IS OA E?

Since n = 100, then the product is:

. . . . .h(100) = 2 * 4 * 6 * ... * 100

Take out the factors of 2:

. . . . .h(100) = 250(1 * 2 * 3 * ... * 50)

Now add the 1:

. . . . .h(100) + 1 = 250(1 * 2 * 3 * ... * 50) + 1 will be odd.

we can use the technique

For example

Take 2*4*6*8*10 + 1 = 3841

Now this number is NOT divisible by any number less than 5.

Take 2*4*6*8*10*12 + 1 = 46081

Now this number is NOT divisible by any number less than 6.

agrwal12 wrote:For every even integer n, the function h(n) is defined to be the product of all 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) above 40

How to solve this problem in shortest possible way ?

Thanks,

h(100) = 2*4*6*8*10*...*100

Since h(100) is a multiple of every even integer from 1 to 100, it will also be a multiple of every odd integer from 1 to 50 (we can just divide all the even factors by 2).

So, h(100) will be a multiple of every prime < 50.

If h(100) is a multiple of every prime < 50, h(100) + 1 CANNOT be a multiple of any of those primes.

For example:

3 goes into h(100), so the next number that will be a multiple of 3 is h(100) + 3

5 goes into h(100), so the next number that will be a multiple of 5 is h(100) + 5

Therefore, the smallest prime that will be a factor of h(100) + 1 must be greater than 49: choose (e).

Thanks for such a nice explaination ..

what level of difficulty is this question ?
500-600
600-700
or 700 + ?

Hi Stuart ... as always you provide great explanations ... but I am suffering brian-block today and am having trouble getting to grasps with this:

"Therefore, the smallest prime that will be a factor of h(100) + 1 must be greater than 49"

Sorry for the basic question ... would appreciate some more clarification behind this. Thanks.

I too dont understand the explanation

'Since h(100) is a multiple of every even integer from 1 to 100, it will also be a multiple of every odd integer from 1 to 50 (we can just divide all the even factors by 2). '

help

kumard24 wrote:I too dont understand the explanation

'Since h(100) is a multiple of every even integer from 1 to 100, it will also be a multiple of every odd integer from 1 to 50 (we can just divide all the even factors by 2). '

help

Double down, I have the same problem. Can you explain this part of it?? specifically, "we can just divide all the even factors by 2", how do you get odd integrers from 1 to 50 when you divide the even numbers from 1-100, should fo example 8/2=4 which is not an odd number!

thanks

I think it should be 750 difficult level.

Math is so beautiful. Sorry guys, I am struggling with the GMAT too. Largely on the numbers properties which is why I'm here but when I think of the explanations I am just amazed by people whose brains can unravel these puzzles. It is awesome. I don't know if it is a matter of nurture vs. nature, I bet it is a bit of both but I guess beat the gmat is about the idea that nurture can work depending on how long you want to work. Sorry for the preaching but felt the need to say this. I've been at this for 3 months with hours and hours of doing these problems and after fixing one problem I'm finding other problems. My score is pretty much the same even though I understand the answers more and I've come to the conclusion that it is based on these numbers problems that the test administrators keep saying is worth 600 - 700. I solve 800 non-number properties problems easier than this. The issue is getting your head around these concepts. So I'm looking for as many sample numbers properties questions as I can get if anyone has it.

N e case, an explanation of the answer is this:

The multiples of this function IS every odd integer under 50 because every odd integer under 50 multiplied by 2 is an even integer under 100; e.g., 49*2 = 98 and 47*2=95, etc. This is what is meant by divide all the even factors by 2. 98/2 = 49; 94/2 = 47. More accurately stated, it is divide every other even number starting with 98 by 2 because the numbers in between are 2 multiplied by another even number (96/2 = 48); but that explanation would probably be confusing.

51*2= 102. 102 is greater than 100 (the question says, the function is the product of every even integer from 2 to n and we are told that n is 100); therefore 102 cannot be one of the multiples of this function and any odd number above 50 doesn't have to be considered.

Now you know that every odd number below 50 is a factor in the final product of the function. You also know that all prime numbers except for 2 are odd. So that means you can now consider what the possibility is that P, the prime number is one of those numbers under 50.

Well, we know all the prime numbers under 50 starting with 2, 3, 5, 7, 11, etc. Since the function is h(100) + 1 and we know that P is the lowest prime that divides h(100) + 1 evenly, we know that 3, 5, 7, 11 ... will not divide evenly into h(100) +1 because IF h(100) is a multiple of the prime then the next multiple of the prime will include an addition of that prime.

For example, 14 is a multiple of 7 the next multiple of 7 is 14 + 7 or 21. Similarly, 6 is a multiple of 3 the next multiple of 3 is 6 + 3 or 9. So that means that if h(100) were 14 then h(100) + 1 = 15, which is not a multiple of 7. Or if h(100) were 6 then h(100) + 1 would be 7, which is not a multiple of 3. Therefore, we know that P can't be any of the prime numbers under 50 and it must be one that is above 50.

Remember that this is a very large number - the product of every even integer from 2 - 100. There is a possibility that there is a prime above 50 that we don't want to and don't have to calculate above 40 that is a factor of this large number.

I hoped this helped. Explaining helps you learn too so this was good for me.

I just found this site that has an ebook that seems to be good for number properties questions. I have just bought the ebook and so far it seems good with the explanations but I can't say for sure yet. The one good thing about this book is it has many sample questions so at the very least you'll be able to practice, practice, practice the number properties problems which is so integral in GMAT.

Sorry forgot the site address: https://questionbank.4gmat.com/mba_prep_ ... r_systems/

On the bottom you can download and buy the ebook for $6.99. You will need to be able to download files with the DNL extension so if your computer doesn't have that capability you will need to go get a free software that does this.

One thing about the ebook. You can't print it and it is only good for the computer that you download on so you can't forward it to yourself because you can only use it when you are online. They do this to try to prevent piracy. So don't download and pay for this ebook on a computer you don't have continuous and constant access to.

I think 4GMAT also ship their materials to those outside of India. I had taken their contact classes in Chennai, India. I have a score of over 700. You guys should try their materials....