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Digits 1-9 (Posted on 2002-06-05) Difficulty: 3 of 5
Can you arrange the digits 1-9 in any order so that:
  • The number formed by the first two digits is divisible by 2.
  • The number formed by the first three digits is divisible by 3.
  • The number formed by the first four digits is divisible by 4
  • and so on up to nine digits...
  • See The Solution Submitted by Rhonda Wendel    
    Rating: 3.1429 (7 votes)

    Comments: ( Back to comment list | You must be logged in to post comments.)
    Hints/Tips Eliminate more; Test less(Part 1) | Comment 5 of 13 |
    (In reply to re: First thoughts (again) by Nick Reed)

    (The system refused my post. I assume it was because of its length. So I am posting in parts)The number must meet the following constraints:

    1) the fifth digit must be 5

    2)the even digits mus be even, and
    2a) the fourth and eigth digits must be 2 and 6 in some order, so
    2b) the second and sixth digits must be 4 and 8 in some order, and
    2c) the first, third, seventh and ninth must be 1,3,7and 9 in some order

    3) [This is a consequence of the original constraints that the first 3, 6, and 9 digits are divisible by the specified number of digits] Breaking up the number into three three-digit numbers, the three numbers must each be divisble by 3.

    Consider the middle number (digits 4,5,6) By constraint 2, it must be 254, 258, 654, or 658, but by constraint 3 it cannot be 254 or 658. Furthermore, once digits 4 and six are decided, digits 2 and eight are determined. So the number looks like either this: x4x258x6x or this: x8x654x2x


      Posted by TomM on 2002-06-07 14:53:20

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