damilolaamele wrote:Please what's the most effective way to approach this kind of question?
For every 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, then p is
a) Between 2 and 10
b) Between 10 and 20
c) Between 20 and 30
d) Between 30 and 40
e) Above 40
[spoiler]OA: E[/spoiler]
Important Concept: If k is a positive integer that’s greater than 1, and if 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)
Factor to get: h(100) = 2[(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,
