You have 100 doors in a row that are all initially closed. you make 100 passes by the doors starting with the first door every time. the first time through you visit every door and toggle the door (if the door is closed, you open it, if its open, you close it). the second time you only visit every 2nd door (door #2, #4, #6). the third time, every 3rd door (door #3, #6, #9), etc, until you only visit the 100th door.

What state are the doors in after the last pass? Which are open?

 

Doors 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be open. All others will be closed.

 

[QUOTE]Originally posted by magician
[B]Doors 1, 4, 9, 16, 25, 36, 49, 64, 81, and 100 will be open.

 

That was a good one!

The answer was all the numbers that had an odd number of factors right? And all numbers have factors in pairs of unique numbers, except square numbers, which have one extra pair of equal factors ... or something like that.

 

I think magician also recognizes the real answer to these riddles:

"If boka can find a riddle on the internet, we can find the answer on the internet."

- Warren

 

LOL!

 

Or maybe he simply knows something about tau(n).

 

Yeah, but where is the fun in that?? I IM'd this one to Josh yesterday and he wrote a perl script to do it for him(he is a huge nerd). That was impressive!! From now on, if you've seen it before or found the answer on the internet, wait a while and let ppl try to figure it out...

 

You have a point... cheaters never win. Or rather, they usually win, but shouldn't. Or something.

- Warren

 

Don't mess with magician. He's not only a magician, but he teaches math at a local college.