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Our aim is to find the smallest prime factor in (2*4*6*8*10*........100)+1. (2*4*6*8*10*........100)+1 =[2(1) * 2(2) *2(3)......2(50)]+1 =[2^50 * 50!]+1

Now , Take a simpler example....

5! is divisible by prime numbers less than or equal to 5 (2,3,5)

You can see that 5!+1 is not divisible by prime numbers less than or equal to 5 (2,3,5) [5!+1 = 121 is divisible by the prime number 11]

Take another example, 6!.

6! is divisible by prime numbers less than or equal to 6 (2,3,5)

In the same way , in above example , 50! will be devisible be prime numbers smaller than 50 ; however 50!+1 will not be. So answer is 5. greater than 40.
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Too good. But is there any way that we can actually find the prime number?

amolsk11 wrote:

Our aim is to find the smallest prime factor in (2*4*6*8*10*........100)+1. (2*4*6*8*10*........100)+1 =[2(1) * 2(2) *2(3)......2(50)]+1 =[2^50 * 50!]+1

Now , Take a simpler example....

5! is divisible by prime numbers less than or equal to 5 (2,3,5)

You can see that 5!+1 is not divisible by prime numbers less than or equal to 5 (2,3,5) [5!+1 = 121 is divisible by the prime number 11]

Take another example, 6!.

6! is divisible by prime numbers less than or equal to 6 (2,3,5)

In the same way , in above example , 50! will be devisible be prime numbers smaller than 50 ; however 50!+1 will not be. So answer is 5. greater than 40.

Too good. But is there any way that we can actually find the prime number?

As it's said, all we need to know is that this number is going to be grater than 50. ))) But, yeah, do the prime factorisation for 2^50*50! + 1 = 34243224702511976248246432895208185975118675053719198827915654463488000000000001, I have no idea how to find it mathematically though, (if there is such a way at all) other than dividing that number by primes starting with 53 and looking if there is a remainder. By the way the smallest prime factor is 79 in this case. I wrote simple program in LISP to find it )), took me under 2 min's. Why can't I bring my lappie to the test?
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Too good. But is there any way that we can actually find the prime number?

No, not without computer assistance. In general, if a very large number only has large prime factors, it can take eons to factorize it, even with a computer. This fact is actually the basis of all internet cryptography - internet encryption can be broken, but it would take too much computing time to do it in any reasonable amount of time. So while relatively few people are aware of the fact, prime number theory is the reason you can do banking or shopping on the internet, or send private emails. Without it, the internet would be a very different thing.
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If you are looking for online GMAT math tutoring, or if you are interested in buying my advanced Quant books and problem sets, please contact me at ianstewartgmat at gmail.com

Hi guys I was wonderig if anyone could offer some suggestions on how to solve this problem:

For every positive integer n, the function h(n) is defined by the roduct of all even integers from 2 to N inclusve. If p is the smallest prime factor of h(100)+1, then p is

a. between 2 and 10 b. between 10 and 20 c. between 20 and 30 d. between 30 and 40 e. greater than 40

the equation works out to be h(n) = n(n-2)(n-4)(n-6)(n-8)....(n-(n-2))

so h(100) = 100*98*96*94*92.....4*2

which is a gigantic even number, with the smallest prime factor of 2

however, when you add one, it becomes prime itself (or safe to assume so), so the smallest prime factor could be insinuated to be itself, or a gigantic random number way over 40.

Hi guys I was wonderig if anyone could offer some suggestions on how to solve this problem:

For every positive integer n, the function h(n) is defined by the roduct of all even integers from 2 to N inclusve. If p is the smallest prime factor of h(100)+1, then p is

a. between 2 and 10 b. between 10 and 20 c. between 20 and 30 d. between 30 and 40 e. greater than 40

This question is testing the concept of coprimes. 2 positive integers are coprime when their greatest common factor (their only common factor) is 1. Now note that two different prime numbers are always coprime. For example, 3 and 7 are coprime. So are 13 and 19. But the two integers need not be prime numbers in order to be coprime. For example, 4 and 9 are coprime (1 is their only common factor). Also, Important! Two consecutive integers are always coprime. The question is testing you on this concept. Let's solve it now: h(n) = 2*4*6*....................*n. ----n is even. h(100) = 2*4*6*........ 94*96*98*100. h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100). Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47. Clearly, this prime factor is greater than 40.

By the way, you cannot assume that h(100) + 1 is a prime number. Just use the concept of coprimes.

That said, this is a previous post of mine in this forum. I was then TheGMATDoctor. You can check out my posts by doing a search. --------------------------------------- This was today's GMAT math recipe from the GMAT Chef.
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For every positive integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100)+1, then p is... A) between 2 and 10 B) between 10 and 20 C) between 20 and 30 D) between 30 and 40 E) greater then 40

I would really appreciate a response to this question. I feel like I am missing a fundamental principle/trick. This question has been bugging me for about a week.

h(100) = 100x99x98....3x2 h(100)+1= 100x99x98....3x2 + 1 So it can not be divided by any number b/w 2 and 100. e.g. it can not be divided by 91 because h(100)= y.91 (y is the multiplication of other numbers b/w 2 and 100) so h(100)+1=y.91 + 1 so this number is equal to 1 in (mod 99). For any number b/w 2 and 100 this holds. So the greatest prime number will be h(100)+1 or a number that is b/w 100 and h(100)+1. Both are greater than 40

h(100) and h(100)+1 are relatively prime, so they cannot have a prime factor in common other than 1.

(for any integer n, n and n+1 are relatively prime)

so we have h(100)= 100x98x.....x4x2 = 2(1x2x3x4x.....x49x50) so we know that any number from 1 to 50 is a factor of h(100) and therefore cannot be a factor of h(100)+1. the smallest factor of h(100)+1 will be greater than 50.

Sorry. I did not see "EVEN". But that does not change the solution.

Quote:

h(100)+1 are relatively prime

Not necessarily. For example, h(7) = 2 x 4 x 6 = 48 (not prime), h(7) + 1 = 49 (not prime)

I was talking about "relatively prime numbers" (or coprime) not "prime numbers" this is the definition "Two integers are relatively prime if they share no common positive factors (divisors) except 1" so 48 and 49 are relatively prime.

check on wiki for more info. clearly I don't know if we can do this question w/o the notion of coprime.