For the sum of all positive integers x, the occurrence of x > 0
$$such\ \frac{24}{12-x}is\ a\ natural\ number$$
A natural number is a set of whole non-negative numbers
Therefore, the expression $$\frac{24}{12-x}$$ where x > 0 must produce a positive, whole number and 12 - x must be a number that can divide 24 without remainder i.e factors of 24.
Testing all factors of 24; factors of 24 = 1, 2, 3, 4, 6, 8, 12, 24
For 12 - x = 1; +x = 1 - 12 = + 11; x = 11
For 12 - x = 2; +x = 2 - 12 = + 10; x = 10
For 12 - x = 3; x = 9
For 12 - x = 4; x = 8
For 12 - x = 6; x = 6
For 12 - x = 8; x = 4
For 12 - x = 24; x = -12
From the test above 12 and 24 wil be discarded as x cannot be 0 and it cannot be negative.
Possible positive values of x includes, 11, 10, 9, 8, 6, 4. Check if thse possitive values will make the expression $$\frac{24}{12-x}$$ a natural number.
$$When\ x\ =\ 11;\frac{24}{12-x}=\frac{24}{12-11\ }=\frac{24}{1}=24$$
$$When\ x\ =\ 10;\frac{24}{12-x}=\frac{24}{12-10\ }=\frac{24}{2}=12$$
$$When\ x\ =\ 9;\frac{24}{12-x}=\frac{24}{12-9\ }=\frac{24}{3}=8$$
$$When\ x\ =\ 8;\frac{24}{12-x}=\frac{24}{12-8\ }=\frac{24}{4}=6$$
$$When\ x\ =\ 6;\frac{24}{12-x}=\frac{24}{12-6\ }=\frac{24}{6}=4$$
$$When\ x\ =\ 4;\frac{24}{12-x}=\frac{24}{12-4\ }=\frac{24}{8}=3$$
Given that all the test cases yielded natural numbers, then all positive integers of x that satisfies the given expression = 11, 10, 9, 8, 6 and 4
The sum of all these values = 11 + 10 + 9 + 8 + 6 + 4 = 48
Answer = C
