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If k is a positive integer and (30 - k)/(k - 10) is a prime number 19 Sep 2023, 09:19
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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

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If k is a positive integer and (30 - k)/(k - 10) is a prime number Updated on: 19 Sep 2023, 09:57
Originally posted by chetan2u on 19 Sep 2023, 09:35.
Last edited by chetan2u on 19 Sep 2023, 09:57, edited 1 time in total.
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Bunuel
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
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\(\frac{30-k}{k-10}\)

K cannot be less than 11 as the expression will become negative or undefined at 0.
Also, at k > 20, the expression will become less than 1.

So k can be between 10 and 20.
At k = 11, \(\frac{30-11}{11-10}=19\)……Yes
At k = 12, \(\frac{30-12}{12-10}=9\)……No
At k = 13, \(\frac{30-13}{13-10}=17/3\)….No
At k = 14, \(\frac{30-14}{14-10}=4\)….No
At k = 15, \(\frac{30-15}{15-10}=3\)….Yes
At k = 16, \(\frac{30-16}{16-10}=7/3\)…..No
At k = 17, \(\frac{30-17}{17-10}=13/7\)…No
13/7<2 and further numbers will keep becoming lesser.

Only possibilities are 19 and 3, so median of prime numbers is 11.
k is 11 and 15, so median of values of k is 13..

C
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If k is a positive integer and (30 - k)/(k - 10) is a prime number 19 Sep 2023, 09:38
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chetan2u
Bunuel
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

\(\frac{30-k}{k-10}\)

K cannot be less than 11 as the expression will become negative or undefined at 0.
Also, at k > 20, the expression will become less than 1.

So k can be between 10 and 20.
At k = 11, \(\frac{30-11}{11-10}=19\)……Yes
At k = 12, \(\frac{30-12}{12-10}=9\)……No
At k = 13, \(\frac{30-13}{13-10}=17/3\)….No
At k = 14, \(\frac{30-14}{14-10}=4\)….No
At k = 15, \(\frac{30-15}{15-10}=3\)….Yes
At k = 16, \(\frac{30-16}{16-10}=7/3\)…..No
At k = 17, \(\frac{30-17}{17-10}=13/7\)…No
13/7<2 and further numbers will keep becoming lesser.

Only possibilities are 19 and 3, so median is 11.

E
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The question asks about the median of all possible values of k, not the possible value of prime numbers.
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If k is a positive integer and (30 - k)/(k - 10) is a prime number 19 Sep 2023, 09:53
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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.
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If k is a positive integer and (30 - k)/(k - 10) is a prime number 19 Sep 2023, 09:56
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unraveled
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.
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The question asks about the median of all possible values of k.
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If k is a positive integer and (30 - k)/(k - 10) is a prime number 19 Sep 2023, 10:27
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Bunuel
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

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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
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If k is a positive integer and (30 - k)/(k - 10) is a prime number 26 Oct 2024, 06:04
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