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The sum of 99 integers is 9999. What is the greatest integer in the se

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The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 17 Apr 2015, 12:45
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The sum of 99 consecutive integers is 9999. What is the greatest integer in the set?

1. 100
2. 120
3. 149
4. 150
5. 151
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 17 Apr 2015, 15:00
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Hi Turkish,

This question is based on a few Number Properties (and you probably already know what they are, you're just not used to thinking about then when dealing with a large group of numbers).

To start, we're told that a group of 99 CONSECUTIVE integers has a sum of 9999. This tells us a couple of things:

1) The AVERAGE of the group will be the "middle" number (meaning the 50th number) in the sequence.
2) The AVERAGE IS 9999/99 = 101

We're asked for the LARGEST number in the sequence. Since we know.....

the 50th number is 101.....

and there are 49 consecutive numbers "below it" and 49 consecutive numbers "above it"...

The LARGEST number must be 101 + 49 = 150

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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 30 Jul 2015, 08:20
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another way : uses sequence:
sum of n numbers = (n(2a+(n-1)d))/2 where n = no of terms a = 1st term , d - difference between two terms (ie d= a2-a1)

here sum = 9999 , n= 99 d = 1 (since consecutive numbers)
9999= (99/2) * (2a+(99-1)1)
from this 'a' (ie the 1st term) = 52

nth term in a sequence : nth term = a + (n-1) d
here a = 52 , n = 99 , d = 1
So nth term = 52 + (99-1)*1

therefore Nth term = 150
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 03 Jun 2016, 07:01
To do the long approach:

a + a+1 + a+2 +.... + a+98 = 9999

99a+ (sum of numbers 1 to 98) = 9999

99a + 99*98/2 = 99a + 99*49 = 99a + 4851 = 9999

99a = 5148, a = 52.

a + 98 = 150.

Answer: D
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 03 Jun 2016, 07:45
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Turkish wrote:
The sum of 99 consecutive integers is 9999. What is the greatest integer in the set?

1. 100
2. 120
3. 149
4. 150
5. 151


Let's suppose first integer is n and 99th integer is n+98

Since these are consecutive integers, Average of the set will be= 1st integer+ last integer/2

n+n+98/2= 9999/99

n+49= 101
n+98= 150

D is the answer
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 03 Jun 2016, 20:59
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Since there are 99 integers, and they are all consecutive, we can find the 50th term in the set by finding the average.

Average = Sum of Consecutive Integers/Number of Integers = 9999/99 = 101

Since 101 is the 50th term, and all of the terms are consecutive, then the largest term, or the 99th term, is equal to 101+49 = 150
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se  [#permalink]

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New post 16 Apr 2018, 00:30
There are a total of 99 numbers. The sum is 9999. Let the first number be a. The sum is given by (99/2)(2a+98)=9999.
Solving we get a=52. Last number is a+98 i.e., 150. Option D.
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Re: The sum of 99 integers is 9999. What is the greatest integer in the se &nbs [#permalink] 16 Apr 2018, 00:30
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