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        riddles >> putnam exam (pure math) >> Free Throws
        
(Message started by: THUDandBLUNDER on Dec 7^th, 2004, 12:42am)

Title: Free Throws
Post by THUDandBLUNDER on Dec 7^th, 2004, 12:42am

Shaq's coach has been keeping a record of the number S(N) of successful free throws he has made in his first N attempts of the season. Earlier in the season S(N) was less than 80% of N, but later in the season S(N) was more than 80% of N. Was there necessarily a time in between when S(N) was exactly 80% of N?

Title: Re: Free Throws
Post by Icarus on Dec 7^th, 2004, 12:40pm

Yes: [hide]Let S, N be the values immediately before reaching or passing the 80% mark. Then 4N - 1 <= 5S < 4N. Since 5S is an integer, it must be the only integer in that range. I.e. 5S = 4N-1. So S+1 = (1/5)(4N-1) + 1 = 1/5 (4N-1+5) = 4/5 (N+1). [/hide]

Title: Re: Free Throws
Post by Foolish on Dec 7^th, 2004, 4:46pm

Why does this work for 80% and not for all values?
35% for example 3/9 -> 4/10 skips over that value.
40% however, it seems to hold for. Is it the fact that .35 isn't made of prime/prime?

There must be something underlying that I am missing.

Title: Re: Free Throws
Post by THUDandBLUNDER on Dec 7^th, 2004, 9:22pm

on 12/07/04 at 16:46:08, Foolish wrote:

Why does this work for 80% and not for all values?

It works for all percentages of the form 1 - 1/k, where k is integer.

Title: Re: Free Throws
Post by Foolish on Dec 7^th, 2004, 9:35pm

on 12/07/04 at 21:22:25, THUDandBLUNDER wrote:

It works for all percentages of the form 1 - 1/k, where k is integer.

Ah, that makes sense. Thanks :)

Title: Re: Free Throws
Post by Barukh on Dec 8^th, 2004, 2:08am

on 12/07/04 at 21:22:25, THUDandBLUNDER wrote:

It works for all percentages of the form 1 - 1/k, where k is integer.

... and I think it has something to do with continued fractions.