The Answer is E.

=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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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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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

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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