If k divides 25 what is prob(k odd)?

[jbriggs444]

[12] (2*6) * 9

Six is not a prime number.

I am not at all sure what that nine is doing there.

 

[Chris Miller]

Right. Thanks! Should be 2*2*3, raising the number of 2s to 22 (9 was the 2 count at that step, and should be 10) making P=1/(22*2) = 1/44. Really appreciate your help.

 

[Chris Miller]

Went to bed. Thought about it. Got back up. For every odd factor (including 1) there are t even factors. So the odds (not probability) of picking an odd one at random are 1 to t. So the probability is 1/(t+1). So for 25! it's 1/(22+1) or 1/23. Now going back to bed, and hopefully not having to get back up again. Tonight.

 

[ddddd28]

The number of factors can be simply described as the product:
(q1+1)(q2+1) ... (qn+1). q1, q2 ... qn are the number of times a prime appears in the factorization.
To find the required probability is to calculate the product of 25! and the product of 25! divided by 2^q.

 

[jbriggs444]

The number of factors can be simply described as the product:
(q1+1)(q2+1) ... (qn+1). q1, q2 ... qn are the number of times a prime appears in the factorization.
To find the required probability is to calculate the product of 25! and the product of 25! divided by 2^q.

You may wish to read the whole thread and refine that answer a bit. What you have written seems to claim that the probability is $\frac{25!\ 25!}{2^{22}}$. In addition to being incorrect, that number is larger than one while probabilities are always less than or equal to one.

 

[ddddd28]

You misunderstood me. I meant that the number of factors of 25! is 23*11*7*4*3*2*2*2*2.
Since we also need to find how many odd factors there are, we don't need to include 23 in the product of 25!/2^22.

 

[ddddd28]

To make my point fully clear, take 60. The factorization is 2^2*3*5. Therefore, there are 3*2*2= 12 factors :
1,2,3,4,5,6,10,12,15,20,30,60. Similarly, there are only 2*2= 4 odd factors: 1,3,5,15. So, the probability in this case is one third.

 

[PeroK]

To make my point fully clear, take 60. The factorization is 2^2*3*5. Therefore, there are 3*2*2= 12 factors :
1,2,3,4,5,6,10,12,15,20,30,60. Similarly, there are only 2*2= 4 odd factors: 1,3,5,15. So, the probability in this case is one third.

You don't need to calculate the number of factors to get the answer. That's the point.

 

[ddddd28]

You are right, but it still works. Of course, the amount of 2s only matters, because the rest of the factorization is reduced anyway.

 

[Chris Miller]

Yeah, really all you need to do is count the number of times 2 appears in its prime factorization. So no need to factor the odd numbers (1,3,5,7...25) at all.

 

[jbriggs444]

Yeah, really all you need to do is count the number of times 2 appears in its prime factorization. So no need to factor the odd numbers (1,3,5,7...25) at all.

There is a shortcut for that part as well.

How many times does an even number occur in the set {1, 2, 3, ... 25} ? Answer: 25 divided by 2 and rounded down = 12.
Divide each of those even numbers in half.
How many times does an even number occur in the set {1, 2, 3, ...12} ? Answer: 12 divided by 2 and rounded down = 6.
Divide each of those even numbers in half.
How many times...

so all you have to do is add up 12 + 6 + 3 + 1 = 22 occurrences of 2 in the prime factorization of 25!

If you refer back to post #7 by @Vanadium 50, you will see a result obtained for 1000 factorial. Using this trick, it is fairly easy to verify the result obtained there.

[You might even note that 1000 is close to 1024 and economize even more on the arithmetic].

 

[ddddd28]

Your shortcut works well with factorials, what about the rest of the numbers?

 

[jbriggs444]

Your shortcut works well with factorials, what about the rest of the numbers?

It does not work for determining the number of 2's in the prime factorizations of numbers that are not factorials.

Edit: Argument about whether this is a bug or a feature removed.

 

[ddddd28]

not important in test taking.

Are we taking a test right now?
I believe that, when not tested, one should not be satisfied with a sufficient solution, because the most important thing is not to provide a solution, but to learn something from the attempts and to improve oneself for the next problem