Rotations divisible by 37

HardCombinatorics Brainteasers~10m

Let $x = \overline{37abc}$ be a $5$ -digit number (digits $a, b, c \in \{0,\dots,9\}$ ). How many such $x$ have the property that $\overline{37abc}$ , $\overline{37bca}$ , and $\overline{37cab}$ are all divisible by $37$ ?

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$1000\equiv1\pmod{37}$ , so the leading $37$ drops out: each number $\equiv$ its last three digits mod $37$ .
You get a cyclic system $26a+10b+c\equiv0$ , and its two rotations. Count digit-triples satisfying all three.

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Rotations divisible by 37

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