All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars
 perplexus dot info

 Digits 0-9 In Reverse (Posted on 2008-11-07)
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?

Note: Try to derive a non computer assisted method, although computer program/spreadsheet solutions are welcome.

 Submitted by K Sengupta No Rating Solution: (Hide) The 3-digit multiple of G which fails to occur is 960. For an explanation, refer to the solution submitted by Dej Mar in this location. Charlie extends the problem in this location giving, amongst other things, various examples of N with different end digits.

 Subject Author Date examples (spoiler) Charlie 2008-11-07 13:39:24 Solution Dej Mar 2008-11-07 11:43:09

 Search: Search body:
Forums (0)