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Conference rusure::math

Title:Mathematics at DEC
Moderator:RUSURE::EDP
Created:Mon Feb 03 1986
Last Modified:Fri Jun 06 1997
Last Successful Update:Fri Jun 06 1997
Number of topics:2083
Total number of notes:14613

1991.0. "IMO 95 Q6" by JOBURG::BUCHANAN () Tue Aug 29 1995 16:53

    	Let p be an odd prime. How many different subsets A of the set
    {1,2,...,2p} are there which satisfy:
    
    	(i) |A| = p
    	(ii) sum(a in A) == 0 mod p
    
    cheers,
    Andrew.
T.RTitleUserPersonal
Name
DateLines
1991.1HANNAH::OSMANsee HANNAH::IGLOO$:[OSMAN]ERIC.VT240Wed Aug 30 1995 15:162
What does |A| mean in this context
1991.2WRKSYS::ROTHGeometry is the real life!Wed Aug 30 1995 15:593
   I'd exepct that |A| means the cardinality of the set A

   - Jim
1991.3RUSURE::EDPAlways mount a scratch monkey.Thu Aug 31 1995 09:448
    See also topic 1930.
    
    
    				-- edp
    
    
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1991.4DECADA::YODERMFYThu Aug 31 1995 12:2617
Consider the permutation u which rotates P={1,2,...,p} forward by 1 and rotates
Q={p+1,p+2,...,2p} backward by 1.  If a set A has k elements in P and p-k in Q,
and its sum mod p is c, call (k,p-k,c) the signature of A.  Then Au will have
the signature (k,p-k,c+2k); and conversely, if Au has this signature, A has
signature (k,p-k,c).  For k=0 or p this isn't too interesting, but otherwise,
since p is an odd prime, adding 2k repeatedly will cycle through all the values
mod p; so u, iterated if necessary, provides a 1-1 map between sets of signature
(k,p-k,a) and (k,p-k,b) for any a and b.

There is one set with signature (0,p,0) and one with signature (p,0,0): the
set's sum in both cases is == p(p+1)/2 == 0 (mod p).  All other sets of
cardinality p divide according to their sum mod p, by using the above argument,
into equinumerous classes of size (C(2p,p) - 2)/p.  So the answer is

  2 + (C(2p,p) - 2)/p = (C(2p,p)+2p-2)/p

where C(a,b) = a!/(b!(a-b)!) is the combinations function.