document.write( "Question 877456: Find A, B and C in the multiplication table.\r
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Algebra.Com's Answer #529369 by Edwin McCravy(20054) $\"\"$ $\"About$

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document.write( "A B C x A A =  A C 6 C\r\n" );
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document.write( "A must be 1 because if it were as much as 2, \r\n" );
document.write( "the smallest ABC could be would be 201. Then\r\n" );
document.write( "we'd have\r\n" );
document.write( "201 x 22 = 4422 but the product would then have to \r\n" );
document.write( "start with A=2.\r\n" );
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document.write( "So A=1\r\n" );
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document.write( "So if we perform the multiplication we'd have this:\r\n" );
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document.write( "          1BC\r\n" );
document.write( "          x11\r\n" );
document.write( "          1BC\r\n" );
document.write( "         1BC0\r\n" );
document.write( "         1C6C\r\n" );
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document.write( "From the 2nd from the right column in the \r\n" );
document.write( "addition part, either B+C=6 or B+C=16\r\n" );
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document.write( "if B+C=6 then the 2nd column from the left would be 1+B=C\r\n" );
document.write( "So we'd have the system of equations \r\n" );
document.write( " But that has solution B=5/2, C=7/2,\r\n" );
document.write( "which aren't digits.\r\n" );
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document.write( "So B+C=16  and the only digits that have sum 16 are 7 and 9\r\n" );
document.write( "Then there'd be 1 to carry to the 2nd column from the left,\r\n" );
document.write( "so 1+1+B=C or 2+B=C, so B=7 and C=9.\r\n" );
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document.write( "So the solution is:\r\n" );
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document.write( "          179\r\n" );
document.write( "          x11\r\n" );
document.write( "          179\r\n" );
document.write( "         1790\r\n" );
document.write( "         1969\r\n" );
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document.write( "Edwin

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