document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=262014>Question 262014</A>: <I><SPAN STYLE=\"word-break: break-all;\"> A three-digit number, which is divisible by 10, has a hundreds digit that is one less than its tens digit. The number also is 52 times the sum of its digits. Find the number. </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #195300 by </B><A HREF=/tutors/aboutme.mpl?userid=dabanfield><B>dabanfield(803)</B></A><img src=\"/images/stars/stars-11.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=dabanfield><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=195300\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A three-digit number, which is divisible by 10, has a hundreds digit that is one less than its tens digit. The number also is 52 times the sum of its digits. Find the number.\r<BR>\n" );
document.write( "\n" );
document.write( "Since the number is divisible by 10 the ones digit is 0. Let x be the tens digit and y be the hundreds digit. The number then is 100*y + 10*x + 0*1 Then we have:\r<BR>\n" );
document.write( "\n" );
document.write( "1.) y = x - 1 and<BR>\n" );
document.write( "2.) 100*y + 10*x + 0*1 = 52*(x+y+0)\r<BR>\n" );
document.write( "\n" );
document.write( "Simplifying 2.) we have:<BR>\n" );
document.write( "3.) 100*y + 10*x = 52*(x+y)\r<BR>\n" );
document.write( "\n" );
document.write( "Substituting x - 1 for y from equation 1.) in equation 3.) we have:\r<BR>\n" );
document.write( "\n" );
document.write( "100*(x-1) + 10*x = 52*(x+(x-1))\r<BR>\n" );
document.write( "\n" );
document.write( "Solve the above for x and then calculate y = x-1 </SPAN>\n" );
document.write( "</TD></TR></TABLE>\n" );