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        riddles >> easy >> Private Club Password
        
(Message started by: Jonsey on Nov 10^th, 2003, 3:35pm)

Title: Private Club Password
Post by Jonsey on Nov 10^th, 2003, 3:35pm

Can anyone help solve this riddle?

A man wants to get into a private club in which he is NOT a member. He listens to the first member as he knocks on the door.. door opens and asks for a password after he gives the number 12. 1st member answers 6 and is allowed to enter. The 2nd member knocks and the number given him is 6, he responds with the answer 3 and is also allowed to enter. Now our man thinks he has heard enough and knocks on the same door. It opens with the number 10, his response is then 5, but it is wrong and he is not allowed in.

WHAT THEN SHOULD HIS ANSWER HAVE BEEN??????

Title changed to be meaningful and moved to Easy forum by Icarus.

Title: Re: Private Club Password
Post by Icarus on Nov 10^th, 2003, 4:34pm

I moved this problem to the Easy forum not because I have the answer, but because it is most similar to other problems in this forum. A look at other threads in the Putnam Forum should make it abundantly clear that this puzzle is not the sort that belongs there.

The problem with this riddle is that there are too many possible answers. All one has to do is dream up a formula that carries 12 to 6 and 6 to 3 but does not take 10 to 5. (Other possibilities also exist, but I will assume that the numbers are dependent.)

The easiest formula that takes 12 to 6 and 6 to 3 is of course y=x/2, so this is the one the guy guesses and fails with.

But give me any other value that "should have been" his answer, and I could give you any of an uncountably infinite number of formulas that he "should have used".

There is simply not enough information given to tell "what his answer should have been".

My guess however is that the INTENDED answer to this ill-posed riddle was that he should have answered "3".

As a hint for why here are 3 other challenges and answers:

20 [to] 6
7 [to] 5
4 [to] 4

Title: Re: Private Club Password
Post by Eigenray on Nov 10^th, 2003, 9:29pm

Icarus is obviously thinking of the function f(n) = -513/116480 n⁵ + 6073/26880 n⁴ - 17897/4160 n³ + 3338519/87360 n² - 22163/140 n + 12885/52 . . .

...Oh. But that doesn't satisfy f(10x+y)=f(10x)+f(y) for 1<x<10, 1[le]y<10, does it?

A related sequence: 4,5,3,1,11,23,...?
[edit]Fixed typo cause I can't read my own handwriting[/edit]

Title: Re: Private Club Password
Post by BNC on Nov 10^th, 2003, 11:16pm

on 11/10/03 at 16:34:17, Icarus wrote:

4 [to] 4

Hehe. There's a whole thread dedicated to numbers like that ;D

Title: Re: Private Club Password
Post by harpanet on Nov 11^th, 2003, 2:56am

This works with the data given.
Take the challenge number and sum its digits. Then double the result and sum its digits to give the response. So

12 -> 3 -> 6 -> 6
6 -> 6 -> 12 -> 3

therefore

10 -> 1 -> 2 -> 2

Title: Re: Private Club Password
Post by Guestmess on Feb 18^th, 2005, 3:54pm

three

Title: Re: Private Club Password
Post by eva on Feb 21^st, 2005, 12:39am

the answer is 3...i guess. bcoz when 12 was said,6 was the reply.12 has six letters in it. when 6 was said, someone answered 3 bcoz it contains 3 letters.therefore,3 must be the answer...hmmmmmmmm :P :P :P