document.write( "Question 1209434: A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C \n" ); document.write( "

Algebra.Com's Answer #848915 by MathTherapy(10555) $\"\"$ $\"About$

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document.write( "A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C\r\n" );
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document.write( "4AB = CA\r\n" );
document.write( "Obviously, AB and CA are 2-digit numbers\r\n" );
document.write( "As such. AB = 10A + B, and CA = 10C + A. So, 4AB = CA becomes:\r\n" );
document.write( "4(10A + B) = 10C + A	\r\n" );
document.write( "40A + 4B = 10C + A	\r\n" );
document.write( "4B = 10C + A - 40A	\r\n" );
document.write( "4B = 10C - 39A	\r\n" );
document.write( "B = \r\n" );
document.write( "Now, we can see that C - 39A is a MULTIPLE of 4, so A CANNOT be an ODD DIGIT. Furthermore, the digit A MUST be less than 3,\r\n" );
document.write( "as ANY digit greater than 3 would yield a negative (< 0) numerator. And, neither can digit A be 0 because, while the units\r\n" );
document.write( "digit in CA can be 0, the tens digit in AB CANNOT.\r\n" );
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document.write( "Having said that, digit A being less than 3, not ODD, and NOT 0 makes it 2\r\n" );
document.write( "B = \r\n" );
document.write( "B =  ----- Substituting 2 for A\r\n" );
document.write( "B = \r\n" );
document.write( "At this point, 10C MUST be greater than 78, which means that C can be 8 or 9. But, if C = 8, then numerator becomes 2, and 2 is\r\n" );
document.write( "NOT a MULTIPLE of 4. So, C MUST be 9\r\n" );
document.write( "B =  ------ Substituting 9 for C\r\n" );
document.write( "B = \r\n" );
document.write( "B = \r\n" );
document.write( "B = 3	\r\n" );
document.write( "Therefore, (A, B, C) = (2, 3, 9), and A + B + C = 2 + 3 + 9 = 14	\r\n" );
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document.write( "You can do the CHECK!!

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