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Math Help - find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)

  1. #1
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    find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)

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  2. #2
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    Re: find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)

    Hello, dave52!

    \text{(a) Find }\,2^{3^{100}}\text{ (mod 5)}

    \text{(b) Find the last digit of }\,2^{3^{100}}

    Here is the first one . . .


    Consider the first few powers-of-2:

    . . \begin{array}{ccc} 2^1 &\equiv& 2\text{ (mod 5)} \\ 2^2 & \equiv& 4\text{ (mod 5)} \\ 2^3 &\equiv& 3\text{ (mod 5)} \\ 2^4 &\equiv& 1\text{ (mod 5)} \\ 2^5 &\equiv& 2 \text{ (mod 5)} \\ \vdots && \vdots \end{array}

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


    Now what about 3^{100}\,?

    We find that: . 3^{10} \:=\:59,\!049 \:\equiv\:1\text{ (mod 4)}

    . . . . . . And:. \left(3^{10}\right)^{10} \:\equiv\:1^{10} \:\equiv\:1\text{ (mod 4)}

    . . . . Hence: . 3^{100} \:=\:4k+1, for some integer k.


    So we have: . 2^{3^{100}} \;=\;2^{4k+1} \;=\;2^{4k}\cdot2^1 \;=\;(2^4)^k\cdot 2

    Therefore: . (2^4)^k\cdot2 \;\equiv\;1^k\cdot2\text{ (mod 5)} \;\equiv\;2\text{ (mod 5)}

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  3. #3
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    Re: find 2^(3)^(100) mod 5 and the last digit of 2^(3)^(100)

    There are two interpretations of 2^3^100

    1. As above and with most calculators 2^{3^{100}}; 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. 2^{3\cdot 100} -- 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.
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