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E
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Difficulty:
55%
(hard)
Question Stats:
60% (01:24) correct
40%
(01:44)
wrong
based on 1398
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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
(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.
23 Jul 2013, 07:57
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Bookmarks
hb
\(x=2*4*6*...*50=(2*1)*(2*2)*(2*3)*...*(2*25)=2^{25}(1*2*3*...*25)=2^{25}*25!\). 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.
23 Jul 2013, 08:00
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Bookmarks
hb
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
General Discussion
29 Nov 2015, 10:59
Kudos
Bookmarks
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:
GMAT assassins aren't born, they're made,
Rich
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:
GMAT assassins aren't born, they're made,
Rich
as I posted this ...
