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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 05 Feb 2019, 04:51
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D
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54% (02:24) correct 46% (02:32) wrong based on 125 sessions

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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
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D
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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 Updated on: 05 Feb 2019, 06:18
Originally posted by Archit3110 on 05 Feb 2019, 05:24.
Last edited by Archit3110 on 05 Feb 2019, 06:18, edited 1 time in total.
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Bunuel
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


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.
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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 05 Feb 2019, 06:12
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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
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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 05 Feb 2019, 19:44
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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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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 05 Feb 2019, 22:02
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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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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 30 Jan 2024, 00:17
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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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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 30 Jan 2024, 07:08
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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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If 1 ≤ n ≤ 100, and n + 7/2 is a multiple of 4 but not a multiple of 3 30 Jan 2024, 07:28
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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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