factors = (8+1)*(4+1)*(2+1)*(1+1)=270

Total odd factors = (4+1)*(2+1)*(1+1)=30

Probability of odd factors = 30/270 = 1/9

Answer B.

Is this a valid way to get to the answer? The explanation seems a bit complicated for my tiny brain to grasp while this makes more sense to me haha.

Shivang29 · Answer

The way I did this question is as follows:

10! = 2^8*3^4*5^2*7

This means that there are 270 factors in 10! (8+1)*(4+1)*(2+1)*(1+1)

of these the ones that account for odd numbers is (4+1)*(2+1)*(1+1) = 30

keep in mind that we need to do the above step because odd*odd = odd and we could have any number of the odd factors together.

this gives us 30/270 = 1/9

Sanjanaad · Answer

So did I!  Bunuel  is this also a right way to solve it?

Bunuel · Answer

_____________________________
Yes.

Goldenfuture · Answer

But this not coveres types like 15 extra - I don't get the logic

Bunuel · Answer

Here is alternative discussion maybe it could help: https://gmatclub.com/forum/12-days-of-c ... 23300.html

M40-44

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