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Math Help - show that 11^33-11^31 is divisible by 5

  1. #1
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    show that 11^33-11^31 is divisible by 5

    show that 11^{33}-11^{31} is exactly divisible by 5
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    Quote Originally Posted by Punch View Post
    show that 11^{33}-11^{31} is exactly divisible by 5
    Easy with modular arithmetic

    11^{33}-11^{31} \equiv 1^{33}-1^{31} \equiv 1 - 1 \equiv 0 \pmod{5}
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  3. #3
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    LOL i didnt understand a thing....

    how about using simple methods, methods that are below or equal to o levels requirement
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    Quote Originally Posted by Punch View Post
    LOL i didnt understand a thing....

    how about using simple methods, methods that are below or equal to o levels requirement
    All right how's this: We know that 11 raised to any exponent n > 0 will end in a 1. So 11^n - 11^k for n,k>0 must end in a 0, hence be divisible by both 2 and 5.
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    i still dont understand... why does 11 raised to any exponent n>0 will end in a 1?
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    Quote Originally Posted by Punch View Post
    i still dont understand... why does 11 raised to any exponent n>0 will end in a 1?
    Consider the way you do long multiplication on paper. The units digit of 11 is a 1, so when you multiply it by 11 you get units digit 1, and when you multiply that number (121) bu 11 you again get units digit 1, etc. Try it and see.
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    If you are looking for an alternative way (I do not know what counts as an o level):

    Use the exponent laws and distributive law to rewrite:

    11^{33} - 11^{31} = 11^{31}11^{2} - 11^{31} = \cdots

    Can you continue?
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  8. #8
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    11^{31}[11^2-11]
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    Quote Originally Posted by undefined View Post
    Consider the way you do long multiplication on paper. The units digit of 11 is a 1, so when you multiply it by 11 you get units digit 1, and when you multiply that number (121) bu 11 you again get units digit 1, etc. Try it and see.
    so u mean that each time i multiply by 11, i get an 1 on the end of the number and when i minus another multiple of 11, i am deducting that 1 and so the number becomes even and is divisible by 5
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    Quote Originally Posted by Punch View Post
    11^{31}[11^2-11]
    Should be 11^{31}[11^2-1]. Now evaluating 11^2 - 1 will tell you something...
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  11. #11
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    Quote Originally Posted by Punch View Post
    so u mean that each time i multiply by 11, i get an 1 on the end of the number and when i minus another multiple of 11, i am deducting that 1 and so the number becomes even and is divisible by 5
    You mean power of 11, not multiple of 11. Remember ordinary elementary school subtraction.. start at the units digit... so if the top number has a 1 and so does the bottom number, the difference must have 0 as units digit..... so it's a multiple of 10, as you say even and divisible by 5.
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