document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=374811>Question 374811</A>: <I><SPAN STYLE=\"word-break: break-all;\"> how many 13 digit numbers are possible by using the digits 1,2,3,4,5 which are divisible by 4 if repetition of digits is allowed?   </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #269959 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=269959\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> The last two digits must be a multiple of 4. This means that the last two digits must be either 12, 24, 32, 44, or 52 (as these are the only two digit multiples of 4 possible with these digits)\r<BR>\n" );
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document.write( "There are five ways to choose one of these 2-digit combinations. The other 11 digits don't matter, and are mutually exclusive to each other so there are 5^11 ways to choose these 11 digits.\r<BR>\n" );
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document.write( "Therefore the total number of ways is 5*(5^11) = 5^12 = 244,140,625 ways. </SPAN>\n" );
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