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Question

Let n(>1) be a positive integer, then the largest integer m such that (nm+1) divides (1+n+n2+....+n127) is 
  1. 63
  2. 32
  3. 64
  4. 127


Solution

The correct option is C 64
Since (nm+1) divides (1+n+n2+....+n127)
Therefore 1+n+n2+...+n127nm+1 is an integer
1n1281n×1nm+1 is an integer
(1n64)(1+n64)(1n)(nm+1)
is an integer when largest m=64.
 

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