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n is a positive integer, what is the tens' digit of 7^n?

Last Two digits of 7 follow following pattern

  • Last Two digits of 7^1 = 07
  • Last Two digits of 7^2 = 49
  • Last Two digits of 7^3 = 49*7 = 43
  • Last Two digits of 7^4 = 43*7 = 01
  • Last Two digits of 7^5 = 01*7 = 07
  • Last Two digits of 7^6 = 07*7 = 49

=> If the power is divisible by 4 then tens digit = tens digit of of \(7^{Cycle}\) = Tens digit of \(7^4\) = 0

(1) n is divisible by 4.
SUFFICIENT as we know that tens' digit will be equal to 0

(2) n is divisible by 3.
NOT SUFFICIENT as we do not whether the tens is 3, 6, 9, etc.

So, Answer will be A
Hope it helps!

Link to Theory for Last Two digits of exponents here.

Link to Theory for Units' digit of exponents here.
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