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1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!.

The question should be: is the prime number p a factor of n!, where n is a nonnegative integer.

(Basically the question asks whether \(p\leq{n}\))

(1) p is a factor of \(n!+(n+2)!\) --> \(p\) is a factor of \(n!(1+(n+1)(n+2))\).

Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\).

But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\).

Two different answer, hence not sufficient.

(2) p is a factor of \(\frac{(n+2)!}{n!}\) --> \(p\) is a factor of \((n+1)(n+2)\).

Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient.

(1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)+1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Actually there is a pattern in divisibility. Let's say a+b is divisible by c then we are not sure if a is divisible by c. But if b is divisible by c then a is divisible by c. Hence combining 1) + 2) works. I don't have see a better to solve this in 60 secs. But will love to hear from others.

1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!.

The question should be: is the prime number p a factor of n!, where n is a nonnegative integer.

(Basically the question asks whether \(p\leq{n}\))

(1) p is a factor of \(n!+(n+2)!\) --> \(p\) is a factor of \(n!(1+(n+1)(n+2))\).

Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\).

But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\).

Two different answer, hence not sufficient.

(2) p is a factor of \(\frac{(n+2)!}{n!}\) --> \(p\) is a factor of \((n+1)(n+2)\).

Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient.

(1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)+1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.

Hi Bunuel,

Can u please explain this part- [color=#ffff00]then it should be a factor of the first multiple of this expression: n![/color]

Last edited by zanaik89 on 17 Oct 2017, 18:58, edited 1 time in total.

1. Prime number p divides n! + (n+2)!. 2. The prime number p divides (n+2)!/n!.

The question should be: is the prime number p a factor of n!, where n is a nonnegative integer.

(Basically the question asks whether \(p\leq{n}\))

(1) p is a factor of \(n!+(n+2)!\) --> \(p\) is a factor of \(n!(1+(n+1)(n+2))\).

Now, if \(p=3\) and \(n=0\) (\(n!+(n+2)!=1+2=3\)) then the answer will be NO: \(p=3\) is not a factor of \(n!=0!=1\).

But if \(p=2\) and \(n=2\) (\(n!+(n+2)!=2+24=26\)) then the answer will be YES: \(p=2\) is a factor of \(n!=2!=2\).

Two different answer, hence not sufficient.

(2) p is a factor of \(\frac{(n+2)!}{n!}\) --> \(p\) is a factor of \((n+1)(n+2)\).

Now, if \(p=2\) and \(n=0\) then the answer will be NO but if \(p=2\) and \(n=2\) then the answer will be YES. Not sufficient.

(1)+(2) Now, \((n+1)(n+2)\) and \((n+1)(n+2)+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1. So, as from (2) \(p\) is a factor of \((n+1)(n+2)\) then it can not be a factor of \((n+1)(n+2)+1\), thus in order \(p\) to be a factor of \(n!*((n+1)(n+2)+1)\), from (1), then it should be a factor of the first multiple of this expression: \(n!\). Sufficient.

Answer: C.

Hi Bunuel,

Can u please explain this part- [color=#ffff00]then it should be a factor of the first multiple of this expression: n![/color]

p is NOT a factor of (n+1)(n+2)+1. For p to be a factor of n!*((n+1)(n+2)+1), it must be a factor of n!
_________________

(1) Prime number p divides n! + (n+2)!. (2) The prime number p divides (n+2)!/n!.

P divides N only if p<=n

from 1

p divides n! [1+(n+1)(n+2)] thus p divides either n! or divides [1+(n+1)(n+2)) ... insufficient

from 2

p divides (n+1) (n+2) ... no mention whether it divides n! or not ... insuff

from both

1+(n+1)(n+2) and ( n+1) ( n+2) are co primes , i.e. no common divisor except 1

and since from 1 p divides n! [1+(n+1)(n+2)] and divides (n+1)(n+2) then it must divide n! ... suff

C

I solve the problem in the same way. I will see my interpretation of result from solution of the statement1 that statement 1 has two answer but it answers the question that p divides n!

So please explain that how come your answer is C not A.

In my option, with above solution answer will be option A.

Last edited by Adityagmatclub on 22 Oct 2017, 11:05, edited 1 time in total.