Find the last two digit of 42^166

To find the last two digits of 42^166, we need to find 42^166 mod 100.

Step 1: We know that 42^166 = (40+2)^166.

Step 2: By the binomial theorem, (40+2)^166 = 40^166 + C(166, 1)40^1652 + C(166, 2)40^1642^2 + ... + 2^166.

Step 3: Notice that all terms in the expansion except the last one are divisible by 100. So, they will not contribute to the last two digits.

Step 4: The last term, 2^166, will determine the last two digits.

Step 5: To find the last two digits of 2^166, we need to find 2^166 mod 100.

Step 6: We know that 2^10 = 1024, which ends in 24.

Step 7: So, 2^20 = (2^10)^2 ends in 76 (because 24^2 = 576).

Step 8: Similarly, 2^40 = (2^20)^2 ends in 76, 2^80 = (2^40)^2 ends in 76, and 2^160 = (2^80)^2 ends in 76.

Step 9: So, 2^166 = 2^160 * 2^6 ends in 76 * 64 = 84 (because 76*64 = 4864).

Therefore, the last two digits of 42^166 are 84.