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Going Cyclic With Geometric (Posted on 2008-08-27) Difficulty: 3 of 5
Three distinct 3-digit positive decimal (base 10) integers P, Q and R, having no leading zeroes and with P > Q > R, are such that:

(i) P, Q and R (in this order) are in geometric sequence, and:

(ii) P, Q and R are obtained from one another by cyclic permutation of digits.

Find all possible triplet(s) (P, Q, R) that satisfy the given conditions.

Note: While the solution may be trivial with the aid of a computer program, show how to derive it without one.

  Submitted by K Sengupta    
Rating: 4.0000 (3 votes)
Solution: (Hide)
(P, Q, R) = (432, 324,243), (864, 648, 486) are the only possible solutions.

For an explanation, refer to the respective methodologies submitted by Steve Herman and Charlie in the comments.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
re: confession, solution (computer) and an observationSteve Herman2008-08-27 14:10:19
Solutionconfession, solution (computer) and an observationCharlie2008-08-27 12:18:46
Solution? (spoiler?)Steve Herman2008-08-27 12:01:52
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