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Math Help - Anyone know how to solve this problem?

  1. #1
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    Anyone know how to solve this problem?

    can you show the work to so i can understand it. thanks

    how many digits does the number 8^41* 5^116 have. justify your answer algebraically
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  2. #2
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    Quote Originally Posted by mcdaking84 View Post
    can you show the work to so i can understand it. thanks

    how many digits does the number 8^41* 5^116 have. justify your answer algebraically
    A number N has n digits where n = floor(log_10(N))+1 (floor the greatest integer less than or equal to its argument)

    8^41* 5^116 = [10^(log_10(8))]^41 * [10^(log_10(5))]^116

    .................. = 10^(41*log_10(8)+log_10(5)*116) ~= 10^(118.107)

    so floor(log_10(8^41* 5^116))+1 = 119, so 8^41* 5^116 has 119 digits.

    RonL
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  3. #3
    Senior Member ecMathGeek's Avatar
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    Quote Originally Posted by CaptainBlack View Post
    A number N has n digits where n = floor(log_10(N))+1 (floor the greatest integer less than or equal to its argument)

    8^41* 5^116 = [10^(log_10(8))]^41 * [10^(log_10(5))]^116

    .................. = 10^(41*log_10(8)+log_10(5)*116) ~= 10^(118.107)

    so floor(log_10(8^41* 5^116))+1 = 119, so 8^41* 5^116 has 119 digits.

    RonL
    Can you prove that theorem?
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  4. #4
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    Quote Originally Posted by ecMathGeek View Post
    Can you prove that theorem?
    Is the Pope Jewish?

    If N has 1 digit 1=<N<10, so 0<=log_{10}(N)<1 so floor(log_{10}(N))=0
    and n=floor(log_{10}(N))+1 is the number of digits of N.

    If N has n digits then 10^{n-1}<=N<10^n, so n-1<=log_{10}(N)<n, so
    floor(log_{10}(N))=n-1, and hence n=floor(log_{10}(N))+1.

    RonL
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  5. #5
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    Note that (8^41)=(2^123). Hence (8^41)(5^116) has 116 trailing zeros.
    123-116=7 and 2^7=128, so it does indeed have 119 digits.
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  6. #6
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    Hello, mcdaking84!

    It's easier than you think . . .


    How many digits does the number N .= .(8^41)(5^116) have?
    Note that: .8^41 .= . (2^3)^41 .= .2^123 .= .(2^7)(2^116)


    We have: .N .= .(2^7)(2^116)(5^116) .= .(2^7)(10^116) .= .128 x 10^116


    Therefore, N has 119 digits.


    Oops . . . exactly what Plato said . . .
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  7. #7
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    Quote Originally Posted by Soroban View Post
    Hello, mcdaking84!

    It's easier than you think . . .
    It depends on what you know. Anyone who was brought up with log books
    would know the relationship between the number of digits in front of a
    decimal point and the integer part of the log of a number >=1 virtualy
    instinctivly.

    Also the log_{10} method is more universal (it does not depend on
    special numbers).

    RonL
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