- Find the smallest 10 digit number that uses all the digits 0-9, and the first N digits (taken as an N-digit number) are divisible by N.
- Now cycle the division requirement. Find the smallest 10 digit number that uses all the digits 0-9, and the first N digits are divisible by (N+k mod 10 + 1) for k = 0 to 8. (Sorry that expression looks worse than it is.)
- For example, for k=0, The first digit is divisible by 2, the first two digits are divisible by 3, ..., all 10 are divisible by 1.
- For k=7, The first digit is divisible by 9, the first two digits are divisible by 10, ..., all 10 are divisible by 8.

Source: Original, based on a previous puzzle from Martin Gardner.

Solutions were received from Kirk Bresniker, Joseph DeVincentis, Jozef Hanenberg. There may be others. I lost all saved email received prior to 7/12/05. Jozef's summary is below.

The original problem (i.e. k=9) has only one
solution: 3816547290

k=0 has only one solution: 8165432709

k=1 has only one solution: 3654729018

k=5 has two solutions: 7290183654 and
7290381654

k=8 has only one solution: 0381654729 (if a
leading zero is allowed, otherwise there's no solution)

k=2, 3, 4, 6 or 7 have no solutions

Notice the similarity between the solutions
for k=1 and the first for k=5 and also between the second for k=5 and
the solutions for k=8 and k=9

The method used is the following: look first
where 0 and 5 must be placed, next look for the places where the even
and odd digits must come: they come in an alternating pattern. Then
look for divisibility by 3 or 9, combine it with divisibility by 4 and 8
and finally you look for divisibility by 7.

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