valleeny wrote:The solution on MGMAT states "No consecutive integers share the same prime factors".
In fact, can I also say "No consecutive integers share the same factor, irregardless if its prime or non prime, except 1" ?
Can someone confirm?
Yes. Consecutive integers are
co-primes: they share no factors other than 1. Here is my explanation:
If x is a positive integer, the only factor common both to x and to x+1 is 1. They share no other factors.
Let's examine why:
If x is a multiple of 2, the next largest multiple of 2 is x+2.
If x is a multiple of 3, the next largest multiple of 3 is x+3.
Using this logic, if we go from x to x+1, we get only to the next largest multiple of 1. So 1 is the only factor common both to x and to x+1. They share no other factors. (As noted above, integers that share no factors other than 1 are called
coprimes.)
Thus, in the problem above, we know that 1 is the only factor common both to h(100) and to h(100) + 1. They share no other factors.
h(100) = 2 * 4 * 6 *....* 94 * 96 * 98 * 100
Factoring out 2, we get:
h(100) = 2^50 (1 * 2 * 3 *... * 47 * 48 * 49 * 50)
Looking at the set of parentheses on the right, we can see that every prime number between 1 and 50 is a factor of h(100). This means that NONE of the prime numbers between 1 and 50 is a factor of h(100) + 1, because h(100) and h(100) + 1 share no factors other than 1.
So the smallest prime factor of h(100) + 1 must be greater than 50.
The correct answer is
E.
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