Manhattan:Divisibility of primes

[shubhamkumar]

If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10

(2) x is even

OA:E.
Found this excellent question in Mgmat CAT.

[Pharo]

Wow seriously good

The only way I can think of is actually getting your hands dirty and doing it. (You will only need fast calculation skills )

Let's say f(x) = x! + (x+1)
f(2) = 2! + (2+1) = 5
f(3) = 3! + (3+1) = 10; at this stage we know that if x is odd, we know that f(x) does not output a prime for all x.
f(4) = 4! + (4+1) = 29; now if x is even, it's looking good.
; i am skipping odd numbers since 3 gave me a non prime.
f(6) = 6! + (6+1) = 727 ; not divisible by 2,3,4,5,7 etc.. So might be a prime.
f(8) = 8! + (8+1) = 40329 ; divisible by 3 .. So it's not a prime.
; Now i know that f(x) does not ALWAYS spit out prime numbers.

So the answer is E -- If you are fast with manual calculations you can solve this question in less than 2 minutes. Even not, i think this is somewhat of an hard question so if you are given this question it is likely that you saved some time from earlier questions so it's all good

[Mike@Magoosh]

Hi, there. I'm happy to help with this.

Prompt: If x is a positive integer, is x! + (x + 1) a prime number?

In general, without restrictions, that would be a hard question to answer. One thing is clear ---- positive integer x goes into (x! + x) so it's not going to go into (x! + x + 1). That much is clear.

Also, just to be clear
1! = 1
2! = 2
3! = 3*2*1 = 6
4! = 4*3*2*1 = 24
5! = 5*4*3*2*1 = 120
6! = 6*5*4*3*2*1 = 720
after that, it gets bigger than I'd like to think about without a calculator.

Statement #1: x < 10

Well,
x = 1, then x! + x + 1 = 3, which is prime
x = 2, then x! + x + 1 = 5, which is prime
x = 3, then x! + x + 1 = 10, which is not prime

Therefore, knowing x < 10 is not sufficient to determine a definitive answer to the question. Statement #1, by itself, is not sufficient.

Statement #2: x is even

x = 2, then x! + x + 1 = 2 + 2 + 1 = 5, which is prime
x = 4, then x! + x + 1 = 24 + 4 + 1 = 29, which is prime
x = 6, then x! + x + 1 = 720 + 6 + 1 = 727

The number 720 has lots of factors of 2's and 3's in it, and a factor of 5, but no factor of 7. 720 is not divisible by 7, and therefore 727 is not divisible by 7. That leaves the harder question: is 727 prime? I'm going to say that, without a calculator, most people will not be able to answer that question in a reasonable amount of time. Let's just skip it, and hope that the problem works out without determining that. (PS, as it happens, 727 is prime.)

x = 8, then x! + x + 1 = 8! + 8 + 1 = 8! + 9

Now, 8! = 8*7*6*5*4*3*2*1, so 8! has two factors of 3 (one from the 3 and one from the 6) --- therefore, 8! is divisible by 3*3 = 9, and therefore 8! + 9 is divisible by 9. If it's divisible by 9, then it's not a prime number. Knowing x is even does not determine an answer to the prompt question. Statement #2, by itself, is not sufficient.

Combined statements:
Notice that x = 2 & x = 4 satisfy both conditions and give a "yes" answer, but x = 8 satisfies both conditions and gives a "no" answer. Even knowing x satisfies both conditions is not enough to answer the question. Combined, both statements are still insufficient.

Answer = E

Notice, in the reasoning in statement #2, we go to a point where we couldn't easily make a determination about x = 6, so we simply moved on. In the most challenging questions, sometimes you will have to do this. In the finely constructed questions of the GMAT (and MGMAT has awfully good questions too!), you have to trust that if one case is too hard to figure out, there's going to be another case that is much more accessible. There will always be a way to solve the problem quickly and efficiently, and it's important to trust that when you are considering whether or not to abandon the analysis of one particular case because it's gotten too hairy.

Does all this make sense?

Here's a similar DS, for more practice:
https://gmat.magoosh.com/questions/864

Let me know if you have any questions about what I've said.

Mike

[Pharo]

Notice, in the reasoning in statement #2, we go to a point where we couldn't easily make a determination about x = 6, so we simply moved on. In the most challenging questions, sometimes you will have to do this. In the finely constructed questions of the GMAT (and MGMAT has awfully good questions too!), you have to trust that if one case is too hard to figure out, there's going to be another case that is much more accessible. There will always be a way to solve the problem quickly and efficiently, and it's important to trust that when you are considering whether or not to abandon the analysis of one particular case because it's gotten too hairy.

That is an awesome tip! Thank you Mike