For every positive even integer n, the function h(n) is defined to be

I'm not taking credit for the following solution. Found it in another forum.

Two consecutive integers cannot be divisible by the same integer greater than 1.

So if we can prove that h(100) is divisible by every number smaller than 50, we proved that h(100)+1 is NOT divisible by any number smaller than 50 (besides 1).

2*4*6*...*98*100=2*(1*2*3*...*50)=2*(50!)
Hence, h(100) is divisible by every number smaller than 51. So the smallest prime factor of h(100)+1 is at least 53.

Here is some theory first:
Pick any two consecutive numbers. Can they both be even i.e. have 2 as a factor?
e.g. (5,6) or (101, 102) or (999, 1000) etc .. Only one number in these pairs will have 2 as a factor.
Now think: If you pick any two consecutive integers, can they both have 4 as a factor? or 7 as a factor? or 99 as a factor? No!
(For that matter, once you get one multiple of 99, you will not get another one in the next 98 numbers. The next multiple will appear when you add 99 to this multiple. e.g. you pick 99. Can 100, 101, 102... be multiples of 99? The next one is 198 so numbers from 100 to 197 are not multiples of 99.)

We can say that consecutive numbers will not have any common factor other than 1. (1 is a factor of every number.)
e.g. if N = 2*3*5*7*11,

Think consecutive numbers now:
(N - 4), (N - 3), (N - 2), (N - 1), N, (N + 1), (N + 2), (N + 3), (N + 4)...

(N + 1) and (N - 1) will not have any of 2, 3, 5, 7 and 11 as factors. (Since they are consecutive with N)
(N + 2) and (N - 2) will not have 3, 5, 7 and 11 as factors. (2 will be a factor of both these numbers since 2 is a factor of N)
(N + 3), ( N + 4), (N -3 ), (N - 4) will not have 5, 7 and 11 as factors. (But (N + 3) and (N - 3) will have 3 as a factor. (N + 4) and (N - 4) will have 2 as a factor)
and so on...

Let's get back to the original question:
2*50! = 2*1*2*3*4*5*6*7*8*9*10*11.....*50

This number has all numbers till 50 as its factors. So 2*50! + 1 cannot have all these numbers as its factor.
Therefore the smallest prime number that will be the factor of 2*50! + 1 is greater than 50.

A trick question can be "What is the smallest factor of (2*50! + 1)?"
The smallest factor will be 1.

Stiv,

Here are some other official GMAT questions that are conceptually related:

1) GMAT Prep: https://gmatclub.com/forum/does-the-inte ... 26735.html


Video explanation: https://gmatquantum.com/official-guides ... h-edition/

3) Video explanation to the original problem(For every positive even integer n, the function h(n) is....): https://gmatquantum.com/gmatprep-number ... n-defined/

Dabral

Video solution from Quant Reasoning starts at 19:52
Subscribe for more: https://www.youtube.com/QuantReasoning? ... irmation=1

Bunuel - how did you get h(100) = 2^50 * 50! ?? Sorry for been a pain.

Thanks Bunuel. It was very helpful. This was my 3rd question on the practice test, I spent a minute without even knowing where to start from, so I made a random guess and moved on. I got 13 incorrect questions out of the 37, but I managed to score 48. thanks again

That is an impressive solution. <-- understatement

Check out part i of the link given above as well. That should cover the theory that is useful for such questions. As for examples, you can search for 'factors consecutive integers' in the Quant forum and you should hit quite a few questions based on these concepts.
I got one in one of my posts: if-n-is-a-positive-integer-and-r-is-the-remainder-when-119518.html?hilit=factor%20consecutive%20integers

Hey Bunuel, I used the following logic, and got E.
(2*4*6*8.... 100) +1 is of the format 6p +1. Therefore, its a prime number and has only two factors, one and itself. So, its definitely greater than 40.

So, E

Is it correct?

Every prime number greater than 3 is of one of the two forms: (6a -1), (6a + 1)
e.g. 5 = 6*1 - 1; 7 = 6*1 + 1; 11 = 6*2 - 1; 13 = 6*2 + 1 etc

But every number of the form (6a -1) or (6a + 1) is not prime.
e.g. 25 = 6*4 + 1 i.e. it is of the form 6a+1. But 25 is not prime.

So just because a number is of the form 6a + 1, we cannot say that it is definitely prime.

Got this one on the 9th question of my GMATPrep test. I actually reached a good point on this problem.

I got to the point that the product of 2*4*6*....*100 = (2^50)*50!

At this point I lost the plot and developed a weird approach here.

Since the last number in the given h(n) was 100 so the actual value could be like xxx00. So now, h(n) + 1 would be some value like xxx01.

So the possible prime factors are effectively 3, 7, 11, 13 etc.

I simply gambled on one of these till 20 being a factor and chose A. It was wrong though. The correct answer is indeed E.

The explanations above specially the one by Bunuel where the point of co-prime numbers not having any common factors but 1 was particularly good stuff and I had not read it before. This information will surely help the next time provided I am able to fully recognize it down.

h(n) = the product of all even integers from 2 to n inclusive
=> h(100) + 1 = (2 * 4 * 6 * 8 * ....... * 100) + 1
= 2^50 * (1*2*3*4*......*50) + 1

As all the numbers from 2 to 50 are factors of 2^50 * (1*2*3.....*50), none of these can be factors of 2^50 * (1*2*3*4*......*50) + 1 (i.e. of h(100)+1)

Therefore the smallest prime factor of h(100) + 1 is greater than 50.

This corresponds to option (E).

This is a long explanation just so you can understand it fully from different angles. Most of my explanations are MUCH shorter than this! =)

Any time you have these examples where they give you a hypothetical situation with a large number like 100---recreate the question at a smaller scope.
So instead of doing h(100), try h(10).
h(10) = 2 * 4 * 6 * 8 * 10

You stop at 10, since n=10.
What's the smallest prime factor of h(10)? Well, that's easy--it's 2.
But the question is asking what is the smallest prime of h(n) + 1. So what is the smallest prime of h(10)+1?

Well, it turns out all of the factors of h(n) cannot also be factors of h(n)+1 (except 1).
This is because if you try to divide any of the factors of h(10) into h(10) +1, you'll be off by one.
So all factors (except 1) of h(10) cannot go into h(10)+1.
It's the same reason why the factors of 20 will not go into 21.


You see, h(10) = 2 * 4 * 6 * 8 * 10
h(10) = (2 * 1) * (2 * 2) * (2 * 3) * (2 * 4) * (2 * 5)
= (2^5) (1*2*3*4*5)


So the factors of h(10) include 1, 2, 3, 4, and 5.
Since factors are all even, the largest factor will be n/2, or in this case 10/2 = 5
Every number less than five is a factor. When multiplied by 2, it is one of the original factors of h(10)
The prime factors of h(10) are 2, 3, and 5.

But what are the prime factors of h(10) +1 ?
Well, none of the factors we mentioned above will be factors of h(5)+1.

Again, it's the same reason why the factors of 20 will not go into 21.

We mentioned all the numbers from 1-5. So it must be true that, anything 5 and below cannot be a factor. So it must be something bigger than 5. It must be something bigger than the halfway point of n. All numbers up to the midpoint when multiplied by two are one of the original factors of h(10). Since n =10, and we know it's not going to be any number less than 5, then it must be bigger than 5. We don't know exactly what...the smallest prime can be that very number itself (h(10)+1). If not, it is at least some number bigger than 5.


So if n=100 (as in this question), then it must be bigger than 50.


The factors we are talking about are the ones up to the halfway point of 100--which is 50.

So the factors going up to 50 will not also be factors of h(100) + 1.

The question is asking what the smallest prime is. Well, you know the smallest prime has to be greater than the midpoint of 50. So the only answer choice that encompasses that is answer E.



Hope that helps.

Zeke

This is a great question. Let me try to post a simple solution!

Let's divide the solution into two parts:

Part (1):

We need to simplify h(100) first.

h(100) = 2*4*6*8*....*98*100.

Consider that 2 = 2*1, 4 = 2*2, etc. up to 100 = 2*50.

h(100) then simplifies to \(h(100) = 2*1 * 2*2 * 2*3 *.... * 2*49 * 2*50 = 2^{50} * 1*2*3*4*... *50\) (A)

Part (2):

Consider the second part of the question. We need to find a bound for the smallest prime factor of h(100)+1. Any number p that is a factor of h(100)+1, leaves a remainder of (p-1) when it divides h(100).

remainder(h(100)/p) = p-1.

Another way of saying this is "p is a prime number that is not a divisor of h(100)" (B)

From (A):

Clearly h(100) is divisible by all prime numbers less than 50 (as h(100) is divisible by 50!). The smallest prime that is not a divisor of h(100) should therefore be greater than 50.

Looking at the choices, the most appropriate choice is:

(e) greater than 40.

Hope that Helps!

Responding to a pm:



I am not sure where you got this question since the original question here talks about h(100) + 1. Even if we do assume h(50) + 1, note that h(2) = 2
h(4) = 2*4
h(6) = 2*4*6
....
So I am not sure how you got average 26.
The correct answer here is (E) as explained in the link given in my post above.

hi

i did the same thing as karishma and when i got 2551, i thought that cannot be correct as the answer mentions above 40, so my solution must be wrong .. anyway i clicked e and it turned out right.

Bunnel & Karishma: how can u guys solve all qs so easily?

Merging similar topics. Please refer tot he discussion on page 1.

Check questions about various functions in Special Questions Directory

Operations/functions defining algebraic/arithmetic expressions
Symbols Representing Arithmetic Operation
Rounding Functions
Various Functios

Hope it helps.