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the first part has all numbers (including prime) as it's factors. First part+1 is a consecutive number so they do not share any common factors (other than 1, which neither prime nor composite). Therefore the smallest prime factor for h(100)+1 is greater than 50.

Answer : E
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For every positive even integer n, the function h(n) is defined to be [#permalink]

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26 Nov 2016, 07:47

1

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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) +1, than 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

Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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26 Nov 2016, 13:43

Top Contributor

blazov 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) +1, than 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

Important Concept: If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1 For example, since 7 is a factor of 350, we know that 7 is not a factor of (350+1) Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313

Now let’s examine h(100) h(100) = (2)(4)(6)(8)….(96)(98)(100) = (2x1)(2x2)(2x3)(2x4)....(2x48)(2x49)(2x50) Factor out all of the 2's to get: h(100) = [2^50][(1)(2)(3)(4)….(48)(49)(50)]

Since 2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is not a factor of h(100)+1 (based on the above rule)

Similarly, since 5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is not a factor of h(100)+1 (based on the above rule)

. . . . Similarly, since 47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is not a factor of h(100)+1 (based on the above rule)

So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.

Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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05 Jan 2017, 22:39

H(100) will be a multiple of all of the prime factors below 50 because it is the product of all of the primes below 50 multiplied by 2 (94 is 47 x 2, 62 is 31 x 2, etc. If h(100) is a multiple of each of these, then when we add 1 it will throw us off of being a multiple of all of these, so the greatest prime factor will have to be greater than 50, which is greater than 40.
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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21 Apr 2017, 11:22

One of the best question and best explanation by Bunuel.

Bunuel wrote:

enigma123 wrote:

h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40 < 30 > 40 < 10 Indeterminate

Below is the proper version of this question:

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) +1, then p is? A. between 2 and 20 B. between 10 and 20 C. between 20 and 30 D. between 30 and 40 E. greater than 40

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

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

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01 May 2017, 05:36

Bunuel 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) +1, then p is?

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

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

Answer: E.

Dear Bunuel, if the question changed to positive ODD integer, will the answer be the same?

I think answer choice A should be between 2 and 10.
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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) +1, then p is?

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

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

Answer: E.

Dear Bunuel, if the question changed to positive ODD integer, will the answer be the same?

I think answer choice A should be between 2 and 10.

Yes. 49th odd number is 97, which is prime.
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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13 Jun 2017, 21:49

Hello, as the question about odd number modification came up,I would like to know how to solve if the question was modified to deal with only odd numbers, as hazelnut and Buñuel discussed.

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) +1, then p is?

A. Between 2 and 20 B. Between 10 and 20 C. Between 20 and 30 D. Between 30 and 40 E. Greater than 40

We are given that h(n) is defined to be the product of all the even integers from 2 to n inclusive. For example, h(8) = 2 x 4 x 6 x 8.

We need to determine the smallest prime factor of h(100) + 1. Before determining the smallest prime factor of h(100) + 1, we must recognize that h(100) and h(100) + 1 are consecutive integers, and consecutive integers will never share the same prime factors.

Thus, h(100) and h(100) + 1 must have different prime factors. However, rather than determining all the prime factors of h(100), let’s determine the largest prime factor of h(100). Since h(100) is the product of the even integers from 2 to 100 inclusive, let’s find the largest prime number such that 2 times that prime number is less than 100.

That prime number is 47, since 2 x 47 = 94, which is less than 100. The next prime after 47 is 53, and 2 x 53 = 106, which is greater than 100.

Therefore, 47 is the largest prime number that is a factor of h(100). In fact, all prime numbers from 2 to 47 are included in the prime factorization of h(100). Since we have mentioned that h(100) + 1 will not have any of the prime factors of h(100), all the prime factors in h(100) + 1, including the smallest one, must be greater than 47. Looking at the answer choices, only choice E can be the correct answer.

Answer: E
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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26 Jul 2017, 02:51

Bunuel wrote:

enigma123 wrote:

h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40 < 30 > 40 < 10 Indeterminate

Below is the proper version of this question:

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) +1, then p is? A. between 2 and 20 B. between 10 and 20 C. between 20 and 30 D. between 30 and 40 E. greater than 40

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

Answer: E.

Hi Bunuel,

Thanks for the explanation! It was helpful. I tried to apply the same logic with simple numbers. Consider we have 15. So if try to find out the least prime factor of 15 , we must take 14 and check for its prime factors. They are 2 and 7. So according to the above theory, will the least prime factor of 15 be > 7??? It should not! ( here, do we need to check >2 or >7 ??)

h(n) is the product of the even numbers from 2 to n, inclusive, and p is the least prime factor of h(100)+1. What is the range of p?

< 40 < 30 > 40 < 10 Indeterminate

Below is the proper version of this question:

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) +1, then p is? A. between 2 and 20 B. between 10 and 20 C. between 20 and 30 D. between 30 and 40 E. greater than 40

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

Answer: E.

Hi Bunuel,

Thanks for the explanation! It was helpful. I tried to apply the same logic with simple numbers. Consider we have 15. So if try to find out the least prime factor of 15 , we must take 14 and check for its prime factors. They are 2 and 7. So according to the above theory, will the least prime factor of 15 be > 7??? It should not! ( here, do we need to check >2 or >7 ??)

Where am I going wrong??

Thanks, Uma

Two consecutive numbers are co-prime, so 14 and 15 are also co-prime, they do not share any common factor but 1. The factors of 14 are 1, 2, 7, and 14 and the factors of 15 are 1, 3, 5, and 15. As you can see no common factors but 1 (including no common primes).

The reasons you are confused is that \(h(100)=2^{50}*50!\) has ALL prime numbers from 1 to 50 as its factors, thus \(h(100)+1=2^{50}*50!+1\) won't have ANY prime factor from 1 to 50.

But 14 does NOT have ALL primes from 1 to 14, so 15 might have some primes from 1 to 14 (but 14 won't have the same primes as 15).
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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02 Aug 2017, 06:37

Can we solve in below method

h(100) = 2*4*6*...*100 From the function we can find the number of factors as (100/2)+(100/4)+(100/8)+(100/16)+(100/32)+(100/64) = 50+25+12+6+3+1 = 97 which is the least prime for the given function.

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