document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=167371>Question 167371</A>: <I><SPAN STYLE=\"word-break: break-all;\"> Three exteenagers find the product of their ages is 26,390. Find the sum of their ages? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #123324 by </B><A HREF=/tutors/aboutme.mpl?userid=Alan3354><B>Alan3354(69443)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=Alan3354><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=Alan3354><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=123324\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Three exteenagers find the product of their ages is 26,390. Find the sum of their ages?<BR>\n" );
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document.write( "Since they're \"exteenagers\" they're all older than 19.  The problem is to factor 26390 into 3 factors that are greater than 19.<BR>\n" );
document.write( "There's only one 2 factor in 26390, so only one factor is an even number.<BR>\n" );
document.write( "One approach is to factor it completely and the assemble 3 numbers greater than 19.<BR>\n" );
document.write( "26390 = 2 x 5 x 7 x 13 x 29<BR>\n" );
document.write( "29 is OK by itself, so make 2 numbers >19 out of the remaining 4.<BR>\n" );
document.write( "2, 5, 7 and 13<BR>\n" );
document.write( "2 times 5 or 7 is less than 20, so that's no good.<BR>\n" );
document.write( "2*13 = 26 and 5*7 = 35 works, so it's:<BR>\n" );
document.write( "26, 35 and 29<BR>\n" );
document.write( "BTW, there's no algebraic solution to this problem.  It's trial and error.<BR>\n" );
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