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Thread: 'simple' proofs

  1. #1
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    'simple' proofs

    What are the last two digits of 3^1234
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by anncar View Post
    What are the last two digits of 3^1234
    $\displaystyle 3^{1234}=(3^2)^{617}=9^{617}$

    if you perform the multiplication at least 6 or 10 powers of 9 (i.e up to $\displaystyle 9^{10}$, you should notice that the last digit is 9 for odd powers and 1 for even powers; and the second to the last digit increases from 0, 2 ,4, 6, 8 then 0 again, from one odd power to the next odd power starting from 9^1, while it decreases from 8,6,4,2,0 then 8 again, from one even power to the next even power starting from 9^2..
    hence, $\displaystyle 3^{1234}=(3^2)^{617}=9^{617}$ has 69 as the last two digits..
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  3. #3
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    Hello, anncar!

    Another approach . . .


    What are the last two digits of $\displaystyle 3^{1234}$
    We have: .$\displaystyle 3^{1234} \:=\:3^{2\cdot617} \:=\:(3^2)^{617} \:=\:9^{617}\:=\:(10 - 1)^{617}$


    Then we have:

    . . $\displaystyle \underbrace{10^{617} - {617\choose1}10^{616} + {617\choose2}10^{615} + \cdots - {617\choose2}10^2}_{\text{These do not affect the last two digits}} + {617\choose616}10 - 1$

    Therefore, the number ends with: .$\displaystyle {617\choose616}10 - 1 \;=\;6170 - 1 \;=\;61{\color{blue}69} $

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  4. #4
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    Yet another approach is to use mod 100. This just depends on what course is asking you the question. If it's a number theory course, try this technique:

    Determine $\displaystyle 3^{1234} \mod 100$.

    Note that $\displaystyle \phi(100) = \phi(2^2)\cdot\phi(5^2) = 2 \cdot 20 = 40$, so $\displaystyle 3^{40} \equiv 1 \mod 100$.

    Now let's determine $\displaystyle 1234 \mod 40$. You can do this by straight pen and paper division to get $\displaystyle 1234 = 30 \cdot 40 + 34$, so $\displaystyle 1234 \equiv 34 \mod 40$.

    So now we know that $\displaystyle 3^{1234} \equiv 3^{34} \mod 100$, which is a much simpler problem to solve.

    We know $\displaystyle 3^5 = 243 \equiv 43 \mod 100$, so the problem again reduces to $\displaystyle 43^6 \cdot 3^4 \mod 100$. Note $\displaystyle 43^2 \equiv 49 \mod 100$, so we reduce further to $\displaystyle 49^3 \cdot 81 \mod 100$, and since $\displaystyle 49^2 \equiv 1 \mod 100$, then the problem simplifies to solving $\displaystyle 49 \cdot 81 \mod 100$

    Working out this calculation, you get $\displaystyle 49 \cdot 81 = 3969 \equiv 69 \mod 100$.

    So the final answer is 69.. again. --- this is a good problem showing how different areas of math can be used to solve a single problem.
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