Sum of the pandigital numbers

MediumCombinatorics Brainteasers~7m

Consider every $9$ -digit integer that uses each of the digits $1$ through $9$ exactly once. Their total sum can be written as $k \cdot a! \cdot (10^{b} - 1)$ for integers $k, a, b$ with $a$ and $b$ as large as possible. Find $a\cdot b\cdot k$ .

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Each digit occupies each of the $9$ positions in $8!$ of the numbers, so each position's digit-total is $8!\cdot45$ .
Weight by place values $1+10+\dots+10^8=(10^9-1)/9$ ; the $45/9$ gives $k=5$ , $a=8$ , $b=9$ .

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Asked at: Jane Street, Two Sigma

Sum of the pandigital numbers

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