Vincen wrote: ↑Fri Jul 09, 2021 10:40 am
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?
A. Between \(2\) and \(20\)
B. Between \(10\) and \(20\)
C. Between \(20\) and \(30\)
D. Between \(30\) and \(40\)
E. Greater than \(40\)
Answer:
E
Source: GMAT Prep
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
