For every positive even integer n, the function h(n) is defined to be the product of all even integers from 2 to n, inclusive. If p is the smallest prime factor of h(100) + 1, the p is
A: Between 2 & 10
B: Between 10 & 20
C: Between 20 & 30
D: Between 30 & 40
E: Greater than 40
Since the difference between them is 1, h(100) and h(100)+1 are consecutive integers.
Consecutive integers are COPRIMES: they share no factors other than 1.
Let's examine why:
If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.
Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1.
So 1 is the only factor common both to x and to x+1.
In other words, x and x+1 are COPRIMES.
Thus, in the problem above, h(100) and h(100)+1 are coprimes. They share no factors other than 1.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)
Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100).
Since h(100) and h(100)+1 are coprimes, NONE of the prime numbers between 1 and 50 can be a factor of h(100)+1.
Thus, the smallest prime factor of h(100) + 1 must be greater than 50.
The correct answer is
E.
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