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

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 [#permalink] New post 05 Jun 2007, 08:22
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ummm on the gmat i probably would have looked at this for 2 seconds and guessed.
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 [#permalink] New post 05 Jun 2007, 08:31
priyankur_saha@ml.com wrote:
It's E ....

Must be 47 or higher!!


i think ive seen thsi problem before, but can you explain?
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 [#permalink] New post 05 Jun 2007, 09:29
Correct the answer is E but how do you solve the problem??
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 [#permalink] New post 05 Jun 2007, 09:49
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This is a repost:

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....6*4*2) + 1 can be simplified to (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

hope this will help.

:-D
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 [#permalink] New post 05 Jun 2007, 10:42
YOur explanation helps a lot. I was not aware of that factorial rule.
I also didn't think of 2^50*50!

Thanks very much,

Eric
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 [#permalink] New post 05 Jun 2007, 19:12
here's how i did it-

h(100) = 2*4*6* .... 98*100

h(100) = 2*1 * 2*2 * 2*3 * .... 2*49 * 2*50

h(100) = 2^50 * 1*2*3 * ... 49*50

so primes 43 and 47 are the largest prime factors

but they wont be prime factors of h(100) + 1, since they wont divide into a number one larger than that. Therefore, the largest prime factor must be larger than 50, so E.
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 [#permalink] New post 15 Jun 2007, 22:14
r019h,

Can you explain why did you chose the largest prime factors 43 and 47 while the question is about the smallest prime factor?
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Smallest Prime Factor [#permalink] New post 07 Jul 2007, 11:49
For Every Positive Integer n, the function h(n) is defined to be product of all even integers from 2 to n.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 was able to find answer using extrapolation but would like to know what techniques can be used for these types of questions.
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 [#permalink] New post 07 Jul 2007, 13:41
KillerSquirrel wrote:
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

the answer is (E)


well done and excellent work KillerSquirrel. I never understood this question. now you made it crystal clear for me.

you are the real Gmat Math mastero!!!!!!!! :flower
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GmatPrep-PS [#permalink] New post 09 Jul 2007, 05:30
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 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. over 40


OA later. Thanks .
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 [#permalink] New post 09 Jul 2007, 08:28
Solution is E.
H(100)+1=2*4*...100+1=2^50*(1*2*3...*47*48*49*50)+1
From above, it is seen that first part is divisible by all prime numbers ranging from 2 to 47. But 1 is divisible by none of them. Hence, if it is ever divisible by any prime number, then the divisor must be greater than 47.
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Better explanation? [#permalink] New post 09 Jul 2007, 12:05
I think I see the concept, but could someone verbalize the answer a bit more for clarity?

Thanks!
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 [#permalink] New post 09 Jul 2007, 14:14
h (100)= 2 x 4 x 6 x...x 100= 2^50 x 50!.

As h(100) is a multiple of every positive integer from 2 to 50, h(100) + 1 is not a multiple of any positive integer from 2 to 50. Thus, the smallest prime factor of h(100) +1 must be greater than 50.
Note: h(100) + 1, when divided by any positive integer from 2 to 50, will yield a remainder of 1, and thus is not a multiple of that integer.
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 [#permalink] New post 10 Jul 2007, 02:01
OA---E.

YOU are all GREAT. Thank u very much.
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FUNC 10 h (n) [#permalink] New post 29 Jul 2007, 11:08
For every positive 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
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Re: FUNC 10 h (n) [#permalink] New post 29 Jul 2007, 17:39
nfa1rhp wrote:
For every positive 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


would the answer be A? Here is my logic but I am not sure:

for h(100)+1, we know the answer will be odd
since n = 100 and we are multiplying the even integers 2*4*6......*100 and adding 1 to the product.

this is a sequence so we can say that 2 * (difference of the first and last term) * # of terms = 2 * 98 * 50 = 9800 + 1 = 9801

Hence, the smallest prime factor p will be 3
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Re: FUNC 10 h (n) [#permalink] New post 29 Jul 2007, 20:40
nfa1rhp wrote:
For every positive 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


The answer is E and this has been discussed many times in the last few weeks. Do search, and you will get the solution.
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 [#permalink] New post 30 Jul 2007, 18:05
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from gmatprp [#permalink] New post 10 Aug 2007, 16:56
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, the p is...

between 2 and 10

between 10 and 20

between 20 and 30

between 30 and 40

greater than 40

please explain in details if possible
thanks
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Re: from gmatprp [#permalink] New post 10 Aug 2007, 21:30
Ravshonbek 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, the p is...

between 2 and 10

between 10 and 20

between 20 and 30

between 30 and 40

greater than 40

please explain in details if possible
thanks




probably the most discussed problem: http://www.gmatclub.com/phpbb/viewtopic ... t=smallest
Re: from gmatprp   [#permalink] 10 Aug 2007, 21:30
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