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

Can someone walk me through this one?

Guest, this is definitely a difficult number properties question. Let's first consider the prime factors of h(100). According to the given function,
h(100) = 2*4*6*8*...*100

By factoring a 2 from each term of our function, h(100) can be rewritten as
2^50*(1*2*3*...*50).

Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1.

Since the smallest prime number that can be a factor of h(100) + 1 has to be greater than 50, The correct answer is E.

Hope that helps
-Dan

I got stuck on this question as well. can you please tell me how 2^50 can be factored? I'm not sure I'm following the math.

Thank you!

the number h(100) is 2 x 4 x 6 x 8 x ... x 100
which is (2 x 1) x (2 x 2) x (2 x 3) x ... x (2 x 50)

if you pull out all those 2's and send them all to the front of the expression - something you're allowed to do, because you can multiply a group of numbers in any order you please - you get 2^50 in front, because there are fifty 2's.

thanks, ron!

Nice job.

Ron,

I'm missing something in the following step

"Thus, all integers up to 50 - including all prime numbers up to 50 - are factors of h(100).

Therefore, h(100) + 1 cannot have any prime factors 50 or below, since dividing this value by any of these prime numbers will yield a remainder of 1. "

So because h(100) can be factored by every integer up to 50 means that h(100)+1 can't have a prime factor below 50? Not quite getting this one... Please flush out.


Thanks-

Halo3,

h(100)=2^50 * 50! = 2^50 * (1*2*3*...*50)

Here h(100) is divisible by any numer between 2 and 50. Right?
So h(100)+1 can't be divisile by any number between 2 and 50. h(100)+1 will always give a reminder of 1 when u divide the no by any no between 1 to 50.

So h(100)+1 doesnot have any factor between 2 and 50. Hence h(100)+1 doesnot have any prime factor between 2 and 50.

Cheers,

i'll assume you mean "flesh out", unless you want this whole thread destroyed (which you presumably don't).
heh heh

the idea is this: if a number is divisible by some prime p, then the next multiple of p will be p units bigger. for instance, 75 is divisible by 5. this means that the next greatest multiple of 5 is 80, which is 5 units away.
hopefully, this fact is clear. once you realize this, it follows that consecutive integers can't share ANY primes, because they're only 1 unit apart (too close together to work for any common factor except 1, which is trivially a factor of any integer at all, anywhere).
that's the basis for saying h(100) + 1 has no prime factors below 50. if it did, then you'd have two multiples of the same prime 1 unit apart, and that's impossible.

Makes sense now - appreciate the help!

We're glad it helped.