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].
(In reply to
solution by Charlie)
Precisely how I attacked the problem, with of course, the same result.
I've never seen the word "prepend" before, so I guess I've even learned something in the solution!
re: solution
