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Is the average of n consecutive integers equal to 1 ? [#permalink]
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Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
Bunuel wrote:
abhisheknandy08 wrote:
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,


Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.


KarishmaB

I hope that you are having a wonderful week :)
I was confused as to how you can just divide the whole inequality by n for statement 2. I have only seen manipulation of inequalities when you divide the middle term in the inequality (e.g., in which the inequality would be divided by s in this case --> 0/s < 1 < n/s). To clarify, you can divide by any term in the inequality as long as you consistently divide each variable or expression in the inequality by that term (except I guess in this example you cannot divide the inequality by the first term because it is a 0).

Many thanks :)
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Is the average of n consecutive integers equal to 1 ? [#permalink]
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woohoo921 wrote:
Bunuel wrote:
abhisheknandy08 wrote:
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,


Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.


KarishmaB

I hope that you are having a wonderful week :)
I was confused as to how you can just divide the whole inequality by n for statement 2. I have only seen manipulation of inequalities when you divide the middle term in the inequality (e.g., in which the inequality would be divided by s in this case --> 0/s < 1 < n/s). To clarify, you can divide by any term in the inequality as long as you consistently divide each variable or expression in the inequality by that term (except I guess in this example you cannot divide the inequality by the first term because it is a 0).

Many thanks :)


You can divide an inequality by any positive number without impacting the inequality.
You might have seen division by middle term (provided we know the term is positive) because in that question it would have helped simplify. Otherwise, you can divide by any positive number.

a < b
a/2 < b/2 (valid)

a < b < c
a/10 < b/10 < c/10 (valid)

a < b < c
1 < b/a < c/a (valid if a is known to be a positive number)

etc.
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Is the average of n consecutive integers equal to 1 ? [#permalink]
Bunuel wrote:
abhisheknandy08 wrote:
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,


Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.


Bunuel
I thought that we can only work with manipulating the inner term (e.g., s) in this case per my question above for Karishmab

In other words, if I have: 5 < x+2 < 9

Can you do the following for division?

Dividing all terms by 5 --> 1 < (x+2)/5 < 9/5

Dividing all terms by x+2 --> x/(x+2) < 1 < 9/(x+2)

Dividing all terms by 9 --> 5/9 < (x+2)/9 < 1


Can you do the following for subtraction?

Again, if I have: 5 < x+2 < 9
Subtracting 9 from all terms --> -4 < x-7 < 0
Subtracting x+2 from all terms --> -x+3 < 0 < -x-2

and vice versa for multiplication?

Many thanks :) KarishmaB

Originally posted by woohoo921 on 29 Dec 2022, 11:37.
Last edited by woohoo921 on 01 Jan 2023, 09:35, edited 1 time in total.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
I think Statement 1 is not sufficient: assume -1;0;1;2 here average is 1, 1;2;3;4 here average is 5. Isn't it insufficient
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
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woohoo921 wrote:
Bunuel wrote:
abhisheknandy08 wrote:
Hi Bunuel ,

Could you post the detail solution for this question .

Regards,


Is the average of n consecutive integers equal to 1 ?

Any set of consecutive integers is an evenly spaced set, where mean = median. Thus for the average to be 1 we should have one of the following types of sets:
{1}
{0, 1, 2}
{-1, 0, 1, 2, 3}
So, basically there should be equal number of elements less than 1 and greater than 1. You can notice that in this case the number of elements turns out to be odd.

(1) n is even --> n is not odd --> 1 cannot be the average. Sufficient.

(2) if S is the sum of the n consecutive integers, then 0 < S < n --> divide by n: 0 < S/n < 1. Since S/n (the sum of the elements divided by the number of elements) is the average, then we directly got that the average is not 1, it's < 1. Sufficient.

Answer: D.

Hope it's clear.


Bunuel
I thought that we can only work with manipulating the inner term (e.g., s) in this case per my question above for Karishmab

In other words, if I have: 5 < x+2 < 9

Can you do the following for division?

Dividing all terms by 5 --> 1 < (x+2)/5 < 9/5

Dividing all terms by x+2 --> x/(x+2) < 1 < 9/(x+2)

Dividing all terms by 9 --> 5/9 < (x+2)/9 < 1


Can you do the following for subtraction?

Again, if I have: 5 < x+2 < 9
Subtracting 9 from all terms --> -4 < x-7 < 0
Subtracting x+2 from all terms --> -x+3 < 0 < -x-2

and vice versa for multiplication?

Many thanks :) KarishmaB



We can add or subtract the same term on both sides of the inequality without changing the sign of the inequality. We can multiply/divide the same term on both sides without impacting the inequality sign provided the term is positive.
Here, since (x + 2) lies between 5 and 9, we know it must be positive. So we can multiply or divide the inequality by (x + 2) and the inequality sign remains the same.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
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MacroMX wrote:
I think Statement 1 is not sufficient: assume -1;0;1;2 here average is 1, 1;2;3;4 here average is 5. Isn't it insufficient




When we have even number of consecutive integers, average will be the average of the two middle integers so it will not be an integer.

Avg of -1, 0, 1, 2 is the average of 0 and 1 which is 0.5.
Avg of 1, 2, 3, 4 is the average of 2 and 3 which is 2.5.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
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KarishmaB

I thought that you can't multiply nor divide inequalities together, or does that rule only apply to when you have two separate inequalities?
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
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woohoo921 wrote:
KarishmaB

I thought that you can't multiply nor divide inequalities together, or does that rule only apply to when you have two separate inequalities?


We can multiply and divide inequalities as long as both sides are non negative in both inequalities (and there is no 0 in the denominator in case of division).
We can multiply when both inequalities have the same inequality sign and we can divide when they have opposite inequality signs.

A < B
C < D
-----
AC < BD - valid if A, B, C and D are all non negative


A < B
C > D
-----

A/C < B/D - valid if A and B are non negative and C and D are positive.
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Re: Is the average of n consecutive integers equal to 1 ? [#permalink]
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