On 22-9-2009 19:10, Warp wrote:
>   Some relatively easy math problems for your consideration and enjoyment:
> 
> 
>   1) The classical proof that there are infinitely many primes is a proof by
> contradiction: Let's assume that there's a largest prime. If we multiply
> all the primes up to that largest prime and add 1, we get a number which
> is not divisible by any of the primes, and thus the assumption we made is
> false: There was a prime which is larger than the one we assumed was the
> largest.
> 
>   However, consider this: 2*3*5*7*11*13 + 1 = 30031, which is a composite
> number.
> 
>   Isn't this a contradiction to the proof? It clearly doesn't hold that the
> product of the first n primes plus 1 is a prime.
> 
>   How to explain this apparent contradiction?

The claim is not that the products of primes plus one is a prime, just 
that it has a prime divisor that is not one in the product.

>   2) Assume you have an array of 24 integers. Each element of that array
> can get a value between 0 and 7. Thus the total amount of different contents
> for such as an array is 8^24, which is approximately 4*10^21.
> 
>   Now assume that you fill the array with some values and then calculate
> all the possible permutations of that array. The amount of permutations
> for 24 elements is 24!, which is approximately 6*10^23.
> 
>   Now here's the apparent paradox: The total amount of different contents is
> about 4*10^21, and naturally all those permutations should be among them as
> well. How come the total amount of permutations, 6*10^23, is way larger
> than the total amount of possible different array contents?

8 is less than 24

>   3) Assume two people, person A and person B, who want to decide who gets
> a price by tossing a coin.
> 
>   Person A is a bad loser and a bully, so if he loses he says "I said it's
> two out of three". So they play it like that. If A loses again, he says
> "I said it's three out of five", and so on, until he wins.
> 
>   How many tosses is this game expected to last, in average?

same answer as: what will the average number if boys be if every pair of 
parents continues getting children untill they get a boy. (assuming 50% 
chance of a boy, which is not exactly correct, and that every pair that 
gets one child can in principle get an infinite number)

>   4) By their nature, factorials tend to accumulate trailing zeroes. For
> example, 5! (120) has one trailing zero, 10! (3628800) has two trailing
> zeroes, 25! (15511210043330985984000000) has six trailing zeroes, etc.
> 
>   Give a (non-recursive) mathematical function which, for an integer n,
> gives the number of trailing zeroes in n!.

just count the number of 5s, 25s, 125s etc

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