document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=26602>Question 26602</A>: <I><SPAN STYLE=\"word-break: break-all;\"> You have 13 coins, which all look alike. 12 coins weigh exactly the same, but the other is heavier. You have a pan balance.  Tell how to find the heavier coin in just 3 weighings.  Assume you can't tell just by holding the coins which on is heavier.\r<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #14452 by </B><A HREF=/tutors/aboutme.mpl?userid=kev82><B>kev82(151)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=kev82><IMG SRC=http://www.algebra.com/images/aboutme-small.gif BORDER=0 alt=\"About Me\" VALIGN=MIDDLE></A>&nbsp;<IMG SRC=http://www.algebra.com/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=14452\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi,\r<BR>\n" );
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document.write( "This is quite a famous problem. If you searched the web, you might find a better explanation than mine, but here goes.\r<BR>\n" );
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document.write( "1) Split the coins into 3 piles, two piles of six, and one pile of one. Weigh the two piles of six, if they are equal then the heavier coin must be the other one and you're done. If they're not equal then the heavier pile contains the coin.\r<BR>\n" );
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document.write( "2) Take the pile of six coins that contain the heavier one, and split it into three piles of two. Weigh two of these piles. If they are the same then the heavier coin is in the third pile, else it is in the heavier pile.\r<BR>\n" );
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document.write( "3) You should now know that the heavier coin is one of possibly two. Weigh them and the heaviest is your coin.\r<BR>\n" );
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document.write( "Hope that helps,<BR>\n" );
document.write( "Kev </SPAN>\n" );
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