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].
For the last two digits to be divisible by 10 the number must end in zero. The third and second from the right must be divisible by 9, but the first digit must be even to accommodate the next pair's (to the left) being divisible by 8. The possibilities are
Prepending digits to complete multiples of 8 gives
(multiple-of-8 '72' left out as that involves repetition of the 7)
Prepending a digit to make a multiple of 7, without producing a duplication of digits leaves
The next digit on the left must be the end of a multiple of 5 but can't be 0, to avoid repetition and so must be 5. 56 and 59 are not multiples of 6, so the rightmost 6 digits must be
Multiples of 4 that don't repeat digits already present are 12, 72 and 28, giving us
For the last one, beginning with 2, only 12 and 72 fit as multiples of 3, but the first digit of each is odd, preventing a multiple of 2 for the leftmost pair, so that last sequence is out.
The only possible ways of fitting the remaining digits work out as
these last being the two answers.
Posted by Charlie
on 2003-05-27 03:39:19