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

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 [#permalink] New post 10 Oct 2006, 12:11
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
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For every positive integer n... [#permalink] New post 02 Nov 2006, 19:12
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
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 [#permalink] New post 02 Nov 2006, 19:47
i for E. posted before, but i forgot the explanation.
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 [#permalink] New post 03 Nov 2006, 17:52
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.
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 [#permalink] New post 03 Nov 2006, 22:41
E was the OA - thank you for the explanation!
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PS - Prime Factor [#permalink] New post 18 Nov 2006, 11:58
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
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 [#permalink] New post 18 Nov 2006, 17:11
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.
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PS: smallest prime factor (gmatprep) [#permalink] New post 16 Jan 2007, 09:10
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
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 [#permalink] New post 16 Jan 2007, 11:20
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.
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 [#permalink] New post 16 Jan 2007, 16:38
sorry, A is not the right answer. h(100) is not equal to 201. read the question again carefully
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 [#permalink] New post 16 Jan 2007, 18:57
OA is E. can someone please explain this?
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 [#permalink] New post 16 Jan 2007, 18:59
Has been discussed many times, could be searched.
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Difficult prime number question [#permalink] New post 20 Jan 2007, 15:48
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
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 [#permalink] New post 21 Jan 2007, 16:29
just use the search option.... question discussed many times on many threads... answer is E.
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Function problem [#permalink] New post 05 Feb 2007, 19:58
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
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Re: Function problem [#permalink] New post 06 Feb 2007, 00:16
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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


:twisted: 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.

:wink:
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Smallest prime factor problem [#permalink] New post 06 Feb 2007, 16:28
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
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 [#permalink] New post 06 Feb 2007, 17:30
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
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Re: Smallest prime factor problem [#permalink] New post 06 Feb 2007, 17:36
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 :P .

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.
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gmat prep question [#permalink] New post 02 Mar 2007, 09:31
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, 09:31
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