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

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

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New post 24 Mar 2013, 06:30
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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) +2, then p is?

A. between 2 and 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. 2
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 24 Mar 2013, 06:38
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skamal7 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) +2, then p is?

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


\(h(100)+2=2*4*6*...*100+2\). Notice that we can factor out 2 from h(100)+2, thus the smallest prime factor of h(100)+2 is 2: \(h(100)+2=2*(4*6*...*100+1)\).

Answer: E.

Similar question to practice: for-every-positive-even-integer-n-the-function-h-n-is-126691.html

Hope it helps.
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 26 Mar 2013, 14:19
skamal7 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) +2, then p is?

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


h(100) will have 0 in its unit digit because while doing the calculation of the function we are multiplying by 10 (and 100). So h(100)+2 will have 2 in the units digit so smallest prime factor is 2.
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 03 Apr 2017, 10:45
The answer will be even and thus the SMALLEST prime factor would be 2 (even).
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 22 Aug 2019, 01:48
h(100) = (100)(98)(96)...(6)(4)(2)=(2)(50)(2)(49)(2)(48)...(2)(3)(2)(2)(2)(1)=2^50*50!
h(100)+2= 2^50*50!+2
2 is the smallest prime number for which h(100)+2 results in a remainder of 0. Therefore E is the correct answer.
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 26 Aug 2019, 04:01
coylahood wrote:
h(100) = (100)(98)(96)...(6)(4)(2)=(2)(50)(2)(49)(2)(48)...(2)(3)(2)(2)(2)(1)=2^50*50!
h(100)+2= 2^50*50!+2
2 is the smallest prime number for which h(100)+2 results in a remainder of 0. Therefore E is the correct answer.


Hi,

Can you elaborate this a little bit . I am confused between how do we decide smallest /largest prime no. in such cases .
I HAD ATTEMPT A PREVIOUS question by your method and I thought 2 is smallest even prime no. but that case turned out to be different.

below is the link of that question

https://gmatclub.com/forum/for-every-po ... 12521.html
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Re: For every positive even integer n, the function h(n)  [#permalink]

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New post 26 Aug 2019, 05:11
gmatway wrote:
coylahood wrote:
h(100) = (100)(98)(96)...(6)(4)(2)=(2)(50)(2)(49)(2)(48)...(2)(3)(2)(2)(2)(1)=2^50*50!
h(100)+2= 2^50*50!+2
2 is the smallest prime number for which h(100)+2 results in a remainder of 0. Therefore E is the correct answer.


Hi,

Can you elaborate this a little bit . I am confused between how do we decide smallest /largest prime no. in such cases .
I HAD ATTEMPT A PREVIOUS question by your method and I thought 2 is smallest even prime no. but that case turned out to be different.

below is the link of that question

https://gmatclub.com/forum/for-every-po ... 12521.html


h(100)+2 is an even number and every even number must be divisible by 2. According to number theory, 2 is the only even prime number and it is also the smallest prime number. Since h(100)+2 is divisible by 2, then 2 must be the smallest prime factor of h(100)+2.
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Re: For every positive even integer n, the function h(n)   [#permalink] 26 Aug 2019, 05:11
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