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

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Discussion Overview

The discussion centers around the probability of selecting an odd factor from the factorial of 25, denoted as 25!. Participants explore various methods for determining this probability, including prime factorization and combinatorial reasoning, while addressing the challenges posed by the large number of factors in 25!. The conversation includes elements of mathematical reasoning and competition problem-solving.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest using prime factorization to analyze the factors of 25!, while others express uncertainty about how to apply this method effectively.
  • There is a debate about the feasibility of listing all factors of 25! due to its size, with some arguing that it is impractical.
  • One participant mentions that the problem is from a local mathematics competition and emphasizes the need for a more efficient approach than listing factors.
  • Another participant points out that the method of choosing a factor at random is crucial to determining the probability, suggesting that the problem statement should clarify this aspect.
  • Some participants propose that counting the factors of a factorial can be simplified through recursive reasoning, while others challenge the validity of this approach.
  • There are discussions about the arithmetic involved in calculating the number of factors for larger factorials, with some participants providing examples and corrections to earlier claims.
  • One participant shares a prime factorization for 25! and invites others to use this information to determine the number of factors.
  • There is a suggestion to visualize the number of factors through a mathematical function, although the idea remains underdeveloped.
  • Another participant speculates that the number of odd factors might equal the number of even factors, hinting at a potential probability of 0.5.

Areas of Agreement / Disagreement

Participants express a range of views on the methods to solve the problem, with no consensus on a single approach. Disagreement exists regarding the necessity of listing factors versus using prime factorization, as well as the implications of how factors are chosen at random.

Contextual Notes

Participants note the complexity of the problem due to the size of 25! and the implications of prime factorization. Some discussions highlight the need for clarity in the problem statement regarding the selection of factors.

Who May Find This Useful

This discussion may be useful for students and enthusiasts interested in combinatorial mathematics, probability theory, and problem-solving strategies in competitive mathematics contexts.

  • #31
Chris Miller said:
[12] (2*6) * 9
Six is not a prime number.

I am not at all sure what that nine is doing there.
 
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  • #32
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.
 
  • #33
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.
 
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  • #34
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.
 
  • #35
ddddd28 said:
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.
 
  • #36
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.
 
  • #37
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.
 
  • #38
ddddd28 said:
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.
 
  • #39
You are right, but it still works. Of course, the amount of 2s only matters, because the rest of the factorization is reduced anyway.
 
  • #40
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.
 
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  • #41
Chris Miller said:
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].
 
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  • #42
Your shortcut works well with factorials, what about the rest of the numbers?
 
  • #43
ddddd28 said:
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.
 
Last edited:
  • #44
jbriggs444 said:
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
 

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