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    Numerical problem


    Guys, please help me in the following problem.

    How many bits are required to represent a 3 digit pin?
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    1. How many possible "digit"s are there?
    2. How many possible bits are there?
    3. What's the lowest power of (answer from #2) that is at least as much as (answer to #1)?
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    A digit can be from 0 to 9. Total no of digits are 3.
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    So that's 10^3=1000 combinations, right? What's the lowest power of 2 that's >= 1000? And how many bits is that?
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    2 raised to the power 10 is 1024.
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    Originally Posted by sargun70
    2 raised to the power 10 is 1024.
    Yes there are 1000 combinations.
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    Seems I have to be more and more exact... I meant > not >=.

    Have you figured out the answer yet? Do you know why the answer is not 11?
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    Yes I get it. The answer is 10.

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