document.write( "<A HREF=/cgi-bin/jump-to-question.mpl?question=1133861>Question 1133861</A>: <I><SPAN STYLE=\"word-break: break-all;\"> a>b>c, a+b+c=de, de+d+e=fg, fg+f+g=hi. use 1 to 9 only </SPAN>\n" );
document.write( "</I><HR><TABLE width=100%><TR><TD><B><A HREF=http://www.algebra.com/>Algebra.Com's</A> Answer #751315 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=751315\">put this solution on YOUR website!</A><BR><SPAN STYLE=\"word-break: break-all;\"> <br><BR>\n" );
document.write( "a+b+c = de<br><BR>\n" );
document.write( "The sum of three 1-digit integers is a 2-digit integer.  The tens digit of that 2-digit integer MIGHT be 2; but it is much more likely to be 1.  So let's assume d=1.<br><BR>\n" );
document.write( "de+d+e = fg<br><BR>\n" );
document.write( "If d=1, then this becomes 1e+1+e = fg, or 11+2*e = fg.  That means f has to be 2.  So now we know 11+2*e = 2g.<br><BR>\n" );
document.write( "fg+f+g = hi<br><BR>\n" );
document.write( "Since f=2, this becomes 2g+2+g = hi, or 22+2*g = hi.  That means h has to be 3.  So now we know 22+2*g = 3i.<br><BR>\n" );
document.write( "So digits 1, 2, and 3 are d, f, and h; so a, b, c, e, g, and i must be, in some order, digits 4 through 9.<br><BR>\n" );
document.write( "Now we now 11+2*e = 2g; and g can't be 1, 2, or 3.  That means e has to be 7, 8, or 9.<br><BR>\n" );
document.write( "And we know 22+2*g = 3i; and i can't be 1, 2, or 3.  That means g has to be 6, 7, 8, or 9.<br><BR>\n" );
document.write( "That's all the logical analysis I can see that we can do.  Now we just need to try sets of digits for a, b, and c for which the sum a+b+c is greater than 13 and less than 20; and see where that leads us.  For each set of digits we choose for a, b, and c, we stop looking at that case when a letter needs to be a digit that has already been used.<br><PRE STYLE=\"white-space: pre-wrap; white-space: -moz-pre-wrap; white-space: -pre-wrap; white-space: -o-pre-wrap; word-wrap: break-word;\">\r\n" );
document.write( "   a, b, c    e (units digit  g (units digit  i (units digit\r\n" );
document.write( "                 of a+b+c)       of 11+2*e)      of 22+2*g)\r\n" );
document.write( "  -----------------------------------------------------------\r\n" );
document.write( "    456              5 X\r\n" );
document.write( "    457              6               3 X\r\n" );
document.write( "    458              7               5 X\r\n" );
document.write( "    459              8               7               6  << YES; this should be the solution; but let's keep checking\r\n" );
document.write( "    467              7 X\r\n" );
document.write( "    468              8 X\r\n" );
document.write( "    469              9 X\r\n" );
document.write( "    478              9               9 X\r\n" );
document.write( "    567              8               7 X\r\n" );
document.write( "    568              9               9 X<br></PRE><BR>\n" );
document.write( "There is indeed a single solution:  (a,b,c,d,e,f,g,h,i) = (9,5,4,1,8,2,7,3,6)<br><BR>\n" );
document.write( "a>b>c?  9>5>4?  YES<BR>\n" );
document.write( "a+b+c = de?  9+5+4 = 18?  YES<BR>\n" );
document.write( "de+d+e=fg?  18+1+8 = 27?  YES<BR>\n" );
document.write( "fg+f+g=hi?  27+2+7 = 36?  YES </SPAN>\n" );
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