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

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Manager
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08 Jun 2008, 12:21
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, the 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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08 Jun 2008, 12:35
ldpedroso 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, the 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

before posting a question..please search for it on this forum...this question has been solved 1000 times here..
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Function h(n) Gmat prep question [#permalink]

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07 Jul 2008, 21:04
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 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

OA TO FOLLOW
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Re: Function h(n) Gmat prep question [#permalink]

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07 Jul 2008, 21:14
neeraj.kaushal 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 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

OA TO FOLLOW

My choice is E.
h(100)= 2*4*6....*100 so it contains all the prime number from 2 to 49 so h(100)+1 is not divisible by 2, 3,..., 49 so its smallest prime factor must be greater than 49.
Is it right?
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Re: Function h(n) Gmat prep question [#permalink]

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07 Jul 2008, 21:20
refer 7-t64451
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Re: Function h(n) Gmat prep question [#permalink]

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12 Jul 2008, 08:09
There is something I don't understand here. Just because the largest prime factor of h(100) is 47, shouldn't mean that the smallest prime factor of h(100)+1 will be bigger than 47. Let me give you an example:

9 = 3*3, the largest prime number here is 3, so when we look at the next higher number:

10=2*5, so the smallest prime factor here is 2, which is not bigger than 3. So how can we say that the bigger number's smallest prime factor will be bigger than the biggest prime factor of the number just under it??? Can someone please explain this concept to me? thanks
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Re: Function h(n) Gmat prep question [#permalink]

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12 Jul 2008, 21:30
tarek99 wrote:
There is something I don't understand here. Just because the largest prime factor of h(100) is 47, shouldn't mean that the smallest prime factor of h(100)+1 will be bigger than 47. Let me give you an example:

9 = 3*3, the largest prime number here is 3, so when we look at the next higher number:

10=2*5, so the smallest prime factor here is 2, which is not bigger than 3. So how can we say that the bigger number's smallest prime factor will be bigger than the biggest prime factor of the number just under it??? Can someone please explain this concept to me? thanks

False analogy
h(100) is divisible by all prime numbers from 2, ..., 47 while 9 is not divisible by all prime numbers from 2 to 3. If 9 were, 10 would never divisible by 2.
If A is divisible by p and A+1 is divisible by p too, 1 is divisible by p. It's wrong.
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Re: Function h(n) Gmat prep question [#permalink]

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13 Jul 2008, 11:48
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GMAT Prep: Function - part 2 [#permalink]

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23 Jul 2008, 17:47
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

OA is e.
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Re: GMAT Prep: Function - part 2 [#permalink]

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23 Jul 2008, 20:24
h (100) would mean 2*3*4*5*6...............................100 = X
Now X has the following properties:

1) it's an even number.
2) it ends in more than three zeros thus a multiple of 5,25, 125 etc.
3) its a multiple of all the prime factors till 100.

Now X+1
would not be an even number.
would not have zeros at the end
would not be a multiple of any of the prime numbers till 100.

Thus there are two possibilities:
1) Either X+1 is a prime number
2) X+1 is a composite number.

Now for 1) there's no option to proove its validity.
For 2) the only option one can pick up is e. Coz' from the above we can see that X+1 would not be a multiple of any number from 2 to 100.
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13 Aug 2008, 23:19
Any mathematical way to get the answer? I tried the lengthy way of finding pattern...but unsuccessful.

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) greather than 40
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Re: GMATPresp: Smallest Prime Factor [#permalink]

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14 Aug 2008, 04:15
scthakur wrote:
Any mathematical way to get the answer? I tried the lengthy way of finding pattern...but unsuccessful.

PS take the trouble of typing the question mate..

Anyways..

h(100) = 2*4*6*....*100 = 2^50 * 50!

Now, a general rule. If x>1 is a factor of a number n, then x will not a factor of (n+1). This can be tried out with a few examples.
Factors of 15 are 3,5,15. None of these will be factors of 16. Factors of 16 are 2,4,8,16. None of these are factors of 17.. so on and so forth.

So none of the numbers 1-50 will be a factor of h(100)+1. The prime factor has to be greater than 50. So E
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Re: GMATPresp: Smallest Prime Factor [#permalink]

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14 Aug 2008, 16:16
h(100)= 2*4*6*8*....*100 = 2^50(1*2*3*4*5*6* ... *50) = 2^50*50!

Notice that h(100) is divisible by every integer from 1 to 50 because of the 50!

Adding one implies that you're adding a "remainder" of 1. Therefore, the result must be divisible by a number greater than 50.

E.
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Re: GMATPresp: Smallest Prime Factor [#permalink]

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14 Aug 2008, 16:50
It's amazing how you guys are able to see this:

h(100)= 2*4*6*8*....*100 = 2^50(1*2*3*4*5*6* ... *50) = 2^50*50!

I never would have been able to see this pattern.
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30 Sep 2008, 03:45
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

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30 Sep 2008, 06:53
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My answer will be A .

h(100) = 2*4*6*...100
= 2 (1*2*...50)
= 2*50!
last digit of this series will be 0

now P is smallest prime factor of h(100)+ 1 = 2*50! + 1.

so this must be divisible by 2,3,5 or 7 : Answer A or

P's value must be greatar than 50 ..means greater than 40 , Anwer will be E

I just tried..not sure
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30 Sep 2008, 07:12
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h(100) + 1 = 2^50 * 50! + 1

let's assume x = 2^50 * 50!

now x is a multiple of all prime numbers between 1 and 50, thus x + 1 cannot be a multiple of any of those numbers. Thus E.
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30 Sep 2008, 07:16
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vr4indian, thanks a lot

OAE is E, it is from practice test mba.com,
to be honest, even now I don't understand how it goes.....
thanks

brgds
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30 Sep 2008, 07:21
aim2010, thanks a lot

now, trying to understand all this....
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30 Sep 2008, 10:25
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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.

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