20150401, 23:29  #1 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2·2,897 Posts 
Factorial puzzle
I just noticed that 10!=6!*7!.
Are there any other instances where x!=y!*z! with y and z > 1? If not can anyone prove this? 
20150401, 23:41  #2 
Just call me Henry
"David"
Sep 2007
Cambridge (GMT/BST)
2×2,897 Posts 
Just realised that there is an infinite number of examples where (n!)!=n!*(n!1)!
Are there any more apart from this trivial form? 
20150402, 00:22  #3 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}×131 Posts 
well one condition on this occuring is that one of the factorials involved has to involve the last prime below the factorial being divided by other factorials so for example the condition for example the reason 10!=6!*7! works is because one of the factorials involves 7 a prime number no solution can leave this out. also a key part is that the other factorial has to include possible factorizations of the rest so 8*9*10 = 8*90 = (2*4)*(5*6*3) which means it can barely fall below sqrt(number) when trying to find solutions to number! okay I made an error but I was trying to limit the cases.
Last fiddled with by science_man_88 on 20150402 at 00:53 
20150402, 02:27  #4 
"Forget I exist"
Jul 2009
Dumbassville
2^{6}·131 Posts 

20150402, 12:02  #5 
Einyen
Dec 2003
Denmark
3·17·59 Posts 

20150402, 12:58  #6 
"Forget I exist"
Jul 2009
Dumbassville
10000011000000_{2} Posts 
using pari I've checked for solutions using up to the last 10000 factorials before a factorial up to 30000! I think:
Code:
factorfactorial(e=0)={ a=parvector(10000,n,(n+e)!); parfor(x=10,#a, d=precprime(x);xd, c, if(c==0c>=10000,next()); b=setintersect(vector(c1,t,a[x]/a[x+t(c+1)]),a); if(b==[],, print(x","b) ) ); e=e+#a; print(e); factorfactorial(e) }; Last fiddled with by science_man_88 on 20150402 at 13:58 
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