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 A versatile number (Posted on 2014-10-13)
Find the biggest integer N such that:
1. All its digits are distinct.
2. Each of these digits divides N.

Only a p&p solution requested.

Bonus: What would be the answer, if one of N"s digits were exempted from being its divider?

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 Bonus only--not the actual puzzle. | Comment 1 of 6

To be divisible by 8 we need to change the last 3 digits to 120.

Divisibility by 8 assures divisibility by 2 and 4. Divisibility by 5 is obvious. As the sum of digits is 45, it's divisible by 3, 6 and 9.

What's left is 7. The number is 3 mod 7.

We need to swap two digits in the hope of reducing this by 3. We can subtract 9 times a power of ten to swap two digits.

900 mod 7 is 4
9000 mod 7 is 5
90000 mod 7 is 1
900000 mod 7 is 3

We can subtract 900000, getting 9875643120.

Unfortunately we forgot one thing: it's not divisible by zero--nothing is; we can't have a zero, and therefore can't have a number that's divisible by both 2 and 5. So I've solved only the bonus, not the puzzle itself.

For the puzzle itself, since we  can't have a number that's divisible by both 2 and 5, it's better to get rid of the 5, as 4 and 8 depend on the 2. It's hard for me to figure that without a computer.

 Posted by Charlie on 2014-10-13 08:57:02

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