For every positive even integer n, the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive. if p is the smallest prime factor of h(100) + 1, then p is

More than 50

Step-by-step explanation:

To solve, we need to first see that the function is h(n). Picking main points from the question statement:

From here, we can write h(100) as:

h(100) =

h(100) = =

so,

h(100)+1 =

Now two numbers,

h(100) and h(100)+1 are consecutive integers and since they are consecutive so they are co-prime. Hence they only have common factor of 1. Example, 13 and 14 have only common factor of 1

As h(100) has all prime numbers from 1 to 50 and according to above statement h(100)+1 won't have any prime factor from 1 to 50, so the smallest prime factor p is greater than 50.

omg what happend to your question? if you could edit this question i can

step-by-step explanation:

the answer is a. q=6,r=6√2

step-by-step explanation: