The smallest prime factor of h(100) + 1

Hello I searched this question about a week ago and below is the answer I found. You can find more explanations by doing a search of the problem either on this forum or many others . Good Luck!

Clearly we cannot calculate h(100), and the smallest prime factor is not trivial. However, we can say that it is even, and therefore divisible by 2. Also, every even number between 2 and 100 is included in the factors, and
so is every prime number less than 50 because 100 = 2*50. But none of these prime numbers can be a factor of h(n)+1 because they are factors of h(n) and the are larger than 1.

Therefore the smallest factor must be larger than the largest prime less than 50 (which is 43), and the smallest factor must be greater than 40. (e).

Generally, the smallest prime factor of h(n)+1 is larger than n/2.

Questions on Divisibility and Primes
https://www.testsandtutors.com/course/qu ... and-Primes

IMO..we can solve the same even more simply...even without bothering about finding the prime numbers less than 50.....

The property of numbers that we need to use is

"No two consecutive positive integers(n, n+1) are ever divisible by same number except 1."

Now if we see that h(n) =(2^50) * 50!.

so it is divisible by all numbers from 2-50....

so the minimum number(including prime/non prime etc etc..) that will divide h(n) will be 51......

Thus it leads to E.

What am I missing in understanding this question? I thought the question is asking for the smallest factor not the biggest? So the smallest prime factor of h(100) is 2? I don't understand.

Yes, but 2 is not a factor of f(100)+1, since f(100)+1 is odd

This is also discussed at https://www.beatthegmat.com/numbers-t61737.html and https://www.beatthegmat.com/maths-coprim ... 58941.html with an alternative approach used to solve.

This is a killer of a question but there are lots of discussions about it on the forum. The key to solving this question is knowing that the best way to prove that an expression is divisible by n is to demonstrate that the expression can be written as the product of n * integer. For example, 4!+6 is divisible by 3 because it can be written as the product of 3 * integer >> 4! + 6 = 3*(4*2*1 + 2)

On the other hand, one way to prove that an expression is not divisible by n is to demonstrate that the expression can be written as the sum of (multiple of n) + (non-multiple of n). For instance, 3p+7 cannot be a multiple of 3 (assuming p is an integer) because it equals a multiple of 3 + a non multiple of 3.

We can use this to prove that all numbers from 1 to 40 cannot be factors of h(100)+1.

The answer is E. You can see a detailed explanation and step-by-step video to solve this question at GMATPrep question 1282. To practice similar questions, set topic='Number Properties' and difficulty='700+' in the Drill Generator

Hope that helps,
-Patrick

On lines similar to that discussed in this post.

There are no common factors between two consecutive apart from 1. For example take 9 and 10, there are no mutual factors among them except for 1. Similarly consider 5! and 5!+1. The factors of 5! are 2, 3, 4, 5 ... and in the case of 5!+1=121 the factors are 11, 121.

Therefore in this question, the factors of 100! include 2, 3, 4, 5, 6 .... 100

so the consecutive number to 100!+1 plus will not have any of the above mentioned factors apart from 1, it may have factors that are greater than 100.

And the option which offers that answer is E. Prime number is greater than 40