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

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

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27 Aug 2005, 04:55
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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

OPEN DISCUSSION OF THIS QUESTION IS HERE: for-every-positive-even-integer-n-the-function-h-n-is-126691.html
[Reveal] Spoiler: OA
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Re: PS - factors (from GmatPrep) [#permalink]

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27 Aug 2005, 08:56
yaron wrote:
Here is a PS from GmatPrep, Can you explain?

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

Thanks, Yaron

First thing I think there are some mistakes in the question I think the bolded parts above should be "function f(n)" and f(100) + 1

Solution

f(100) + 1 = 2(1x2x3...50) + 1

2^50x 50! +1

So the number should be something like xyz....0000001

So the prime factor should be much higher than 40...infact over 1000 probably...

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29 Aug 2005, 14:49
E is it

f(100) = 2* 4 * 6------100 + 1
or
2^50(1* 2* 3* 4.... *46 * 47 * 48 * 49 * 50) + 1

who knows what will divide this number..
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29 Aug 2005, 15:46
We had this question or a very similar one, two weeks ago !
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30 Aug 2005, 15:09
f(100) = 2* 4 * 6------100 + 1
or
2^50(1* 2* 3* 4.... *46 * 47 * 48 * 49 * 50) + 1

Why wud it be 2^50? Shudnt it be 2(1.2.3.4...50)+1?
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12 Oct 2005, 00:39
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 0 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
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12 Oct 2005, 03:47
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rahulraao 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

A. Between 0 and 20
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40

h(100)+1= 2.4.6.8......100 + 1
for every prime factor from 2, 3, to 47 (47 is the largest prime factor here coz 47*2=94<100. Remember all numbers here are even so to find the prime factor we have to, at least, devide each of them by 2). As we see the product contains prime factor from 2 to 47 ----> these prime numbers can't be factors of h(100)+1!. Thus the possible smallest prime factor of h(100)+1 must be greater than 47. E is correct.
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23 Oct 2005, 18:11
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
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23 Oct 2005, 20:23
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rigger 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 2 and 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

h(100)= 2*4*6*.....*100 = (2*1)*(2*2)*(2*3) .....*(2*47)*(2*48)*(2*49)*(2*50)
As we observe h(100) is divisible by primes like 2,3,5......,47 ---> h(100)+1 is not divisible by these primes, in other words, these primes can't be factor of h(100)+1. Thus, the smallest prime factor of h(100)+1 must be greater than 47 . E is my choice.
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26 Jan 2006, 19:27
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

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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26 Jan 2006, 19:48
2 * 4 * 6 * 8........94....* 100

2*1 *2*2 *2*3....2*47 *2*50

In this list 47 is the biggest prime which does not divide h(100) + 1. So the prime must be bigger than 47. E.
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26 Jan 2006, 20:12
h(100)= 2(1*2*3*4*5...*50) -> Largest prime number 47
h(100)+1 should result in another prime number, which I guess will not be lower than 47..

Going with E
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27 Jan 2006, 13:36
I don't agree. 47 is not a factor of h(100)+1 but of h(100).

I don't know thw answer though.
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27 Jan 2006, 13:40
Also, the question asks for the smallest and not the largest prime factor...
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27 Jan 2006, 16:05
mbadownunder, could you please indicate what is the correct answer, even if we don't have the correst reasoning ? It may help us find the solution...

Thanks.
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PS: Prime factor (from GMAT Prep) [#permalink]

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16 Jul 2006, 12:05
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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

Thanks
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16 Jul 2006, 12:18
E.

(1) Note first:Let x, y be two pos. integers then xy+1 is not divisible by x or y.
(2) Therefore H(100)+1 = 2*4*6*....*96*98*100 + 1 cannot be divisible by any odd number k smaller than 50 (because k*2 is a factor of H(100)).
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16 Jul 2006, 12:33
Thanks game over. That is the correct answer. Initially I did not follow your line number 2, but I think I follow now.

Is your reasoning that if K*2 can divide the number n, then it cannot possiblly divide the number n + 1.
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16 Jul 2006, 12:48
Quote:
Is your reasoning that if K*2 can divide the number n, then it cannot possiblly divide the number n + 1

[If your statement is correct, then 30 cannot be divisible by 6, because it is divisible by 5]

Example for (1)

Since 30 = 5*6 is divisible by 5 and 6, 30+1= 31 cannot be divisible by 5 and 6.

Example for (2):

Is H(100)+1 divisible by 37?

We know that H(100) = 2*4*6* .... * 74 * ... * 92*96*98*100.
Hence H(100) is divisible by 74 and therefore divisible by 37.

Using (1), we know that H(100)+1 cannot be divisible by 37.
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17 Jul 2006, 13:40
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H(100) + 1 = 2 * 4 * 6 * 8 * 10 * 12 ..* 100 + 1

Note that n is not the number of terms, the function is defined as the product of all even integer up to n.

Therefore, h(100) + 1 = (taking 2 as the common factor) 2^50 * ( 1 * 2 * 3 * 4 * â€¦* 50) + 1

Note also that the problem is not asking for a value of P, it is only asking you what might be P.

Note that a number divisible by 2, can be written as 2k where k is an integer, similarly, a number divisible by 3 can be written as 3i, where i is an integer.

Note also that h(100) is divisible by all the numbers 2,3,4,5,6,7,...50.

Hence, when h(100) + 1 is divided by the numbers 2,3,4,5,6,7,..50, the remainder is 1.

Hence, Note that h(100) cannot be divisible by any number that is less than or equal to 50. Hence, the smallest number that is a factor of h(100) is 51. Answer E.

-mathguru
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17 Jul 2006, 13:40

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