Best Answer Prime, 29 January 2013 - 06:46 AM

But they don't need to be integers, those 11 periodic values.

Welcome to BrainDen.com - Brain Teasers Forum. Like most online communities you must register to post in our community, but don't worry this is a simple free process. To be a part of BrainDen Forums you may create a new account or sign in if you already have an account. As a member you could start new topics, reply to others, subscribe to topics/forums to get automatic updates, get your own profile and make new friends. Of course, you can also enjoy our collection of amazing optical illusions and cool math games. If you like our site, you may support us by simply clicking Google "+1" or Facebook "Like" buttons at the top. If you have a website, we would appreciate a little link to BrainDen. Thanks and enjoy the Den :-) |

Guest Message by DevFuse

Started by bushindo, Jan 27 2013 04:53 PM

Best Answer Prime, 29 January 2013 - 06:46 AM

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Go to the full post
10 replies to this topic

Posted 27 January 2013 - 04:53 PM

This is based on a previous puzzle.

Find a closed form expression for function f(N) such that

f( N ) = 0 for 1 <= N <= 11

f( N ) = 1 for 12 <= N <= 22

f( N ) = 2 for 23 <= N <= 33

f( N ) = 3 for 34 <= N <= 44

f( N ) = 4 for 45 <= N <= 55

...

No modulus or rounding operators are allowed.

Posted 27 January 2013 - 08:38 PM

Spoiler for How about

*Vidi vici veni.*

Posted 28 January 2013 - 02:36 AM

Spoiler for How about

Comments

Spoiler for

Posted 28 January 2013 - 10:14 PM

I am assuming N must be an integer.

As for how to slice two pi into 11 pieces, which would yield the required progression, I am stumped.

I just want to throw a different thought into this subject:

Spoiler for multiples

Past prime, actually.

Posted 29 January 2013 - 02:26 AM

Spoiler for My 2 cents.

Posted 29 January 2013 - 06:46 AM Best Answer

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Past prime, actually.

Posted 29 January 2013 - 07:24 AM

Wouldn't a polynomial be continuous?

*Vidi vici veni.*

Posted 29 January 2013 - 07:37 AM

Oops, messed up a little...

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Spoiler for CORRECTION

**Edited by Prime, 29 January 2013 - 07:37 AM.**

Past prime, actually.

Posted 29 January 2013 - 07:44 AM

Wouldn't a polynomial be continuous?

No. We have exactly 11 pairs of points (x_{k},y_{k}). Where x_{k} is one of the tan(...) values and y_{k} is one of the integer values 0 through 10. It should be big and messy, but a regular polynomial of 10th degree.

And another correction to start output with zeros instead of ones, just subtract another 1 from that whole thing.

**Edited by Prime, 29 January 2013 - 07:50 AM.**

Past prime, actually.

Posted 29 January 2013 - 06:08 PM

But they don't need to be integers, those 11 periodic values.

Spoiler for one messy solution

Excellent answer

Spoiler for

0 members, 0 guests, 0 anonymous users