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Math Help - An interesting Korean math olympiad problem about exponents

  1. #1
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    An interesting Korean math olympiad problem about exponents

    How can I find the last two numbers of 2^2005-2^1999 ??

    I tried everything I can but I can't think of how to approach this problem

    Any help is appreciated.,,,,,
    Last edited by FutureKSAStudent; January 13th 2010 at 02:06 AM.
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  2. #2
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    2^{2005}-2^{1999}

    =2^{1999}(2^{6}-1)
    =2^{1999}(63)

    Now we need to find the last two digits in 2^{1999}, and the problem is solved.

    The given number is divisible by 4, hence its last two digits are multiples of 4.

    So the digits are either 04,08,12,16,......96.

    We can see that the last digit in 2^{1999} clearly is 8.



    Now since the above number is divisble by four, the, possible values for the last two digits are 08,28,48,68,88.


    Now, consider the powers of 2 which have the last digit 8

    8
    128
    2048
    32768
    524288
    8388608

    We can see, that the digits 08,28,48,68,88 repeat at regular intervals


    Hence, we can obtain the last two digits in the number 2^{1999}

    which are 88

    So the last two digits in the number 2^{2005}-2^{1999} are the last two digits in the number 63X88,

    i.e. 44
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  3. #3
    Senior Member Dinkydoe's Avatar
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    We can calculate it by looking at residu-classes modulo 100
    Observe that
    2^{10}\equiv 24 mod 100
    2^{20}\equiv 24^2 \equiv 76mod 100
    2^{40}\equiv 76^2 \equiv 76 mod 100

    Thus we observe that  2^{20k + a} \equiv 2^{a}(76)mod 100

    Thus 2^{2005}-2^{1999} \equiv (2^5-2^{19})76 = 2^5(1-2^{14})76 \equiv 32(1-84)76\equiv -56\cdot 76 \equiv 44 mod 100

    Thus 44 are the last 2 digits.
    Last edited by Dinkydoe; January 13th 2010 at 04:55 AM.
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  4. #4
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    Quote Originally Posted by Dinkydoe View Post
    We can calculate it by looking at residu-classes modulo 100
    Observe that
    2^{10}\equiv 24 mod 100
    2^{20}\equiv 24^2 \equiv 76mod 100
    2^{40}\equiv 76^2 \equiv 76 mod 100

    Thus we observe that  2^{20k + a} \equiv 2^{a}(76)mod 100

    Thus 2^{2005}-2^{1999} \equiv (2^5-2^{19})76 = 2^5(1-2^{14})76 \equiv 32(1-84)76\equiv -56\cdot 76 \equiv 44 mod 100

    Thus 44 are the last 2 digits.
    This is a more formal method I guess...
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  5. #5
    Senior Member Dinkydoe's Avatar
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    This is a more formal method I guess...
    Indeed: Namely for any n we write (\mathbb{Z}/n\mathbb{Z})=\left\{\overline{0},\overline{1},\cdo  ts \overline{n-1}\right\} for the set residu-classes modulo n.
    A class \overline{a} consists of all elements in \mathbb{Z} that have the same residu by dividing through n. One may write the class    \overline{a} = a+ n\mathbb{Z}= \left\{a+ nb| b\in \mathbb{Z}\right\}

    Then for numbers k_1,k_2 we write k_1\equiv k_2 mod n when they appear in the same class, that is: have the same residu modulo n.

    One can prove that you can sum and multiply residuclasses. Thus \overline{a}+\overline{b} = \overline{a+b}. And \overline{a}\cdot\overline{b} = \overline{ab}

    What this means is: given numbers d_1, d_2. Say that d_1,d_2 divided by n have residu r_1,r_2 respectively, then d_1+d_2 has residu r_1+r_2 mod n when divided by n. Likewise: d_1d_2 has residu r_1r_2mod n when divided by n.

    So, one can do many handy calculations by looking at residu-classes.
    Last edited by Dinkydoe; January 13th 2010 at 07:32 AM.
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