Yes: !x (mod a) stops changing at x=a or sooner, since y! = 0 when y>=a. So !c (mod a) = !a (mod a) = k (mod a), and ditto for b replacing a. Therefore, !c = k both (mod a) and (mod b), whence mod LCM(a,b). qed. Rich PS: a=0 needs to be treated as a special case, since congruence mod 0 is a tad squirrely, and LCM(0,x)=0 violates the usual rule LCM(a,b)>=max(a,b), which is implicitly used in the above proof. --R ----- Quoting David Wilson <davidwwilson@comcast.net>:
Let !n = 0! + 1! + ... + n! = A003422(n)
For integer k, is it true that
!a == k (mod a)
and
!b == k (mod b)
together imply
!c == k (mod c)
where
c = lcm(a, b)
?
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