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E
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Difficulty:
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55% (01:56) correct
45%
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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 10
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
1.jpg [ 49.61 KiB | Viewed 87719 times ]
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.
h(100) = 2 * 4 * 6 ****************100
Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.
100 = 2 + (n-1) 2
n = 50
Product of terms = Average * number of terms
Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.
H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
A. Between 2 and 10
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
Attachment:
1.jpg [ 49.61 KiB | Viewed 87719 times ]
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.
h(100) = 2 * 4 * 6 ****************100
Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.
100 = 2 + (n-1) 2
n = 50
Product of terms = Average * number of terms
Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.
H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
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, then, according to the 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 \(h(100)+1\) must be more than 50.
Answer: E.
A. between 2 and 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
\(h(100)+1= \)
\( =2*4*6*...*100+1= \)
\( =2^{50}*(1*2*3*..*50)+1= \)
\(= 2^{50}*50!+1\)
\( =2*4*6*...*100+1= \)
\( =2^{50}*(1*2*3*..*50)+1= \)
\(= 2^{50}*50!+1\)
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, then, according to the 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 \(h(100)+1\) must be more than 50.
Answer: E.
28 Jan 2012, 18:08
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enigma123
Given that the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive, so: \(h(100)=2*4*6*...*100=(2*1)*(2*2)*(2*3)*(2*4)*...*(2*50)\) --> factor out all 50 2's: \(h(100)=2^{50}*(1*2*3*..*50)=2^{50}*50!\).
Hope it's clear.
enigma123
Here is some theory first:
Pick any two consecutive numbers. Can they both be even i.e. have 2 as a factor?
e.g. (5,6) or (101, 102) or (999, 1000) etc .. Only one number in these pairs will have 2 as a factor.
Now think: If you pick any two consecutive integers, can they both have 4 as a factor? or 7 as a factor? or 99 as a factor? No!
(For that matter, once you get one multiple of 99, you will not get another one in the next 98 numbers. The next multiple will appear when you add 99 to this multiple. e.g. you pick 99. Can 100, 101, 102... be multiples of 99? The next one is 198 so numbers from 100 to 197 are not multiples of 99.)
We can say that consecutive numbers will not have any common factor other than 1. (1 is a factor of every number.)
e.g. if N = 2*3*5*7*11,
Think consecutive numbers now:
(N - 4), (N - 3), (N - 2), (N - 1), N, (N + 1), (N + 2), (N + 3), (N + 4)...
(N + 1) and (N - 1) will not have any of 2, 3, 5, 7 and 11 as factors. (Since they are consecutive with N)
(N + 2) and (N - 2) will not have 3, 5, 7 and 11 as factors. (2 will be a factor of both these numbers since 2 is a factor of N)
(N + 3), ( N + 4), (N -3 ), (N - 4) will not have 5, 7 and 11 as factors. (But (N + 3) and (N - 3) will have 3 as a factor. (N + 4) and (N - 4) will have 2 as a factor)
and so on...
Let's get back to the original question:
2*50! = 2*1*2*3*4*5*6*7*8*9*10*11.....*50
This number has all numbers till 50 as its factors. So 2*50! + 1 cannot have all these numbers as its factor.
Therefore the smallest prime number that will be the factor of 2*50! + 1 is greater than 50.
A trick question can be "What is the smallest factor of (2*50! + 1)?"
The smallest factor will be 1.
.. anyway i clicked e and it turned out right.
