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

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02 Mar 2007, 16:43
Stormgal wrote:
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

h(n) = 2*(n/2)!
h(100) = 2*(50)!

since h(100) is divisible by every number from 1 to 50, h(100) wont be divisible by any those numbers (i think), so E would be the answer.

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05 Mar 2007, 02:45

I couldn't explain it better myself so I won't.
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05 Mar 2007, 13:41
E....you know from the function that the number is going to be VERY large so E is the most likely answer...but cookie's answer explains in detail.
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Function & prime factor Qs [#permalink]

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24 Mar 2007, 17:02
1) For every positive even 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
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24 Mar 2007, 21:10
I go with D

h(n) = 2 x 4 x 6 ... 100

each of the prime factors within the ranges given in the choices will factor out h(n)

say for eg 30-40, guess the biggest prime factor is 37

37 x 2 already exists in h(n)

but for a prime number > 40, well chances are it might factor out h(n) + 1

not sure if my logic is right, whats the OA ?
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18 Apr 2007, 19:33
For every positive interger n, the function h(n) is def. 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 bet. 2 and 10
b. bet. 10-20
c. bet. 20-30
d. bet 30-40
e. greater than 40

I'm not quite sure how to approach this question and I'm not sure what they're asking. Any help is appreciated. OA will follow.
TIA
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18 Apr 2007, 20:03
ayl989 wrote:
For every positive interger n, the function h(n) is def. 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 bet. 2 and 10
b. bet. 10-20
c. bet. 20-30
d. bet 30-40
e. greater than 40

I'm not quite sure how to approach this question and I'm not sure what they're asking. Any help is appreciated. OA will follow.
TIA

I am taking a shot. Others can chime in

H(100) = 2 x 4 x 6 x ... x 100
= 2(1 x 2 x 3 x ... x 50)

so H(100) + 1 = 2(1 x 2 x 3 x ... x 50) + 1

Now any divisor of the above expression must divide both terms. Any prime factor below 50 will always divide the first term completely but it cant divide 1. So I think that required prime factor should be > 50

I would chose E.
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Help with a GMAT question [#permalink]

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21 Apr 2007, 10:24
Can someone help me with this question? I felt a tickle in my brain when I saw this

For every positive even integer n, 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 know the answer is E, but I haven't a clue as to why or how. HELP!
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Re: Help with a GMAT question [#permalink]

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22 Apr 2007, 11:51
afiggy1 wrote:
Can someone help me with this question? I felt a tickle in my brain when I saw this :oops:

For every positive even integer n, 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 know the answer is E, but I haven't a clue as to why or how. HELP! :cry:

I am using . as multiplication.
h(100) = 2 . 4 . 6 . ... . 98 . 100
or h(100) = 2(1 . 2. 3. 4. ... . 49.50)

Now h(100) + 1 = 2(1 . 2. 3. 4. ... . 49.50) + 1
now every number from 2 t0 50 will divide the first term but not the second term so the divisor has to be > 50

I am not sure if this is the most elegant answer but it makes sense to me. Any comments?
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Re: Help with a GMAT question [#permalink]

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22 Apr 2007, 12:44
techjanson wrote:
afiggy1 wrote:
Can someone help me with this question? I felt a tickle in my brain when I saw this

For every positive even integer n, 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 know the answer is E, but I haven't a clue as to why or how. HELP!

I am using . as multiplication.
h(100) = 2 . 4 . 6 . ... . 98 . 100
or h(100) = 2(1 . 2. 3. 4. ... . 49.50)

Now h(100) + 1 = 2(1 . 2. 3. 4. ... . 49.50) + 1
now every number from 2 t0 50 will divide the first term but not the second term so the divisor has to be > 50

I am not sure if this is the most elegant answer but it makes sense to me. Any comments?

Hey Tech -

Thanks for the explanation ... what puzzles me about this question is the part where it mentions p as being "the smallest prime factor"; intuitively, doesn't that sound like p would be one of the first 10 primes? Maybe my thinking is screwed up, but that's what I initially gravitate towards.

In any event, I appreciate your help, Tech.

Regards,
Alex
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23 Apr 2007, 10:00
This was the third question of my practice GMAT that I downloaded from the site. It must be pretty easy if you know a certain rule, but I am stumped...

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 than 40

Thanks for the help...
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Re: Obviously Missing Something Here... [#permalink]

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23 Apr 2007, 12:44
rfraser4 wrote:
This was the third question of my practice GMAT that I downloaded from the site. It must be pretty easy if you know a certain rule, but I am stumped...

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 than 40

Thanks for the help...

Wilfred has a posted a very nice explanation ;
http://www.gmatclub.com/phpbb/viewtopic ... actor++100
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Re: Obviously Missing Something Here... [#permalink]

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23 Apr 2007, 13:49
Thanks Goalsnr. Wilfred's explanation was a little confusing for me, but it got me on the right track and I get it now. I posted the solution in my own words to help out anyone else confused by Wilfred's...
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24 Apr 2007, 17:02
h(100) = [ 2 x 4 x 6 ... x 100 ] + 1

h(100) is a multiple of all even numbers from 2 to 100, so it's also a multiple of all numbers from 1 to 50 (Dividing the even numbers by 2).

h(100) + 1 is a number next to h(100), so it can't have a factor from 2 to 50. So, the smallest prime factor should be greater than 50. Hence the answer is E.

Edit: I seem to have repeated the solution already provided.
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24 Apr 2007, 17:33
the person who posted that h(100) + 1 = 2(1....50) + 1 is wrong.

If you factor 2 out of 2*4, it is not 2(1*2). It will be 2^2 * (1*2).

Thus it would look more like 2^50(1....50) + 1.

However, I still didn't know how to solve this problem.

So I factored it further. I might be missing some factors but this is what I got:

2^109 * 3^20 * 5^12 * 7^6 * 11^4 * 13^2 17^2 * 19^2 * 23^2 + 1.

this is where i am stuck.
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GmatPrep 1: Real tough problem [functions] [#permalink]

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01 Jun 2007, 07:24
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 100
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

I have no clue how to approach this problem.
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01 Jun 2007, 07:32
In order to better understand this problem , I suggest to plug in some values.

h(n) where n is positive and even (e.g 2,4,6) and h(n) is the prouduct of all even numbers from 2 to n.

n=10

h(10) = 10*8*6*4*2

so after we understand what this problem wants , we can now solve it !

h(100)+1 = 100*98*96*94....4*2 + 1 (can be simplified to (2^50)*50!+1)

h(100)+1 = (100*98*96*94....6*4*2) + 1 = (2^50)*50!+1

what is the smallest prime factor ?

For any factorial + 1 the smallest factor (apart from 1) is greater then any of the members of the factorial

2! + 1 = 3: smallest factor is 3
3! + 1 = 7: smallest factor is 7
4! + 1 = 25: smallest factor is 5
5! + 1 = 121: smallest factor is 11
6! + 1 = 721: smallest factor is 103

50! + 1 = a number greater then 50

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01 Jun 2007, 07:44
OMG, that's a killer. There is no way I will be able to solve such a problem in less than 2 minutes.

But thanks for the solution.
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01 Jun 2007, 08:07
Mishari wrote:
OMG, that's a killer. There is no way I will be able to solve such a problem in less than 2 minutes.

But thanks for the solution.

I think you don't have to solve this at all. the trick is to know how to factor 100*98*97.... into 2^50*50!+1.

and to apply common sense afterwards.

thats all

you don't get points on the way only on the outcome !

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05 Jun 2007, 08:01
How do you solve this? It's from the gmat prep practice exam.

For every positve 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
Prime number problem   [#permalink] 05 Jun 2007, 08:01

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