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Divisible by 2007 (Posted on 2007-03-11)

Prove that A= 3410^n +3000^n -2964^n -1439^n is a multiple of 2007 for all positive integer values of n.

See The Solution Submitted by Dennis

Rating: 3.3333 (3 votes)

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Note that 223*9 = 2007.

So we need to show that A is separately divisible by 223 and 9 irrespective of the value of n.

Posted by K Sengupta on 2007-03-11 11:36:50

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