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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 Full solution | Comment 5 of 7 |
OK, yes, I made a mistake in my analysis.  Here is a revised solution:

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 between 100000 and 999999
    And qn is between 1000001 and 1000009
   
   It follows that n must be between 2 and 10
   
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
   
5) This leaves a small number of possibilities, when k divides 1000000 and n divides (1000000 + k):

   k 1000000 + k   n  q       p
   --   --------------  --- ---------   ---------
   2 1000002      3   333334 166667   <-- only solution
   2 1000002      6   166667  83333.5     <-- p is not integral
   4 1000004      4   250001  62500.25   <-- p is not integral
   5 1000005      5   200001  40000.2     <-- p is not integral
   8 1000008      3   333336   41667      <-- p too small
   8 1000008      4   250002  31250.25   <-- p is not integral
   8 1000008      6   166668  20833.5     <-- p is not integral
   8 1000008      8   125001  15625.125 <-- p is not integral
   8 1000008      9   111112   13889      <-- p too small

So, Dej Mar's solution is the only one

  Posted by Steve Herman on 2010-06-29 17:04:49
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