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

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Homework Help Overview

The problem involves finding the sum of numbers between 200 and 800 that are multiples of 6 but not multiples of 9. Participants are exploring the arithmetic properties of these numbers and the implications of divisibility.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss identifying multiples of 6 within the specified range and calculating their sum using arithmetic series. There is also a focus on how to exclude multiples of 9 from this sum. Questions are raised about the least common multiple of 6 and 9 and the conditions for divisibility.

Discussion Status

The discussion is ongoing, with participants attempting to clarify their calculations and reasoning. Some have pointed out potential errors in previous calculations, and there is a shared interest in understanding the relationship between the multiples of 6 and 9.

Contextual Notes

Participants are working under the constraints of the problem statement, which specifies the range and the conditions for divisibility. There is an indication that some calculations may need to be revisited for accuracy.

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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