Man, I hate this problem! I saw it a few weeks ago and it took me over an hour of research (particularly on BTG) to finally understand it. I'm praying I get something like this on my GMAT, though, as I get it inside and out now!
Anyhow, the single most valuable piece of info you need to understand is the concept of co-primes (and a well deserved shout-out to a post by Mitch which broke this down and finally made it click for me).
Co-primes are any two integers that share no common positive factors other than 1. And that is the definition of any two consecutive numbers.
Once you get your head around that, a question like this can be broken down really easily.
We know that h(100) is the sum of the evens between 2 and 100, inclusive. And we know that once we know that number, we just add 1 to get the sum of h(100)+1.
Look at the pattern in the formula for h(100):
h(100) = 2 * 4 * 6 * 8 * 10.... etc. They're all increasing by an increment of 2, so we can extract a 2 from this and we get:
h(100) = 2^50(1*2*3*4*5...*48*49*50)
And that's it. The greatest prime number in h(100) is, therefore, 47. And that means the SMALLEST prime in h(100)+1 cannot be less than or equal to 47.
And a quick look at the answer choices reveals E.
But since we only need to know the smallest prime number of h(100)+1, all we have to do is find the LARGEST prime of h(100), because we'll know that it can't be equal t
PRIME FACTOR
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mk101
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I guess the key to solving this question is to understand two thingsearnest10 wrote:
1) The definition of h(n) = 2.4.6.8..............n where n is an even no. Please note that this function is valid only for even numbers.
There fore h (100) = 2.4.6.8..........100 = 2.(2.2)(2.3)(2.4)..........(2.50)
taking 2 common from all the above terms = 2 ^50 . (1.2.3.4.5...........50)
thus
This number is divisible by all numbers from 1 to 50
2 Than no two consecutive numbers can ever have a common factor. For example 21,22.... any number or factor that divides21, can not divide 22 or for that matter ......50 and 51 ... any number that divides 50 can not divide 51.
Since we have already proved that h(n) is already divisible by all numbers from 1 to 50, none of the numbers from 1 to 50 should divide h(n) + 1. Hence the least prime number that divides h(n) + 1 must be greater than 50...
Now take a look at the options and immediately get the answer as "e"
