Find all ten digit numbers each containing the digit from 0 to 9 once and only once, with the property that the successive pairs of digits from left to right are divisible by 2,3,4,5,6,7,8,9 and 10.

[In other words: The two-digit number formed with the 'N'th digit and the '(N+1)'st digit is divisible by '(N+1)', where N = 1,2,3,4,....,9].

1) Digits 5 and 10 must be 5 and 0 respectively.

2) Digits 2, 4, 6, and 8 are even
2a) Digits 2 and and 6 are 4 and 8 in some order
2b) Digits 4 and 8 are 2 and six in some order

3) Digits 5 and 6 add to a multiple of three, so by 2a, digit 6 must be 4

x8xe54xex0

4) Digit 7 must be 2 or 9 and must be odd, so it is 9;

x8xe549ex0

5) Digit 8 must be 6 (96 is the only number between 90 and 99 divisible by 8); digit 4 then is 2

x8x25496x0

6) Digits 2 and 3 add up to a multiple of 3, so digit 3 must be 1 or 7;

x8125496x0 or x8725496x0

7) Digits 8 and 9 must add up to 9; Digit 9 is 3

7812549630 or 1872549630

Posted by TomM on 2003-05-27 07:51:13