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Math Help - Winning ticket (Challenging)

  1. #1
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    Winning ticket (Challenging)

    Professor Gamble buys a lottery ticket, which requires that he pick six different integers from 1 to 46 inclusive. He chooses his numbers so that the sum of the base-ten logarithm of his six numbers is an integer. It so happens that the integers on the winning ticket have the same property- the sum of the base -ten logarithms is an integer. What is the probability that Professor Gamble holds the winner ticket?
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  2. #2
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    Hello, acc100jt!

    A fascinating problem!
    I have a game plan and a weak finish.

    Well, maybe not . . .

    Professor Gamble bought a lottery ticket; he picked six different integers from 1 to 46 inclusive.
    He chose his numbers so that the sum of the base-ten logs of his six numbers is an integer.

    It so happens that the integers on the winning ticket have the same property -
    the sum of the base-ten logarithms is an integer.
    What is the probability that Professor Gamble holds the winning ticket?

    Let his six numbers be: . a,b,c,d,e,f.

    If the sum of the logs of his six numbers is an integer,
    . . then the product of the 6 numbers is a power-of-10.

    \log a + \log b + \log c + \log d + \log e + \log f \:=\:n \quad\Rightarrow\quad \log(a\!\cdot\!b\!\cdot\!c\!\cdot\!d\!\cdot\!e\!\c  dot\!f) \:=\:n

    . . Hence: . a\!\cdot\!b\!\cdot\!c\!\cdot\!d\!\cdot\!e\!\cdot\!  f \:=\:10^n


    His product ranges from a minimum of: . 1\!\cdot\!2\!\cdot\!3\!\cdot\!4\!\cdot\!5\!\cdot\!  6 \:=\:720

    . . to a maximum of: . 41\!\cdot\!42\!\cdot\!43\!\cdot\!44\!\cdot\!45\!\c  dot\!46 \:=\:6,\!744,\!109,\!680

    Hence, his product is one of: . 10^3,\;10^4,\;10^5\;\hdots \;10^9

    . . We have only 7 cases to consider.



    Can his product be 10^3 = 1000? . . . No

    1000 \;=\;1\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5

    . . which cannot be partitioned into 6 distinct factors.



    Can his product be 10^4 = 10,\!000? . . . Yes

    10,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5\c  dot5

    . . which factors into: . 1\cdot2\cdot4\cdot5\cdot10\cdot25



    Can his product be 10^5 = 100,\!000? . . . Yes

    100,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot5\cdot5\c  dot5\cdot5\cdot5

    . . which factors into: . 1\cdot4\cdot5\cdot10\cdot20\cdot25



    Can his product be 10^6 = 1,\!000,\!000? . . . Yes

    1,\!000,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot5\c  dot5\cdot5\cdot5\cdot5\cdot5

    . . which factors into: . 1\cdot5\cdot10\cdot20\cdot25\cdot40



    Here's where it gets interesting.

    Can his product be 10^7 = 10,\!000,\!000 or larger? . . . No

    10,\!000,\!000 \;=\;1\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\cdot2\:  \cdot\:\underbrace{5\cdot5\cdot5\cdot5\cdot5\cdot5  \cdot5}_{\text{seven factors of 5}}


    The only numbers from 1 to 46 of the form . 2^m5^n

    . . are: . 1, 2, 4, 5, 8, 10, 16, 20, 25, 32, 40

    The ones containing factors-of-5 are: . \begin{Bmatrix} 5 &=& 5 \\ 10 &=& 2\cdot 5 \\ 20 &=& 2^2\cdot5 \\ 25 &=& 5^2 \\ 40 &=& 2^3\cdot5 \end{Bmatrix}

    Using all of them, we can account for six factors-of-5 . . . and no more.

    Hence. 10^7 cannot be factored into six distinct factors from 1 to 46.

    . . The same holds true for: . 10^8\,\text{ and }\,10^9



    I found three outcomes that satisfy the log/sum/integer constraint.

    Therefore, the probability that Professor Gamble wins is \frac{1}{3}

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  3. #3
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    Brilliant!!! Thanks a lot!
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