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chetan2u
souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32


This will happen when N is prime number..
Till 100, we have 25 prime numbers...

But there is one more 4, because 3! has only 2*3..
So 25+1=26
C


I was able to understand uptill this part
"Till 100, we have 25 prime numbers...

can you please ellaborate further thank you.
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chetan2u
souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32


This will happen when N is prime number..
Till 100, we have 25 prime numbers...

But there is one more 4, because 3! has only 2*3..
So 25+1=26
C


I was able to understand uptill this part
"Till 100, we have 25 prime numbers...

can you please ellaborate further thank you.

If you count the number of prime numbers upto 100, it will be 25.
Its better to remember this.
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chetan2u
souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32


This will happen when N is prime number..
Let's see WHY?
(N-1)! Means 1*2*3....(N-2)(N-1)..
So all numbers till (N-1) are factors of (N-1)!...
Also a prime number does not have any factor less than N except 1..
So N will not divide (N-1)! As they do not have any common factor.


Till 100, we have 25 prime numbers...

But there is one more : 4, because 3! = 2*3=6

So 25+1=26
C

hero_with_1000_faces I have elaborated the answer here itself..
Pl revert incase of any other problem


chetan2u can you please provide any references for concept related to this question, then Ill try to re-read your explanation as now I am unable to understand the questions and thus your explanation. Thanks a lot.
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souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32

1≤N≤100
N=1: (1-1)=0!/1=divisible
N=2: (2-1)=1!/2≠divisible
N=3: (3-1)=2!/3≠divisible
N=4: (4-1)=3!/4≠divisible

All prime numbers ≤100 are possible values of N
primes: [2,3,5,7,11…97]=25
Also, N=4 is a possible value
Total=25+1=26 values

Ans (C)
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rahul16singh28
souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32

It will include all the Primer Numbers from 1 - 100 (25 Prime Numbers) and 4.
How 4? please explain.
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rahul16singh28
souvonik2k
N is a positive integer at most equal to 100. If (N-1)! is not divisible to N, how many different values can N assume?
A) 22
B) 25
C) 26
D) 30
E) 32

It will include all the Primer Numbers from 1 - 100 (25 Prime Numbers) and 4.
How 4? please explain.

Because 4 is not divisible by 3! = 6.
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