document.write( "Question 982668: How many three-digit numbers are there in which one of the digits is the sum of the other two? \n" ); document.write( "

Algebra.Com's Answer #603492 by Edwin McCravy(20059) $\"\"$ $\"About$

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How many three-digit numbers are there in which one of the digits is the sum of
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document.write( "Case 1. Those which contain one 0, such as 404. (None can contain more than \r\n" );
document.write( "one 0).  The other two digits are the same. We can choose that non-zero digit 9\r\n" );
document.write( "ways.  The non-zero digit must come first. So for each of those there are\r\n" );
document.write( "2! = 2 ways the other two digits can be arranged.  So Case 1 contains 2∙9 = 18\r\n" );
document.write( "ways.   \r\n" );
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document.write( "Case 2.  Those which have two digits alike.\r\n" );
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document.write( "They can only be like 112, 224, 336, 448.  Each of those 4 can be arranged in \r\n" );
document.write( "3!/2! = 6/2 = 3 ways. That's 12 for Case 2.\r\n" );
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document.write( "Case 3.  Those which have no digits alike.  We list the ways where the digits\r\n" );
document.write( "are ascending.\r\n" );
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document.write( "123, 134, 145, 156, 167, 178, 189\r\n" );
document.write( "235, 246, 257, 268, 279\r\n" );
document.write( "347, 358, 369\r\n" );
document.write( "459\r\n" );
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document.write( "That's 16.  Each can be arranged 3!=6 ways.  16∙3! = 16∙6 = 96 ways for Case 3.\r\n" );
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document.write( "Total for all three cases = 18+12+96 = 126.\r\n" );
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document.write( "Edwin

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