document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=442481>Question 442481</A>: <I><SPAN STYLE=\"word-break: break-all;\"> show that a number and it's cube leaves the same remainder when divided by 6<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 #305217 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=305217\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> If x^3 and x leave the same remainder upon dividing by 6, then x^3 &#8801; x (mod 6) --> x^3 - x &#8801; 0 (mod 6).\r<BR>\n" );
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document.write( "The left expression factors to\r<BR>\n" );
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document.write( "x(x^2 - 1) = x(x+1)(x-1). At least one of these numbers must be even, and exactly one of them is divisible by 3, so x(x+1)(x-1) &#8801; 0 (mod 6), so our original claim is true. </SPAN>\n" );
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