Twisty Multiples of Seven

Suppose we have a number $abcd$ (concatenated digits) that has the property given in the problem. Then we can assume this number obeys the two following congruences: $10^3 a + 10^2b + 10c + d \equiv 0 \mod 7 \\ 10^3 b + 10^2 a + 10d + c \equiv 0 \mod 7$ And as $10 \equiv 3 \mod 7$ we may simplify these into: $6 a + 2b + 3c + d \equiv 0 \mod 7 \\ 6 b + 2a + 3d + c \equiv 0 \mod 7$ .

Add twice the second congruence to the first to get $3a + 5c \equiv 0 \mod 7 \implies c \equiv 5a \mod 7$ . Now plug this result into either of the previous congruences to get $2b + d \equiv 0 \mod 7 \implies d \equiv 5b \mod 7$ . We obtained basically the same result for the numbers $c,a$ and the numbers $d,b$ so here is when I start computation: Find all pairs of digits $x,y$ such that $y \equiv 5x \mod 7$ .

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The way this table works is if you choose one $x$ then choosing any of the $y$ in the following row (and of the same column) will yield a valid solution. For $b,d$ we have no restrictions and we see there are $14$ ways of picking values for them. But for $a,c$ we have the restriction that we cannot pick $a=0$ so we can't count the last column. If we ignore the last column we have just $12$ ways of picking. This gives us a total of $14 \times 12 = 168$ ways of picking.