MHB How to find smallest prime factor?

[highschoolmath]

I was given this problem.

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 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Ideas?

 

[I like Serena]

I was given this problem.

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 10
b) between 10 and 20
c) between 20 and 30
d) between 30 and 40
e) greater than 40

Ideas?

Hi high schoolmath! Welcome to MHB! (Smile)

Here are my ideas:

1. What is the general form of h(n)?
2. What are its divisors?
3. What happens to those divisors if we add 1 to h(n)?

(Thinking)

 

[highschoolmath]

Unfortunately, that's all the info I have! I wondering if the solution has to do with changing the equation? If we know h(100)= 2x4x6x...x98x100, then this should equal (2^50)(50!) right? But then the problem is to find the smallest prime factor of (2^50)(50!)+1.

I think there must be an easy solution to this somehow!

 

[I like Serena]

Unfortunately, that's all the info I have! I wondering if the solution has to do with changing the equation? If we know h(100)= 2x4x6x...x98x100, then this should equal (2^50)(50!) right? But then the problem is to find the smallest prime factor of (2^50)(50!)+1.

Good! (Nod)

I think there must be an easy solution to this somehow!

You're almost there.
Note that all numbers up to 50 are dividers of (2^50)(50!).
Will they also divide (2^50)(50!) + 1? (Wondering)

 

[highschoolmath]

Well, I know that (2^50)(50!) + 1 is an odd number, so none of the even numbers from 1 to 50 will divide into it. But I'm not sure how I can figure out whether the odd numbers can divide into it, let alone the prime numbers.

50! will have I imagine at least 5 zeroes because it's multiplying 10 five times. So, I might be wrong, but I might guess that the number will end in "00001".

Is there another rule I should know? I think it's getting close...

 

[Jameson]

Hmm, I think I see where ILS is going with this. We are looking at $$\frac{2^{50}50!+1}{p}$$, where $p$ is prime.

What if we rewrote this as $$\frac{2^{50}50!}{p}+\frac{1}{p}$$?

This is just using a property of fractions. Now, in order for the whole thing to be a whole number, either both terms must be whole numbers or both terms must be fractions and add together in such a way that they become a whole number. If we got a whole number plus a fraction, that would clearly not be a whole number right? :)

 

[I like Serena]

Let's pick a number that is divisible by 5.
Say 45, which is also divisible by 3.
Is 46 divisible by 5? Or by 3?
Why not? (Wondering)

 

[highschoolmath]

I get it!

If the prime number were less than 50, then it would divide into the left part of the expression, resulting in a whole number, and then adding 1/p would result in a fraction.

But since we know p divides into the entire expression resulting in a while integer, then p must be greater than 50. Right?!

Is this how you both would have reasoned it out? I'd appreciate hearing your thought processes for learning purposes. In retrospect, this seems so obvious!

 

[I like Serena]

I get it!

If the prime number were less than 50, then it would divide into the left part of the expression, resulting in a whole number, and then adding 1/p would result in a fraction.

But since we know p divides into the entire expression resulting in a while integer, then p must be greater than 50. Right?!

Is this how you both would have reasoned it out? I'd appreciate hearing your thought processes for learning purposes. In retrospect, this seems so obvious!

That is correct and indeed the way to reason it out. (Mmm)

 

[highschoolmath]

Awesome- thanks to you both!