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

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

by imane81 » Wed Feb 17, 2010 12:04 pm
Thank you for your input on this one

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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by komal » Wed Feb 17, 2010 12:15 pm
imane81 wrote:Thank you for your input on this one

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
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
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by harsh.champ » Thu Feb 18, 2010 3:01 pm
komal wrote:
imane81 wrote:Thank you for your input on this one

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
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
Hey komal,
I didn't get this :-2*50!+1 will not be divisible by any no <50
Even if we take 50 or 51 how can we confidently say that 2*50!+1 will be divisible by it.The remainder 1 will be there then also.
Kindly clarify.
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by Stuart@KaplanGMAT » Thu Feb 18, 2010 7:00 pm
imane81 wrote:Thank you for your input on this one

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
This is probably the most asked question on this site (or close to it); do a search on "h(100)" and you'll see lots of explanations.
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by vicksikand » Sun Oct 17, 2010 10:30 am
harsh.champ wrote:
komal wrote:
imane81 wrote:Thank you for your input on this one

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
source : testmagic
h(100)= 2*4*6............100
h(100)+1=2*4*6............100 +1
2(1*2*3.......50)+1

so when h(100)+1 is divided by any of the nos between 2 and 50 will leave a remainder 1.

2*50!+1 will not be divisible by any no <50

answer E
Hey komal,
I didn't get this :-2*50!+1 will not be divisible by any no <50
Even if we take 50 or 51 how can we confidently say that 2*50!+1 will be divisible by it.The remainder 1 will be there then also.
Kindly clarify.
Just FYI:
h(100) = 2^50.50! and not 2.50!
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