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From: eklhad@ihnet.UUCP (K. A. Dahlke)
Newsgroups: net.math
Subject: Re: a hard one!?!?  (not a spoiler)
Message-ID: <153@ihnet.UUCP>
Date: Mon, 20-Aug-84 15:58:43 EDT
Article-I.D.: ihnet.153
Posted: Mon Aug 20 15:58:43 1984
Date-Received: Tue, 21-Aug-84 00:34:29 EDT
Organization: AT&T Bell Labs, Naperville, IL
Lines: 16

The cheap way.

main(){
register i,j;
for(i=1;;++i){
for(j=2;j<11;++j) if(i%j!=j-1) break;
if(j==11){ printf("%d\n",i); exit(0); }
}
}

If you are not into computers, try the chinese remainder theorem.
x%5=4 x%7=6 x%8=7 x%9=8
bc(1) or a calculator makes short work of this.
-- 

Karl Dahlke    ihnp4!ihnet!eklhad