Hmmm (1 Viewer)

[ND]

Can anyone figure out how this works?

http://digicc.com/fido/

 

[redslert]

having maths withdraw ND?

 

[J0n]

The puzzle is based on 2 provable facts:

1. An integer is divisible by 9 IFF the sum of its digits is divisible by 9.

2. If you take any 2 permutations of the same set of digits, and 
subtract one from the other, the result is always divisible by 9. 



Given these facts, the puzzle operation is simple. After subtraction you
 have a number divisible by 9.

You tell me all the digits except 1. I add them up in modulo 9, giving X, where X is a digit 0..9.

The missing digit D is given by 9 - X.  If I get X = 3, the missing digit must be a 6.

You are asked not to omit a zero, because I can't tell then whether you omitted a 0 or a 9.


Proofs:

   1. Divisibility by 9

      Any integer M is a sum of terms a(n) * 10^n,  n = 0, 1, 2 ....
         where each a(n) is a digit 0..9

      So we write
                   M = Sum[ a(n) * 10^n ]   (for each n = 0, 1 , 2, etc)
      So,    M mod 9 = Sum[ (a(n) * 10^n) mod 9 ] 

      But 10^n mod 9 = 1 for any n, so
             M mod 9 = Sum[ a(n) mod 9 ] 
                     = Sum[ a(n) ] mod 9

      QED


   2. Subtraction of any 2 permutations of an integer is divisible by 9

      Given M as defined above, if M2 is a permutation of M, then 

                  M2 = Sum[ a(n) * 10^m(n) ]   (for each n = 0, 1 , 2, etc)

      Notice I've just changed the power of 10, each m(n) is also 0, 1, 2  etc
         but in some other order.

      Example: 4321 =   1     +  2x10    + 3x100 + 4x1000
               2134     1x100 +  2x1000  + 3x10  + 4

      Subtraction gives
              
          (M - M2) mod 9 = Sum[ a(n) * (10^n - 10^m(n)) ] mod 9


      But we know ANY power of 10 = 1 mod 9, so

                         = Sum[ a(n) * (1 - 1) ] mod 9
                         = 0 mod 9

      QED

 

[KeypadSDM]

Remainder of any number when divided by 9:

Let the digits be A, B, C, ... n
:. Remainder = A * 10^n + B*10^(n-1) + ... + n * 10^0 (Mod 9)
= A * 1^n + B*1^(n-1) + ... + n * 1^0 (Replacing 10's with 1's (Same value in Mod 9))
= A + B + ... + n (Mod 9)
Hence IFF the sum of the digits is divisible by 9, then the number itself is divisible by 9

 

[ND]

No wonder i couldn't figure it out; i don't know any modular arithmetic. Thanks.

Originally posted by redslert
having maths withdraw ND?

Heheh something like that.

 

[KeypadSDM]

Originally posted by ND
No wonder i couldn't figure it out; i don't know any modular arithmetic. Thanks.

Well it's not NECESSARY to figure it out. With my proof, you could just assume that 10^n - 1 = 9B where B is an integer. It's pretty obvious 10^n - 1 = 9999...999 To n-1 places.

 

[flyin']

ND, no modular arithmatic for first year Actuarials unless you do 135, btw.

 

[ND]

Originally posted by KeypadSDM
Well it's not NECESSARY to figure it out. With my proof, you could just assume that 10^n - 1 = 9B where B is an integer. It's pretty obvious 10^n - 1 = 9999...999 To n-1 places.

To be honest, i didn't even really read through the proofs, i just saw "mod 9" and thought "ugh, modular arithmetic". I gotta learn me that someday... (i've never done anything outside of 4u, and that includes most yr7-10 work too)

ND, no modular arithmatic for first year Actuarials unless you do 135, btw.

But i thought that 133 was higher than 135? or is modular arithmetic too easy for 133? (i remember spice girl saying he learned it in yr7)

 

[flyin']

133 is the big brother of 136. And 132 is the big brother of 135... Well, knowing some modulo is useful for Group Theory.

 

[Grey Council]

isn't spice girl a girl? uh, don't tell me im wrong...

 

[evilc]

you are wrong

 

[Dumbarse]

yeh i always thought spicegirl was some super maths freak chick, then i found out it was just a guy with a woosie name