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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
A. 3
B. 5
C. 7
D. 101
E. 505
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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
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
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.
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.
and so discarded all higher values. 
