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Math Help - Lock Combos

  1. #1
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    Lock Combos

    A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?

    Do I use 3C9?
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  2. #2
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    Quote Originally Posted by magentarita View Post
    A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?
    If the product is even, then at least one of the digits must be even.
    There are 10^3 possible combinations. WHY?
    There are 5^3 possible combinations having only odd digits. WHY?
    So what is the answer?
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  3. #3
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    sorry...

    Quote Originally Posted by Plato View Post
    If the product is even, then at least one of the digits must be even.
    There are 10^3 possible combinations. WHY?
    There are 5^3 possible combinations having only odd digits. WHY?
    So what is the answer?
    I don't get it.
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  4. #4
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    Quote Originally Posted by magentarita View Post
    A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?

    Do I use 3C9?
    <br />
\text{\#  of combinations where order matters: }^{\text{10}} \text{P}_\text{3}  = 720<br />

    <br />
\text{We have 5 odd numbers i}\text{.e}\text{. 1,3,5,7,9}<br />

    <br />
\text{\#  of combos of only odd numbers }^\text{5} \text{P}_\text{3}  = 60<br />

    <br />
\text{Hence \#  of even combos: }^{\text{10}} \text{P}_\text{3} \text{  -  }^\text{5} \text{P}_\text{3}  = 720 - 60 = 660.<br />

    Last edited by Math's-only-a-game; May 28th 2009 at 07:22 AM.
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  5. #5
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    Quote Originally Posted by Math's-only-a-game View Post
    <br />
\text{\#  of combinations where order matters: }^{\text{10}} \text{P}_\text{3}  = 720<br />

    <br />
\text{We have 5 odd numbers i}\text{.e}\text{. 1,3,5,7,9}<br />

    <br />
\text{\#  of combos of only odd numbers }^\text{5} \text{P}_\text{3}  = 60<br />

    <br />
\text{Hence \#  of even combos: }^{\text{10}} \text{P}_\text{3} \text{  -  }^\text{5} \text{P}_\text{3}  = 720 - 60 = 660.<br />
    Math's-only-a-game, Why do you assume that the three numbers are all different?
    Surely 525 could be the code to a combination lock.
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  6. #6
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    Quote Originally Posted by Plato View Post
    Math's-only-a-game, Why do you assume that the three numbers are all different?
    Surely 525 could be the code to a combination lock.
    Yep, you are right. I'll try again. Thanks for your correction.
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  7. #7
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    Smile ok

    I want to thank all who took time to reply, especially Plato.
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  8. #8
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    Quote Originally Posted by magentarita View Post
    A combination lock uses three digits from 0-9. If the product of the digits are even, how many lock combos are possible?

    Do I use 3C9?
    <br />
\text{Basics:}<br />

    <br />
\text{If the order doesn't matter, it is a Combination}\text{.}<br />

    <br />
\text{If the order does matter it is a Permutation}\text{.}<br />

    <br />
\text{In this case it is a permutation where repetitions are allowed}\text{.}<br />

    <br />
\text{Permutations with Repetition:}<br />

    <br />
\text{If you have n things to choose from, and you choose r of them, }<br />
    <br />
\text{then the permutations are:}<br />

    <br />
\text{n x n x }...\text{ (r times)  =  nr}<br />

    <br />
\text{(Because there are n possibilities for the first choice, }<br />

    <br />
\text{THEN there are n possibilites for the second choice, and so on}\text{.)}<br />

    <br />
\text{Hence as plato said there are }10^3  \text{lock permutations and as there are 5 odds digits }<br />
    <br />
\text{we have }5^3 \text{odd digit permutations}\text{.}<br />

    <br />
\text{Answer: }10^3  - 5^3  = 1000 - 125 = \text{875 permutations having even digits }<br />
    <br />
\text{and hence their product is even}\text{.}<br />

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