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Bunuel
If x>100, what is the smallest value of x for which (3+x)/12 is an integer.

A. 102
B. 103
C. 104
D. 105
E. 106

12*9 = 108

Now, to make the numerator 108 , we must add 105 to 3

Answer, must be (D) 105.
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Bunuel
If x>100, what is the smallest value of x for which (3+x)/12 is an integer.

A. 102
B. 103
C. 104
D. 105
E. 106

Since (3 + x)/12 must be an integer and x >100, we must find the smallest value of x greater than 100 that makes 3 + x a multiple of 12.

Since (105 + 3)/12 = 108/12 = 9, 105 is the smallest possible value of x.

Answer: D
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To be divisible by 12 (2^2*3), (3+x) should be even and therefore x must be odd (odd+odd=even).
105 fits all the requirements.

105+3= 108

It's divisible by 3 (1+0+8=9)
Also 108 is a multiple of 12.
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Here's another way of looking at it :

(3+x)/12 => 3/12 + x/12

3/12 leaves a remainder of 3. Hence for the whole thing to be divisible by 12, the second part needs to leave a reminder 9.

105 leaves a remainder of 9 when divided by 12.
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Bunuel
If x>100, what is the smallest value of x for which (3+x)/12 is an integer.

A. 102
B. 103
C. 104
D. 105
E. 106

Very simple- 12 x 9 = 108- this is the smallest number that satisfies x>100 so

3 + 105 /12 = 1

Thus
"D"
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It's possible to use the following approaching:

You need the least multiple of 12 greater than 100.

You can get this quickly, using easy numbers:

12x5 = 60x2 = 120

120-12=108

3+x = 108
x = 105
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It should be odd+odd=even to be divisible by an even so A,C,E eliminated now put values
Answer is: D is it right approach?
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Bunuel
If x>100, what is the smallest value of x for which (3+x)/12 is an integer.

A. 102
B. 103
C. 104
D. 105
E. 106
if x>100 than (3+x)>103
for integer, it shall be multiple of 12 and >103
the smallest value is 108

3 + x = 108
x = 105

Ans D
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