Number Trouble

Let $A = \overline{abc} = 100a+10b+c$ and $B = \overline{def} = 100d + 10e +f$ , then we have:

$\begin{aligned} 6(1000A+B) & = 1000B + A \\ (6000-1)A & = (1000-6)B \\ 5999A & = 994B \\ 7\dot{} 857\dot{} A & = 2\dot{} 7 \dot{} 71 \dot{} B \\ 7\dot{} 857\dot{} (2\dot{} 71) & = 2\dot{} 7 \dot{} 71 \dot{} (857) \\ \Rightarrow A = \overline{abc} & = 2\dot{} 71 = 142 \\ \Rightarrow B = \overline{def} & = 857 \\ \Rightarrow a+b+c+d+e+f & = 1+4+2+8+5+7 =\boxed{27} \end{aligned}$ .

According to the divisibility rule for 3, the sum of the digits have to be divisible by 3, and according to the divisibility rule for 9, the sum of the digits have to be divisible by 9. abcdef and defabc are just the same digits. Since defabc is a multiple of 6, it has to be divisible by 3. Since defabc's digits sum up to a multiple of 3, then abcdef's digits also sum up to a multiple of 3, and abcdef is also divisible by 3. Since abcdef is a multiple of 3, after being multiplied by 6, then defabc is divisible by 9. Using the same logic, that means that abcdef is also divisible by 9. 27 is the only answer that is a multiple of 9.