20070720, 15:24  #1 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Factorial
To me this is a tricky one. I think its for people of Alpertron's and Maxal's calibre. Anyway its open for one and all! Show that 61! + 1 ==0 and 63! + 1 == 0 mod 71. Take a shot at it though. Mally Last fiddled with by mfgoode on 20070720 at 15:34 Reason: wrong thread 
20070720, 18:34  #2 
Sep 2002
Vienna, Austria
3×73 Posts 
61!=(9!)^(1) (mod 71) 63!=(7!)^(1) (mod 71) 7!=1 (mod 71) 9!=72*7!=1 (mod 71) 
20070720, 20:26  #3 
Aug 2002
Buenos Aires, Argentina
2×17×43 Posts 
More verbose solution:
According to Wilson's theorem (p1)!+1 is divisible by p. Since 71 is prime, we have: 70!+1 = 0 (mod 71), which implies 70! = 1 (mod 71). Now 61! = 70!/(70*69*68*67*...*61) But 70 = 1, 69 = 2, 68 = 3, etc. (mod 71) so 70 * 69 * 68 *...*62 = 9! (mod 71) (because the number of factors is odd). 61! = 70!/(9!) > 61! = 1/9! (mod 71) In the same way: 63! = 1/7! (mod 71) Now notice that 7! = 5040 = 71^21 = 1 (mod 71). So 63! = 1/(1) = 1, then 63!+1 = 0 (mod 71) 9! = 9*8*7! = 72*7! = 1*7! = 1 (mod 71) So 61! = 1/(1) = 1, then 61!+1 = 0 (mod 71) Last fiddled with by alpertron on 20070720 at 20:37 
20070721, 15:42  #4 
Bronze Medalist
Jan 2004
Mumbai,India
2^{2}·3^{3}·19 Posts 
Right proof!
Yeah wpolly and Alpertron thats about it! Good work, Thank you. Mally 
20070723, 23:22  #5 
Feb 2006
Denmark
2·5·23 Posts 
By the way, there are more congruent factorials modulo 71 than modulo any smaller prime. See http://primepuzzles.net/problems/prob_027.htm for more primes with many congruent factorials.
n! == 8978998 (mod 10428007), for n = 816488, 1251081, 3384225, 4112650, 4237275, 4431559, 4467010, 4835062, 7328694, 7385077, 7415726, 8460938, 8689396, 9295594, 9661614. I used a computer for that! 
20070724, 08:36  #6  
(loop (#_fork))
Feb 2006
Cambridge, England
2·7·461 Posts 
Quote:


20070724, 14:24  #7  
Feb 2006
Denmark
2·5·23 Posts 
Quote:


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