Question 1209434
<pre>
A, B and C are distinct digits and 4 * AB = CA. Find the sum A+B+C

4AB = CA
Obviously, AB and CA are 2-digit numbers
As such. AB = 10A + B, and CA = 10C + A. So, 4AB = CA becomes:
4(10A + B) = 10C + A	
40A + 4B = 10C + A	
4B = 10C + A - 40A	
4B = 10C - 39A	
B = {{{(10C -  39A)/4	}}}
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,
as ANY digit greater than 3 would yield a negative (< 0) numerator. And, neither can digit A be 0 because, while the units
digit in CA can be 0, the tens digit in AB CANNOT.

Having said that, digit A being less than 3, not ODD, and NOT 0 makes it 2
B = {{{(10C -  39A)/4	}}}
B = {{{(10C -  39(2))/4}}} ----- Substituting 2 for A
B = {{{(10C -  78)/4	}}}
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
NOT a MULTIPLE of 4. So, C MUST be 9
B = {{{(10(9) -  78)/4}}} ------ Substituting 9 for C
B = {{{(90 -  78)/4}}}
B = {{{12/4}}}
B = 3	
Therefore, (A, B, C) = (2, 3, 9), and A + B + C = 2 + 3 + 9 = 14	

You can do the CHECK!!</pre>