- #1
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 = 6n → n1 = ##\frac{200}{6}## = 33.3, so the first multiple of 6 would be at n1 = 34.
800 = 6n → n2 = ##\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?