If x is a positive integer greater than 1, is x! + x + 1 a p

Question

If x is a positive integer greater than 1, is x! + x + 1 a prime number?

(1) x < 10

(2) x is odd

OE
 (1) INSUFFICIENT
If x = 2: 2! + (2 + 1) = 5, which is prime.
If x = 3: 3! + (3 + 1) = 6 + (3 + 1) = 10, which is not prime.

(2) SUFFICIENT:
If x = 3: 3! + (3 + 1) = 10, which is not prime.
If x = 5: 5! + (5 + 1) = (5 × 4 × 3 × 2 × 1) + 6.
This expression must be divisible by 3, since both of its terms are divisible by 3.
Furthermore it must be even, because both terms are even.
Therefore, it is not a prime number.

Bunuel · Accepted Answer

If x is a positive integer greater than 1, is x! + x + 1 a prime number?

(1) x < 10.

If x=2, then x! + x + 1 = 5 = prime.
If x=3, then x! + x + 1 = 10, not a prime.

Not sufficient.

(2) x is odd. So, x is an odd integer greater than 1: 3, 5, 7, ... In this case x! + x + 1 = even + odd + odd = even > 2, not a prime. Sufficient.

Answer: B.

australia123 · Answer

Let X<10, then
for the x=2
x! + x + 1 = 5 = prime number
& for the x=3
x! + x + 1 = 10, not a prime number
So it is not possible if x is a positive integer greater than 1, is x! + x + 1 a prime number..

him1985 · Answer

Is there any other way , apart from testing number. I solved it correctly but took more than 2 min.

gianghoang217 · Answer

(1) x < 10.

If x=2, then f(x)= 5 => prime
If x=3, then f(x) = 10 => not prime

Not suff.

(2) x is odd.

x! + x + 1 = x(x-1)(x-2)(x-3)...(x-n) + x + 1 (with n always is even)

=> x  + 1

We have (x-1) always even, (x-2) odd, (x-3) even, ... (x-n)odd => (x-1)(x-2)(x-3)...(x-n) = even

=>   = odd (1+ even)
=> x  = odd (odd * odd)
=> x  + 1 = even (odd + 1)

=> suff.

I just write in detail, it would be very quick to infer that.

mvictor · Answer

I got to B in ~1 minute..
1. x<10 -> 1<x<10
suppose x=2.
2+2+1 = 5
suppose x=3
3! = 6.
6+3+1 = 10 - not prime.

2 outcomes - insufficient.

2. x is odd.
since x can't be 1, x! of any odd number will always be even.
since x is odd, x+1 will always be even
now..x!(even) + x+1(even) = even.
even number can't be a prime number.

mkumar26 · Answer

definitely no. this is why option b ? please help anyone :roll:

Bunuel · Answer

This is an YES/NO data sufficiency question. A statement is sufficient if you can get a definite YES or definite NO answer to the question asked.

So, (2) is sufficient because it gives a definite NO answer to the question: x! + x + 1 is NOT a prime.

mkumar26 · Answer

thank bunuel sir ! :-D

LakerFan24 · Answer

The main takeaways from this question is this:  2 is the ONLY number (apart from 1) that, when expressed as a factorial, will give you a prime # (when you add "x" and "1") . Why, you ask? B/c  any number > 2 will include "2" in its prime factorization . Meaning, if you have 3! = 3x2; or 5! = 5x4x3x2.

ALSO, any number multiplied by an even will be EVEN. and the ONLY EVEN PRIME = 2.

Therefore, (1) = insufficient b/c x<10 includes 2 (prime) as well as 3, 4, 5, 6, 7, 8, 9 (which are all NOT prime).
(2) is sufficient b/c we know "1" cannot be included here, so the only numbers to play with are: 3, 5, 7, 9 -- which are all NOT prime. This is sufficient b/c we can safely say "x! + x + 1 is NOT PRIME" under these conditions

Archit3110 · Answer

given x>1
is x! + x + 1 a prime number
#1
x < 10
at x=2 we get yes and we get no at all odd numbers as e+o+o = even
insufficient
#2
x is odd
always no to get prime number as e+o+o = even and all number will be >2
sufficient
option B

achloes · Answer

Any factorial greater than or equal to 2 will always be even.

S2 tells us that x is odd. So x! (even) + x (odd) + 1 (odd) = even. Because x>1, the min value it can take on is 3, we can be sure that all of the values generated will be even values that are greater than 2, which is the only even prime. Therefore, using this statement, we can definitively answer no to the question.

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If x is a positive integer greater than 1, is x! + x + 1 a p

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If x is a positive integer greater than 1, is x! + x + 1 a p

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