document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=4067>Question 4067</A>: <I><SPAN STYLE=\"word-break: break-all;\">  Show that the sum of any 5 consecutive counting numbers must have a factor of 5 </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #1795 by </B><A HREF=/tutors/aboutme.mpl?userid=khwang><B>khwang(438)</B></A><img src=\"/images/stars/stars-10.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=khwang><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=1795\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\">  Let the median(3rd number) of the five consecutive numbers be n, <BR>\n" );
document.write( " then the 5 numbers must be n-2,n-1,n,n+1,n+2.<BR>\n" );
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document.write( " We see that their sum S= n-2 + n-1 +n + n+1+ n+2 = <BR>\n" );
document.write( "  n-2 + + n+2 + n-1 + n+1 + n = 5n \r<BR>\n" );
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document.write( " In other words, 5 is a factor of the sum S. <BR>\n" );
document.write( " This completes the proof.\r<BR>\n" );
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document.write( " In general,for 2k+1 (odd) consecutive numbers, their sum must be<BR>\n" );
document.write( " 2k+1 times of the median.\r<BR>\n" );
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document.write( " How about the sum of even consecutive numbers?\r<BR>\n" );
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document.write( " Kenny </SPAN>\n" );
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