Think of this problem as a reverse of
Digits 19.
N is a 10digit positive integer which contains each of the
ten base10 digits
0 to 9 exactly once, such that:
(i) The number formed by the
last two digits is divisible by 2.
(ii) The number formed by the
last three digits is divisible by 3.
(iii) The number formed by the
last four digits is divisible by 4.
and so on
up to ten digits...
We will now consider the last three digits of each of the values of N that satisfy all the given conditions and regard them as 3 digit numbers. Let us denote the
greatest common divisor of all these 3digit numbers as G.
It will be observed that a certain 3digit multiple of G (with no leading zero),
each of whose digits are distinct, will
never occur as the last three digits of any of the possible values of N.
What is that 3digit multiple of G which fails to occur?
Note: Try to derive a non computer assisted method, although computer program/spreadsheet solutions are welcome.
The last three digits of each base10 tendigit pandigital number is either 120, 240, 360, 480, 720, or 840. The greatest common divisor of these threedigit numbers is 120. Dividing each by this GCD results in 1, 2, 3, 4, 6, and 7, respectively. 5 is skipped as 120 x 5 equals 600. This GCD multiple is excluded, as not all digits are distinct.
Others have noted that all tendigit pandigital numbers are divisible by 9. Therefore, in order by the last nine digits of a tendigit base10 pandigital number to be divisible by 9 – a condition required of this problem  the first digit will need be a 9. Thus, the last remaining
3digit multiple of the GCD, 8 x 120, with the product 960, can not occur as the 9 is delegated as the first digit of the tendigit pandigital number; and, the 9, being required as distinct in the tendigit pandigital number, can not also be located within the pandigital number’s last three digits.
The 3digit multiple of G, with distinct digits, that fails to occurs is, then,
960.

Posted by Dej Mar
on 20081107 11:43:09 