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What is the sum of the integers from 100 to 200 inclusive which are divisible from 6 but not from 12?

1) 1200
2) 1350
3) 1850
4) 2550
5) 2650

This means we need to find the sum of the odd multiples of 6 from 100 to 200, inclusive, since even multiples of 6 will be divisible by 12. The first odd multiple of 6 within the range is 6 x 17 = 102, the next one is 6 x 19 = 114, etc. The last one is 6 x 33 = 198. So the odd multiples of 6 within the range are: 102, 114, 126, …, 198.

This is an arithmetic sequence of integers with a common difference of 12, so the number of integers is

(198 - 102)/12 + 1 = 96/12 + 1 = 9

The average of the integers is (198 + 102)/2 = 300/2 = 150. Thus, the sum is

150 x 9 = 1350

Answer: B/2
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Chethan92
What is the sum of the integers from 100 to 200 inclusive which are divisible from 6 but not from 12?

1) 1200
2) 1350
3) 1850
4) 2550
5) 2650

If you're unfamiliar with the formulas for arithmetic series, you can also answer the question this way...

We want the sum: 102 + 114 + 126 + . . . + 186 + 198
If we add the terms in PAIRS (starting at each end), we get: (102 + 198 ) + (114 + 186 ) + . . .
Simplify to get: 300 + 300 + ....

The question now is, "How many terms are in the sum 102 + 114 + 126 + . . . 186 + 198?

Notice that 102 = (17)(6) = (2 x 8 + 1)(6)
and 114 = (19)(6) = (2 x 9 + 1)(6)
and 126 = (21)(6) = (2 x 10 + 1)(6)
.
.
.
and 126 = (33)(6) = (2 x 16 + 1)(6)

We can see that the number of terms = the number of integers from 8 to 16 inclusive

A nice rule says: the number of integers from x to y inclusive equals y - x + 1

So, the number of terms = 16 - 8 + 1 = 9
So, there are 9 terms, which means we have 4 pairs that add to 300, and 1 value, which we can deduce must be HALF of 300

So, the SUM = (4)(300) + 150 = 1350

Answer: B

Cheers,
Brent
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