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Question

If p(n) = 10n+3×4n+1+p is divisible by 3 for all values of n  N. Then find the least positive value of p for which p(n) is true. 
  1. 2
  2. 3
  3. 5
  4. 1


Solution

The correct option is A 2
Since p(n) = 10n+3×4n+1+p is divisible by 3 for all values of n  N, this implies that it is true for n = 1.
 Now, 101+3×41+1+p      = 58+p
Nearest value of 58 + p to be divisible by 3 and for which p is positive is 60.
Therefore, p = 2.

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