For every positive even integer n, the function h(n)

[RonPurewal]

so does this mean...that the next factor is greater than 50 ?

still not getting this :(

i'm not quite sure what you mean by "the next factor". could you explain?

[gkhan]

Ron,

Can your or someone from the MGMAT staff, point me to a good source to learn functions? I am struggling with every single middle to upper level function question there is, I have also implemented your strategy and I still seem to get confused by plugging in numbers for a or b for f(x).

Thanks,

Shaun

[RonPurewal]

Ron,

Can your or someone from the MGMAT staff, point me to a good source to learn functions? I am struggling with every single middle to upper level function question there is, I have also implemented your strategy and I still seem to get confused by plugging in numbers for a or b for f(x).

Thanks,

Shaun

if you are looking to "learn functions" in general, then you might want to try one of the following two more general resources:
(1) an algebra book, perhaps from the local public library;
(2) web searches for terms such as "function notation".

you may want to try the latter of these first, since you can execute the search without having to go chase down books. if you fail to find something sufficiently tailored to your level, then go find a book.

[matva88]

i like to explain problems step-by-step in my own words to help me understand them. Can someone tell me if this explanation is correct?
this is the way my thought process brings me through the problem. pretty bad, i know, but i'm just starting preparation.

Steps:
(1) h(100)=2*4*6*8...*100 <--- easy first step though the way they describe this function is kind of confusing.

(2) Recognize that there are 50 even #s between 2 and 100 so this pattern will go on 50 times until you get to 100.

(3) At this point you still have no clue what h(100) will equal, since it is too long to multiply out. Looking at the answer choices, you realize you don't need an exact #.

(4) Try to manipulate the #s to find patterns/reveal something.
Try factoring out:
h(100)=2*4*6..=(2*1)*(2*2)*(2*3) ...
Right there you'll notice a pattern that continues as (2*1)*(2*2)*(2*3)*(2*4)... 50 times which, according to the pattern and step 2, will end in (2*50). this is essentially h(100) factored out, which means that every # 1-50 is a factor of h(100). every even # 2-100 will also be a factor of h(100), but we're not concerned with evens, only primes.

(5) if every #1-50 is a factor of h(100) then every # 1-50 can't be a factor of h(100)+1 which means the factor we're looking for must be greater than 50.

it is testing your knowledge of primes. The way to find primes is to factor out! so that's what you gotta do. Once you start factoring out you'll pick up the pattern easy since its 1,2,3,4,5.
It is also testing your ability to understand functions, but to a lesser extent. I kind of wish they made the function clearer cause it left me unsure, and the problem is hard enough even when you know what the function is.

[tejkumar.m]

Hi Manhattan Staff,

That was a good explanation by you. Thanks, it helped me understand..

[akhp77]

h(100) + 1 = (2 x 4 x ...100 + 1) = (2^50) * (1 x 2 x...50) +1


Now if we divide this number by any number lesser than 50 you will get a reminder of 1, you should always use a number greater than 50 to completely divide this number, hence the least prime factor will have to be greater than 50 to satisfy the condition.

[StaceyKoprince]

The smallest prime factor has to be greater than 50, yes.

[maaz_gmat]

can we have a different explanation on this one. It seems simple, but I still just don't get it..

I ain't sleepy, just stupid.

[tanishsr1013]

Hi all,

Good question and nice explanation.


Can I generalize the following quote (from Ron reply) for all numbers?

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.

Let say, if A is divisible by B, that is A/B = I, where I is an integer. The next number divisible by B has to A+B, otherwise any number between A and A+B will yield 1 to B-1 as remainder when divided by B.

-regards,
Tanish S.

[RonPurewal]

so does this mean...that the next factor is greater than 50 ?

still not getting this :(

we proved that there aren't any factors that are below 50 (other than 1, which is not a prime number). there's no such thing as the "next" factor -- we just ruled out everything 50 and below, so we're left with factors that are greater than 50.

[varun.pendharkar]

I understand this problem but I do not understand why the correct answer choice 'E' says greater than 40. Should it not be greater than 50 instead?

[RonPurewal]

I understand this problem but I do not understand why the correct answer choice 'E' says greater than 40. Should it not be greater than 50 instead?

all numbers that are greater than 50 are also greater than 40.

analogy:
joe is 46 and his wife is 41.
are they both over 40? --> yes
are they both over 25? --> also yes
same thing here.

[varun.pendharkar]

Hi ,

I agree with the analogy. But greater than 40 also includes numbers that are not greater than 50 and those numbers will not satisfy the condition.
Hence the question regarding choice E.

It is saying that if eqn X satisfies x> 15 it also satisfies x>10. I think that is incorrect assumption.

Please let me know your thoughts

[jnelson0612]

Hi ,

I agree with the analogy. But greater than 40 also includes numbers that are not greater than 50 and those numbers will not satisfy the condition.
Hence the question regarding choice E.

It is saying that if eqn X satisfies x> 15 it also satisfies x>10. I think that is incorrect assumption.

Please let me know your thoughts

Again, if I have values that are above 50, such as 51, 52, 53, 54, etc., each of those answers also MUST be above 40. Numbers between 40 and 50 are not relevant here and should not be considered; the correct p must be above 50, as we have learned, so it must also be above 40. The GMAT is adding on an additional layer of difficulty by not giving an answer choice of p>50.

[stephen.fotiu]

Halo 3,


For example, say we have the number 20......1, 2, 4, 5, 10 are all factors of 20. Adding 1 to 20 gives 21. Those factors are not factors of 21. There is a remainder of one. 2 was the least prime factor, but 3 is now the least prime factor for 21.

A number as large as 50x2x50! Includes way more prime factors than 20 does. So once 1 is added to that number, it cancels out all of those prime factors, because those prime factors now result in a remainder of one when divided into the number 50x2x50!+1. The least prime factor is going to be larger than any prime factor of 50x2x50!. Definitely greater than 47, so the answer to the question would be E, greater than 40.