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
Important Concept:
If integer k is greater than 1, and k is a factor (divisor) of N, then k is not a divisor of N+1
For example, since 7 is a factor of 350, we know that 7 is not a factor of (350
+1)
Similarly, since 8 is a factor of 312, we know that 8 is not a factor of 313
Now let's examine h(100)
h(100) = (2)(4)(6)(8)....(96)(98)(100)
= (2x
1)(2x
2)(2x
3)(2x
4)....(2x
48)(2x
49)(2x
50)
Factor out all of the 2's to get: h(100) = [2^50][
(1)(2)(3)(4)....(48)(49)(50)]
Since
2 is in the product of h(100), we know that 2 is a factor of h(100), which means that 2 is
not a factor of h(100)
+1 (based on the above rule)
Similarly, since
3 is in the product of h(100), we know that 3 is a factor of h(100), which means that 3 is
not a factor of h(100)
+1 (based on the above rule)
Similarly, since
5 is in the product of h(100), we know that 5 is a factor of h(100), which means that 5 is
not a factor of h(100)
+1 (based on the above rule)
.
.
.
.
Similarly, since
47 is in the product of h(100), we know that 47 is a factor of h(100), which means that 47 is
not a factor of h(100)
+1 (based on the above rule)
So, we can see that none of the primes from 2 to 47 can be factors of h(100)+1, which means the smallest prime factor of h(100)+1 must be greater than 47.
Answer =
E
Cheers,
Brent
