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Last edited by Nevernevergiveup on 10 Jan 2016, 09:42, edited 3 times in total.

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10 (B) Between 11 and 15 (C) Between 15 and 20 (D) Between 20 and 25 (E) Greater than 25

My Question: Please provide an explanation on how to arrive at the answer.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 02 Chapter: Homework Topic: Arithmetic Question: 105 Question: Page 251 Solution: PDF Page 20 of 32

\(x=2*4*6*...*50=(2*1)*(2*2)*(2*3)*...*(2*25)=2^{25}(1*2*3*...*25)=2^{25}*25!\). This number is obviously divisible by each prime less than 25.

Now, x and x+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.

Since x has all prime numbers from 1 to 25 as its factors, according to above x+1 won't have ANY prime factors from 1 to 25. Hence the smallest prime factor of x+1 will be greater than 25.

Re: x is the product of all even numbers from 2 to 50, inclusive [#permalink]

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23 Jul 2013, 08:00

4

This post received KUDOS

hb wrote:

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 02 Chapter: Homework Topic: Arithmetic Question: 105 Question: Page 251 Solution: PDF Page 20 of 32

Questioin 105. x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be (A): Between 1 and 10 (B): Between 11 and 15 (C): Between 15 and 20 (D): Between 20 and 25 (E): Greater than 25

My Question: Please provide an explanation on how to arrive at the answer.

SOME THEORY FIRST: HAVE A LOOK OF THIS PATTERN.

Factorial 2 +1=3 smallest factor is 3==>this is greater than 2 factorial 3 + 1=7 smallest factor is 7==>this is greater than 3 factorial 4 + 1 = 25 smallest factor is 5==>which is greater than 4 factorial 5 + 1= 121 smallest factor is 11==>which is greater than 5 (note: i am excluding 1 as a smallest factor)

now by seeing this pattern you can figure out that.. (factorial x + 1)==>smallest factor will always be greater than x

now coming to our problem all even no.s between 2 to 50 2*4*6*8......*48*50 now take 2 from each number common. 2^25(1*2*3*4*5*......25) or 2^25*factorial 25 now x+1= 2^25*factorial 25 + 1===>clearly smallest factor will be greater than 25(as proved above)

hence E

hope it helPS
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The above prompt is essentially just a 'lift' of the following GMAC question (but the concept is exactly the same):

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:

1. between 2 and 10 2. between 10 and 20 3. between 20 and 30 4. between 30 and 40 5. greater than 40

The main idea behind this prompt is:

"The ONLY number that will divide into X and (X+1) is 1."

In other words, NONE of the factors of X will be factors of X+1, EXCEPT for the number 1.

Here are some examples: X = 2 X+1 = 3 Factors of 2: 1 and 2 Factors of 3: 1 and 3 ONLY the number 1 is a factor of both.

X = 9 X+1 = 10 Factors of 9: 1, 3 and 9 Factors of 10: 1, 2, 5 and 10 ONLY the number 1 is a factor of both. Etc.

Since the H(100) is (100)(98)(96)....(4)(2)....we can deduce.... 1) This product will have LOTS of different factors 2) NONE of those factors will divide into H(100) + 1.

H(100) contains all of the primes from 2 through 47, inclusive (the 47 can be "found" in the "94"), so NONE of those will be in H(100) + 1. We don't even have to calculate which prime factor is smallest in H(100) + 1; we know that it MUST be a prime greater than 47....and there's only one answer that fits.

x is the product of all even numbers from 2 to 50, inclusive [#permalink]

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12 Oct 2017, 13:27

Bunuel wrote:

hb wrote:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10 (B) Between 11 and 15 (C) Between 15 and 20 (D) Between 20 and 25 (E) Greater than 25

My Question: Please provide an explanation on how to arrive at the answer.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 02 Chapter: Homework Topic: Arithmetic Question: 105 Question: Page 251 Solution: PDF Page 20 of 32

\(x=2*4*6*...*50=(2*1)*(2*2)*(2*3)*...*(2*25)=2^{25}(1*2*3*...*25)=2^{25}*25!\). This number is obviously divisible by each prime less than 25.

Now, x and x+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.

Since x has all prime numbers from 1 to 25 as its factors, according to above x+1 won't have ANY prime factors from 1 to 25. Hence the smallest prime factor of x+1 will be greater than 25.

Re: x is the product of all even numbers from 2 to 50, inclusive [#permalink]

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12 Oct 2017, 13:51

gmatcracker2017 wrote:

Bunuel wrote:

hb wrote:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10 (B) Between 11 and 15 (C) Between 15 and 20 (D) Between 20 and 25 (E) Greater than 25

My Question: Please provide an explanation on how to arrive at the answer.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 02 Chapter: Homework Topic: Arithmetic Question: 105 Question: Page 251 Solution: PDF Page 20 of 32

\(x=2*4*6*...*50=(2*1)*(2*2)*(2*3)*...*(2*25)=2^{25}(1*2*3*...*25)=2^{25}*25!\). This number is obviously divisible by each prime less than 25.

Now, x and x+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.

Since x has all prime numbers from 1 to 25 as its factors, according to above x+1 won't have ANY prime factors from 1 to 25. Hence the smallest prime factor of x+1 will be greater than 25.

hi Bunuel since x is the product of all even integers from 2 to 50 inclusive,

x = (2 * 4 * 6 * 8 * 10 * 12 * 14.....* 50) which can be rewriten as

2( 1 * 2 * 3 * 4 * 5 * 6 * 7 *........* 25) so, x is equal to 2 * 25!

please say to me why this is not okay ...

thanks in advance, man

gmatcracker2017:

I believe you are mixing the rule up with a question where we are talking about the sum of all even integers between 2 and 50. Then it would be 2+3+6...+50 which can factor out just one 2, but since it's the product, you have to factor out all 25 "2"s. I believe that is where your thought processes is getting confused, but mine gets confused often, so Bunuel may be best to confirm.

Re: x is the product of all even numbers from 2 to 50, inclusive [#permalink]

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12 Oct 2017, 14:13

gmatcracker2017 wrote:

Bunuel wrote:

hb wrote:

x is the product of all even numbers from 2 to 50, inclusive. The smallest prime factor of x+1 must be

(A) Between 1 and 10 (B) Between 11 and 15 (C) Between 15 and 20 (D) Between 20 and 25 (E) Greater than 25

My Question: Please provide an explanation on how to arrive at the answer.

Disclaimer: I have used the Search Box Before Posting. I used the first sentence of the question or a string of words exactly as they show up in the question below for my search. I did not receive an exact match for my question.

Source: Veritas Prep; Book 02 Chapter: Homework Topic: Arithmetic Question: 105 Question: Page 251 Solution: PDF Page 20 of 32

\(x=2*4*6*...*50=(2*1)*(2*2)*(2*3)*...*(2*25)=2^{25}(1*2*3*...*25)=2^{25}*25!\). This number is obviously divisible by each prime less than 25.

Now, x and x+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.

Since x has all prime numbers from 1 to 25 as its factors, according to above x+1 won't have ANY prime factors from 1 to 25. Hence the smallest prime factor of x+1 will be greater than 25.