Does the integer K have

[Guest]

Does the integer k have a factor p such that 1<p<k?

1. k>4!
2. 13!+2(<=)k(<=)13!+13

This is from GMAT Prep software (It is also in OG but since the sum is exposed in Prep software I think it should be fine to post this)

The above question has been answered several times by MGMAT... I have a different question though..

From what I understand "K" could range from 13!+2 to 13!+13, which is from 6227020802 to 6227020813 (Calculated).. And the quesion stems asks whether K has a afactor P such that P ranges from 1 to 6227020813.. Right ?

What if P is 6227020801, why is this is a factor ? I know that something is wrong with the way I interpret this sentence but not sure what.. Please explain

[Guest]

Anyone ?

[Guest]

I think the range HAS to be less than k, given "such that 1<p<k"

[RonPurewal]

the question is asking whether k has a factor that is greater than 1, but less than itself.
if you're good at these number property rephrasings, then you can realize that this question is equivalent to "is k non-prime?", which, in turn, because it's a data sufficiency problem (and therefore we don't care whether the answer is "yes" or "no", as long as there's an answer), is equivalent to "is k prime?".
but let's stick to the first question - "does k have a factor that's between 1 and k itself?" - because that's easier to interpret, and, ironically, is easier to think about (on this particular problem) than the prime issue.

--

key realization:
every one of the numbers 2, 3, 4, 5, ..., 12, 13 is a factor of 13!.

this should be clear when you think about the definition of a factorial: it's just the product of all the integers from 1 through 13. because all of those numbers are in the product, they're all factors (some of them several times over).

--

consider the lowest number allowed by statement 2: 13! + 2.
note that 2 goes into 13! (as shown above), and 2 also goes into 2. therefore, 2 is a factor of this sum (answer to question prompt = "yes").

consider the next number allowed by statement 2: 13! + 3.
note that 3 goes into 13! (as shown above), and 3 also goes into 3. therefore, 3 is a factor of this sum (answer to question prompt = "yes").

etc.
all the way to 13! + 13.
works the same way each time.
so the answer is "yes" every time --> sufficient.

--

by the way, the gmat will NEVER, EVER, EVER require you to calculate a value that's even half as ugly as the one that you worked out for this problem (i sincerely hope that you didn't work out that value by hand). if you're even considering doing a calculation like that, run for your life, and try to think about another method.

[Guest]

Ron,

Thanks for the elabrote reply. However I still am not convinced.. we have sucessfully proved that if p ranges from 2 to 13 then yes I agree that p is a factor of k such that 13!+13 > K >13!+2.. What if P exceeds 13 ? Let us consider only the question stem and the 2nd statement.

Stem says 2 things:
1) K is an integer
2) P can range from 1 to K

2nd statement says:
"K" can be any integer from 13!+2(<=)k(<=)13!+13 so in all there are 12 possible values for K

Now it is nowhere mentioned that P cannot exceed 13, what if it does ?.. the highest possible value for P is 13!+12, since it is less than 13!+13 and the lowest possible value for P is 2 as it is greater than 1.. A data sufficiency question is sufficeint only if I can prove that P is always a factor of K...Right ? So what am i missing ?

If it can be proven that every value of P, which ranges from 1 to K is a factor of K, only then it becomes sufficient..Thnx

[RonPurewal]

Ron,

Thanks for the elabrote reply. However I still am not convinced.. we have sucessfully proved that if p ranges from 2 to 13 then yes I agree that p is a factor of k such that 13!+13 > K >13!+2.. What if P exceeds 13 ? Let us consider only the question stem and the 2nd statement.

Stem says 2 things:
1) K is an integer
2) P can range from 1 to K

2nd statement says:
"K" can be any integer from 13!+2(<=)k(<=)13!+13 so in all there are 12 possible values for K

Now it is nowhere mentioned that P cannot exceed 13, what if it does ?.. the highest possible value for P is 13!+12, since it is less than 13!+13 and the lowest possible value for P is 2 as it is greater than 1.. A data sufficiency question is sufficeint only if I can prove that P is always a factor of K...Right ? So what am i missing ?

If it can be proven that every value of P, which ranges from 1 to K is a factor of K, only then it becomes sufficient..Thnx

before attacking the most difficult data sufficiency problems - such as this one - you should take the time to fully grasp the way data sufficiency works. it doesn't appear that you have yet reached that level of understanding.

remember, as soon as you can ascertain that the prompt question has a definitive answer, then that's "sufficient". there is no need to proceed any further.

in this problem, the prompt asks, "Is there a factor p such that...?"
this means that, if you can show that there is even one such factor, then it's "sufficient" and you are DONE.
we have ascertained that every one of the "k"s in that range has at least one such factor.
to wit, 13! + 2 has the factor 2; 13! + 3 has the factor 3; ...; 13! + 13 has the factor 13.
that's all we need to know.
sufficient.

you are right that it's difficult to ascertain whether numbers greater than 13 are factors of these "k"s. luckily, we don't have to care about that.

[crissro]

Since 2 and 7 are factors of k isn't 14 a factor as well?
I know this is irrelevant for solving the posted question, just a thought while going through the comments
Thank you

Ron,

Thanks for the elabrote reply. However I still am not convinced.. we have sucessfully proved that if p ranges from 2 to 13 then yes I agree that p is a factor of k such that 13!+13 > K >13!+2.. What if P exceeds 13 ? Let us consider only the question stem and the 2nd statement.

Stem says 2 things:
1) K is an integer
2) P can range from 1 to K

2nd statement says:
"K" can be any integer from 13!+2(<=)k(<=)13!+13 so in all there are 12 possible values for K

Now it is nowhere mentioned that P cannot exceed 13, what if it does ?.. the highest possible value for P is 13!+12, since it is less than 13!+13 and the lowest possible value for P is 2 as it is greater than 1.. A data sufficiency question is sufficeint only if I can prove that P is always a factor of K...Right ? So what am i missing ?

If it can be proven that every value of P, which ranges from 1 to K is a factor of K, only then it becomes sufficient..Thnx

before attacking the most difficult data sufficiency problems - such as this one - you should take the time to fully grasp the way data sufficiency works. it doesn't appear that you have yet reached that level of understanding.

remember, as soon as you can ascertain that the prompt question has a definitive answer, then that's "sufficient". there is no need to proceed any further.

in this problem, the prompt asks, "Is there a factor p such that...?"
this means that, if you can show that there is even one such factor, then it's "sufficient" and you are DONE.
we have ascertained that every one of the "k"s in that range has at least one such factor.
to wit, 13! + 2 has the factor 2; 13! + 3 has the factor 3; ...; 13! + 13 has the factor 13.
that's all we need to know.
sufficient.

you are right that it's difficult to ascertain whether numbers greater than 13 are factors of these "k"s. luckily, we don't have to care about that.

[RonPurewal]

Since 2 and 7 are factors of k isn't 14 a factor as well?
I know this is irrelevant for solving the posted question, just a thought while going through the comments

we try not to include comments that are, as you said, irrelevant to the question. but, yes, if 2 and 7 are factors then 14 will be a factor as well.

[administrator]

I often use smaller sets of data to mimic large sets presented in GMATPrep questions. I understand now that the set of 12 numbers from 13! + 2 to 13! + 13 will each have a factor that is >1 and < k, so no problem with the actual question. However, I was surprised to see that answer because I used a sample set of 4!+2 to 4!+13 which results in the 12 members 26, 27, 28, 29, 30....37. This smaller set contains prime numbers 29, 31, and 37 which ONLY have factors of 1 or itself. For example, 4! + 5 = 29 and 29 is NOT divisible by 5. Hummmm... what am I missing? Does something magical happen to primes > 6 billion such as 13!?

[jlucero]

I love when students use smaller numbers to help them with larger concepts, but you make one small mistake in your logic- the question talks about numbers between 13!+2 through 13!+13. Your smaller number should examine numbers between 4!+2 through 4!+4, all of them being non-prime. Here's why:

13! = 2x3x4...x12x13

Notice this big ol' factorial is made up of a lot of smaller numbers multiplied together. Because I multiplied by 2 (and b/c I multiplied by some other even numbers), 13! must be divisible by 2. Therefore, if I had 2 to an even number, 13!+2 must be even. So it's non-prime. For the same reason, 13!+3 is divisible by 3, 13!+4 is divisible by 4, etc.

In your analogy 4!+5, there's no factor that 4! and 5 have in common, so you aren't able to definitively prove that 4!+5 is a non-prime number. (side note: you also can't prove that it's a prime number as 4!+11= 35 which is non-prime)

To stretch this concept to all numbers, for all numbers where n>2, all numbers between n!+2 and n!+n will be non-prime.

[administrator]

Excellent; thanks.

[RonPurewal]

Excellent; thanks.

sure

[geezer0305]

Ron,

Thanks for the elabrote reply. However I still am not convinced.. we have sucessfully proved that if p ranges from 2 to 13 then yes I agree that p is a factor of k such that 13!+13 > K >13!+2.. What if P exceeds 13 ? Let us consider only the question stem and the 2nd statement.

Stem says 2 things:
1) K is an integer
2) P can range from 1 to K

2nd statement says:
"K" can be any integer from 13!+2(<=)k(<=)13!+13 so in all there are 12 possible values for K

Now it is nowhere mentioned that P cannot exceed 13, what if it does ?.. the highest possible value for P is 13!+12, since it is less than 13!+13 and the lowest possible value for P is 2 as it is greater than 1.. A data sufficiency question is sufficeint only if I can prove that P is always a factor of K...Right ? So what am i missing ?

If it can be proven that every value of P, which ranges from 1 to K is a factor of K, only then it becomes sufficient..Thnx

Hi,


This question indirectly asks us to check whether k is a prime number (if it is, then there is no factor p such that 1<p<k and if k isn't a prime number, then a factor p exists)

Statement 1 is insufficient: As it only tells us that k>4!--> k could be a prime number greater than 4! (in which case a factor p, such that 1<p<k, would not exist)

Statement 2: 13!+2(<=)k(<=)13!+13

Consider 13! -- it is divisible by 13,12,11,10,9,8,7,6,5,4,3,2,1
Also, since it is a multiple of 10, it ends in 0 (therefore is even)

Now, consider the inequality:

k can be any of the following numbers:-
13!+2 => Multiple of 2 + 2 --> p can be 2
13!+3 => Multiple of 3 + 3 --> p can be 3
13!+4 => Multiple of 4 + 4 --> p can be 4,2
13!+5 => Multiple of 5 + 5 --> p can be 5
13!+6 => Multiple of 6 + 6 --> p can be 6,2,3
13!+7 => Multiple of 7 + 7 --> p can be 7
13!+8 => Multiple of 8 + 8 --> p can be 2,4,8
13!+9 => Multiple of 9 + 9 --> p can be 3,9
13!+10 => Multiple of 10 + 10 --> p can be 10,2,5
13!+11 => Multiple of 11 + 11 --> p can be 11
13!+12 => Multiple of 12 + 12 --> p can be 12,2,3,4,6
13!+13 => Multiple of 13 + 13 --> p can be 13

Thus, in all of the above cases a factor p exists when k satisfies the given equality. Therefore, statement 2 is sufficient

Regards,

[tim]

thanks; let us know if there are any further questions on this one..