wu :: forums - Print Page


    
      
        wu :: forums
        (http://www.ocf.berkeley.edu/~wwu/cgi-bin/yabb/YaBB.cgi)
      

        riddles >> medium >> Enigma Number 1606 
        
(Message started by: grad on Sep 11^th, 2010, 6:44pm)

Title: Enigma Number 1606
Post by grad on Sep 11^th, 2010, 6:44pm

Hey, does anyone know the answer to this:
http://www.newscientist.com/article/mg20727710.900-enigma-number-1606.html

or what to do to solve it?

Title: Re: Enigma Number 1606
Post by TenaliRaman on Sep 14^th, 2010, 10:44am

As it stands, I am not sure enough information is available. I am getting at least two answers. And if one of those is the expected answers, that lock can be broken by simply trial and error (without resorting to any math).

So yeah, I am finding the question ambiguous.

-- AI

Title: Re: Enigma Number 1606
Post by grad on Sep 14^th, 2010, 5:34pm

So, can you tell me how you get those two answers?

Did you use a program or what?

Title: Re: Enigma Number 1606
Post by towr on Sep 15^th, 2010, 1:01am

Finding the answer is easy: http://www.newscientist.com/article/mg20727771.000-enigma-number-1612.html
But why that's the answer, I don't know.
I can't make enough sense of the problem to start with.

Title: Re: Enigma Number 1606
Post by towr on Sep 15^th, 2010, 1:40am

Bart Schipperijn's explanation of the problem in the comments seems to be the right approach.

The weight of the balls is one of [hide]4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209, 2809, 3481, 3721[/hide]
The weights next to the scale have a weight from [hide]4000, 2000, 1000, 800, 500, 400, 250, 200, 160, 125, 100, 80, 50, 40, 32, 25, 20, 16, 10, 8, 5, 4, 2, 1[/hide]
There's just one value for the weight that singles out one total weight for the balls.

Title: Re: Enigma Number 1606
Post by Immanuel_Bonfils on Sep 16^th, 2010, 4:03pm

I found rather amusing (should I say annoying?) the Enigma Number 1602

http://www.newscientist.com/article/mg20727670.800-enigma-number-1602.html

since it seems to have no unique solution.

Title: Re: Enigma Number 1606
Post by towr on Sep 17^th, 2010, 2:13am

For 1602
[hide] 21-29 are superfluous, because if one of them were wrong, then either one of 1-9 or 20 would be wrong as well. So given that only one is wrong, it's in 10-19.
It's easily deduced that T+E+E+N=10, so we can scrap 14,16,17,19. Because if one of them were wrong, 4,6,7 or 9 would also be wrong. That leaves just 10,11,12,13,15,18.
If we assume T+E+N =/= 10, then we run into a problem, because we have F+I+F=5 implying I is odd, but T+H+R+E+E=T+H+I+R implies I is even, so therefore T+E+N=10=T+E+E+N ===> E=0.
Equation 13 then implies 9 is even, so 13 has to be wrong.

Then, from eight and eighteen we can deduce N=10;
N+I+N+E=9 ===> I = -11
T+H+I+R+T+E+E+N = 3 + 10 + -11 = 2

E=T=0
N=10
F=V=8
I=S=-11
O=-9
W=11
Y=-1
L=-7
X=28

H+R=3
U+R=5
G+H=19
[/hide]

Title: Re: Enigma Number 1606
Post by Immanuel_Bonfils on Sep 17^th, 2010, 6:13am

[hide]Sure! Didn't notice that 18 is also an exception in the "teen"s ...[/hide]

edit: hide

Title: Re: Enigma Number 1606
Post by Immanuel_Bonfils on Sep 17^th, 2010, 6:32am

[hide]Think would be easier to get E = 0 from 14, 16, 17,19 minus 10. [/hide]

Title: Re: Enigma Number 1606
Post by towr on Sep 17^th, 2010, 6:44am

on 09/17/10 at 06:32:21, Immanuel_Bonfils wrote:

[hide]Think would be easier to get E = 0 from 14, 16, 17,19 minus 10. [/hide]

[hide]You can't from the outset assume 10 is correct. 4/14, 6/16, 7/17, 9/19 tells you T+E+E+N=10, but if T+E+N isn't 10, then E isn't 0 either. So you have to prove that E=/=0 leads to a contradiction.[/hide]