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We need to find the tens digit of 36^10

36^10 = (6^2)^10 = 6^(2*10) = 6^20

Last Two digits of 6 follow following pattern

  • Last Two Digits of 6^1 = 06
  • Last Two Digits of 6^2 = 36
  • Last Two Digits of 6^3 = 36*6 = 16
  • Last Two Digits of 6^4 = 16*6 = 96
  • Last Two Digits of 6^5 = 96*6 = 76
  • Last Two Digits of 6^6 = 76*6 = 56
  • Last Two Digits of 6^7 = 56*6 = 36 [ same as 6^2 ]
  • Last Two Digits of 6^8 = 36*6 = 16 [ same as 6^3 ]

=> So, from 6^2 cycle of tens' digit repeats as 3, 1, 9, 7 , 5, 3, 1 9, 7 , 5
=> We have a cycle of 5 from the 2nd term

=> Tens' digit of 6^20 = Now 20 = 1(6^1 not part of cycle) + 15 (cycle of 5*3) + 4 (as cycle starts from 2nd term) = Fourth term of the cycle = 7

So, Answer will be D
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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Look for the cyclicity patterns in this question.
6,36,216,1296,7776, ---56, ---36,---1,...
here leaving first number 6 , cycle repeats as 3,1,9,7,5 (5 nos)
so we have 6^10
means we have total 9 powers (leaving first number)
Remainder of 9/5 = 4
4th tens number is 7
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We can check the pattern for tenth digit for increasing powers of 6.

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Harsha
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