document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=377809>Question 377809</A>: <I><SPAN STYLE=\"word-break: break-all;\"> If 45 is added to a two-digit number, the result is a number with the same digits, but in reverse order.  The sum of the digits is 11.  What is the original number?<BR>\n" );
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document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #268589 by </B><A HREF=/tutors/aboutme.mpl?userid=richard1234><B>richard1234(7193)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=richard1234><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=richard1234><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=268589\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> Assume the original number is 10a+b where a,b are digits. Since the order of the digits is reversed upon adding 45, we have\r<BR>\n" );
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document.write( "10a + b + 45 = 10b + a\r<BR>\n" );
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document.write( "9a - 9b = -45\r<BR>\n" );
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document.write( "a - b = -5\r<BR>\n" );
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document.write( "We are also given a + b = 11. Adding the last two equations together we have 2a = 6 --> a = 3. From this we can plug into either of the last two equations to obtain b = 8. Therefore the original number is 38. </SPAN>\n" );
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