document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=13233>Question 13233</A>: <I><SPAN STYLE=\"word-break: break-all;\"> How many integers between 10 and 100 have exactly one digit equal to 3? </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #6728 by </B><A HREF=/tutors/aboutme.mpl?userid=rapaljer><B>rapaljer(4671)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=rapaljer><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=rapaljer><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=6728\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Ooops!!  In the posted solution for this, smik seems to have missed a few of the possibilities.  \r<BR>\n" );
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document.write( "These were listed:13, 23, 30, 43, 53, 63, 73, 83, 93 .  \r<BR>\n" );
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document.write( "Also, the numbers 31, 32, 34, 35, 36, 37, 38, and 39 should have been listed.  There is a formula of sorts, there are 9 possibilities for the units digit, and 9 possibilities for the tens digit to be a 3, and then subtract one for the case in which both digits are 3, since that case was counted twice.  So there should be 9+9-1 = 17 possibilities.  Right?\r<BR>\n" );
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document.write( "R^2 at SCC </SPAN>\n" );
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