document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=562260>Question 562260</A>: <I><SPAN STYLE=\"word-break: break-all;\"> the digit in the ones place is the sum of the digits in a dozen\r<BR>\n" );
document.write( "\n" );
document.write( "please help </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #364524 by </B><A HREF=/tutors/aboutme.mpl?userid=issacodegard><B>issacodegard(60)</B></A><img src=\"/images/stars/stars-08.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=issacodegard><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=issacodegard><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=364524\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Hi, i think the problem is to find a positive integer x such that the ones place of x is the sum of the digits of 12x.  Denote this as ones(x)=dsum(12x).\r<BR>\n" );
document.write( "\n" );
document.write( "You can check that x=9 works, but i'll try to give some explanation for the problem.\r<BR>\n" );
document.write( "\n" );
document.write( "We know that the ones place of x can be 0,1,2,3,4,5,6,7,8, or 9<BR>\n" );
document.write( "If ones(x)=0, then dsum(12x)=0, which can't be true<BR>\n" );
document.write( "If ones(x)=1, then dsum(12x)>=ones(12x)=ones(12)*ones(x)=2*1=2, so it can't equal 1<BR>\n" );
document.write( "If ones(x)=2, then dsum(12x)>=ones(12x)=2*2=4, so it can't equal 2<BR>\n" );
document.write( "If ones(x)=3, then dsum(12x)>=ones(12x)=2*3=6, so it can't equal 3<BR>\n" );
document.write( "If ones(x)=4, then dsum(12x)>=ones(12x)=2*4=8, so it can't equal 4\r<BR>\n" );
document.write( "\n" );
document.write( "This at least eliminates half of the possible x's, since it shows that ones(x) must be 5,6,7,8, or 9. </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );