document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=313391>Question 313391</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many three digit numbers are such that the product of their digits is 120? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #224067 by </B><A HREF=/tutors/aboutme.mpl?userid=toidayma><B>toidayma(44)</B></A><img src=\"/images/stars/stars-07.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=toidayma><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=toidayma><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=224067\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Firstly, we have 120 = 2^3*3*5<BR>\n" );
document.write( "since each digit is smaller than 10, therefore, one digit must be 5. We have 2 digits left with product of 2^3*3<BR>\n" );
document.write( "there are only two case for one digit that has the factor of 3: 3, 3*2 (3*2^2 is larger than 9)<BR>\n" );
document.write( "So there are totally two 3-digit sets that has the product of 120: (5,3,8) and (5,6,4)<BR>\n" );
document.write( "With each set, we have 3! different digits that satisfied the product of the three digits is 120.<BR>\n" );
document.write( "Thus, there are totally 2*3! = 12 such digits. </SPAN>\n" );
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