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2014-08-18T00:29:46+08:00

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S_n =  \frac{n}{2}(a_1+a_n)

a_1=16  since it is the first multiple of 8 after 9.
a_n = 192  since it is the last multiple of 8 before 199.
n = 23  because there are a total of 23 multiples of 8 between 9 and 199.

S_{23} = \frac{23}{2}(16+192)=\frac{23}{2}(208)=23(104)=\boxed{2392}

Therefore, the sum of all he integer multiples of 8 between 9 and 199 is 2,392.

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