If k is a positive integer and (30 - k)/(k - 10) is a prime number

If k is a positive integer and (30 - k)/(k - 10) is a prime number, what is the median of all possible values of k?

A. 15
B. 14
C. 13
D. 12
E. 11

Since k > 0 and an integer
(30 - k)/(k - 10) gives us that
k = 11 minimum

Thus,
(30 - k)/(k - 10) = 19/1 = 19
Next, prime number less than 19 is 17, let's check
(30 - k)/(k - 10) = 17
30 - k = 17k - 170

This leaves us with k being not an integer

Trying further with prime number 13
(30 - k)/(k - 10) = 13
30 - k = 13k - 130
Again not an integer

(30 - k)/(k - 10) = 11
30 - k = 11k - 110

(30 - k)/(k - 10) = 7
30 - k = 7k - 70

(30 - k)/(k - 10) = 5
30 - k = 5k - 50

(30 - k)/(k - 10) = 3
30 - k = 3k - 30
k = 15

(30 - k)/(k - 10) = 2
30 - k = 2k - 20
k is not an integer

We only have two values of k i.e. 11 and 15

(30 - k)/(k - 10) = 23
30 - k = 23k - 230

(30 - k)/(k - 10) = 29
30 - k = 29k - 290

... and so on

Now, if k = 20

(30 - k)/(k - 10) = 1 which is not possible since (30 - k)/(k - 10) is a prime number. So it has to be at least 2 or more.
Now that we have checked all possibilities of (30 - k)/(k - 10)
k has only two values i.e. 11 and 15.

I think something is missing.

The question asks about the median of all possible values of k.

Given:

\(\frac{(30 - k)}{(k - 10)} = n\) ⇒ n is prime number

\(30 - k = nk - 10n\)

\(30 + 10n = (n+1)k\)

\(\frac{30 + 10n}{(n+1)} = k\)

n = 2

\(k = \frac{30 + 10*2}{(2+1)} = \frac{50}{3} \) -- We have to disregard this value as k is an integer

n = 3

\(k = \frac{30 + 10*3}{(3+1)} = \frac{60}{4} = 15 \) -- Valid

n = 5

\(k = \frac{30 + 10*5}{(5+1)} = \frac{80}{6}\) -- We have to disregard this value as k is an integer

n = 7

\(k = \frac{30 + 10*7}{(7+1)} = \frac{100}{8}\) -- We have to disregard this value as k is an integer

n = 11

\(k = \frac{30 + 10*11}{(11+1)} = \frac{140}{12}\) -- -- We have to disregard this value as k is an integer

n = 13

\(k = \frac{30 + 10*13}{(13+1)} = \frac{160}{14}\) -- We have to disregard this value as k is an integer

n = 17

\(k = \frac{30 + 10*17}{(17+1)} = \frac{200}{18}\) -- We have to disregard this value as k is an integer

n = 19

\(k = \frac{30 + 10*19}{(19+1)} = \frac{220}{20} = 11\) -- Valid

n = 23

\(k = \frac{30 + 10*23}{(23+1)} = \frac{260}{24} \) -- We have to disregard this value as k is an integer

n = 29

\(k = \frac{30 + 10*29}{(29+1)} = \frac{320}{30}\) -- We have to disregard this value as k is an integer

n = 31

\(k = \frac{30 + 10*31}{(31+1)} = \frac{340}{32}\) -- We have to disregard this value as k is an integer

We see a pattern for value of \(n \geq 31\)

Remainder(\(\frac{30 + 10*n}{(n+1)}\))

= Remainder(\(\frac{30}{(n+1)}\)) + Remainder(\(\frac{10n}{(n+1)}\))

\(= 30 + 10 * (-1) \)

\(= 20\)

Hence, k will never be equal to an integer for \(n \geq 31\). We can stop here.

Set = {11,15}

Median = \(\frac{11 + 15 }{ 2}\) = 13

Option C