document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=919569>Question 919569</A>: <I><SPAN STYLE=\"word-break: break-all;\"> in a three-digit number, the hundreds digit is equal to the tens digit and is 2 more than the ones digit. The number formed by reversing the digits is 19 times the sum of the digits. Find the original number </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #557807 by </B><A HREF=/tutors/aboutme.mpl?userid=JoelSchwartz><B>JoelSchwartz(130)</B></A><img src=\"/images/stars/stars-09.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=JoelSchwartz><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/images/nopicture.jpg><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=557807\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> h=hundreds digit<BR>\n" );
document.write( "t=tens digit<BR>\n" );
document.write( "u=ones digit<BR>\n" );
document.write( "h=t<BR>\n" );
document.write( "h=2+u<BR>\n" );
document.write( "19(t+h+u)=100u+10t+h<BR>\n" );
document.write( "19(2+u+2+u+u)=100u+10(2+u)+2+u<BR>\n" );
document.write( "19(3u+4)=100u+20+10u+2+u<BR>\n" );
document.write( "57u+76=111u+22<BR>\n" );
document.write( "54=111u-57u<BR>\n" );
document.write( "54=54u<BR>\n" );
document.write( "u=1<BR>\n" );
document.write( "h=3<BR>\n" );
document.write( "t=3<BR>\n" );
document.write( "19(7)=100+30+3<BR>\n" );
document.write( "133=133<BR>\n" );
document.write( "100h+10t+u<BR>\n" );
document.write( "300+30+1=331<BR>\n" );
document.write( "The original number is 331 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );