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Thread: [SOLVED] calcul riddle

  1. #1
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    Exclamation [SOLVED] calcul riddle

    I have to find a little riddle , but i don't find a résulat.

    can you help me please ?!








    thx you!
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  2. #2
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    C'est quoi la question?
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  3. #3
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    il faut résoudre A
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  4. #4
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    Solve for A?!
    That makes no sense.
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  5. #5
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    Tu ne peux pas trouver un résultat sans avoir une question ... quelle est-elle donc ?

    You can't find a solution without having a question ... so what is it ?
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  6. #6
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    Quote Originally Posted by Bacterius View Post
    Tu ne peux pas trouver un résultat sans avoir une question ... quelle est-elle donc ?

    You can't find a solution without having a question ... so what is it ?
    The question probably is to simplify the given expression. I don't have time but someone else might chip in.
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  7. #7
    Super Member Bacterius's Avatar
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    I'll give a clue (don't have time either) :

    4^3 = 8^2

    Which makes me think this expression might simplify to a constant.
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  8. #8
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    Quote Originally Posted by Wilmer View Post
    Solve for A?!
    That makes no sense.
    It does make sense, actually; it is simply a hint that tells us the result is constant:

    1. (8^{n+1} + 8^n)^2 = ((2^3)^{n+1} + (2^3)^n)^2 = (2^{3n+3} + 2^{3n})^2 = (2^{3n}(1 + 8))^2 = \boxed{2^{6n}\cdot 81}

    2. (4^{n}-4^{n-1})^3 = ((2^2)^n - (2^2)^{n-1})^3 = (2^{2n}-2^{2n-2})^3 = (2^{2n}(1 - \frac{1}{4}))^3 = \boxed{2^{6n}(\frac{27}{64})}

    Therefore A = \frac{2^{6n}\cdot 81}{2^{6n}\cdot \frac{27}{64}} = 81 \cdot \frac{64}{27} = 3 \cdot 64 = 192

    Therefore \boxed{A \equiv 192}

    You can plug in any value of n\in \mathbb{N}, n >1 and you will see that this is what you'll get.
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  9. #9
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    Yes you've got to simplify the expression and you will find all the n's evenyuallu cancel out. I got 216 as the value for A.
    Start by writing 8 as 2^3 and 4 as 2^2.
    Then use your index laws and factorising.
    A few hints: (2^3)^(n+1) = 2^(3n+3) = 2^(3n) x 2^3.
    Give it a try....I'll check in again later if you need more help. Good luck.
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  10. #10
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    Quote Originally Posted by Debsta View Post
    Yes you've got to simplify the expression and you will find all the n's evenyuallu cancel out. I got 216 as the value for A.
    Start by writing 8 as 2^3 and 4 as 2^2.
    Then use your index laws and factorising.
    A few hints: (2^3)^(n+1) = 2^(3n+3) = 2^(3n) x 2^3.
    Give it a try....I'll check in again later if you need more help. Good luck.
    Sorry - did the calculation at the end wrong. The answer is 192.
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