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Math Help - Proving that dividing by 5 is possible

  1. #1
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    Proving that dividing by 5 is possible

    Prove that 2^{60}+2^{59}-2^{58} can be divided by 5.
    Obviously if a number can be divided by five it either ends with a 0 or a 5. But how do you ADD numbers with exponents? I know how to divide, multiply, but not add or subtract. And my calculator doesn't fit that many numbers.
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  2. #2
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    e^(i*pi)'s Avatar
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    Quote Originally Posted by Evaldas View Post
    Prove that 2^{60}+2^{59}-2^{58} can be divided by 5.
    Obviously if a number can be divided by five it either ends with a 0 or a 5. But how do you ADD numbers with exponents? I know how to divide, multiply, but not add or subtract. And my calculator doesn't fit that many numbers.
    Use that fact that a^{b+c} = a^ba^c

    2^{60} = 2^{58 + 2} = 4 \cdot 2^{58}

    2^{59} = 2^{58+1} = 2 \cdot 2^{58}

    You can then factor out 2^{58} as a common factor: 2^{58}(4+2-1) = 5 \cdot 2^{58}

    Obviously there being a 5 means it is divisible by 5
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  3. #3
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    Quote Originally Posted by e^(i*pi) View Post
    Use that fact that a^{b+c} = a^ba^c
    But is that the same as a^b+a^c?
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  4. #4
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    e^(i*pi)'s Avatar
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    No! It most definitely isn't. For example 2^6 \neq 2^3 + 2^3 \implies 64 \neq 8+8. You may have to (re)visit the laws of exponents if you don't understand



    What I did was to say that 2^{60} = 2^{58+2} which is easy enough to verify.

    It is a standard law of exponents that a^{b+c} = a^b \times a^c. If you think about the example above: 2^6 = 2^3 \times 2^3 \implies 64 = 8 \times 8 which is fine.

    In other words 2^{60} = 2^{2+58} = 2^2 \cdot 2^{58} Likewise 2^{59} = 2^{1+58} = 2^1 \times 2^{58}

    I chose 2^{58} because that is the greatest common factor of the original expression.

    Rewriting the original equation using my subsitution gives:

    2^2 \cdot 2^{58} + 2^1 \cdot 2^{58} - 2^{58}

    Exponents behave in the same way as other numbers when it comes to factorising so you can factor out 2^{58}

    2^{58}(2^2+2^1-1)

    The part in the brackets can be simplified to 5 meaning this expression is divisible by 5
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  5. #5
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    Ok, thanks
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  6. #6
    Super Member Bacterius's Avatar
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    Or even more easily, if you're familiar with number theory :

    \varphi{(5)} = 4, Euler's Theorem states a^b \equiv a^{b \mod{\varphi{(n)}}} \pmod{n} if \gcd{(a, n)} = 1.

    2^{60}+2^{59}-2^{58} \equiv 2^{60 \mod{4}} + 2^{59 \mod{4}} - 2^{58 \mod{4}}

    Which gives : \equiv 2^0 + 2^3 - 2^2 \equiv 1 + 8 - 4 \equiv 5 \equiv 0 \pmod{5}

    Hence 2^{60}+2^{59}-2^{58} is divisible by 5.
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  7. #7
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    That was supposed to be an exercise for 14 year olds , so I don't think a 14 year old knows about number theory
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  8. #8
    Super Member Bacterius's Avatar
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    There is no age to mathematics
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