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Given n+7/2 = 4x
so n+7=8x
Meaning that n+7 is some numbers which can be divided by 8.
then N could be an odd number.
III. is out.

Try check other numbers.
From 8, 16, 24,32 ,40, 48, 56, 64, 72, 80, 88, 96, 104
cut off the numbers which are divided by 3.
We will have N+7 as 8, 16,32,40,56,64,80,88,104

We will have N as 1,9,25,33,49,57,73,81,97

9 could be divided by 3
and
73 is a prime number.

So D is the answer.
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given n+7/2 is a multiple of 4
thus \(4k =\frac{n+7}{2}\)
hence n is of form 8k-7
now this expression will never be divisible 4. .... statement III out
k can take any number for which n will be prime such as k =3 = 24-7 =17.
Hence statement I is possible.
8k-7 can be re-written as\(8k-4-3= 4(2k-1) -3\)
this is divisible by 3 for k = 2,
Hence statement II is possible.


Thus only I and II statements are possible.
Thus option D
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globaldesi
given n+7/2 is a multiple of 4
thus \(4k =\frac{n+7}{2}\)
hence n is of form 8k-7
now this expression will never be divisible 4. .... statement III out
k can take any number for which n will be prime such as k =3 = 24-7 =17.
Hence statement I is possible.
8k-7 can be re-written as\(8k-4-3= 4(2k-1) -3\)
this is divisible by 3 for k = 2,
Hence statement II is possible.


Thus only I and II statements are possible.
Thus option D



24 will not be allowed as it is divisible by 3 however the working theory is correct 👍 prime 73 fits the solution
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(n+7)/2 = 4k

=> n+7 = 8k => n= 8k-7 where, (1<=n<=100)
=> n has to be odd
possible values from this equation {1,9,17,25,33,41,49,57,65,73,81,89,97} for k=1 to13

(n+7)/2 cannot be a multiple of 3

=> above values minus multiples of 12 (multiples of both 4 and 3)

=> (n+7)/2 = 12l

=> n = 24b-7
here n takes 4 values {17, 41, 65,89}

Hence all values that n can take= multiples of 4 set minus minus multiples of 12 set,

= {1,9,25,33,49,57,73,81,97}

Hence,
n can be a multiple of 3 and can be a prime.
and also from before n has to be odd.

Answer: D
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n=8a-7 but not 6b-7

n=1,9,17,25,33,41,49,57,65,73.......

n is not -1,5,11,17,23, 29,35,41,..

n=1,9(mul of 3),25,33,49,..73(prime)...

Hence I & II are correct
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