document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=266643>Question 266643</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Cory and Melissa are racing electronic cars around a circular track. They begin at the same time going in the same direction. Cory's car completes a revolution in 40 seconds, while Melissa's car completes a revolution in 35 seconds. How long will it take them before both cars reach the starting point again simultaneously? <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 #195925 by </B><A HREF=/tutors/aboutme.mpl?userid=ptaylor><B>ptaylor(2198)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=ptaylor><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/cgi-bin/show-face.mpl?user=ptaylor><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=195925\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> OK.  We need to find the lowest number that is divisible by both 40 and 35---in other words, the Least Common Multiple (LCM)<BR>\n" );
document.write( "Here's one way to do that(reduce each to prime numbers):<BR>\n" );
document.write( "40=2*20=2*5*4=2*5*2*2<BR>\n" );
document.write( "35=7*5<BR>\n" );
document.write( "Since both contain 5, we need only to use 5 once.<BR>\n" );
document.write( "so, we have 7*5*2*2*2=280 sec<BR>\n" );
document.write( "In 280 sec, Cory's car completes exactly 8 revolutions<BR>\n" );
document.write( "In 280 sec, Melissa's car completes exactly 7 revolutions<BR>\n" );
document.write( "Therefore, after 280 sec, they have both returned to the starting point at the same time.\r<BR>\n" );
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document.write( "Hope this helps--ptaylor </SPAN>\n" );
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