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Math Help - number theory

  1. #1
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    number theory

    What is the last two digits of this number ?

    16^198
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  2. #2
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    Quote Originally Posted by mathaddict View Post
    What is the last two digits of this number ?

    16^198
    16^198 = (2^4)^198 = 2^792.

    Now consider powers of 2 and look for the pattern in the last two digits ....
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    sorry , i still don get u
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    Quote Originally Posted by mathaddict View Post
    sorry , i still don get u
    2^2 =
    2^3 =
    2^4 =
    2^5 =
    2^6 =
    2^7 =
    2^8 =
    2^9 =
    2^10 =
    2^11 =
    2^12 =
    2^13 =
    2^14 =

    Look for a pattern in the last two digits of the value. Use that pattern to answer your question.
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  5. #5
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    Hello,

    Here is another way :
    6^1=6
    6^2=36
    6^3=216
    etc...

    More generally :
    (10k+6)^1=10k+6
    (10k+6)^2=10(10k^2+2k)+6
    etc...

    Any power of something ending with a 6 ends with a 6

    So 16^{n} ends with a 6, whatever positive integer n \ge 1 is.


    Now, what are the two last digits ?
    16 is divisible by 4. So 16^(198) is obviously divisible by 4.
    We know that a number is divisible by 4 iff its two last digits form a multiple of 4.
    There are only 5 two-digit numbers ending with a 6 and divisible by 4 :
    16,36,56,76,96.
    That's for giving you an idea.

    16^1=\bold{1}6
    16^2=2\bold{5}6
    16^3=40\bold{9}6
    16^4=16 \times 4000+16 \times 96=16 \times 4000+1536=\dots\bold{3}6
    16^5=\dots=\dots \bold{7}6
    16^6=\dots=\dots \bold{1}6
    16^5=\dots=\dots \bold{5}6

    etc...

    You can see that it repeats every 5 powers. So if the power is in the form 5k+1, the last 2 digits are 16.
    If in the form 5k+2, it'l be 56 and so on

    198=5 \times 39+3
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    Wonderful , thank you for the very precise explaination
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