| Author |
Message |
|
Guest
|
Post subject: For every positive even integer n, the function h(n)  Posted: Tue Sep 04, 2007 1:55 am |
|
|
|
|
For every positive even 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 than 40
Can someone walk me through this one?
|
|
  |
|
 |
|
dbernst
|
Post subject: Posted: Thu Sep 06, 2007 3:51 pm |
|
 |
| ManhattanGMAT Staff |
|
|
Posts: 304
|
Guest, this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100
By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).
Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.
Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.
Hope that helps
-Dan
Quote: For every positive even 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 than 40
|
|
| |
|
|
|
guest612
|
Post subject: Posted: Sat Apr 19, 2008 10:45 pm |
|
|
|
|
I got stuck on this question as well. can you please tell me how 2^50 can be factored? I'm not sure I'm following the math.
Thank you!
|
|
| |
|
|
|
RonPurewal
|
Post subject: Posted: Thu Apr 24, 2008 3:27 am |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 7146
|
guest612 wrote: I got stuck on this question as well. can you please tell me how 2^50 can be factored? I'm not sure I'm following the math.
Thank you!
the number h(100) is 2 x 4 x 6 x 8 x ... x 100
which is (2 x 1) x (2 x 2) x (2 x 3) x ... x (2 x 50)
if you pull out all those 2's and send them all to the front of the expression - something you're allowed to do, because you can multiply a group of numbers in any order you please - you get 2^50 in front, because there are fifty 2's.
|
|
| |
|
|
|
guest612
|
Post subject: great Posted: Fri Apr 25, 2008 10:06 pm |
|
|
|
|
| |
|
|
|
rfernandez
|
Post subject: Posted: Fri May 02, 2008 3:38 pm |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 386
|
|
| |
|
|
|
Halo3
|
Post subject: Posted: Sun Aug 03, 2008 5:15 pm |
|
|
|
|
Ron,
I'm missing something in the following step
"Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1. "
So because h(100) can be factored by every integer up to 50 means that h(100)+1 can't have a prime factor below 50? Not quite getting this one... Please flush out.
Thanks-
|
|
| |
|
|
|
John
|
Post subject: Posted: Sun Aug 03, 2008 6:18 pm |
|
|
|
|
Halo3,
h(100)=2^50 * 50! = 2^50 * (1*2*3*...*50)
Here h(100) is divisible by any numer between 2 and 50. Right?
So h(100)+1 can't be divisile by any number between 2 and 50. h(100)+1 will always give a reminder of 1 when u divide the no by any no between 1 to 50.
So h(100)+1 doesnot have any factor between 2 and 50. Hence h(100)+1 doesnot have any prime factor between 2 and 50.
Cheers,
|
|
| |
|
|
|
RonPurewal
|
Post subject: Posted: Tue Aug 12, 2008 4:30 am |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 7146
|
Halo3 wrote: Ron,
I'm missing something in the following step
"Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1. "
So because h(100) can be factored by every integer up to 50 means that h(100)+1 can't have a prime factor below 50? Not quite getting this one... Please flush out.
Thanks-
i'll assume you mean " flesh out", unless you want this whole thread destroyed (which you presumably don't).
heh heh
the idea is this: if a number is divisible by some prime p, then the next multiple of p will be p units bigger. for instance, 75 is divisible by 5. this means that the next greatest multiple of 5 is 80, which is 5 units away.
hopefully, this fact is clear. once you realize this, it follows that consecutive integers can't share ANY primes, because they're only 1 unit apart (too close together to work for any common factor except 1, which is trivially a factor of any integer at all, anywhere).
that's the basis for saying h(100) + 1 has no prime factors below 50. if it did, then you'd have two multiples of the same prime 1 unit apart, and that's impossible.
|
|
| |
|
|
|
Halo3
|
Post subject: Thanks Posted: Tue Aug 12, 2008 6:46 pm |
|
|
|
|
Makes sense now - appreciate the help!
|
|
| |
|
|
|
rfernandez
|
Post subject: Posted: Fri Aug 22, 2008 1:59 am |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 386
|
|
| |
|
|
|
supratims
|
Post subject: Re: For every positive even integer n, the function h(n) Posted: Tue Aug 04, 2009 3:31 pm |
|
|
| Students |
|
|
Posts: 30
|
|
nice explanation. Thanks !
|
|
| |
|
|
|
Ben Ku
|
Post subject: Re: For every positive even integer n, the function h(n) Posted: Thu Aug 13, 2009 4:01 pm |
|
|
| ManhattanGMAT Staff |
|
|
Posts: 823
|
|
You're welcome!
_________________
Ben Ku
Instructor
ManhattanGMAT
|
|
| |
|
|
|
hugolinares
|
Post subject: Re: For every positive even integer n, the function h(n) Posted: Fri Aug 21, 2009 7:06 am |
|
|
| Forum Guests |
|
|
Posts: 2
|
|
Great explanation Ron, Thanks!
|
|
| |
|
|
|
anoo.anand
|
Post subject: Re: Posted: Sat Oct 03, 2009 1:23 am |
|
|
| Students |
|
|
Posts: 73
|
RonPurewal wrote: Halo3 wrote: Ron,
I'm missing something in the following step
"Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).
Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1. "
So because h(100) can be factored by every integer up to 50 means that h(100)+1 can't have a prime factor below 50? Not quite getting this one... Please flush out.
Thanks- i'll assume you mean " flesh out", unless you want this whole thread destroyed (which you presumably don't). heh heh the idea is this: if a number is divisible by some prime p, then the next multiple of p will be p units bigger. for instance, 75 is divisible by 5. this means that the next greatest multiple of 5 is 80, which is 5 units away. hopefully, this fact is clear. once you realize this, it follows that consecutive integers can't share ANY primes, because they're only 1 unit apart (too close together to work for any common factor except 1, which is trivially a factor of any integer at all, anywhere). that's the basis for saying h(100) + 1 has no prime factors below 50. if it did, then you'd have two multiples of the same prime 1 unit apart, and that's impossible. so does this mean...that the next factor is greater than 50 ? still not getting this :(
|
|
| |
|
|
|
