Author 
Message 
TAGS:

Hide Tags

Director
Joined: 17 Jul 2006
Posts: 706

yezz wrote: For every positive integer n, the function h(n) is defined to be the product of all 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
eg:
5! = 5*4*3*2*1 = 120 largest prime factor is 5
120+1 = 121 the smallest prime factor is 11 bigger than the largest prime of 120
h(100) = 2*4*6*8*.......100 = 2^50(1*2*3*4*........47*48*49*50)
47 is the largest prime in h(100) thus sure h(100) + 1 has smallest prime greater than 47
E is the answer
I HOPE THIS HELPS
Yezz great explanation. I like it. YOU Really ROCK



Director
Joined: 11 Sep 2006
Posts: 514

For every positive integer n... [#permalink]
Show Tags
02 Nov 2006, 20:12
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
between 2 and 10
between 10 and 20
between 20 and 30
between 30 and 40
greater than 40
_________________
...there ain't no such thing as a free lunch...



VP
Joined: 25 Jun 2006
Posts: 1166

i for E. posted before, but i forgot the explanation.



Manager
Joined: 01 Nov 2006
Posts: 70

If an integer > 1 divides x then it doesn't divide x + 1.
Any prime factor p less than 50 divides h(100) because 2*p is in the product h(100). Thus, the smallest prime factor that divides h(100) + 1 must be greater than 50.



Director
Joined: 11 Sep 2006
Posts: 514

E was the OA  thank you for the explanation!
_________________
...there ain't no such thing as a free lunch...



Intern
Joined: 02 Mar 2006
Posts: 14

PS  Prime Factor [#permalink]
Show Tags
18 Nov 2006, 12:58
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



Manager
Joined: 10 Jul 2006
Posts: 72

Fight hard,
I believe this question had been covered recently. If you'd like some explaination on this question, I think doing a search for keywords such as prime would get you the answer.



Intern
Joined: 09 Dec 2006
Posts: 8

PS: smallest prime factor (gmatprep) [#permalink]
Show Tags
16 Jan 2007, 10:10
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



Senior Manager
Joined: 01 Feb 2005
Posts: 271

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
n = 100. Therefore h(n) = 2n = 200. therefore H(100) + 1 = 201.
201 =3X67. 3 being the smallest prime number, I got A.



Intern
Joined: 09 Dec 2006
Posts: 8

sorry, A is not the right answer. h(100) is not equal to 201. read the question again carefully



Intern
Joined: 09 Dec 2006
Posts: 8

OA is E. can someone please explain this?



Senior Manager
Joined: 24 Oct 2006
Posts: 339

Has been discussed many times, could be searched.



Intern
Joined: 11 Oct 2006
Posts: 21
Location: Indianapolis

Difficult prime number question [#permalink]
Show Tags
20 Jan 2007, 16:48
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) betwen 2 and 10
B) between 10 and 20
C) between 20 and 30
D) between 30 and 40
E) greater than 40



Senior Manager
Joined: 23 Jun 2006
Posts: 387

just use the search option.... question discussed many times on many threads... answer is E.



Manager
Joined: 05 May 2005
Posts: 78

Function problem [#permalink]
Show Tags
05 Feb 2007, 20:58
How would you solve?
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



Senior Manager
Joined: 04 Jan 2006
Posts: 279

Re: Function problem [#permalink]
Show Tags
06 Feb 2007, 01:16
2
This post received KUDOS
above720 wrote: How would you solve?
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
question!
h(2) = 2 = (2x1)
h(4) = 2 x 4 = (2x1) x (2x2) = (2^2) * 2!
h(6) = 2 x 4 x 6 = (2x1) x (2x2) x (2x3) = (2^3) * 3!
Therefore,
h(n) = (2^(n/2) * (n/2)!
h(100) = (2^50) * 50!
If p is the smallest prime factor of h(100) + 1, then p is?
Keep dividing the term above with each prime number. 2, 3, 5, 7, 11, 13, 17, 19, 23, ... and see what number can divide it evenly.
Assuming p = 2; (h(100) + 1)/2 = (2^50) * (3x4x5...x50) + 1/2
This means that by using 2, the remainder = 1/2
Assuming p = 3; (h(100) + 1)/3 = (2^50) * (2 x 4 x 5 x 6... x50) + 1/3
This means that by using 3, the remainder = 1/3
We can conclude that All prime number from 2 to 47 will give the remainder of 1/2, 1/3, 1/5, 1/11, ..., 1/47
Therefore, p is greater than 40.
E. is the answer.



Intern
Joined: 09 Jan 2007
Posts: 24

Smallest prime factor problem [#permalink]
Show Tags
06 Feb 2007, 17:28
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.
between 2 and 10
between 10 and 20
between 20 and 30
between 30 and 40
above 40
The answer is above 40, but why? From the problem I figure that
h(100)+1 is a really large odd number, but where do you go from there?
Thanks



Intern
Joined: 13 Jun 2005
Posts: 28

Here's my cut on this
h(10) = 2x4x6x8x10 = 2^5 (1x2x3x4x5)
The smallest prime factor of h(10) + 1 needs to be greater than 5
cannot be 1 or 2 or 3 or 4 or 5
Similarly h(50) = 2^25 (1X2x....X25)
smallest prime factor of h(50)+1 need to be > than 25
h(100) = 2^50 (1x2x3....50)
For h(100)+1, smalles prime factor needs to > than 50



Senior Manager
Joined: 04 Jan 2006
Posts: 279

Re: Smallest prime factor problem [#permalink]
Show Tags
06 Feb 2007, 18:36
frankmay32780 wrote: 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.
between 2 and 10 between 10 and 20 between 20 and 30 between 30 and 40 above 40
The answer is above 40, but why? From the problem I figure that h(100)+1 is a really large odd number, but where do you go from there?
Thanks
I have answered this question before .
Here is the solution.
h(2) = 2 = (2x1) = (2^1)*(1!)
h(4) = 2 x 4 = (2x1) x (2x2) = (2^2) * (2!)
h(6) = 2 x 4 x 6 = (2x1) x (2x2) x (2x3) = (2^3) * (3!)
Therefore; h(n) = (2^(n/2)) * (n/2)!
h(100) = (2^50) * (50!)
The important clue here is that h(100) has 50! as part of its factor. 50! = 1 x 2 x 3 x 4 x ... x 49 x 50.
If we pick any prime number from 2 to 50. we can safely say that the remainder will be zero.
From the question, p is the smallest prime factor of h(100)+1.
Check by using number 2; [h(100) + 1]/2 = (Integer1) + 1/2
using number 3; [h(100) + 1]/3 = (Integer2) + 1/3
using number 5; [h(100) + 1]/5 = (Integer3) + 1/5
.
.
.
using number 47; [h(100) + 1]/47 = (Integer bla bla bla) + 1/47
Clearly there are no prime numbers from 2 to 47 that can divide h(100) + 1 evenly.
E. is the answer.



Intern
Joined: 30 Dec 2004
Posts: 25

gmat prep question [#permalink]
Show Tags
02 Mar 2007, 10:31
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 2and 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40




gmat prep question
[#permalink]
02 Mar 2007, 10:31



Go to page
Previous
1 2 3 4 5 6 7 8 9 10 11 ... 14
Next
[ 273 posts ]





Similar topics 
Author 
Replies 
Last post 
Similar Topics:


705


For every positive even integer n, the function h(n) is defined to be

enigma123 
49 
20 Jun 2017, 07:37 

15


For every positive even integer n, the function h(n) is defined to be

fla162 
5 
24 May 2016, 03:26 

9


For every positive even integer n, the function h(n)

skamal7 
5 
03 Apr 2017, 10:45 

28


For every positive even integer n, the function h(n) is defi

GODSPEED 
7 
05 Jul 2014, 05:23 

27


For every positive even integer n, the function h(n) is defi

joyseychow 
8 
15 Nov 2013, 15:06 



