document.write( "Question 1202988: If A, B, and C are different digits in the addition problem below and A < B \n" ); document.write( "\n" ); document.write( " BC
\n" ); document.write( "+BC
\n" ); document.write( "1AB \n" ); document.write( "

Algebra.Com's Answer #838201 by greenestamps(13200) $\"\"$ $\"About$

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document.write( "   BC\r\n" );
document.write( " + BC\r\n" );
document.write( " ----\r\n" );
document.write( "  1AB

\n" ); document.write( "(1) In the units column, 2 times C yields final digit B; that means B must be even

\n" ); document.write( "(2) 2 times a 2-digit number yields a 3-digit number, so the 2-digit number is greater than 50, so B is greater than or equal to 5.

\n" ); document.write( "(3) (1) and (2) together mean B is either 6 or 8. That gives us 4 possibilities:

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document.write( "   63        84        68        89\r\n" );
document.write( " + 63      + 84      + 68      + 89\r\n" );
document.write( " ----      ----      ----      ----\r\n" );
document.write( "  126       168       136       178

\n" ); document.write( "In the first two cases, B > C, so those are not solutions.

\n" ); document.write( "In both of the other cases, the condition A < B < C is satisfied, so those are both solutions to the problem.

\n" ); document.write( "ANSWERS:

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document.write( "    68        89\r\n" );
document.write( "  + 68      + 89\r\n" );
document.write( "  ----      ----\r\n" );
document.write( "   136       178

\n" ); document.write( "Note we could have ruled out the first two cases above by further logical reasoning; however, with the possibilities reduced to only four, it was faster and easier simply to see which of the cases satisfied the conditions.
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