document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=527848>Question 527848</A>: <I><SPAN STYLE=\"word-break: break-all;\"> I have tried many different ways, but i cannot figure out the answer, i have even asked friends and family to help out!\r<BR>\n" );
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document.write( "\"In the following addition problem, the letters A, B, and C stand for three different digits. What is the sum of the digits that A, B, and C represent?\"<BR>\n" );
document.write( "                                 ABC<BR>\n" );
document.write( "                               + ACB<BR>\n" );
document.write( "                               ______<BR>\n" );
document.write( "                                 CBA\r<BR>\n" );
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document.write( "I have tried certain combinations such as  123+132 = 255 , but the \"C\"'s,\"A\"'s, and \"B\"'s do not match up :(\r<BR>\n" );
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document.write( "P.S. Thank you in advance </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #349076 by </B><A HREF=/tutors/aboutme.mpl?userid=KMST><B>KMST(5328)</B></A><img src=\"/images/stars/stars-13.gif\" alt=\"\" >&nbsp;<A HREF=/tutors/aboutme.mpl?userid=KMST><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=KMST><BR>You can <A HREF=\"/cgi-bin/embed-solution.mpl?show=1&solution=349076\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> A is 4 or less because A+A<10 (because the sum has only 3 digits, not 4).<BR>\n" );
document.write( "When you add C+B in the ones' place the answer ends in A, but it must be 10+A, because the sum of the tens place (including a 1 possibly carried over) does not en in A, but ends in B.<BR>\n" );
document.write( "So B+C=10+A for the ones<BR>\n" );
document.write( "and B+C+1=10+B for the tens<BR>\n" );
document.write( "That means that B=A+1<BR>\n" );
document.write( "and A+A+1(carried over)=C<BR>\n" );
document.write( "A=4, gives you B=5, and C=9, and that is one solution.<BR>\n" );
document.write( "Is there another one?<BR>\n" );
document.write( "If A<4, then B=A+1<4+1=5, and C=A+A+1=not more than 3+3+1=7<BR>\n" );
document.write( "B+C<7+5=12=10+A, which would make A = 2 or less, but that would make<BR>\n" );
document.write( "B=A+1<4, C=A+A+1=5 or less, and then B+C<9, which does not work.<BR>\n" );
document.write( "So there is only one solution, the one listed above. </SPAN>\n" );
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