For Every Positive Even Integer n, the Function h(n) is Defined to be GMAT Problem Solving

Question: 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?

1. Between 2 and 10
2. Between 10 and 20
3. Between 20 and 30
4. Between 30 and 40
5. Greater than 40

Solution and Explanation:

Approach Solution 1:

h(100)+1=2∗4∗6∗...∗100+1=2^50∗(1∗2∗3∗..∗50)+1=2^50∗50!+1
Now, two numbers h(100)=2^50∗50! And h(100)+1=2^50∗50! + 1 are consecutive. When two successive integers are co-prime, they only have one factor in common. For instance, since 20 and 21 are two consecutive integers, the only element they have in common is 1.
As h(100)=2^50∗50! has ALL prime numbers from 1 to 50 as its factors, then, according to the above, h(100)+1 = 2^50∗50!+1
From 1 to 50, above will not have ANY prime factors. Accordingly, p (>1), the least prime factor of h(100)+1 must be more than 50.
E is the correct answer.

Correct Answer: E

Approach Solution 2:

Let's break down the answer into two parts:
Part (1):
First, we must simplify h(100).
h(100) = 2*4*6*8*....*98*100.
Consider that 2 = 2*1, 4 = 2*2, etc. up to 100 = 2*50.
h(100) then simplifies to h(100) = 2∗1∗2∗2∗2∗3∗....∗2∗49∗2∗50 = 2^50∗1∗2∗3∗4∗...∗50h(100)=2∗1∗2∗2∗2∗3∗....∗2∗49∗2∗50=250∗1∗2∗3∗4∗...∗50 (A)
Part (2):
Think about the second component of the query. We must establish a limit for h(100)+1's smallest prime factor. Any number p that divides h(100)+1 by itself leaves a residue of (p-1) (100).
rest (h(100)/p) = p-1
"p is a prime number that is not a divisor of h(100)" is another way to state this (B)
From (A):
Given that h(100) is divisible by 50, it follows that h(100) is divisible by all prime numbers smaller than 50. Therefore, the smallest prime that is not h(100 divisor )'s should be bigger than 50.
Looking at the choices, the most appropriate choice is:
(E) greater than 40.

Correct Answer: E

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