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

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To find the sum of numbers between 200 and 800 that are multiples of 6 but not multiples of 9, the first step is identifying the range of multiples of 6. The first multiple of 6 in this range is 204 (from n=34) and the last is 792 (from n=133). The total count of these multiples is 99, leading to an initial sum of 49,302. To exclude multiples of 9, the least common multiple of 6 and 9 must be calculated, and the relevant multiples subtracted from the total. The discussion highlights the importance of accurately determining the number of terms and the correct arithmetic series sum.
cmkluza
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Homework Statement


Find the sum of the numbers between 200 and 800 inclusive, which are multiples of 6, but not multiples of 9.

Homework Equations

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?
 
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cmkluza said:

Homework Statement


Find the sum of the numbers between 200 and 800 inclusive, which are multiples of 6, but not multiples of 9.

Homework Equations

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?
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:
cmkluza said:
133 - 34 = 99.
How many numbers in the range 1 to 2?
 
Σ(multiples of six)-Σ(multiples of LCM(6,9))
 
What must be true about n, if 6n is to be divisible by 9 ?
 

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