What are the last two digits of this problem?

Question

I need to find the last two digits of 37^1992. The hint is to divide by 100 and find the remainder.

I've looked at similar questions and answers here and elsewhere, but the numbers are much simpler and no one seems to explain how to solve the problem.

Please help.

Answer 1

the remainders... modulo 100
the first column gives the remainder, the second column gives the power of 37

37 -- 1
69 -- 2
53 -- 3
61 -- 4
57 -- 5
09 -- 6
33 -- 7
21 -- 8
77 -- 9
49 -- 10 ... so 37^10 = 49 modulo 100
13 -- 11
81 -- 12 ... this means 37^12 = 81 modulo 100

also note: 49^2 = 01 modulo 100

thus 37^20 = 01 modulo 100

now 1992 = 12 modulo 20
thus

37^1992 = 37^12 = 81 modulo 100

the last two digits is 81.

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Answer 2

The answer is 81, but I cheated and programmed my computer to calculate it by force.