Thread: find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)

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Hello, dave52!

Here is the first one . . .


Consider the first few powers-of-2:

. .

We see that the powers-of-2 have a cycle of 4 steps.


Now what about

We find that: .

. . . . . . And:.

. . . . Hence: . , for some integer


So we have: .

Therefore: .

There are two interpretations of 2^3^100

1. As above and with most calculators ; mod 5, this is 2 as above. So the last digit d is 2. Note to calculate 3^100 mod 4, it's probably easier to notice 3 is -1 mod 4, so 3^100 is (-1)^100 = 1 mod 4.

2. -- this is treating ^ as associating left to right; i.e. a^b^c is (a^b)^c as opposed to 1 where it's a^(b^c). Here, though, 300 is 0 mod 4 and so 2^{300} is 1 mod 5. Then the last digit d is 0 mod 2, and by the Chinese remainder theorem, d is 6.