Think of this problem as a reverse of Digits 1-9
N is a 10-digit positive integer which contains each of the ten
base-10 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 3-digit numbers as G.
It will be observed that a certain 3-digit 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 3-digit multiple of G which fails to occur?
: Try to derive a non computer assisted method, although computer program/spreadsheet solutions are welcome.
The last three digits of each base-10 ten-digit pandigital number is either 120, 240, 360, 480, 720, or 840. The greatest common divisor of these three-digit 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 ten-digit pandigital numbers are divisible by 9. Therefore, in order by the last nine digits of a ten-digit base-10 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 3-digit
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 ten-digit pandigital number; and, the 9, being required as distinct in the ten-digit pandigital number, can not also be located within the pandigital number’s last three digits.
The 3-digit multiple of G, with distinct digits, that fails to occurs is, then, 960
Posted by Dej Mar
on 2008-11-07 11:43:09