If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3

Question

If 1 ≤ n ≤ 100, and  \frac{n + 7}{2}  is a multiple of 4 but not a multiple of 3, then which of the following could be true?

I. n is prime
II. n is a multiple of 3
III. n is a multiple of 4

A. I only
B. II only
C. III only
D. I and II only
E. I, II and III

Archit3110 · Answer

1 ≤ n ≤ 100, and  \frac{n + 7}{2}  is a multiple of 4
for this relation to be valid
n has to be odd no.

so option 3 are ruled out
n= is prime is correct and n is a multiple of 3 is correct 
IMO D is correct.

ak31 · Answer

Question has asked for COULD BE TRUE scenario.

For the expression to be divisible by 2, n value should be ODD
This rules out 3rd option.

Let's try to plug n=9
so, 9+7/2=8

Let's try to plug n=13
so, 13+7/2 =10

Above 2 scenarios explain that n can be multiple of 3 or prime number. Hence, the answer is D

sirpumpkin · Answer

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.

globaldesi · Answer

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

thisisakr · Answer

24 will not be allowed as it is divisible by 3 however the working theory is correct 👍 prime 73 fits the solution

VerbalFaker · Answer

(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

Oppenheimer1945 · Answer

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

bumpbot · Answer

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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3

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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3

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