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

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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
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
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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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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

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01 May 2017, 05:49
ziyuen wrote:
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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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.

Sent from my Redmi Note 4 using GMAT Club Forum mobile app
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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20 Jun 2017, 07:37
enigma123 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

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.

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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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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
Math Expert
Joined: 02 Sep 2009
Posts: 47037
Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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26 Jul 2017, 04:24
umabharatigudipalli wrote:
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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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.

Hence 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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30 Sep 2017, 20:02
The answer should be much simpler.
we know that the products of n consecutive numbers is divisible by n!
In this case 2*4*6....*100=2*(1*2*3...*50). therefor, (1*2*3...*50) must be divisible by 50!, thus all factors inside 50! (including the prime factors) are factors. 50! includes 41 &43. answer is E.

using this method can help you to identify highest and lowest prime factors too
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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11 Dec 2017, 08:26
Great Question, though hard.
Tks Bunuel for the amazing solution
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Re: For every positive even integer n, the function h(n) is defined to be [#permalink]

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10 Jun 2018, 04:54
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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

Hello Bunuel, please tell me mathematical truth

what is the point of factoring 2, i mean how does it help you to solve this queston ? $$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

thank you
Re: For every positive even integer n, the function h(n) is defined to be   [#permalink] 10 Jun 2018, 04:54

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