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

Question

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

pushpitkc · Accepted Answer

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)

niks18 · Answer

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

chetan2u · Answer

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

MadaraU · Answer

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

chetan2u · Answer

Hi

thanks, I missed out on ODD in 'product of ODD integers", although answer will still remain same.
edited the answer..

Timebomb · Answer

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

chetan2u · Answer

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

Timebomb · Answer

chetan2u . Yes. Got it. Thanks for your reply.

Posted from my mobile device

aragonn · Answer

f(n) is product of all odd integer from 1 to n.
f(n) - 1 will be even integer.

As these two are consecutive integers, there will not be any common factors between the two.So next factor for f(n)-1 after 2 will be more than 71.

ScottTargetTestPrep · Answer

We see that f(n) = 1 x 3 x 5 x … x n. So f(71) = 1 x 3 x 5 x … x 71. However, since f(71) is divisible by all odd numbers from 1 to 71, f(71) - 1 will not be divisible by any odd numbers from 1 to 71. So the lowest odd prime factor of f(71) - 1 must be greater than 71.

Answer: E

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

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