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

[imane81]

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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

[komal]

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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

[harsh.champ]

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.

[Stuart@KaplanGMAT]

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.

[vicksikand]

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!