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# Sum of integers

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Manager
Joined: 29 Apr 2003
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04 Dec 2003, 06:35
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

I got this question, and was told that the both
came out to the same answers. Would anyone
show some simple workings to get the answer
instead of suming up each individual integers.

a. The sum of all integers from 19 to 59, inclusive.

b. The sum of all integers from 22 to 60, inclusive.
Director
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04 Dec 2003, 06:53
You can use arithmetic progression formula, or 19+59=78,20+58=78...or 20 pairs and one single number 39, in the middle, or 20x78+39=1599, same for the next, 19 pairs and one single number 41, in the middle,19x82+41=1599....
Manager
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04 Dec 2003, 07:27
Agree with BG. Good explanation.

Now if you look at it from logical point of view...

First, here is the number series in consideration:
19, 20, 21, 22, ..... 59, 60

It's basically going from 22 to 59 then either adding or subtracting 60 (the integer) itself from your list. If you don't include 60 (integer) then you include three numbers (19, 20, 21) that add up to 60. I hope that helps...
Manager
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04 Dec 2003, 10:03
To count the number of consequtive integers in a range, subtract the smaller from the larger and add 1.

22 - 60 = 38, 38 + 1 = 39
39 integers in above range

To get the average of a range, add the smallest and largest integer in the range and divide that amount by 2.

22 + 60 = 82, 82 / 2 = 41
41 is average value above range

To get the sum of a range, multiple the average by the number of integers contained within that range

41 * 39 = 1599
Director
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04 Dec 2003, 12:41
try:
sum of n natural numbers = n(n+1)/2

take 1-59 first, then subtract 1-18

59*60/2 - 18*19/2 = 1599
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