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Thread: Megadigits

  1. #1
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    Megadigits

    Help me please!

    How many digits are in the number 2^2009 5^2005?

    I've no idea how to begin.
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  2. #2
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    Quote Originally Posted by judocallin02 View Post
    Help me please!

    How many digits are in the number 2^2009 5^2005?

    I've no idea how to begin.
    Take logs to base 10: \log(2^{2009}\times5^{2005}) = 2009\log2 + 2005\log5. My calculator gives that as 2006.2041, which means that 2^{2009} \times 5^{2005} lies between 10^{2006} and 10^{2007}.

    Alternatively, 2^{2009} \times 5^{2005} = 2^4\times(2\times5)^{2005}, from which you can estimate the answer.

    The second method is much cleaner in this case, but the log-to-base-ten method works more generally.
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  3. #3
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    thanks
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  4. #4
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    Hello, judocallin02!

    I recommend Opalg's second method . . .


    How many digits are in the number: .  2^{2009} \cdot5^{2005} ?

    2^{2009}\cdot5^{2005} \;=\;\left(2^4\cdot2^{2005}\right)\cdot5^{2005}

    . . . . . . . = \;2^4\cdot\left(2^{2005}\cdot5^{2005}\right)

    . . . . . . . = \;2^4\cdot(2\cdot5)^{2005}

    . . . . . . . = \;2^4\cdot10^{2005}

    . . . . . . . = \;16\cdot\underbrace{1000\hdots000 }_{\text{2005 zeros}}

    . . . . . . . =\;16\underbrace{000\hdots000 }_{\text{2005 zeros}}


    Therefore, the product has {\color{blue}2007} digits.

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