document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1187151>Question 1187151</A>: <I><SPAN STYLE=\"word-break: break-all;\"> The tens digit of a three-digit number is 0. The sum of the other digits is 6. Interchanged the units and hundreds digits decreases the number by 396. 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 #818114 by </B><A HREF=/tutors/aboutme.mpl?userid=greenestamps><B>greenestamps(13200)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=greenestamps><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=greenestamps><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=818114\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "Given the two 3-digit numbers ABC and CBA, and assuming A is greater than C, then the difference ABC-CBA is equal to 99 times (A-C).<br><BR>\n" );
document.write( "Since that difference is 396, we know A-C is 396/99 = 4.<br><BR>\n" );
document.write( "So the hundreds digit is 4 more than the units digit; and the problem says the tens digit is 0 and the sum of the other two digits is 6.  Then we have<br><BR>\n" );
document.write( "A-C = 4<BR>\n" );
document.write( "A+C = 6<br><BR>\n" );
document.write( "which gives us A=5 and C=1.<br><BR>\n" );
document.write( "ANSWER: The original number is 501<br>\r<BR>\n" );
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