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Math Help - 4 appearing in the integers 1-1000000

  1. #1
    Member oldguynewstudent's Avatar
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    4 appearing in the integers 1-1000000

    You write down all of the integers from 1 to 1,000,000. How many times did you write the digit 4?

    I get \sum_{n=0}^{6}10^{n}

    Is this correct?
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by oldguynewstudent View Post
    You write down all of the integers from 1 to 1,000,000. How many times did you write the digit 4?
    One integer in every ten has a 4 in the units position. That gives 10^5 4s in the units position (in the integers from 1 to 10^6). Ten integers in every hundred have a 4 in the tens position (in other words, each of the integers ending in 40, 41, ..., 49). That gives 10\times10^4 = 10^5 (again) 4s in the tens position. And so on. There will be a total of 10^5 4s in the hundreds position, in the thousands position, in fact in each of the six positions in a six-digit number. So the total number of 4s that you would have to write, if you wrote down all the integers from 1 to a million would be 600,000.
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  3. #3
    Member oldguynewstudent's Avatar
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    Thank you so much. As always I was on the right track but off a little.

    The neurons are willing but the dendrites are weak!
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  4. #4
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    4XXXXX

    X may be any value from 0 to 9.
    This counts the number of 4's in the given position from 1 to a million, which is 10^5

    Y4XXXX

    Y can be any value from 1 to 9, X can be any value from 0 to 9.
    However Y can also be missing, hence it has 10 possibilities.

    The total number of 4's in this position is (9)10^4+10^4=(10)10^4=10^5

    YY4XXX

    Y can be any value from 1 to 9 or missing.
    X can be any value from 0 to 9.

    The number of 4's in this position is 10^210^3=10^5

    YYY4XX

    The number of 4's in this position is 10^310^2=10^5

    YYYY4X

    The number of 4's in this position is 10^410=10^5

    YYYYY4

    The number of 4's in this position is 10^5

    The total number of 4's is (6)10^5
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