4

As proven here

$3816547290$ is the only positive integer in which

  • every digit is used;
  • each digit is used only once;
  • the first $n$ digits are divisible by $n$, for $n=1,...,10$.

Is there a more "analytic" way [that is, one that does not involve guessing an trying] to prove that this number is unique and to find it?

apnorton
  • 17,156
  • 4
  • 47
  • 107
Dal
  • 7,726
  • 5
  • 38
  • 94

1 Answers1

6

You can reduce the possibilities quite quickly. Clearly the final digit has to be $0$, and then divisibility by $10$ is assured as is divisibility by $9$.

You can only have divisibility by $5$ if $5$ is in the fifth place, and divisibility by $2$ only works if the digits in even places are even.

Divisibility by $8$ means the digits in the $7, 8$ places are restricted to $32, 72, 16, 96$ because $0, 5$ are taken and there are no two consecutive even digits.

Divisibility by $4$ has the third and fourth places taken by a choice of $12, 32, 72, 92, 16, 36, 76, 96$

This means that $4, 8$ occupy places $2$ and $6$ in some order.

Then the test for divisibility by $3$ can reduce possibilities still further, and is simply applied, leaving divisibility by $7$ as the final arbiter of the remaining options.

Mark Bennet
  • 96,480
  • 12
  • 109
  • 215
  • 1
    $+1$ Thank you very much for your clear exposition, but I am aware of this way of solving the question: I was searching for a different approach (if it exists). – Dal Dec 22 '14 at 00:33