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Sum (Pair Product) = Sum (Triplet Product) (Posted on 2009-11-08) Difficulty: 2 of 5
Determine the probability that for a positive integer N chosen at random between 1000 and 9999 inclusively, the sum of the products of pairs of digits in N is equal to the sum of products of triplets of its digits.

  Submitted by K Sengupta    
Rating: 4.0000 (1 votes)
Solution: (Hide)
The required probability is 51/9000 = 17/3000 = 0.5667 0.00566 (approx.)

For a detailed explanation, refer to the solution submitted Dej Mar in this location.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionsolutionDej Mar2009-11-08 21:03:18
Solutioncomputer solutionCharlie2009-11-08 16:42:29
Questionquestions?Daniel2009-11-08 14:51:02
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