Luke's Cryptarithm

• $300000A$ + $30000B$ + $3000C$ + $300D$ + $30E$ + $3F$ = $100000B + 10000C$ + $1000D + 100E + 10F + A$
• $299999A = 70000B + 7000C + 700D +70E + 7F$
• $42857A = 10000B + 1000C + 100D + 10E + F$
• $A$ can only be one or two as if $A > 3, B > 10.$
• If $A = 1$ , $B = 4$ , $C = 2$ , $D = 8$ , $E = 5$ , $F = 7$
• If $A = 2, B = 8, C = 5, D = 7, E = 1, F = 4$
• $1 + 4 + 2 + 8 + 5 + 7 = 27$

$\begin{aligned} \overline{A\blue{BCDEF}} \times 3 & = \overline{\blue{BCDEF}A} & \small \blue{\text{Let }N = \overline{BCDEF}} \\ 3\times 10^5 A + 3 \blue N & = 10 \blue N + A \\ 7 N & = 299999 A \\ \implies N & = 42857 A \\ \overline{BCDEF} & = 42857 A \end{aligned}$

This means that $A=1$ or $2$ , because if $A \ge 3$ , $\overline{BCDEF}$ would be $6$ digits. Then we have:

$\begin{cases} A = 1 & \implies \overline{BCDEF} = 42857 \implies \overline{ABCDEF} \times 3 = 142857 \times 3 = 428571 = \overline{BCDEFA} \\ A = 2 & \implies \overline{BCDEF} = 85714 \implies \overline{ABCDEF} \times 3 = 285714 \times 3 = 857142 = \overline{BCDEFA} \end{cases} \\ \implies \begin{cases} A = 1 & \implies A+B+C+D+E+F = 1+4+2+8+5+7 = 27 \\ A = 2 & \implies A+B+C+D+E+F = 2+8+5+7+1+4 = 27 \end{cases}$

The required answer is $\boxed{27}$ .

{ ABCDEF } × 3 = { BCDEFA }

Rewriting the above while giving each digits their own 'house' values, we get :

300000A + 30000B + 3000C + 300D + 30E + 3F = 100000B + 10000C + 1000D + 100E + 10F + A

Rearranging the above to make the same variables to stay on the same sides (with the smaller valued multiples migrating to where the bigger ones' original sides), we get :

(300000 – 1) × A = (10 – 3) × (10000B + 1000C + 100D + 10E + F)
299999A = 7(10000B + 1000C + 100D + 10E + F)
42857A = 10000B + 1000C + 100D + 10E + F

Now, every variable is representing a one digit number, so
A, B, C, D, E & F < 10
and
A > 0 , B > 0

If B < 10,
then A < 10 × 10000 / 42857 = 2.333341

A = 1 will give B = 4, C = 2, D = 8, E = 5 & F = 7
while on the other hand,
A = 2 will give B = 8, C = 5, D = 7, E = 1 & F = 4

Anyway, the answer would be
= A + B + C + D + E + F
= 1 + 4 + 2 + 8 + 5 + 7
= 2 + 8 + 5 + 7 + 1 + 4
= 27