Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 exptools 1/6/84; site ihnet.UUCP Path: utzoo!watmath!clyde!burl!ulysses!mhuxl!houxm!ihnp4!ihnet!eklhad 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