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

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Re: Tough GMAT question. Please help...! THANKS! [#permalink] New post 22 Jul 2009, 06:41
maliyeci and madeinafrica, very good approach.

I too had seen the approach that maliyeci took, but had not thought about madeinafrica's approach. excellent. Thanks for sharing.

+1 for both of u :-D
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Re: Tough GMAT question. Please help...! THANKS! [#permalink] New post 22 Jul 2009, 08:06
Let me dig into the concepts of relative primes, looks interesting.
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function h(n) [#permalink] New post 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] New post 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" :wink: Just one of the discussions is here... smallest-prime-factor-76565.html :wink:
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Re: GMAT Prep Question on Integers [#permalink] New post 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] New post 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] New post 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
Correct answer is e
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Re: GMAT Prep Question on Integers [#permalink] New post 22 Nov 2009, 16:11
thnx alot ..doctor...lernt co-prime concept
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Re: GMAT Prep Question on Integers [#permalink] New post 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] New post 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)

Please help
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Re: Question on Functions [#permalink] New post 24 Jan 2010, 02:24
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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)

Please help


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.

Answer: E.
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Re: GMAT Prep Question on Integers [#permalink] New post 11 Feb 2010, 02:28
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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] New post 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!




Answer: greater than 40
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Re: Math Question [#permalink] New post 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?




Answer: 16
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Re: Math Question [#permalink] New post 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?




Answer: 16

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] New post 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

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Re: Math Question [#permalink] New post 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?

Answer: 16


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] New post 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] New post 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] New post 25 Feb 2010, 07:17
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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!




Answer: greater than 40


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.

Answer: 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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