Is integer k a prime number?

A prime number has only two factors - 1 and itself.
We need to find whether k is prime.

(1) k = 10! + m, where 1 < m < 8
10! has all factors from 1 to 10 because 10! = 1*2*3*4*...*10
m is a value which ranges from 2 to 7. In each one of these cases, you can split k into two factors other than 1 and itself. For example,

k = 10! + 2 = 2 * (1*3*4*5*6*7*8*9*10 + 1) - two factors other than 1 and itself
k = 10! + 3 = 3 * (1*2*4*5*6*7*8*9*10 + 1) - two factors other than 1 and itself
So in all these cases, k will NOT be prime.
Sufficient alone

(2) k is a multiple of 7
k could be a negative multiple of 7, 0, 7 or some other positive multiple of 7.
7 is prime. All others are not prime.
Hence this statement alone is not sufficient.

Answer (A)

I'm wondering why statement #2 is not sufficient, I understand that k being a multiple of 7 could be 7 making it prime or it could be 14,21 or any other non prime multiple of 7. My question is, we know that k=10!+m so how could k=7 if 10! is a large number and m equals a number 2-7.

You cannot use info given in one statement when evaluating another.

VERITAS PREP OFFICIAL SOLUTION:

On a problem such as this, you must use conceptual understanding of the number line and divisibility to answer the question. Algebraic manipulation will not help you. Number picking will not help you. And certainly just doing it will not help you; calculating the number would be impossible!

The goal from each statement is to prove whether k is prime or importantly not prime. It is impossible without computer or calculator assistance to determine whether large number is actually prime (there are too many numbers you would have to check for divisibility), but it is actually quite easy to prove that a number is not prime. If you can prove that k is divisible by anything other than 1 and itself, you have proven that k is not prime. It is on this type of problem that you should be looking to disprove the question and find a no answer.

In statement (1), you learn that k = 10! + either 2, 3, 4, 5, 6, or 7. To review, 10! (10 factorial) represents the product of all positive numbers from 1 to 10, inclusive: 10 x 9 x 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1. As you can see, the number 10! is a multiple of each of 2, 3, 4, 5, 6, and 7. For demonstration, say that m were 7. 10! + 7 will then definitely be divisible by 7. If you find any multiple of 7 on the number line and add another 7 to it, it will always still be divisible by 7. Take the number 63, a multiple of 7. If you add 7 to 63, you get 70, another multiple of 7. This is then true for any potential value of m. If m were 2, then 10! (an even number) + 2 will remain divisible by 2. If m were 3, then 10! (a multiple of 3) + 3 will remain divisible by 3. Statement (1) thus proves that k is not prime, as whatever value of m (2 through 7) we add to 10!, k will remain divisible by that number, so it cannot be a prime number. Statement (1) is sufficient.

Be careful with statement (2). If you carry some information with you from statement (1)—the fact that k is a large multiple of 7—you might think that statement (2) is also sufficient. In other words, you might think, if k is a multiple of 7, then it could never be prime, as it will be divisible by 7. However, remember that there is one multiple of 7 that is prime: 7 itself. Statement (2) is not sufficient, because k could be prime (7) or it could be any of the infinite set of multiples of 7 that are not prime.

The answer to this question is answer choice A.

Why do we overlook the fact that 'm' may not be an integer? I know the that the question stem says "integer k", but that does not imply that 'm' must be an integer value only.

m also must be an integer.

k = 10! + m;
integer = integer + m;
m = integer - integer = integer.

Hope it's clear.

Great Question
here is what i did
We ned to find whether integer k is prime or not
Statement 1

k=10!+m
As k is an integer => m must be an integer too.
m=> (1,8)
m=2=> k is divisible by 2
m=3=> k is divisible by 3
m=4=> k is divisible by 4
m=5=> k is divisible by 5
m=6=> k is divisible by 6
m=7 => k is divisible by 7
hence k is never prime
Statement 2
k=7
k=14
taking the above two cases we can discard this statement
hence A

Hello, does it mean, that first prime number in this series, will be at least 10!+11 or later ? and even 10!+8/9/10 won't ever be prime ?

Hi Shrey9

From the stem we know K is an integer

A) k = 10! + m, where 1 < m < 8

Since K is an Integer M can take only integer values from 2 to 7.
10!=10*9*8*7*6*5*4*3*2*1+ 2/3/4/5/6/7
You can always take one integer common.So K is not prime. Sufficient

To your question if m is 8/9/10 , K would be non prime.
If m= 11, K would be Prime , but you don't need to think of this here.

B) If k is a multiple of 7 K can be 0 , 7 or 14,

We have no idea whether it can be a prime or non prime.

Hope it helps.

While it is possible for a test-taker to prove that a very large integer is not prime -- if the integer is even, if the integer is a multiple of 3, etc. -- it is NOT possible for a test taker to prove that a very large integer IS prime.
Implication:
If a DS problem asks whether a very large integer is prime, the answer must be NO.



Statement 1:
Here, k is a very large integer.
Since the question stem asks whether this very large integer is prime, the answer must be NO.
SUFFICIENT.

Statement 2:
If k=7, the answer to the question stem is YES.
If k=14, the answer to the question stem is NO.
Since the answer is YES in the first case but NO in the second case, INSUFFICIENT.

GMATGuruNY

Logic in highlighted portion is incorrect.

k = 10! + m where 1<m<8
since 10! = 10*9*8*7*6*5*4*3*2*1
10! + 2 = 10*9*8*7*6*5*4*3*2*1 + 2 = 2(10*9*8*7*6*5*4*3*1+1) = multiple of 2
10! + 3 = 10*9*8*7*6*5*4*3*2*1 + 3 = 3(10*9*8*7*6*5*4*2*1+1) = multiple of 3
10! + 5 = 10*9*8*7*6*5*4*3*2*1 + 5 = 5(10*9*8*7*6*4*3*2*1+1) = multiple of 5
10! + 7 = 10*9*8*7*6*5*4*3*2*1 + 7 = 7(10*9*8*6*5*4*3*2*1+1) = multiple of 7
10! + m is NOT a prime number.

Asked: Is integer k a prime number?

(1) k = 10! + m, where 1 < m < 8
k = 10! + m where 1<m<8
since 10! = 10*9*8*7*6*5*4*3*2*1
10! + 2 = 10*9*8*7*6*5*4*3*2*1 + 2 = 2(10*9*8*7*6*5*4*3*1+1) = multiple of 2
10! + 3 = 10*9*8*7*6*5*4*3*2*1 + 3 = 3(10*9*8*7*6*5*4*2*1+1) = multiple of 3
10! + 5 = 10*9*8*7*6*5*4*3*2*1 + 5 = 5(10*9*8*7*6*4*3*2*1+1) = multiple of 5
10! + 7 = 10*9*8*7*6*5*4*3*2*1 + 7 = 7(10*9*8*6*5*4*3*2*1+1) = multiple of 7
10! + m is NOT a prime number.

(2) k is a multiple of 7
If k = 7 ; k is a PRIME number
If k = 14; k is NOT a PRIME number
NOT SUFFICIENT

IMO A

(2) can trick you if you forget that 7 is a multiple of 7 so eliminate only (1) helps and A is correct

This solution is valid.
But the soundness of this approach does not on its own invalidate the soundness of OTHER approaches.
Since Statement 1 indicates that every option for k is a very large integer, the answer to the question stem -- is k prime? -- must be NO.
The GMAT cannot expect a test-taker to prove -- in 2 minutes or less -- that a very large integer is prime.

I think it should be given M is an integer or it should be given K is an integer because we have to find whether k is prime number or not and if "M" is not and intger K will be fractional and it wont be and prime number too . I think we should consider this rather than what you have mentioned.