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Math Help - help on sequence

  1. #1
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    help on sequence

    I cannot get the correct answer of this problem. The answer is 445 but I got 435.


    May be I didn't understand what the last sentence meant.
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  2. #2
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    Quote Originally Posted by shrrikesh View Post
    I cannot get the correct answer of this problem. The answer is 445 but I got 435.


    May be I didn't understand what the last sentence meant.
    It is a trick.
    Find these values: a_{10},~a_{11}~\&~a_{12}.
    Do you now see where the extra 10 comes from?
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  3. #3
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    Quote Originally Posted by shrrikesh View Post
    I cannot get the correct answer of this problem. The answer is 445 but I got 435.


    May be I didn't understand what the last sentence meant.
    I got 435 too - perhaps they're somehow rounding it differently?


    edit: found it... those dastardly so and sos...
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  4. #4
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    Quote Originally Posted by Unenlightened View Post
    I got 435 too - perhaps they're somehow rounding it differently?
    edit: plato - did you get a different answer?
    The answer is 445.
    a_{10}=10,~a_{11}=10,~a_{12}=11
    That is 30 integers 1 to 29 with two 10's.
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  5. #5
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    I feel cheated
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  6. #6
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    445 is correct.
    If n=10: |190/19| = 10
    If n=11: |219/20| = 10

    Missed one of those?

    Your answer of 435 is sum of digits from 1 to 29 ; 29*(30/2) = 435.
    So you're missing a term, since you need 30 terms.
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  7. #7
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    Quote Originally Posted by shrrikesh View Post
    I cannot get the correct answer of this problem. The answer is 445 but I got 435.


    May be I didn't understand what the last sentence meant.
    First, make a simple conversion

    \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor }  = \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{n\left( {n + 9} \right) + 10 - n}}{{n + 9}}} \right\rfloor }  = \sum\limits_{n = 1}^{30} {\left\lfloor {n + \frac{{10 - n}}{{n + 9}}} \right\rfloor }  =

    = \sum\limits_{n = 1}^{30} n  + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor }  = 465 + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor } .

    \frac{{10 - n}}{{n + 9}} = 0 \Leftrightarrow n = 10

    Therefore, we consider two cases:

    1) if n = \left\{ {1,{\text{ }}2,{\text{ }} \ldots ,{\text{ }}10} \right\} then \left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor  = 0.

    2) if n = \left\{ {11,{\text{ 1}}2,{\text{ }} \ldots ,{\text{ }}30} \right\} then \left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor  =  - 1.

    So we have

    \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor }  = \sum\limits_{n = 1}^{10} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor }  + \sum\limits_{n = 11}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor }  = 0 + \left( { - 20} \right) =  - 20.

    Finally we have

    \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{{n^2} + 8n + 10}}{{n + 9}}} \right\rfloor }  = \sum\limits_{n = 1}^{30} n  + \sum\limits_{n = 1}^{30} {\left\lfloor {\frac{{10 - n}}{{n + 9}}} \right\rfloor }  = 465 + \left( { - 20} \right) = 445.
    Last edited by DeMath; July 29th 2009 at 01:55 PM. Reason: refinement
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