document.write( "Question 1151097: A, B, C and D are distinct digits, and 4 x AAB = CDA. If C is less than D, find the sum A+B+C+D. \n" ); document.write( "

Algebra.Com's Answer #772721 by ikleyn(52786) $\"\"$ $\"About$

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document.write( "Since AAB is 3-digit number and CDA is 3-digit number, there are only 2 possibilities for A:  A= 1  or  A= 2.\r\n" );
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document.write( "But A is the last digit of the number CDA, which is multiple of 4; so it leaves only one possibility for A to be 2:  A = 2.\r\n" );
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document.write( "Then  B = 3  or  B = 8.\r\n" );
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document.write( "Case 1).  If B = 3, then  4 * 223 = 892;  so,  C= 8  and  D= 9.\r\n" );
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document.write( "Case 2).  If B = 8, then  4 * 228 = 912;  so,  C= 9  and  D= 1.\r\n" );
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document.write( "The condition  C < D is satisfied in case 1 ONLY;  so, case 2) is denied.\r\n" );
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document.write( "Therefore,  A+B+c+D = 2+3+8+9 = 22.    ANSWER\r\n" );
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