If n is an even positive integer, and n 10, then (n+ 2)! + (n+ 1)! +

Question

If n is an even positive integer, and n > 10, then (n + 2)! + (n + 1)! +n! must be divisible by

I) n^2
II) (n + 1)^2
III) (n + 2)^2

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

aniket16c · Answer

Simplifying the equation:
--> (n+2)!+(n+1)!+n!
--> n!  
--> n!(n+2)^2

1. Option 1: n^2
n! (n+2)^2 / (n^2) 
--> (n-1)!*(n+2)^2 / n
n = positive even integer --> n = 2*odd integer. 
Both will be present in the (n-1)!
--> 1 = correct

2. Option 3: (n + 2)^2 
As seen in above simplification. The simplified equation is easily divisible by (n+2)^2

3. Option 2: (n+1)^2
--> Consider the case that n+1 = odd prime.
--> Eg: n = 16, then n+1 = 17
--> n! (n+2)^2 / (n+1^2) = 16!*(18^2)/(17^2) --> not divisible

Hence answer = D

Russ19 · Answer

MartyTargetTestPrep   BrentGMATPrepNow   chetan2u  
Could you please resolve the problem?

chetan2u · Answer

Let us simplify  (n+2)!+(n+1)!+n! 
 (n+2)(n+1)n!+(n+1)n!+n!=n!((n+2)(n+1)+(n+1)+1)=n!((n+2)(n+1)+(n+2))=n!(n+2)(n+1+1)=n!(n+2)^2

Clearly I and III are correct.

Let us check for (n+1)^2
 n!(n+2)^2. 
Say n+1 is prime. For example n=12
So 12! *14^2  does not have prime number 13. Hence, II need not be true.

D

sanjitscorps18 · Answer

(n + 2)! + (n + 1)! +n!
= n! x (n+2)^2

I) n^2
Since 'n' is an even number and greater than 12 we can safely say that n/2 is also an integer. Hence if we take n as 12 we would also have 6 in 12!. Additionally we would also have 2 in 12!. Hence, we can safely assume that any n^2 would divide the expression

II) (n + 1)^2
By above logic n+1 would be an odd number. To have another n+1! in the factorial we would necessarily require two factors that when multiplied would give n+1, for e.g. 15 is given by 5 and 3. But this cannot be the case for prime numbers and we would never have the second odd number to ensure that (n+1)^2 divides the expression

III) (n + 2)^2
Same logic as n!. This should be able to divide the expression.

Option D

Saya12345 · Answer

*remember this is a MUST be condition, not a COULD be condition.

Simple expand the question stem: 
(n+2)! + (n+1)! + n!

Take n! common from all of these: 
n! *  
n! *  
n! *  ^2

This is obviously divisible by option III

Lets check other options: 
n^2: n! is divisible by n. (Remember that n is even - that means n can be rewritten as n=2k, where k is less than n. And is within the n!)
SO, N! is divisible by n^2 also.

(n+1)^2 
(2k+1)^2 would not be a factor of either n! or n+2

If n is an even positive integer, and n > 10, then (n+ 2)! + (n+ 1)! +

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