document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=572484>Question 572484</A>: <I><SPAN STYLE=\"word-break: break-all;\"> three consecutive numbers whose product is 40 times their sum. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #368413 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=368413\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Let a-1, a, a+1 be the numbers (I chose these instead of a, a+1, a+2 for simplicity). Their sum is 3a, so\r<BR>\n" );
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document.write( "40(3a) = (a-1)(a)(a+1)\r<BR>\n" );
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document.write( "a = 0 works. If a is not zero, then divide through by a:\r<BR>\n" );
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document.write( "120 = (a-1)(a+1) = a^2 - 1\r<BR>\n" );
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document.write( "Therefore a^2 = 121, a = 11 or -11. The numbers are either {-1,0,1}, {10,11,12} or {-12,-11,-10}. </SPAN>\n" );
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