The number of possible divisiors of 2x +12

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Quant Warrior

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If 'x' is a positive integer such that 2x+12 is perfectly divisible by 'x' , then the number of possible values of 'x' is:
a)2
b)5
c)6
d)12

I took random values of x starting from 1 and tried to get the expression ''2x+12' by 'x'. I got the number of values as 6 which is the correct answer. Solution to this question in my book is: the number of factors of 12 would be the number of possible values of 'x' that can divide ''2x+12" perfectly.

Why did they take take number of factors of 12 as the answer ?

Please help.

Regards.
 
Quant Warrior said:
If 'x' is a positive integer such that 2x+12 is perfectly divisible by 'x' , then the number of possible values of 'x' is:
a)2
b)5
c)6
d)12

I took random values of x starting from 1 and tried to get the expression ''2x+12' by 'x'. I got the number of values as 6 which is the correct answer. Solution to this question in my book is: the number of factors of 12 would be the number of possible values of 'x' that can divide ''2x+12" perfectly.

Why did they take take number of factors of 12 as the answer ?

Please help.

Regards.
Click to expand...
(2x + 12)/ x = 2 + (12/x)

Now for 2 + (12/x) to be an integer, the number (12/x) must also be an integer. That can happen only when 'x' is a factor 12. Thus x must be one of (1, 12, 2, 6, 3, 4) 6 numbers (those that are factors of 12).
 
Here's another pproach:

If 2x+12 = xn, then simple rearranging gives x(n-1)=12.

Therefore x divides 12 (and for every divisor x of 12, n-1 will also be a positive integer.)

(x,n) = (1,13), (2, 7), (3,5), (4,4), (6,3), (12,2)
 
Yes, will fix, thanks.

edit:looks like I can't edit, so


If 2x+12 = xn, then simple rearranging gives x(n-2)=12.

Therefore x divides 12 (and for every divisor x of 12, n-2 will also be a positive integer.)

(x,n) = (1,14), (2, 8), (3,6), (4,5), (6,4), (12,3)
 
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