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Thread: 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: .$\displaystyle a,b,c,d,e,f.$

    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.

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


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

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

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

    . . We have only 7 cases to consider.



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

    $\displaystyle 1000 \;=\;1\cdot2\cdot2\cdot2\cdot5\cdot5\cdot5$

    . . which cannot be partitioned into 6 distinct factors.



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

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

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



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

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

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



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

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

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



    Here's where it gets interesting.

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

    $\displaystyle 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 .$\displaystyle 2^m5^n$

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

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

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

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

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



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

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

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