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N is a number where each digit is divisible by its place, for example:

1527

1 has to be divisible by 1, 5 divisible by 2, 2 divisible by 3, and 7 divisible by 4.

Additionally, N must be divisible by all of its digits.

It has to be at least 5 digits long, and have at least 3 different digits.

Does N exist? Is there more than 1 number that fits N? If so is there a pattern? Give a few examples.

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  • 3
    $\begingroup$ The identity that $n$ divides $n$ makes this very, very easy. $\endgroup$ – Aza Nov 2 '14 at 1:37
  • $\begingroup$ @Emrakul Exactly what do you mean by that... $\endgroup$ – warspyking Nov 2 '14 at 9:45
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If such an N exists, it is at least 5 digits long, so it has a 2nd digit and a 5th digit.
The second digit must be divisible by 2, and the 5th digit must be divisible by 5.
N is divisible by both of these, so it must be divisible by 10.
Then the last digit of N is 0.
But no positive number can be divisible by 0, so N cannot exist.

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