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The sum of n consecutive positive integers, where n > 1, is 2020.

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Bunuel
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The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 02:12
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The sum of n consecutive positive integers, where n > 1, is 2020. If n is odd, what is the value of n ?

A. 3
B. 5
C. 7
D. 101
E. 505


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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 03:16
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As \(n\) is odd, \(\frac{2020}{n}\) needs to divide without any remainders and will give us the median number. The only answer choices which divide \(2020\) without a remainder are B. 5 & D. 101.

Choice D. will not work. The median will be \(\frac{2020}{101} = 20\) with 50 integers on either side. This means that the sequence will include zero and negative numbers, while we are told that the sum of n consecutive positive integers = 2020. Therefore, the answer has to be B..

To double check: \(\frac{2020}{5}= 404\). With two numbers on either side, the sequence will be \(402, 403, 404, 405, 406\)

Adding those numbers will give \(2020\).

Answer B
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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 05:22
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As the numbers are consecutive, they are in arithmetic progression.

So the sum of all the terms = number of terms * middle term

We know that number of terms = n

Let's assume that the middle term is x

n * x = 2020

Now n should be a factor of 2020, hence from the given options A and C are out.

Lets evaluate the remaining options

Option E

If n = 505, then x which is the middle term is 4

Now if x = 4, there are 252 numbers to the right of x (i.e. 4) on the number line and there are 252 numbers on the left of x on the number line. But if this were the case, we are also taking negative numbers into account. However the question stem mentions that the numbers are positive.

So eliminate E.

Option D

If n = 101, then x which is the middle term is 20

Now if x = 20, there are 50 numbers to the right of x (i.e. 20) on the number line and there are 50 numbers on the left of x on the number line. But if this were the case, we are also taking negative numbers into account. However the question stem mentions that the numbers are positive.

So eliminate D.

Option B

If n = 101, then x which is the middle term is 404

Hence we have 2 terms on the right of 404 and two terms on the left of 404, and the sum of these 5 terms is 2020.

Ans: Option B

Note: We can optimize this process of elimination by first considering D as the number is between 5 and 505. Based on our finding we can avoid an extra check of option.
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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 05:25
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Bunuel wrote:
The sum of n consecutive positive integers, where n > 1, is 2020. If n is odd, what is the value of n ?

A. 3
B. 5
C. 7
D. 101
E. 505


Somehow, initially I thought the question is asking for minimum value of n, so it is important to read the question clearly.
Edited the solution accordingly

Sum of consecutive numbers = Average*number of terms.
Here, as there are odd number of terms, the average has to be an integer and Sum=n*Average

2020=n*Average=1*2020=5*404 and so on
The above shows n can take any value that is factor of 2020.
Here, B, D and E will fit in.
Let us check each separately-
B) 5: 2020=5*404, so numbers are 402,403,404,405,406…..Fits in perfectly.
D) 101: 2020=101*20. If the median is 20 and terms are consecutive, surely all are not positive terms.
There are only 19 positive terms less than 20, but we require 50 terms less than 20……..Not Possible
E) 505: 2020=505*4. If the median is 4 and terms are consecutive, surely all are not positive terms.
There are only 3 positive terms less than 4, but we require 252 terms less than 4……..Not Possible


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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 06:53
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chetan2u wrote:
Bunuel wrote:
The sum of n consecutive positive integers, where n > 1, is 2020. If n is odd, what is the value of n ?

A. 3
B. 5
C. 7
D. 101
E. 505



Sum of consecutive numbers = Average*number of terms.
Here, as there are odd number of terms, the average has to be an integer and Sum=n*Average

2020=n*Average=1*2020=5*404 and so on
The above shows n can take any value that is factor of 2020.
The smallest odd value is 1 and then 5.
Since n>1, the least value of n is 5.


B


The question does not asks about the least value of n. It asks about the value of n. Not clear how you got 5, and how you discarded say 101.
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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 07:34
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Bunuel wrote:
chetan2u wrote:
Bunuel wrote:
The sum of n consecutive positive integers, where n > 1, is 2020. If n is odd, what is the value of n ?

A. 3
B. 5
C. 7
D. 101
E. 505



Sum of consecutive numbers = Average*number of terms.
Here, as there are odd number of terms, the average has to be an integer and Sum=n*Average

2020=n*Average=1*2020=5*404 and so on
The above shows n can take any value that is factor of 2020.
The smallest odd value is 1 and then 5.
Since n>1, the least value of n is 5.


B


The question does not asks about the least value of n. It asks about the value of n. Not clear how you got 5, and how you discarded say 101.


My mistake. I thought question asking least value of n without reading it completely, :( and so discarded all higher values.
Edited now.
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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   27 Oct 2022, 09:44
The sum of n consecutive positive integers, where n > 1, is 2020. If n is odd, what is the value of n ?

A. 3
B. 5
C. 7
D. 101
E. 505
1. Odd + Odd = Even
2. Odd + Odd + Odd = Odd
3. Odd + Even = Odd
4. Odd + Even + Odd = Even

Case 4. is what we are dealing with here since n is odd numbers of consecutive positive integers.

Testing numbers help but in a longer manner.

A. 3 - Here one of the three numbers must be even. It would be such that 2020 is dived in three ALMOST equal parts. So, 2020/3 =~ 666. Hence 666,667 and 668 but it does not suffice our condition even after checking with more possibilities.
B. 5 - Again similarly, 2020/5 = 404. 402,403,404,405,406 satisfy.
Although we have reached the required answer, do check for other options.
Others, ofcourse, don't satisfy. IMPORTANTLY, best see which of the options divide 2020, leaving an integer.

Answer B.
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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink] New post   18 Apr 2024, 08:23
Take numbers as
x+1,x+2…, x+n

Sum = (n/2)(x+1+x+n)=2020
n(2x+n+1)=4040=2^3.5.101

Take n=101, 2x+102=2^3.5 gives x negative (eliminate)

Take n=505, same as above (eliminate)

Take n=5, you get x=401 (possible)

Hence n=5 and series is 402, 404….

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Re: The sum of n consecutive positive integers, where n > 1, is 2020. [#permalink]
18 Apr 2024, 08:23
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