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

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Re: Hard question: Function and factors [#permalink]

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30 Apr 2011, 23:27
Co-prime numbers concept.
Good question.
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03 May 2011, 03:49
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03 May 2011, 10:33
Answer has to be p > 40. If h(n) = product of all EVEN integers (n included), then the smallest prime factor is necessarily 2. But if you add 1 to any h(n), function, the number becomes odd.

1) So we can multiply to find h(100): 2 x 4 x 6 x 8 x ... x 98 x 100
2) That can be reduced to: 2^1 x 2^2 x 2^3 x 2^4 x... 2^49 x 2^50
This can be further simplified: (2 x 2 x 2 x ... 50 times) x (1 x 2 x 3... x 50)
Or: (2^50) x (50!)

3) Then we add 1 to it >> very large and odd
4) So to get to the lowest common factor, we go back to step (2). The lowest prime factor I can find is 47 among the numbers that reach 50!. That's also > 40.

If there's any hole in my reasoning, I welcome comments. As I can benefit from it too.
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Re: Hard question: Function and factors [#permalink]

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12 Jul 2011, 06:41

=2*4*6*8*10*12*14*...*100
= 2 (1*2*3*4*5*6*7...*50)
47 is the highest prime number before 50..therefore lies greater than 40.
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22 Jul 2011, 11:05
Thanks for you i understood this Q

Bunuel wrote:
mohit005 wrote:
Question is - 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 factor of h(100) + 1, then p is

a) between 2 and 10
b) between 20 and 10
c) between 20 and 30
d) between 30 and 40
e) greater than 40

OA is (e) , but I am not able to figure out why it is (e)

Welcome to the club Mohit.

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

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 numbers from 1 to 50 as its factors, according to above $$h(100)+1=2^{50}*50!+1$$ won't have ANY factor from 1 to 50. Hence $$p$$ ($$>1$$), the smallest factor of $$h(100)+1$$ will be more than 50.

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Smallest prime factor of a factorial [#permalink]

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22 Aug 2011, 05:24
If p is the smallest prime factor of 2(50!) + 1, the 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: Smallest prime factor of a factorial [#permalink]

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22 Aug 2011, 06:22
If you expand the factorial notation, you can see that the expression can't have prime factors below 50, since dividing this value by any of these prime numbers will give a remainder of 1.
So the smallest prime factor has to greater than 50.
Hence E.
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Re: Smallest prime factor of a factorial [#permalink]

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22 Aug 2011, 21:19
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Expert's post
bsaikrishna wrote:
If p is the smallest prime factor of 2(50!) + 1, the 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

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.
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26 Aug 2011, 11:41
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26 Aug 2011, 11:48
Never mind, I got it. Sometimes all it takes is getting up to wash your face and then trying 1 more time.

h(100) = 2*4*6...*100 etc.
[h(100)]/(2^50) = 1*2*3...*50 = 50!
h(100)=(2^50)*50!
which is divisible by all primes up to 50
so when we add 1, the number can't be divisible by any prime less than 50
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28 Sep 2011, 02:54
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Re: What it the lowest prime divisor? [#permalink]

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05 Oct 2011, 00:25
According to the given function,
P(n)=2^50*(1*2*3*...*50).
(Remove 2 from each even factor)

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of P(100).
Hence, P(100) + 1 cannot have any prime factors 50 or below as dividing this value by any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of P(100) + 1 has to be greater than 50, The correct answer is E.
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Re: GMATPrep PS- Function and factors [#permalink]

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06 Oct 2011, 10:30
karishma,

Can you please explain this part .I;m nt abt able to get this

Quote:
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.
Re: GMATPrep PS- Function and factors   [#permalink] 06 Oct 2011, 10:30

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