document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=923015>Question 923015</A>: <I><SPAN STYLE=\"word-break: break-all;\"> What is the smallest natural k for which k! is divisible by 2016?\r<BR>\n" );
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document.write( "The topic is on modulo arithmetic and I really hope you can solve it:) </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #559821 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=559821\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The prime factorization of 2016 is 2^5 * 3^2 * 7.\r<BR>\n" );
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document.write( "Therefore k must be at least 7, but 7! only has four 2's in its prime factorization (2, 2*2, 2*3) but enough 3's. Setting k = 8 gives the desired result, as 8! = 2016*20. </SPAN>\n" );
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