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For an odd integer n, the function f(n) is defined as the product of a

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For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 03 Nov 2017, 04:44
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For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above
[Reveal] Spoiler: OA

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For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 03 Nov 2017, 04:49
nkmungila wrote:
For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above



So F(n) is nothing but 1*3*....*n-2*n
Now n! has all ODD numbers from 1 to n as its factors, so n!-1 will not have any ODD numbers 1 to n as its factors..
so f(n)-1 will be EVEN number and will have 2 as its factor BUT all other prime factors are ODD and will not be a factor of f(n)-1
This basically means f(n)-1 has ONLY 2 as prime factor which is less than n

Therefore f(71)-1 will be 1**3*4*.....*69*71 - 1
and it will have ODD prime factor greater than 71...
So E
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Re: For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 03 Nov 2017, 06:19
chetan2u wrote:
nkmungila wrote:
For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above



So F(n) is nothing but n!, Which is 1*2*3*....*n-1*n
Now n! has all numbers from 1 to n as its factors, so n!-1 will not have any numbers 1 to n as its factors..
This basically means n!-1 is a prime number

Therefore f(71)-1 =71!-1, which will be a prime number and ofcourse much greater than 71. It will be 1*2*3*4*.....*70*71 - 1
So E



f(n) is defined as the product of all odd integers which means F(n) = 1.3.5.7..... . how come is it n! ?
I'm confused
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For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 03 Nov 2017, 08:39
nkmungila wrote:
For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above


Hi MadaraU - the question can be solved as per below process

\(f(71)-1\) & \(f(71)\) are consecutive integers hence will be co-prime. for e.g 2, 3 or 50, 51 are co-prime.

therefore \(f(71)\) and \(f(71)-1\) will not have any common factor.

since \(f(71)\) is the product of all odd numbers less than \(71\)(inclusive) which means all odd numbers less than \(71\) are factors of \(f(71)\), hence \(f(71)-1\) will not have any odd factor less than \(71\)

Option E
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Re: For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 03 Nov 2017, 19:55
MadaraU wrote:
chetan2u wrote:
nkmungila wrote:
For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above



So F(n) is nothing but n!, Which is 1*2*3*....*n-1*n
Now n! has all numbers from 1 to n as its factors, so n!-1 will not have any numbers 1 to n as its factors..
This basically means n!-1 is a prime number

Therefore f(71)-1 =71!-1, which will be a prime number and ofcourse much greater than 71. It will be 1*2*3*4*.....*70*71 - 1
So E



f(n) is defined as the product of all odd integers which means F(n) = 1.3.5.7..... . how come is it n! ?
I'm confused


Hi

thanks, I missed out on ODD in 'product of ODD integers", although answer will still remain same.
edited the answer..
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 17 Feb 2018, 00:18
Hi,. My doubt is will f(71)-1 have any prime number as it's factor apart from 2? I don't think there will be any prime number, even greater than 71 as it's factor chetan2u
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For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 17 Feb 2018, 02:07
nkmungila wrote:
For an odd integer n, the function f(n) is defined as the product of all odd integers from 1 to n. The lowest odd prime factor f(71)-­1 lies between…

A. 3 and 10
B. 11 and 30
C. 31 and 50
D. 51 and 70
E. 71 and above


Function f(n) is defined as the product of all odd integers from 1 to n

If n=7, f(7) = 7*5*3 = 105 | f(7) -1 = 104 = \(2^2\)*13

If n=11, f(11) = 11*9*7*5*3 = 10395 | f(11) - 1 = 10395 - 1 = 10394
Prime factorizing 10394, we will get 2*5197 - a prime number

Extrapolating this we will understand that the lowest odd prime factor for
the function f(n) - 1 must be greater than n.

The lowest prime factor of f(71) - 1 will always be greater than 71(Option E)
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Re: For an odd integer n, the function f(n) is defined as the product of a [#permalink]

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New post 17 Feb 2018, 06:12
Mudit27021988 wrote:
Hi,. My doubt is will f(71)-1 have any prime number as it's factor apart from 2? I don't think there will be any prime number, even greater than 71 as it's factor chetan2u


Hi Mudit27021988,

2 will be a factor but it can and generally should have a ODD factor > 71..
example..
5!-1 = 5*3*1-1=15-1=14... prime factors 2 and 7 > 5
7!-1 = 1*3*5*7 -1 = 104 = 2*2*2*13.. here 13 > 7
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Absolute modulus :http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html


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Re: For an odd integer n, the function f(n) is defined as the product of a   [#permalink] 17 Feb 2018, 06:12
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