gary391 wrote:
If k and n are both integer and k>n>0 is k!/n! divisible by 30 ?
1) K+n =30
2) k-n = 6
what are the factors of\(30 - 2*3*5\)
so k should have these EXTRA factors when compared to n..
lets see the statements:-
1)\(k+n=30\)
we know k >n, so\(k>\frac{30}{2}\) and\(n<\frac{30}{2}\)
so if we try and make n as MAX as possible n will be 14 and thus k will be 30-14=16..\(\frac{16!}{14!}=15*16\) , which is Div by 30
sufficient
2)\(k-n=6\)
this means\(\frac{k!}{n!}\) is PRODUCT of 6 consecutive integers..
ANY 6 consecutive integers will surely have
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