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Difficult Math Problem #73 - Arithmetic, Number Theory

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800guy Master | Next Rank: 500 Posts Default Avatar
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Difficult Math Problem #73 - Arithmetic, Number Theory

Post Fri Dec 08, 2006 10:32 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

    (A) 1050
    (B) 10050
    (C) 5050
    (D) 5000
    (E) 50000

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    projase Junior | Next Rank: 30 Posts Default Avatar
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    Post Fri Dec 08, 2006 6:38 pm
    Those are multiples of 5 and odd numbers...

    The multiples of 5 between 1 and 1000 are 200 numbers, but you must take only the odd numbers...

    5x(1+3+5+...+197+199)=50000.

    Regards,

    PR

    800guy wrote:
    Obtain the sum of all positive integers up to 1000, which are divisible by 5 and not divisible by 2.

    (A) 1050
    (B) 10050
    (C) 5050
    (D) 5000
    (E) 50000

    _________________
    Where there's a will, there's a way...

    800guy Master | Next Rank: 500 Posts Default Avatar
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    Post Mon Dec 11, 2006 5:52 pm
    OA:

    Consider 5 15 25 ... 995
    l = a + (n-1)*d

    l = 995 = last term
    a = 5 = first term
    d = 10 = difference

    995 = 5 + (n-1)*10

    thus n = 100 = # of terms

    consider 5 10 15 20.... 995

    995 = 5 + (n-1)*5

    => n = 199

    Another approach...

    Just add up 995 + 985 + 975 + 965 + 955 + 945 = 5820, so it has to be greater than 5050, and the only possible choices left are 1) & 4)

    Also, series is 5 15 25.... 985 995

    # of terms = 100

    sum = (100/2)*(2*5 + (100-1)*10) = 50*1000 = 50000

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