Recent content by highschoolmath

Awesome- thanks to you both!

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...

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...

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...

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...