Chethan92 wrote:
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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