Answer:
More than 50
Explanation:
To solve, we need to first see that the function is h(n). Picking main points from the question statement:
- h(n) is the product of all even integers (From 2 to n)
- p is the smallest factor of h(100)+1
- h(100)+1 , here n=100
From here, we can write h(100) as:
h(100) =
![2 * 4 * 6 * 8 * ...... * 100](https://img.qammunity.org/2020/formulas/mathematics/high-school/iqqt0rzea52w2lxlyrxq27mhquu5szmdsn.png)
h(100) =
=
![2^(50) * 50!](https://img.qammunity.org/2020/formulas/mathematics/high-school/na285pcpovvywk5tzcpv99nf1h95d20q1b.png)
so,
h(100)+1 =
![(2^(50) * 50! )+1](https://img.qammunity.org/2020/formulas/mathematics/high-school/ov6846rw7j1kaapuxcl8cfa8yyib0rpzz1.png)
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.