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# What is the sum of the integers from 1 to 999 , inclusive?

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What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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22 Nov 2016, 05:50
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What is the sum of the integers from 1 to 999, inclusive?

A. 499000
B. 499500
C. 499999
D. 500500
E. 500999
[Reveal] Spoiler: OA

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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22 Nov 2016, 06:47
Bunuel wrote:
What is the sum of the integers from 1 to 999, inclusive?

A. 499000
B. 499500
C. 499999
D. 500500
E. 500999

B
number of integers = (999-1)+1=999
if you observe 1st term and last term gives you a sum of 1000
there are 999/2= 499.5 499 such pairs
so 499*1000+500= 499500

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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22 Nov 2016, 07:37
Summation of n numbers = n*(n+1)/2

If n= 1000 summation of first 1000 numbers

So 1000*1001/2 = 500500

500500-1000 as we don't need the last term.... 499500

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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22 Nov 2016, 07:58
Bunuel wrote:
What is the sum of the integers from 1 to 999, inclusive?

A. 499000
B. 499500
C. 499999
D. 500500
E. 500999

Sum of the integers from 1 to 999, inclusive = $$\frac{999 ( 999 + 1 )}{2} = 499500$$

Answer will be (B) 499500
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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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03 Jun 2017, 04:05
Bunuel wrote:
What is the sum of the integers from 1 to 999, inclusive?

A. 499000
B. 499500
C. 499999
D. 500500
E. 500999

simplest way i noticed would be:

quantity of numbers: (999-1)+1 = 999 ---> (Last number - First number) + 1

Average: (999+1)/2 = 500 ----> (Last number + First number)/2

now:
Average * quantity
999* 500

for fast calc we notice that 999 i very close to 1,000, So:
500*1000 = 500,000
500,000 - 500 = 499,500

B

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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04 Jun 2017, 16:06
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Hi All,

These types of 'bunching' questions often come down to how you choose to 'organize' the math involved.

Here, we're asked for the sum of all of the integers from 1 to 999, inclusive. We're talking about the first 999 positive integers, but I'm going to make the math a little easier by adding one extra integer that won't affect the sum at all: the number 0.

By including 0, we now have 1,000 total integers and we can 'bunch' them into groups of 2:

0 and 999 = 999
1 and 998 = 999
2 and 997 = 999
Etc.

Since we have 1,000 total numbers, there will be 500 'pairs' that add up to 999. Thus, the sum of those terms is...

(500)(999)

While that calculation might seem a little tough, you can avoid it if you think about the math in a different way...

(500)(1000) = 500,000

999 is "one less" than 1,000 so we just have to subtract "one" of those 500s from 500,000. That gives us:

500,000 - 500 = 499,500

Final Answer:
[Reveal] Spoiler:
B

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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19 Aug 2017, 00:46
Sum = number of terms * mean

- # of terms = (999-1), but we need to add '1' to the resultant as this is an inclusive set, bringing us back to '999'.
- mean of the set = first term+last term divided by 2 which gives us 500 (1+999 = 1000/2 = 500)

Putting it into the original equation:
- sum = 999 (number of terms) * 500 (mean) = 499500

Hope that helps!

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Re: What is the sum of the integers from 1 to 999 , inclusive? [#permalink]

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12 Oct 2017, 14:15
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Number of Integers (Inclusive) = N = (Last # - First #)+1

Sum of Integers = (First # + Last #)*N))/2

1. N= (999-1) + 1 = 999

2. (1+999)*999))/2 = 499,500 B

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Re: What is the sum of the integers from 1 to 999 , inclusive?   [#permalink] 12 Oct 2017, 14:15
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# What is the sum of the integers from 1 to 999 , inclusive?

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