Function h(n) = product of all even numbers from 2 up to n.
if h(100) ... and if p is the small prime factor of h(100) + 1.... what is p?
a) between 10 and 20
b) btwn 20 and 30
c) btwn 30 and 40
d) greater than 40
D is the answer
I took this to mean: h(100) + 1 = [ 2 x 4 x 6 x 8.. 100 ] + 1. Not sure what to do next.
Thanks for your help!
Products of consecutive intervals
This topic has expert replies
There is a simple rule that states that n!+1 cannot be divisible by any number less than or equal to n.
modulo arithmetic:
n! mod x (where x <= n) = 0
n! + 1 mod x = 1, therefore not divisible by x.
In this particular question the function can be written as:
h(100) + 1 = (2^50)*50! + 1
therefore it's primes are > 50
correct answer is D
modulo arithmetic:
n! mod x (where x <= n) = 0
n! + 1 mod x = 1, therefore not divisible by x.
In this particular question the function can be written as:
h(100) + 1 = (2^50)*50! + 1
therefore it's primes are > 50
correct answer is D
Hi Viju
For better understanding of this try to analyse it by putting different values of n. for eg n=5
n!+1 --> 5! +1 =121 which is not divisible by 2,3,4,5 i.e. any number less than or equal to 5 . it is divisible by 11 .
Its a general rule therefore just remember this.
For better understanding of this try to analyse it by putting different values of n. for eg n=5
n!+1 --> 5! +1 =121 which is not divisible by 2,3,4,5 i.e. any number less than or equal to 5 . it is divisible by 11 .
Its a general rule therefore just remember this.