n@resh wrote:winny wrote:someone please explain...
H(n) = 2*4*6*8.....n ( all are even integers)
lets take example:
H(8)+1 = 2.4.6.8 +1 = 2^6(1*2*3)+1 = 385 ; so, least prime factor H(8)+1 is 5 (i.e 5*7*11)
similarly..H(10)+1 = 2*4*6*8*10 +1 = 2^7*(1*2*3*5)+1= 3841;
so, least prime factor of H(10)+1 is 23 (i.e. 23*167)
now come to H(100)+1 = 2^50*(1*2*3*...*50) + 1, will not have prime factors < = 50.
Hence the least prime factor must be >50
Answer E!
I am unable to find any relation between your examples and the solution of the question.
Looks like you jumped to the solution directly (please don't take me wrong).
let's do it with one of your example i.e. h(10)+1 = 2*4*6*8*10+1 = 2^4(1*2*3*4*5)+1
Now as per the remainder theorem if you divide (p+1)by p you will get remainder = 1.
The same logic is applicable here as well.
h(10)+1 = 2^4(1*2*3*4*5)+1 -- means if you divide this number by any integer till 5 you will get remainder = 1.
The same logic is applicable to this question (you don't need to multiply the whole thing and then find the prime factor because that will take lot of time)
h(100)+1 = 2^50 (1*2*.......*50) +1
this number when divided by any number till 50 will give 1 as the remainder. Hence your answer is greater than 50.
Option E 
