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If x is an integer, and 121 < x^2 <= 361, which of the following is

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Bunuel
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If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   19 May 2015, 07:02
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If \(x\) is an integer, and \(121 < x^2 \leq 361\), what is the sum of all possible values of \(x\)?

A. 135
B. 124
C. 116
D. 105
E. None of the above

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Bunuel
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   25 May 2015, 06:57
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If \(x\) is an integer, and \(121 < x^2 \leq 361\), what is the sum of all possible values of \(x\)?

A. 135
B. 124
C. 116
D. 105
E. None of the above


Since \(x\) is squared, for every positive value of \(x\), there will be a corresponding negative value with the same magnitude. Therefore, the sum of all possible values of \(x\) will be 0.


Answer: E
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   19 May 2015, 08:17
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Bunuel wrote:
If x is an integer, and 121 < x^2 <= 361, which of the following is the sum of all possible values of x?

(A) 135
(B) 124
(C) 116
(D) 105
(E) None of the above



The range of x is from 12 .. to 19 and -12 to -19 so the sum is 0 hence ans E, I guess, unless I am missing something!
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   20 May 2015, 01:27
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For every possible positive integer of x there is a negative integer. The sum of the possible integers of x should equal 0.

E
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   07 Jun 2017, 19:15
the range of x can be
-11< x <19
11<x<19
-19<x<-11
11<x<19
I did not understand why only 2 range were considered.
I think we need to consider the entire set of values for x.
Can you please help me understand.
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   07 Jun 2017, 23:12
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Pitabdhi wrote:
the range of x can be
-11< x <19
11<x<19
-19<x<-11
11<x<19
I did not understand why only 2 range were considered.
I think we need to consider the entire set of values for x.
Can you please help me understand.


Hi

We are given 121 < x^2 <=361
So x is an integer whose square is greater than 121 and less than (or equal to) 361.

So if x is a positive integer, it must be greater than 11 and less than/equal to 19.
These integers are 12, 13, 14, ....19

And if x is a negative integer, it should again be such that its square is greater than 11 and less than/equal to 361.
So here, x can take these values: -12, -13, -14, ...-19.

These are the Only and All possible values of x. Adding these you will get '0'.
Please note that the range -11< x <= 19 is NOT correct, and doesn't make sense here. Rather the correct range of x would be

-19<= x < -11 and 11 < x <= 19.
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   07 Jun 2017, 23:50
None of the above because for every possible integer sum is Zero.
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   09 Jan 2020, 04:16
amanvermagmat wrote:
Pitabdhi wrote:
the range of x can be
-11< x <19
11<x<19
-19<x<-11
11<x<19
I did not understand why only 2 range were considered.
I think we need to consider the entire set of values for x.
Can you please help me understand.


Hi

We are given 121 < x^2 <=361
So x is an integer whose square is greater than 121 and less than (or equal to) 361.

So if x is a positive integer, it must be greater than 11 and less than/equal to 19.
These integers are 12, 13, 14, ....19

And if x is a negative integer, it should again be such that its square is greater than 11 and less than/equal to 361.
So here, x can take these values: -12, -13, -14, ...-19.

These are the Only and All possible values of x. Adding these you will get '0'.
Please note that the range -11< x <= 19 is NOT correct, and doesn't make sense here. Rather the correct range of x would be

-19<= x < -11 and 11 < x <= 19.





Hi Bunuel

I'd like to understand why the range -11<x<=19 is NOT correct? I'm unable to grasp the concept why we don't consider the following two cases

-11<x<=19. OR
-19<=x<11
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Bunuel
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   09 Jan 2020, 04:29
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ronak1003 wrote:
amanvermagmat wrote:
Pitabdhi wrote:
the range of x can be
-11< x <19
11<x<19
-19<x<-11
11<x<19
I did not understand why only 2 range were considered.
I think we need to consider the entire set of values for x.
Can you please help me understand.


Hi

We are given 121 < x^2 <=361
So x is an integer whose square is greater than 121 and less than (or equal to) 361.

So if x is a positive integer, it must be greater than 11 and less than/equal to 19.
These integers are 12, 13, 14, ....19

And if x is a negative integer, it should again be such that its square is greater than 11 and less than/equal to 361.
So here, x can take these values: -12, -13, -14, ...-19.

These are the Only and All possible values of x. Adding these you will get '0'.
Please note that the range -11< x <= 19 is NOT correct, and doesn't make sense here. Rather the correct range of x would be

-19<= x < -11 and 11 < x <= 19.





Hi Bunuel

I'd like to understand why the range -11<x<=19 is NOT correct? I'm unable to grasp the concept why we don't consider the following two cases

-11<x<=19. OR
-19<=x<11


The point is when you take the square root from \(121 < x^2 \leq 361 \) you get \(11 < |x| \leq 19 \) (recall that \(\sqrt{x}=|x|\)). This on the other hand translates to \(11 < -x \leq 19 \) and \(11 < x \leq 19 \). Multiply by -1 the first inequality and flip the signs: \(-19 \leq x < -11\). So, finally we get that \(121 < x^2 \leq 361 \) means \(-19 \leq x < -11\) or \(11 < x \leq 19 \).

Hope it's clear.
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   03 May 2021, 02:29
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\(121 = 11^2 \)and \((-11)^2 \)
\(361 = 19^2\) and \((-19)^2 \)

All possible values of 'x' will cancel out each other and the sum will be 0.

Answer E
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Re: If x is an integer, and 121 < x^2 <= 361, which of the following is [#permalink] New post   24 Oct 2023, 13:50
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