If x is an integer, and 121

Question

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

Bunuel · Accepted Answer

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

Anu26 · Answer

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!

9RoKl1 · Answer

For every possible positive integer of x there is a negative integer. The sum of the possible integers of x should equal 0.

E

Pitabdhi · Answer

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.

amanvermagmat · Answer

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.

Simon1994 · Answer

None of the above because for every possible integer sum is Zero.

ronak1003 · Answer

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

Bunuel · Answer

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.

MathRevolution · Answer

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

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