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NUS School Moderator V
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What is the sum of the integers from 100 to 200 inclusive  [#permalink]

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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

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What is the sum of the integers from 100 to 200 inclusive  [#permalink]

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Afc0892 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

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

If you observe the pattern of the numbers that are divisible by 6 but not by 12 in the given range, they lie at a distance of 12 from each other consecutively.

So the series of the terms starts at 102 and ends at 198, they form an AP with common difference of 12.

$$t_{n}=a+(n-1)*d=102+(n-1)*12$$
Or, 198=102+(n-1)*12
Or, n=9

Sum, $$S_{9}=\frac{n}{2}*(a+l)$$=$$\frac{9}{2}*(102+198)=\frac{9}{2}*300=9*150=1350$$

Ans. (B)
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Re: What is the sum of the integers from 100 to 200 inclusive  [#permalink]

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Afc0892 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

We get the following numbers
102, 114,126...

a common difference of 12

a1 =102
tn =200
200 = 102 + (n-1) 12
98 =12n -12
n = 110/12 = 9.6
n = 9

sum = 9/2 ( 2*102 + (8)12) = 150 * 9 = 1350
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Re: What is the sum of the integers from 100 to 200 inclusive  [#permalink]

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Afc0892 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

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

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Re: What is the sum of the integers from 100 to 200 inclusive  [#permalink]

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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

Cheers,
Brent
_________________ Re: What is the sum of the integers from 100 to 200 inclusive   [#permalink] 27 May 2019, 09:35
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