Hi,
h(100) = 2*4*...100 = 2^50(50!)
So, h(100)+1 = 2^50.(50!)+1
As all numbers from 2 to 52 are factors of 50!, this number(h(100)) definitely leaves remainder 1 when it is divided by all numbers upto 52.
So, p > 52.
Such questions are better, with their options mentioned as well.

Hi GMAT Experts,

Please help.

Thanks.

Hey Divya,

I love this question - thanks so much for sharing it. This is one of those questions that just looks awful until you get a feel for what they're testing. In no way, shape, or form should you ever know the actual number created by multiplying 2*4*6*8*.....96*98*100. It's a ridiculously-large number that NASA engineers would round to only a handful of significant digits. You need to recognize that the GMAT isn't testing whether you know this number, but rather "what this number is made of".

One of the best ways to attack a question in which numbers like this are insanely large is to look for patterns, and to do so with smaller numbers so that you can learn why those patterns work. Here, let's skip the large number for a second once we've determined what it really is.

We know that it's 2*4*6*....*98*100. Well, if we want to express it mathematically, and we know that they're asking about prime factors, then one way to simplify it is to note that each number in that set has a prime factor of 2 (they're all even). So if we factor those out, we're left with a 2 for each of the 50 numbers, or 2^50, and then 1*2*3*.....*49*50. So we can call this number:

2^50 * 50!

Now, again, that's an insane number. But if we're looking for prime factors and the 2 is really repetitive, then that 50! becomes more important - there are a lot more unique factors there. So let's focus on that for now, and let's use a factorial that's easier to process: let's just say 3!.

3! has prime factors of 2 and 3. And if we add one to 3!, we get 7, which doesn't have any of the same prime factors in common. Why not? Well, every second number is divisible by 2:

2 4 6 8 10 12 14 16

And every THIRD number is divisible by 3:

3 6 9 12 15 18

And if you add one to any of those numbers, you break that cycle of "every second" number and "every third" number. You're now looking at a number that has none of the same prime factors as the previous number. Add one to any number that's even and the new number is no longer even. Add one to any number that's divisible by 3 and the new number is no longer divisible by 3. So with any number that's a multiple of 3!:

6 12 18 24 30

If you add one:

7 13 19 31 37

The new numbers are not divisible by either 2 or 3. In fact, these first few all happen to be prime (which won't always hold; 54 + 1 = 55, which has a prime factor of 5...but note that it's definitely not even or divisible by 3).



So...if you follow the logic, if we take a factorial like 50! and add one to it, we can prove that it will no longer be divisible by any of factors between 2 and 50. We know it's odd, it's not divisible by 3, it's not divisible by 5, etc. So to answer this question, the lowest prime factor that this new number can have is 53.



I've seen a few different varieties of this same question and I love them all - it's great evidence that the GMAT wants to test your ability to reason through problems that you can't solve by just performing algebra or "crunching numbers". Especially in practice, I think the best way to train yourself to do that is to look for patterns with small numbers that fit the style of the larger numbers that they test, and then ask yourself why those patterns hold. The hardest questions you'll see on test day will force you to find a unique framework to apply to a huge number, so in practice you should train yourself to create and apply those frameworks.

Here is a video solution to this difficult GMAT problem:

https://www.gmatquantum.com/shared-posts ... lem13.html

Dabral