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Sum of range of numbers divisble by 6 but not by 9

  1. Jan 22, 2016 #1
    1. The problem statement, all variables and given/known data
    Find the sum of the numbers between 200 and 800 inclusive, which are multiples of 6, but not multiples of 9.

    2. Relevant equations


    3. The attempt at a solution
    Numbers that are multiples of 6 should be: a = 6n, n ∈ ℤ and a is any multiple of six.
    200 = 6nn1 = ##\frac{200}{6}## = 33.3, so the first multiple of 6 would be at n1 = 34.
    800 = 6nn2 = ##\frac{800}{6}## = 133.3, so the last multiple of 6 would be at n2 = 133.
    I'm then taking the sum of an arithmetic series starting at n1 = 34 and ending at n2 = 133 with a common difference d = 6. If 34 is the starting point then the number of terms I'm looking for, n, is 133 - 34 = 99.

    ##a_1 = 6(34) = 204##
    ##a_{99} = 204 + (99-1)6 = 204 + 588 = 792##
    ##S_{99} = \frac{99(204 + 792)}{2} = 49,302##.

    How would I go about finding the sum for all numbers divisible by both 6 and 9, between 200 and 800, so as to take it out of my current sum, finding the answer?
     
  2. jcsd
  3. Jan 22, 2016 #2

    Samy_A

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    What is the least common multiple of 6 and 9?

    By the way, 49302 is not the sum of all multiples of 6 between 200 and 800. (Doesn't that 792 look suspicious?)
     
    Last edited: Jan 22, 2016
  4. Jan 22, 2016 #3

    haruspex

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    How many numbers in the range 1 to 2?
     
  5. Jan 22, 2016 #4
    Σ(multiples of six)-Σ(multiples of LCM(6,9))
     
  6. Jan 22, 2016 #5

    SammyS

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    What must be true about n, if 6n is to be divisible by 9 ?
     
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