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

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If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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

(1) x < 10

(2) x is even

Originally posted by iamba on 16 Jul 2007, 16:19.
Last edited by Bunuel on 07 May 2015, 08:58, edited 2 times in total.
Renamed the topic, edited the question and added the OA.
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If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post Updated on: 08 May 2015, 03:36
2
4
Radhika11 wrote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10
(2) x is even

Is there any other way to solve this question ?


Dear Radhika11

Here's an alternate solution.

We'll first analyze the question statement and only then go to St. 1 and 2.

Given: x is a positive integer. So, possible values of x: 1, 2, 3, 4 . . .

To find: If x! + x + 1 is Prime

Analysis:

For x = 1, x! = 1
And, x! + x +1 = 1 + 1 + 1 = 3, which is a Prime number

For x > 1,

x! will always be an even number (Because, x! contains the product of consecutive integers. So, this product will have Even AND odd terms. We know that when an even number is multiplied with anything, the product is always even)

So, the sum (x! + x + 1) = (An even number + x + Odd number) = (Odd number + X)

Even for the lowest possible value of x (x = 1), the value of the sum (x! + x + 1) was equal to 3.

So, for x > 1, the value of this sum is definitely going to be greater than 3.

And, we know that all Prime numbers greater than 2 are odd.

So, the sum (Odd number +X) can be prime only if first, this sum is an odd number.

That is, if, X is an even number.

So, for x > 1, x being an even number is a NECESSARY condition for the sum (x! + x + 1) to be prime.

But is it a SUFFICIENT condition? That is, can you say that if x is even, that must mean that the sum (x! + x + 1) will be prime?

Let's see:

x x! + x + 1

2 2! + 2 + 1 = 7 (Prime)
4 4! + 4 + 1 = 29 (Prime)
6 6! + 6 + 1 = 727 (Prime)
8 8! + 9 (both terms in this sum are divisible by 3) (NOT Prime)

Thus, we see that for some even values of x, the sum (x! + x + 1) will be Prime and for others, it will not be.

With this understanding, let's now look at the two statements:

Statement 1: x < 10

As per this statement, x can be even (=> some possibility of the sum (x! + x + 1) to be prime)
or x can be odd (=> NO possibility of the sum (x! + x + 1) to be prime)

Clearly not sufficient.

Statement 2: x is even

As illustrated in our analysis of the question statement, x being Even is not a sufficient condition for the sum to be prime.

Not Sufficient.

Statement 1 + 2

=> x is an even number < 10

As we saw in our analysis above, for x = 8, the sum is not prime. For all other possible values of x, the sum is prime.

Not Sufficient.

Therefore, correct answer: Option E

(Note: Had Statement 1 been: x < 8, then the correct answer would have been Option C)

I hope this helped!

Japinder
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Originally posted by EgmatQuantExpert on 08 May 2015, 01:17.
Last edited by EgmatQuantExpert on 08 May 2015, 03:36, edited 1 time in total.
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 16 Jul 2007, 20:17
iamba wrote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10

(2) x is even


Same here. Got E in 2min and 30sec.

S1: x is less than 10. so I picked 3. 3!+(4)=10 so no. but 2!+3 is a prime number so this stmnt is insuff.

S2: just tried more numbers and came out insuff.
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 16 Jul 2007, 22:04
iamba wrote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10
(2) x is even


Could be B.

if x = 2, x! + (x + 1) = 5 prime
if x = 4, x! + (x + 1) = 29 prime
if x = 6, x! + (x + 1) = 727? probably prime
if x = 8, x! + (x + 1) = 40329? probably prime
if x = 10, x! + (x + 1) = 3628811? probably prime
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 17 Jul 2007, 01:34
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2
Consider 8! + 8 + 1. As both 8! and 8+1 are multiples of 3, so is 8! + 8 + 1. Thus (2) is not sufficient
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 17 Jul 2007, 02:03
'E'.

Stmt2: x! + (x + 1) gives a non prime number if x =8. For x=2,4,6, it gives prime number. So insuff

Taking them together, INSUFF
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 17 Jul 2007, 02:07
kevincan wrote:
Consider 8! + 8 + 1. As both 8! and 8+1 are multiples of 3, so is 8! + 8 + 1. Thus (2) is not sufficient


Yeah True

Stmt 1 is insuff
Stmt 2, we can plug in values....
2,4,6 are looking primes..... than put in 8
8! + 9 = 3(something + 3)..this is clearly not prime!

E
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post Updated on: 17 Jul 2007, 10:06
I get C

Stmt1
x <10> x can be 2 so 2!+3 = 5 - prime
But x can be 3 so 3!+4 = 10 - Non prime
So InSuff

Stmt2
x is even.

x can be 2 so 2!+3 = 5 - prime
But x can be 14 so 14! + 15 is not prime

Combine 1 and 2 you will see x! + x + 1 is prime

Oops I missed 8! + 9. I agree with E

Originally posted by dahcrap on 17 Jul 2007, 09:51.
Last edited by dahcrap on 17 Jul 2007, 10:06, edited 1 time in total.
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 17 Jul 2007, 10:05
Quote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10
(2) x is even


Quote:
get C

Stmt1
x <10> x can be 2 so 2!+3 = 5 - prime
But x can be 3 so 3!+4 = 10 - Non prime
So InSuff

Stmt2
x is even.

x can be 2 so 2!+3 = 5 - prime
But x can be 14 so 14! + 15 is not prime

Combine 1 and 2 you will see x! + x + 1 is prime


I agree with everyone who said E. As was noted upthread, if you combine the 2 statments, you are left with 2, 4, 6, 8 as possibilities for x. However, after Himalayan did the math, it was noted that using x=8 does not produce a prime number.
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If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 16 Mar 2008, 13:58
2
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If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10

(2) x is even
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 16 Mar 2008, 15:15
1
E

1: insuffcieint
2 - 2!+3 = 5 - prime
3 - 3!+4 = 10 - not a prime
8 = 8!+9 - not a prime

2: insuffcient
2 and 8 give different results

Combined - again 2 and 8 give different results - insuffcieint
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 27 Sep 2011, 04:15
we have to pick easy numbers and get the result of a factorial and add that to the sum in the parentheses

1.) Not Sufficient
1! + (1+1) = 1 + 2 = 3 (would be a prime number)
2! + (2+1) =2*1 + 3 = 5 (would be a prime number)
3! + (3+1) = 3*2*1 + 4 = 10 (wouldn't be a prime number)

2.) Not Sufficient
See x = 2 @ First Choice
4! + (4+1) = 4*3*2*1 + 5 = 30 (wouldn't be a prime number)

MGMAT Explanation has place for improvement. YES it SUCKS (Yes u r great with other things Mr and Mrs 69 uhm.. 99 :shock: ).
Q is according to MGMAT 700-800, Divisibility and Primes. Besides I can only recommend the MGMAT online tests (you get the access to 6 online (win and MAC) gmat tests if you buy one book for around $30)
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 06 Oct 2011, 04:07
1
masterrick wrote:
we have to pick easy numbers and get the result of a factorial and add that to the sum in the parentheses

1.) Not Sufficient
1! + (1+1) = 1 + 2 = 3 (would be a prime number)
2! + (2+1) =2*1 + 3 = 5 (would be a prime number)
3! + (3+1) = 3*2*1 + 4 = 10 (wouldn't be a prime number)

2.) Not Sufficient
See x = 2 @ First Choice
4! + (4+1) = 4*3*2*1 + 5 = 30 (wouldn't be a prime number)

MGMAT Explanation has place for improvement. YES it SUCKS (Yes u r great with other things Mr and Mrs 69 uhm.. 99 :shock: ).
Q is according to MGMAT 700-800, Divisibility and Primes. Besides I can only recommend the MGMAT online tests (you get the access to 6 online (win and MAC) gmat tests if you buy one book for around $30)

you'd also need to consider conditions 1+2 together.. which doesn't allow for a counter example until you get to x=8
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 13 Oct 2011, 13:47
igemonster wrote:
masterrick wrote:
we have to pick easy numbers and get the result of a factorial and add that to the sum in the parentheses

1.) Not Sufficient
1! + (1+1) = 1 + 2 = 3 (would be a prime number)
2! + (2+1) =2*1 + 3 = 5 (would be a prime number)
3! + (3+1) = 3*2*1 + 4 = 10 (wouldn't be a prime number)

2.) Not Sufficient
See x = 2 @ First Choice
4! + (4+1) = 4*3*2*1 + 5 = 30 (wouldn't be a prime number)

MGMAT Explanation has place for improvement. YES it SUCKS (Yes u r great with other things Mr and Mrs 69 uhm.. 99 :shock: ).
Q is according to MGMAT 700-800, Divisibility and Primes. Besides I can only recommend the MGMAT online tests (you get the access to 6 online (win and MAC) gmat tests if you buy one book for around $30)

you'd also need to consider conditions 1+2 together.. which doesn't allow for a counter example until you get to x=8



When x=4 4! + (4+1) = 24+5 = 29 (a prime number) not 30
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 13 Oct 2011, 14:04
I picked E.

1.) Not Sufficient
2! + (2+1) =2*1 + 3 = 5 (prime number)
3! + (3+1) = 3*2*1 + 4 = 10 (not a prime number)

2.) Not Sufficient
2! + (2+1) =2*1 + 3 = 5 (prime number)
4! + (4+1) = 4*3*2*1 + 5 = 29 (prime number)
6! + (6+1) = 6*5*4*3*2*1 + 7 = 727 (prime number)
8! + (8+1) = 8*7*6*5*4*3*2*1 + 9 = 40329 (not a prime number)

Combined Statements 1 and 2..not sufficient because when x<10 and x is even you get prime and non-prime numbers

Is there any way to resolve this problem faster???
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 13 Oct 2011, 16:26
My Answer is E.

1: insuffcieint
2 - 2!+3 = 5 - prime
3 - 3!+4 = 10 - not prime
8 = 8!+9 - not prime

2: insuffcient
x= 0, x! + (x + 1) = 1 - Not prime.
x= 2, x! + (x + 1) = 5 - prime.

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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 07 May 2015, 08:49
ishcabibble wrote:
Quote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10
(2) x is even


Quote:
get C

Stmt1
x <10> x can be 2 so 2!+3 = 5 - prime
But x can be 3 so 3!+4 = 10 - Non prime
So InSuff

Stmt2
x is even.

x can be 2 so 2!+3 = 5 - prime
But x can be 14 so 14! + 15 is not prime

Combine 1 and 2 you will see x! + x + 1 is prime


I agree with everyone who said E. As was noted upthread, if you combine the 2 statments, you are left with 2, 4, 6, 8 as possibilities for x. However, after Himalayan did the math, it was noted that using x=8 does not produce a prime number.



Is there any other way to solve this question ?
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 08 May 2015, 03:04
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 08 Feb 2016, 10:19
iamba wrote:
If x is a positive integer, is x! + (x + 1) a prime number?

(1) x < 10

(2) x is even


Is this question representative of an actual GMAT question? As it is not challenging logic wise and takes a lot of listing.I could get to the answer but took me way more than two minutes.
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Re: If x is a positive integer, is x! + (x + 1) a prime number?  [#permalink]

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New post 08 Feb 2016, 18:05
Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

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

(1) x < 10

(2) x is even


In the original condition, there is 1 variable(x), which should match with the number of equations. So you need 1 equation. For 1) 1 equation, for 2) 1 equation, which is likely to make D the answer.
For 1), x=2 -> yes, x=8 -> no, which is not sufficient.
For 2), x=2 -> yes, x=8 -> no, which is not sufficient.
When 1) & 2), x=2 -> yes, x=8 -> no, which is not sufficient.
Therefore, the answer is E.


-> For cases where we need 1 more equation, such as original conditions with “1 variable”, or “2 variables and 1 equation”, or “3 variables and 2 equations”, we have 1 equation each in both 1) and 2). Therefore, there is 59 % chance that D is the answer, while A or B has 38% chance and C or E has 3% chance. Since D is most likely to be the answer using 1) and 2) separately according to DS definition. Obviously there may be cases where the answer is A, B, C or E.
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