If N is a positive integer less than 31, how many values can N take so

Question

If N is a positive integer less than 31, how many values can N take so that (n + 1) is a factor of n! ?

A. 20
B. 18
C. 17
D. 16
E. 12

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nick1816 · Answer

If n+1 is a prime number, n+1 is not a factor of n!.

Number of primes less than 31= 12

Number of values can n take so that (n + 1) is a factor of n! = 30-12=18

GMATinsight · Answer

N < 31

We need to ensure that (n + 1) is a factor of n!

i.e. n+1 must be a composite number i.e .n+1 = prime can be simply avoided

Primes from 1 to 30 are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29,} = 10 Numbers

If n = 1, then (n + 1) is 2 which is NOT a factor of n! i.e. 1!
If n = 2, then (n + 1) is 3 which is NOT a factor of n! i.e. 2!
If n = 3, then (n + 1) is 4 which is NOT a factor of n! i.e. 3!
If n = 4, then (n + 1) is 5 which is NOT a factor of n! i.e. 4!
 If n = 5, then (n + 1) is 6 which is a factor of n! i.e. 5! 
If n = 6, then (n + 1) is 7 which is NOT a factor of n! i.e. 6!
 If n = 7, then (n + 1) is 8 which is a factor of n! i.e. 7! 
 If n = 8, then (n + 1) is 9 which is a factor of n! i.e. 8! 
 If n = 9, then (n + 1) is 10 which is a factor of n! i.e. 9! 
If n = 10, then (n + 1) is 11 which is NOT a factor of n! i.e. 10!
 If n = 11, then (n + 1) is 12 which is NOT a factor of n! i.e. 11! 
If n = 12, then (n + 1) is 13 which is NOT a factor of n! i.e. 12!
 If n = 13, then (n + 1) is 14 which is a factor of n! i.e. 13! 
 If n = 14, then (n + 1) is 15 which is a factor of n! i.e. 14! 
 If n = 15, then (n + 1) is 16 which is a factor of n! i.e. 15! 
If n = 16, then (n + 1) is 17 which is NOT a factor of n! i.e. 16!
 If n = 17, then (n + 1) is 18 which is a factor of n! i.e. 17! 
If n = 18, then (n + 1) is 19 which is NOT a factor of n! i.e. 18!
If n = 22, then (n + 1) is 23 which is NOT a factor of n! i.e. 22!
If n = 28, then (n + 1) is 29 which is NOT a factor of n! i.e. 28!
If n = 30, then (n + 1) is 31 which is NOT a factor of n! i.e. 30!

12 cases are the ones in which (n + 1) is NOT a factor of n!

i.e. 30-12 = 18 cases exist for whom (n + 1) is a factor of n!

Answer: Option B

ArunSharma12 · Answer

N < 31
lets take a few examples
1! = 1, 2 is not a factor
2! = 2, 3 is not a factor
3! = 6, 4 is not a factor
4! =24, 5 is not a factor (why?), because 5 is a prime number. 
5! = 120, 6 is factor.
so a prime numbers will only be a factor of n! if prime number is less than n.

from 5 to 30, there are 25 numbers and out of these 25 numbers 7 are prime numbers.
number of values of N = 25 -7 = 18
Ans: B

chetan2u · Answer

Whenever n+1 is PRIME, it will NOT be a factor of n! as n! is product of all numbers  \leq{n} ...
Now Primes till 31 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31...We include 31 as n+1=31
So there are 11 primes...

Also 2^2 =4, but there is ONLY one 2 in 3! and n+1=4...so NOT a factor..
All numbers after that, be it 3^2 or 2^3 have sufficient number of 3s and 2s in it..
So 30-11-1=18

B

youcouldsailaway · Answer

If we start listing out the numbers, we will see that all the numbers below 5 do not satisfy the condition of  (n+1)  being a factor of  n!  ; we can thus remove them from consideration.

Since n is less than 31, we only have 30 numbers to consider. But even 30 doesn't satisfy the condition since the next consecutive number 31, is not a factor of 30! ; we can thus remove it from consideration.

So from 5 to 29, we have 25 numbers.

Now one thing to note is that if a particular number, let's call it A, is followed by a prime, then A! will not meet the above condition.

From 5 to 29 we have 7 prime numbers (7, 11, 13, 17, 19, 23, 29). Thus we can remove the numbers that come before these primes from consideration too. Now, 25 - 7 = 18, thus 18 numbers remain. Ans is B.

Note: We are not excluding prime numbers themselves from consideration since they do meet the condition above. For example, 11! will include 12 as a factor. We are only using prime numbers to weed out the numbers that don't meet the above condition, which are basically the numbers immediately before the primes.

Kinshook · Answer

If N is a positive integer less than 31, how many values can N take so that (n + 1) is a factor of n! ?

N=0, 1, 2 are not feasible; N+1 can not be prime;

Possible values of N = {3, 5, 7, 8, 9, 11, 13, 14, 15, 17, 19, 20, 21, 23, 24, 25, 26, 27, 29}

N = 3; N! = 6; N+1=4 is NOT a factor of N!
N = 5; N! = 120; N+1=6=2*3 is a factor of N! : Possible case #1
N = 7; N! = 5040; N+1=8=2^3 is a factor of N! : Possible case #2
N = 8; N! = 40320; N+1=9=3^2 is a factor of N! : Possible case #3
N = 9; N! = 9*40320; N+1=10=2*5 is a factor of N! : Possible case #4
N = 11; N! = 11!; N+1=12=2^2*3 is a factor of N! : Possible case #5
N = 13; N! = 13!; N+1=14 is a factor of N! : Possible case #6
N = 14; N! = 14!; N+1=15=3*5 is a factor of N! : Possible case #7
N = 15; N! = 15!; N+1=16=2^3 is a factor of N! : Possible case #8
N = 17; N! = 17!; N+1=18=2*3^2 is a factor of N! : Possible case #9
N = 19; N! = 19!; N+1=20=2^2*5 is a factor of N! : Possible case #10
N = 20; N! = 20!; N+1=21=3*7 is a factor of N! : Possible case #11
N = 21; N! = 21!; N+1=22=2^11 is a factor of N! : Possible case #12
N = 23; N! = 23!; N+1=24=2^3*3 is a factor of N! : Possible case #13
N = 24; N! = 24!; N+1=25=5^2 is a factor of N! : Possible case #14
N = 25; N! = 25!; N+1=26=2*13 is a factor of N! : Possible case #15
N = 26; N! = 26!; N+1=27=3^3 is a factor of N! : Possible case #16
N = 27; N! = 27!; N+1=28=2^2*7 is a factor of N! : Possible case #17
N = 29; N! = 29!; N+1=30=2*3*5 is a factor of N! : Possible case #18

IMO B

If N is a positive integer less than 31, how many values can N take so

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If N is a positive integer less than 31, how many values can N take so

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