The nine-digit number whose every prefix divides

Question

Find a nine-digit number that uses each of the digits 1 to 9 exactly once, such that: - its first digit is divisible by 1, - its first two digits form a number divisible by 2, - its first three digits form a number divisible by 3, - and so on, up to the whole nine-digit number being divisible by 9. Find it, and show how the divisibility rules cut the search down to almost nothing.

Accepted Answer

381,654,729. Position 5 is 5. The first five digits form a multiple of 5, and 0 is not available, so the fifth digit is 5. Even positions hold even digits. The 2-digit, 4-digit, 6-digit and 8-digit prefixes are even, so digits 2, 4, 6 and 8 occupy positions 2, 4, 6 and 8 in some order, and digits 1, 3, 7, 9 occupy positions 1, 3, 7, 9. Rules for 4 and 8. A number is divisible by 4 when its last two digits are; the 4-digit prefix ends in (odd digit, even digit), and an odd ten's digit followed by an even unit's digit is divisible by 4 only if the unit's digit is 2 or 6. So position 4 is 2 or 6…

The nine-digit number whose every prefix divides

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