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x is the product of all even numbers from 2 to 50, inclusive

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

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New post Updated on: 10 Jan 2016, 09:42
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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.

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Originally posted by hb on 23 Jul 2013, 07:44.
Last edited by Nevernevergiveup on 10 Jan 2016, 09:42, edited 3 times in total.
Edited the question and the tags.
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Re: x is the product of all even numbers from 2 to 50, inclusive  [#permalink]

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New post 23 Jul 2013, 07:57
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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.

Answer: E.

Similar questions to practice:
for-every-positive-even-integer-n-the-function-h-n-is-126691.html
for-every-positive-even-integer-n-the-function-h-n-149722.html
if-n-is-a-positive-integer-greater-than-1-then-p-n-represe-144553.html

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

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New post 23 Jul 2013, 08:00
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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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Re: x is the product of all even numbers from 2 to 50, inclusive  [#permalink]

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New post 29 Nov 2015, 10:59
Hi All,

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.

Final Answer:

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

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New post 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.

Answer: E.

Similar questions to practice:
http://gmatclub.com/forum/for-every-pos ... 26691.html
http://gmatclub.com/forum/for-every-pos ... 49722.html
http://gmatclub.com/forum/if-n-is-a-pos ... 44553.html

Hope it helps.


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

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New post 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.

Answer: E.

Similar questions to practice:
http://gmatclub.com/forum/for-every-pos ... 26691.html
http://gmatclub.com/forum/for-every-pos ... 49722.html
http://gmatclub.com/forum/if-n-is-a-pos ... 44553.html

Hope it helps.


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

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New post 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.

Answer: E.

Similar questions to practice:
http://gmatclub.com/forum/for-every-pos ... 26691.html
http://gmatclub.com/forum/for-every-pos ... 49722.html
http://gmatclub.com/forum/if-n-is-a-pos ... 44553.html

Hope it helps.


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


hi zflodeen

oh! I got it ...

I must be sleeping :grin: as I posted this ...

thanks
8-)
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Re: x is the product of all even numbers from 2 to 50, inclusive  [#permalink]

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New post 12 Oct 2017, 14:20
No worries gmatcracker2017. Happens all the time to me!
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Re: x is the product of all even numbers from 2 to 50, inclusive  [#permalink]

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New post 25 Oct 2018, 09:19
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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


Two consecutive integers do not share any common prime factors. Thus, we know that x and x + 1 cannot share any of the same prime factors.

We also see that x, the product of the even numbers from 2 to 50, contains prime factors of 2, 3, 5, 7, 11, 13, 17,19, and 23.

Thus, since x contains the primes from 2 to 23, we see that the smallest prime factor of x + 1 must be at least 29, i.e., greater than 25.

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

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New post 20 Mar 2019, 10:50
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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.


x = (2)(4)(6)....(46)(48)(50)
= (1)(2)(2)(2)(3)(2).....(23)(2)(24)(2)(25)(2)

Notice that:
x is divisible by 2. This tells us that x+1 is 1 greater than a multiple of 2. In other words, x+1 is NOT divisible by 2
x is divisible by 3. This tells us that x+1 is 1 greater than a multiple of 3. In other words, x+1 is NOT divisible by 3
x is divisible by 4. This tells us that x+1 is 1 greater than a multiple of 4. In other words, x+1 is NOT divisible by 4
x is divisible by 5. This tells us that x+1 is 1 greater than a multiple of 5. In other words, x+1 is NOT divisible by 5
.
.
.
x is divisible by 23. This tells us that x+1 is 1 greater than a multiple of 23. In other words, x+1 is NOT divisible by 23
x is divisible by 24. This tells us that x+1 is 1 greater than a multiple of 24. In other words, x+1 is NOT divisible by 24
x is divisible by 25. This tells us that x+1 is 1 greater than a multiple of 25. In other words, x+1 is NOT divisible by 25

We see that x+1 is NOT divisible by 2 to 25
In other words, all integers from 2 to 25 are NOT factors of x+1

So, if a number IS a factor of x+1, that number must be greater than 25

Answer: E

Cheers,
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Re: x is the product of all even numbers from 2 to 50, inclusive   [#permalink] 20 Mar 2019, 10:50
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