Mystery Procedure

7331 /www/nrich/html/content/id/7331/ 1 2011-02-01T00:00:01 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><ul id="stemLinks"> <li><a href="http://nrich.maths.org/5918">Warm-up</a></li> <li><a href="http://nrich.maths.org/5974">Try this next</a></li> <li><a href="http://nrich.maths.org/5928">Think higher</a></li> <li><a href="http://plus.maths.org/content/os/issue36/features/nishiyama/index">Read: mathematics</a></li> <li><a href="http://nrich.maths.org/6025">Read: technology</a></li> <li><a href="http://plus.maths.org/content/brain">Explore further</a></li> </ul> <div> </div> What does this procedure do? Prove it. <mdo:image alt="" height="674" src="procedure.png" width="483"></mdo:image></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0">It is not envisaged that <a href="http://nrich.maths.org/7331">this problem</a> would be used as a class problem. It is more appropriate for an enthusiastic student or small group of students looking for a challenge to work on independently.</mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"> This procedure calculates the highest common factor of A and B. The algorithm swaps over A and B to ensure A > B. We know hcf(A,B) $\leq$ min(A,B) = B in this case. Therefore, the first guess for N = hcf(A,B) is B. It first tests whether N is a factor of B. If that's the case, it then tests whether it's a factor of A. If either of these fail, it decreases N by 1. Eventually, it will either stop at a value of N>1 such that N divides A and B, or will terminate when N=1 (as A/1 and B/1 are both whole numbers). In the latter case, A and B are coprime. </mdoxml> 2 3 0 0 0 1 0 Mystery Procedure Can you work out what this procedure is doing? Applications engineering Applications Technology Admin Individual Information and Communications Technology Programming Applications Maths Supporting SET Admin Stage 5 Decision mapping Applications STEM - design technology Admin No Teachers Notes