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Math Help - The last digit of 7^7^7

  1. #1
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    The last digit of 7^7^7

    I've used modular exponentiation on 7^7 to find the last digit is 3.

    How can I use that fact to find the last digit of 7^7^7? (7 to the power of 7, to the power of 7)
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  2. #2
    Super Member TheChaz's Avatar
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    Re: The last digit of 7^7^7

    Quote Originally Posted by Celcius View Post
    I've used modular exponentiation on 7^7 to find the last digit is 3.

    How can I use that fact to find the last digit of 7^7^7? (7 to the power of 7, to the power of 7)
    I think you need more than (just) the last digit of 7^7.

    Notice that 7 = 7 (mod 10)
    7^2 = 49 = -1 (mod 10)
    7^3 = 7*-1 (mod 10) = -7 (mod 10)
    7^4 = 1 (mod 10)
    7^5 = 7 (mod 10)... now we have repeated.
    So you need to know what 7^7 is (mod 4), since the pattern is length 4. This will require the last TWO digits of 7^7, which are 43 = 3 mod 4. So the last digit of 7^(7^7) should be -7 = 3.

    I've tried to color-code a little so you can see the train of thought
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    Re: The last digit of 7^7^7

    Quote Originally Posted by TheChaz View Post
    I think you need more than (just) the last digit of 7^7.

    Notice that 7 = 7 (mod 10)
    7^2 = 49 = -1 (mod 10)
    7^3 = 7*-1 (mod 10) = -7 (mod 10)
    7^4 = 1 (mod 10)
    7^5 = 7 (mod 10)... now we have repeated.
    So you need to know what 7^7 is (mod 4), since the pattern is length 4. This will require the last TWO digits of 7^7, which are 43 = 3 mod 4. So the last digit of 7^(7^7) should be -7 = 3.

    I've tried to color-code a little so you can see the train of thought
    Not sure, I completely understand what you are saying, also I am colour blind, so I don't think I get that part either.
    Ok, so my first part, was actually using modular exponentiation to find the answer to 7^7 mod 4 = 3
    I know I said last digit, but I mixed that part up with the second part.
    But, I found that the answer is 7^7 mod 4 = 3. Now how does that fact tie into the second part?

    7^1: 7 (mod 10) = 7
    7^2: 49 (mod 10) = 9
    But what is the -1? And the next line 7^3, how did you arrive at that?
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  4. #4
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    Re: The last digit of 7^7^7

    7^1 = 7
    7^2 = 49 = 9 = -1 (mod 10). Using negative numbers isn't necessary.
    7^3 = 49*7 = 9*7 = -1*7 (mod 10) ... here is just multiplies the entire previous line by 7. This simlpifies to
    7^3 = 343 = 63 = -7 (mod 10). Notice that this also equals 3 (mod 10).

    Then 7^4 = 7 times any/all of those = 1 (mod 10
    7^5 = 7 (mod 10)
    etc.

    Since the LAST digit of powers of 7 follows this pattern, we need to know what the exponent of 7 is, mod 4. The exponent is ....43 = 3 mod 4, so

    "The last digit of 7^(7^7) is the same as the last digit of 7^43 and of 7^3. "

    3.
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  5. #5
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    Re: The last digit of 7^7^7

    Hello, Celcius!

    \text{Find the last digit of }7^{7^7}

    Be careful!

    This is not (7^7)^7 \:=\:(823,543)^7 . . . which has only 42 digits.

    Given a "stack" of exponents and no parentheses, we read "from the top down".

    The number is: . 7^{(7^7)} \:=\:7^{823,543} . . . which has nearly 700,000 digits.


    Consider the last digits of powers-of-7:

    . . \begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots && \vdots \end{array}

    We see that the last digits form a 4-step cycle: . 7, 9, 3, 1


    We find that: . 7^7 \:=\:823,543 \:=\:4(205,885) + 3

    Therefore, the last digit of 7^{7^7} is the same as 7^3: .3

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  6. #6
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    Re: The last digit of 7^7^7

    Quote Originally Posted by Soroban View Post
    Hello, Celcius!


    Be careful!

    This is not 823,543)^7 " alt="(7^7)^7 \:=\823,543)^7 " /> . . . which has only 42 digits.

    Given a "stack" of exponents and no parentheses, we read "from the top down".

    The number is: . 7^{(7^7)} \:=\:7^{823,543} . . . which has nearly 700,000 digits.


    Consider the last digits of powers-of-7:

    . . \begin{array}{ccc} 7^1 & \to & 7 \\ 7^2 & \to & 9 \\ 7^3 & \to & 3 \\ 7^4 & \to & 1 \\ 7^5 & \to & 7 \\ \vdots && \vdots \end{array}

    We see that the last digits form a 4-step cycle: . 7, 9, 3, 1


    We find that: . 7^7 \:=\:823,543 \:=\:4(205,885) + 3

    Therefore, the last digit of 7^{7^7} is the same as 7^3: .3

    Thanks for this answer, the only thing is how do I get the deduction that 7^{7^7} is the same as 7^3?
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  7. #7
    Super Member TheChaz's Avatar
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    Re: The last digit of 7^7^7

    Really? We've done this twice!
    The last digit ("last digit" means mod 10) of powers of 7:
    7
    9
    3
    1
    7
    9
    3
    1
    .
    .
    .
    7^3 has last digit 3
    7^7 has last digit 3
    7^11
    7^15
    7^19
    .
    .
    .
    7^10000000000000000000000000000000003 will also have last digit equal to three, since the exponent is congruent to 3 (mod4).
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  8. #8
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    Re: The last digit of 7^7^7

    This problem is not hard.
    We see 7 \equiv -1 \pmod{4} \Rightarrow 7^7 \equiv -1 \pmod{4}.
    Then let 7^7=4k+3.
    So 7^{7^7}=7^{4k+3}=(7^4)^k.7^3.
    From here it is easy to find next.
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