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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.)
    Solution Alternative Methodology Comment 13 of 13 |
    (In reply to Problem Resolution by K Sengupta)

    Let the 9-digit integer represented by pqrstuvwx where each aplhabet represents a unique digit from the set of {1, 2, ...9}.

    Since pqrst must be divisible by 5, t can only be 5 or 0.
     
    However the digits are non-zero, so t = 5. Also, q, s, u, w  are even numbers since they are the last digits of numbers that are divisible by 2, 4, 6, and 8.

    We now investigate the combinations of rs and uvw where rs will be divisible by 4 and fgh will be divisible by 8. This will enable one to conclude that that s is 2 or 6, and w is 2 or 6, since r
    and v cannot be even.
    The possible values for rs are thus  12, 16, 32, 36, 52, 56, 72, 76, 92, and 96.

    The possible values for uvw are 416, 432, 472, 496, 816, 832, 872, 896.

    p+q+r must be divisible by 3 and t+u+v must be divisible by 3.
    The only possibilities for (p, q, r, s, t, u, vwx) are:
    (1,4,7,2,5,8,963), (7,4,1,2,5,8,963), (1,8,3,6,5,4,729); (3,8,1,6,5,4,729); (9,8,1,6,5,4,729);

    Among the five possibilities, only 3816547 is divisible by 7.

    Consequently the reuired number pqrstuvwx is 381654729.


      Posted by K Sengupta on 2007-06-23 05:53:34
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