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

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maliyeci and madeinafrica, very good approach.

+1 for both of u
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Let me dig into the concepts of relative primes, looks interesting.
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function h(n) [#permalink]  13 Sep 2009, 15:59
For every positive even integer n, the function h(n) is defined to be 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
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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Re: function h(n) [#permalink]  13 Sep 2009, 16:23
OMG this question is like a ghost that keeps on appearing here every other week... It has been discussed many many times... just google (or search our forum for "smallest prime factor" Just one of the discussions is here... smallest-prime-factor-76565.html
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Re: GMAT Prep Question on Integers [#permalink]  14 Nov 2009, 07:18
TheGMATDoctor wrote:
This question is testing the concept of coprimes.
2 positive integers are coprime when their greatest common factor (their only common factor) is 1.
Now note that two different prime numbers are always coprime.
For example, 3 and 7 are coprime. So are 13 and 19.
But the two integers need not be prime numbers in order to be coprime.
For example, 4 and 9 are coprime (1 is their only common factor).
Also, Important!
Two consecutive integers are always coprime. The question is testing you on this concept.
Let's solve it now:
h(n) = 2*4*6*....................*n. ----n is even.
h(100) = 2*4*6*........ 94*96*98*100.
h(100) = (2^50)*(1*2*3*.......47*48*49*50). Note:I have pooled together all the 2s from all the even integers from 2 to 100; that's how I got 2^50. Now, the largest prime number involved in the above factorization is 47. All the prime from 2 to 47 are also involved in the above factorization. Actually, 47 is the greatest prime factor of h(100).
Since h(100) and h(100) + 1 are consecutive integers, they are necessarily coprime (see above). h(100) and h(100) + 1 have no common factor except 1, so they have no common prime factor either. The smallest prime factor of h(100) +1 must then be greater than 47.
Clearly, this prime factor is greater than 40.

That's all folks!
Asan Azu, The GMAT Doctor.

Excellent explaination. Thanks
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Re: GMAT Prep Question on Integers [#permalink]  14 Nov 2009, 17:22
Spot on GMAT Doctor +1 Kudos ; I had no idea whatsoever... I was prolly gonna guess E anyways..
but knowing the approach, phew.... great relief knowing what u r doing )
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Pl. provide the solution for the problem? [#permalink]  16 Nov 2009, 12:30
For every +ve even integer n, the function h(n) is defined as the product of all even integer 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) b/w 10 & 20
c) b/w 20 & 30
d) b/w 30 & 40
e) greater than 40

Pl. provide the solution...thanx in advance
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Re: GMAT Prep Question on Integers [#permalink]  22 Nov 2009, 16:11
thnx alot ..doctor...lernt co-prime concept
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Re: GMAT Prep Question on Integers [#permalink]  22 Nov 2009, 20:19
This is how Euclid has proved that there are infinite prime numbers in the number system.

Nice logic +1 to theGMATdoctor
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Hard question: Function and factors [#permalink]  24 Jan 2010, 01:14
Question is - 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 factor of h(100) + 1, then p is

A. between 2 and 10
B. between 20 and 10
C. between 20 and 30
D. between 30 and 40
E. greater than 40

OA is (e) , but I am not able to figure out why it is (e)

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Re: Question on Functions [#permalink]  24 Jan 2010, 02:24
11
KUDOS
Expert's post
mohit005 wrote:
Question is - 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 factor of h(100) + 1, then p is

a) between 2 and 10
b) between 20 and 10
c) between 20 and 30
d) between 30 and 40
e) greater than 40

OA is (e) , but I am not able to figure out why it is (e)

Welcome to the club Mohit.

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

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Re: GMAT Prep Question on Integers [#permalink]  11 Feb 2010, 02:28
1
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brokerbevo wrote:
Wow, lot of info. I understood everything except for this part: h(100) = (2^50)*(1*2*3*.......47*48*49*50)
Could you explain this part a little more clearly? Thanks.

h(100) = 2.4.6.8........100
h(100) = (2*1)(2*2)(2*3)(2*4)........(2*50)

therefore you can factor all 2's = 2^50(1*2*3*....50)
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Math Question [#permalink]  24 Feb 2010, 16:54
For every positive even integer n, the function h(n) is defnied 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

Any help on how to get the correct answer? Thanks!

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Re: Math Question [#permalink]  24 Feb 2010, 17:29
Another math question. Thanks folks!

What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?

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Re: Math Question [#permalink]  24 Feb 2010, 20:57
msv3763 wrote:
Another math question. Thanks folks!

What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?

Since there is a hypotenuse, we are talking of a right isosceles triangle. Let a, b and c be sides of the triangle, where a=b and c is the hypotenuse.
from the prompt, a+b+c=16 + 16√2, and since a=b, it is a+a+c=16 + 16√2
from the pythagorean theorem, a^2+b^2=c^2, but since it is a isosceles and a=b, we re-write this as
a^2+a^2=c^2

Now solve the system:
a+a+c=16 + 16√2
a^2+a^2=c^2

c=16 + 16√2 - 2a
c=a√2

therefore

a√2=16 +16√2 -2a
2a+a√2=16 +16√2
a(2+√2)=16 +16√2
a=(16 +16√2)/(2+√2)

now recall from above that
c=16 + 16√2 - 2a

plug in (16 +16√2)/(2+√2) in place of a to get:
c=16 + 16√2 - 2(16 +16√2)/(2+√2)
c=16 + 16√2 -(32+32√2)/(2+√2)

simplify it further and you will get
c=(32+16√2)/(2+√2)
c=16(2+√2)/(2+√2)
c=16
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Re: Math Question [#permalink]  24 Feb 2010, 21:17
Expert's post
another way:

we have a right isosceles triangle. Let's say hypotenuse is a then perimeter is

P = a/root(2)+a/root(2) + a = 16 + root(2)16

a = 16 * (1+root(2))/(1+root(2)) = 16

Posted from GMAT ToolKit
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Re: Math Question [#permalink]  24 Feb 2010, 21:44
msv3763 wrote:
Another math question. Thanks folks!

What is the value of the hypotenuse of an isosceles triangle with a perimeter equal to 16 + 16√2?

hypotenuse - h
side - s

2s+d = 16+16√2

also in isoceles triangle h^2 = 2s^2 so h=√2s
hence simplyfying main equation we get √2h + h = 16√2 + 16
hence h = 16.
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Re: Math Question [#permalink]  24 Feb 2010, 22:10
Thanks everybody, I really appreciate it. Any luck with the first question? (For every positive even integer n, the function h(n) is defnied 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...)
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Re: Math Question [#permalink]  25 Feb 2010, 02:57
msv3763 wrote:
Thanks everybody, I really appreciate it. Any luck with the first question? (For every positive even integer n, the function h(n) is defnied 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...)

the answer is 79, i did it on PC
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Re: Math Question [#permalink]  25 Feb 2010, 07:17
1
KUDOS
Expert's post
msv3763 wrote:
For every positive even integer n, the function h(n) is defnied 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

Any help on how to get the correct answer? Thanks!

Welcome to the Gmat Club. Below is the solution for your question:

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

Now, two numbers $$h(100)=2^{50}*50!$$ and $$h(100)+1=2^{50}*50!+1$$ are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.

As $$h(100)=2^{50}*50!$$ has all numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

Hope it helps.

P.S. Can you please: post one question per topic, tag the questions you post, and also post the whole questions with answer choices.
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Re: Math Question   [#permalink] 25 Feb 2010, 07:17

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