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Concatenated Numbers and Divisibility (Posted on 2010-06-29) Difficulty: 2 of 5
Each of p and q is a 6-digit base ten positive integer with no leading zero. The 12-digit number that is obtained by writing p and q side-by-side is divisible by the product p*q.

Determine all possible pair(s) (p, q) for which this is possible.

See The Solution Submitted by K Sengupta    
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Solution Improved Solution Comment 7 of 7 |
1) n = (1000000p + q)/pq , where n is a integer
    
2) Multiplying by q,     
   qn = 1000000 + (q/p)
   
   Since qn is an integer, so is q/p.
   
   Let q/p = k, where k must be an integer between 1 and 9.
   
3) Since q is <= 999999
    And qn >= 1000001
   
   It follows that n must be >= 2
   
4) q = (1000000 + k)/n
    p = (1000000 + k)/kn
   
   Since p is integral, k must divide 1000000.
   k must be 1,2,4,5 or 8

   But p <= 999999, so kn must be <10

   But n > 2, so k must be 1,2 or 4
   
5) There are only three possibilities where 
    a) k is 1,2 or 4
    b) n < 10/k, and
    c) n divides (1000000 + k), 
   and only one leads to an integral p:

   k 1000000 + k   n  q       p
   --   --------------  --- ---------   ---------
   2 1000002      3   333334 166667   <-- only solution
   2    1000002        2    500001  250000.5  <-- p not integral
   4    1000004        2    500002  125000.5  <-- p not integral

So, Dej Mar's solution is the only one

  Posted by Steve Herman on 2010-06-30 11:56:23
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