document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1008422>Question 1008422</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The product represented by 553! is a very large number. How many terminating zeros are at the end to the answer to this factorial? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #624315 by </B><A HREF=/tutors/aboutme.mpl?userid=fractalier><B>fractalier(6550)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=fractalier><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=fractalier><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=624315\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> That would be determined by the number of factors of five in 553!.<BR>\n" );
document.write( "There are 110 factors of five, 22 factors of twenty-five (five-squared) and 4 factor of 125 (five-cubed)...thus 553! would end in 136 zeroes. </SPAN>\n" );
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