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Divisible by 2007 (Posted on 2007-03-11) Difficulty: 3 of 5
Prove that A= 3410^n +3000^n -2964^n -1439^n is a multiple of 2007 for all positive integer values of n.

  Submitted by Dennis    
Rating: 3.3333 (3 votes)
Solution: (Hide)
Since (a-b) is a factor of (a^n - b^n), 446 is a factor of 3410^n - 2964^n. Also 1561 is a factor of 3000^n - 1439^n. But 446=2*223 and 1561=7*223 --> 223 is a factor of A.

Similarly, 1971 is a factor of 3410^n - 1439^n and 36 is a factor of 3000^n - 2964^n. Since 36=4*9 and 1971=9*219, 9 is a factor of A. Now 9 and 223 are relatively prime --> 9*223=2007 is a factor of A.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
SolutionSolution : With A Different MethodologyK Sengupta2007-03-14 03:41:32
re: Spoiler Ahead ! (public thank you with full solution)ken2007-03-11 14:07:25
Hints/TipsSpoiler Ahead !K Sengupta2007-03-11 11:36:50
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