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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

For every positive integer n... [#permalink]
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 _________________

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.

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

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.

PS: smallest prime factor (gmatprep) [#permalink]
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

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

Difficult prime number question [#permalink]
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

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

Re: Function problem [#permalink]
06 Feb 2007, 00:16

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

Smallest prime factor problem [#permalink]
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?

Re: Smallest prime factor problem [#permalink]
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 .

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!)

gmat prep question [#permalink]
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

gmatclubot

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

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