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

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Intern
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30 Sep 2008, 10:37
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scthakur wrote:
zaur, as a rule, if n is divisible by a factor, (n+1) will not be divisible by the same factor. You can check this for any number. For example, if 6 is divisible by 2, 3 and 6 then (6+1) will not be divisible by any of 2,3, or 6.

Extending the same logic, h(100) is divisible by all the integers from 1 to 50. Hence, h(100)+1 will not be divisible by any number from 1 to 50 and hence its factor will be greater than 50.

Thanks scthakur! Its much clear now! + 1 kudo for you.
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25 Nov 2008, 21:32
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 an 10
b. between 10 and 20
c. between 20 and 30
d. between 30 and 40
e. greater than 40

Thanks - I am sure its easy I just don't know how to attack it.
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25 Nov 2008, 21:46
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you can se that :h(100)=2*4*6*...*100=(2*1)(2*2)(2*3)...(2*50)
or h(100)=2^50(1*2*3...*50)
h(100)+1=2^50(1*2*3*...*50) + 1
if h(100)+1 is divisible by p then p must be greater than 47.
Here we go:
we can see that h(100) is divisible by 2,3,..47(47 is the greatest prime which is smaller than 50)
so if h(100) +1 is is divisible by p which p <=47 then 1 is divisible by p so p=1 ridiculous
then if h(100) +1 divisible by p, p must be greater than 47 and answer is E
We use the characteristic :
a+ b is divisible by c, and a is divisible by c too. So b is divisible by c
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26 Nov 2008, 10:42
All other prime factors less than 47 have a factor in the term.
so answer is greater than 40.
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26 Nov 2008, 18:34
So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.
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27 Nov 2008, 02:31
my doubt is ques asks the smallest prime factor...?
wouldnt 2 be the samllest prime factor of h(100)?
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27 Nov 2008, 02:32
bertlacy wrote:
So what you are saying is that I should guess and move on? LOL? Can anyone else break that down for the only slightly intelligent people in this forum.

i am one of those slightly intel.

however this is based on a concept that says

if x is devisible by z then sure x+1 is not devisible by z also, the largest prime that devide x has to be less than the largest prime that devide x+1.

x = 50 is devisible by 2,5 ( largest prime is 5)

x+1 = 51 devisble by 3,37 ( largest prime is 37)
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01 Jan 2009, 14:22
pls give an explanation
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03 Jan 2009, 13:39
linau1982 wrote:
pls give an explanation

This is probably the most hardest question i ever encountered.

h (2) = 2
h (4) = 2 x 4 = 2! x 2^2
h (6) = 2 x 4 x 6 = 3! x 2^3
h (8) = 2 x 4 x 6 x 8 = 4! x 2^4
h (10) = 2 x 4 x 6 x 8 x 10 = 5! x 2^5
.
.
.
.
h (100) = 2 x 4 x 6 x .........x 100 = 50! x 2^50 = (1x2x3x4x5x......x40x....48x49x50) (2^50).
The pattren identified is that for every n, h (n) = (n/2)! x 2^(n/2)

Now lets find the p, the smallest prime factor of h (100) + 1.
lets try from the begaining:

if p = 2, [h(100)+1] should be divisible by 2. But it is not.
[h (100) + 1]/2 = h (100)/2 + 1/2 = (50! x 2^50)/2 + 1/2. it doesnot result in an integer. so 2 cannot be p.

if p = 3, [h (100) + 1]/3 = h (100)/3 + 1/3 = (50! x 2^50)/3 + 1/3. it doesnot result in an integer. so 3 cannot be p.
if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 3 cannot be p.
if p = 4, [h (100) + 1]/4 = h (100)/4 + 1/4 = (50! x 2^50)/4 + 1/4. it doesnot result in an integer. so 4 cannot be p.
if p = 5, [h (100) + 1]/5 = h (100)/5 + 1/5 = (50! x 2^50)/5 + 1/5. it doesnot result in an integer. so 5 cannot be p.
.
.
.
.
.
if p = 50, [h (100) + 1]/50 = h (100)/50 + 1/50 = (50! x 2^50)/50 + 1/50. it doesnot result in an integer. so 50 cannot be p.

The finding is that: Something [h(n)] divisible by k is not divisible by k if 1 is added to something [h(n)].
Therefore h (n=100) + 1 has the smallest prime i.e. > 50.
So it is E.
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03 Jan 2009, 15:05
Great explanation.

Just thinking, Is there any number that divides h(100)+1 with the given function?

In other words, How to find the number that divides ( 50! X 2 ^50 ) +1
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03 Jan 2009, 15:39
it is E. This one is from GMATPrep.
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PS -- smallest prime factor [#permalink]

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22 Feb 2009, 12:18
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.

please explain ... this question is from gmatprep1.
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Re: PS -- smallest prime factor [#permalink]

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23 Feb 2009, 11:05
this is gmatprep test # 1 question ... please some one explain it ...
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Re: PS -- smallest prime factor [#permalink]

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23 Feb 2009, 12:03
http://gmatclub.com/forum/7-p556064?t=74417#p556064
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Re: PS -- smallest prime factor [#permalink]

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24 Feb 2009, 08:35
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10 Mar 2009, 15:27
This showed up on my 3rd question on the GMAT Prep.

What the heck >=[
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14 Mar 2009, 16:26
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
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40
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16 Mar 2009, 12:30
nitindas 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
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40

the smallest prime factor of 2*4*6*8...*100 is 2.

2+1=3

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17 Mar 2009, 12:55
This is a question from GmatPrep and answer is
5. greater than 40

The smallest prime factor cannot be 3 bcoz lets say:
we have to find smallest prime factor of 2*4*6 +1

This comes out to be 49

whereas 2 is the smallest prime factor of 2*4*6

but 2+1 =3 is not the factor of 49.

Can somebody explain how to find the smallest prime factor in such a case?
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17 Mar 2009, 14:36
RahlowJenkins wrote:
nitindas 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
1. between 2 and 10
2. between 10 and 20
3. between 20 and 30
4. between 30 and 40
5. greater than 40

the smallest prime factor of 2*4*6*8...*100 is 2.

2+1=3

I gues this is not correct .
You did not take the +1 in consideration , which changes the answer.
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Re: Smallest Prime Factor   [#permalink] 17 Mar 2009, 14:36

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