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

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by BTGmoderatorDC » Tue Mar 12, 2019 9:23 pm

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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

OA B

Source: Veritas Prep
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by GMATGuruNY » Wed Mar 13, 2019 2:46 am
BTGmoderatorDC wrote:What is the sum of the integers from 1 to 999, inclusive?

A. 499000
B. 499500
C. 499999
D. 500500
E. 500999
For any set of consecutive integers:
Count = biggest - smallest + 1.
Average = (biggest + smallest)/2.
Sum = (count)(average).

In the set above:
Count = 999 - 1 + 1 = 999.
Average = (999 + 1)/2 = 500.
Sum = (999)(500) = (1000-1)(500) = 500000-500 = 499500.

The correct answer is B.
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by Scott@TargetTestPrep » Fri Mar 15, 2019 6:51 am
BTGmoderatorDC wrote:What is the sum of the integers from 1 to 999, inclusive?

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

OA B

Source: Veritas Prep
The sum of the integers from 1 to 999, inclusive, is:

sum = quantity x average = 999 x (1 + 999)/2 = 999 x 1000/2 = 999 x 500 = 499,500

Answer: B

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by [email protected] » Fri Mar 15, 2019 9:47 am
Hi All,

We're asked for the sum of the integers from 1 to 999, inclusive. This question can be approached in a number of different ways, including by "bunching"

The sum of 1 and 999 is 1,000
The sum of 2 and 998 is 1,000
The sum of 3 and 997 is 1,000
Etc.

Using the numbers from 1 to 999, inclusive, we'll have 499 'pairs' of 1000 with one extra number 'leftover', the number "in the middle": 500.
Thus, the sum is (499)(1000) + 500 = 499,500

Final Answer: B

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