document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1185885>Question 1185885</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the GCD of three numbers is 30 and their LCM is 900. if two of te numbers are 150 and 60 find the other possible numbers </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #816760 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=816760\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "The LCM of the three number is 900 = (2^2)(3^2)(5^2).<br><BR>\n" );
document.write( "The LCM of 60 and 150 is 300 = (2^2)(3^1)(5^2).<br><BR>\n" );
document.write( "Since the GCD of the three numbers is 30, the third number must be a multiple of 30, and it must contain an additional factor of 3.  So the third number is a multiple of 90.<br><BR>\n" );
document.write( "Finish the problem informally, simply by finding which multiples of 90 make the LCM 900 -- that is, find the multiples of 90 that are factors of 900.<br><BR>\n" );
document.write( "ANSWERS: 90, 180, and 450<br><BR>\n" );
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