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For every positive even integer n, the function h(n) is defined to be the product of all the 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. Greater than 40
1.jpg [ 49.61 KiB | Viewed 88058 times ]
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.
h(100) = 2 * 4 * 6 ****************100
Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.
100 = 2 + (n-1) 2
n = 50
Product of terms = Average * number of terms
Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.
H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
A. Between 2 and 10
B. Between 10 and 20
C. Between 20 and 30
D. Between 30 and 40
E. Greater than 40
Attachment:
1.jpg [ 49.61 KiB | Viewed 88058 times ]
This is how I am trying to solve this. Please help me if you think I am not right. OA is not provided in the book.
h(100) = 2 * 4 * 6 ****************100
Tn = a1 + (n-1) d-----------------------(1) where Tn is the last term, a1 is the first term and d is the common difference of the evenly spaced set.
100 = 2 + (n-1) 2
n = 50
Product of terms = Average * number of terms
Average = (a1+an)/2
Therefore average = 102/2 = 51
Product of the series = 51*50 = 2550.
H(100) + 1 = 2550+1 = 2551 which is prime. And prime numbers have exactly 2 factors 1 and the number itself. Therefore for me D is the answer i.e. < 10
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For every positive even integer n, the function h(n) is defined to be the product of all the 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 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
Now, two numbers \(h(100)=2^{50}*50!\) and \(h(100)+1=2^{50}*50!+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.
As \(h(100)=2^{50}*50!\) has ALL prime numbers from 1 to 50 as its factors, then, according to the above, \(h(100)+1=2^{50}*50!+1\) won't have ANY prime factor from 1 to 50. Hence \(p\) (\(>1\)), the smallest prime factor of \(h(100)+1\) must be more than 50.
Answer: E.
_________________
A. between 2 and 20
B. between 10 and 20
C. between 20 and 30
D. between 30 and 40
E. greater than 40
\(h(100)+1= \)
\( =2*4*6*...*100+1= \)
\( =2^{50}*(1*2*3*..*50)+1= \)
\(= 2^{50}*50!+1\)
\( =2*4*6*...*100+1= \)
\( =2^{50}*(1*2*3*..*50)+1= \)
\(= 2^{50}*50!+1\)
Now, two numbers \(h(100)=2^{50}*50!\) and \(h(100)+1=2^{50}*50!+1\) are consecutive integers. Two consecutive integers are co-prime, which means that they don't share ANY common factor but 1. For example 20 and 21 are consecutive integers, thus only common factor they share is 1.
As \(h(100)=2^{50}*50!\) has ALL prime numbers from 1 to 50 as its factors, then, according to the above, \(h(100)+1=2^{50}*50!+1\) won't have ANY prime factor from 1 to 50. Hence \(p\) (\(>1\)), the smallest prime factor of \(h(100)+1\) must be more than 50.
Answer: E.
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enigma123
Given that the function h(n) is defined to be the product of all the even integers from 2 to n, inclusive, so: \(h(100)=2*4*6*...*100=(2*1)*(2*2)*(2*3)*(2*4)*...*(2*50)\) --> factor out all 50 2's: \(h(100)=2^{50}*(1*2*3*..*50)=2^{50}*50!\).
Hope it's clear.
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enigma123
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.
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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
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
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General Discussion
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
Bunuel
That is an impressive solution. <-- understatement
Stiv
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
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mehulsayani
Joined: 03 Jul 2012
Last visit: 26 Mar 2015
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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?
(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?
mehulsayani
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.
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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.
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.
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.
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).
_________________
=> 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).
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itsmeabhi99
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
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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!
_________________
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!
_________________
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Schedule a free consultation with us: bit (dot) ly (backslash) GM_AugAcd_Profile_Evaluation
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About: We are an India-based MBA admissions consulting firm. We've been helping Indian MBA applicants (students belonging to the most competitive MBA admissions pool) since 2013. Several of our students study at the top 20 schools in the US including Harvard, Booth, Kellogg, MIT, Yale, Ross, and Darden, top European schools including INSEAD, IMD, London Business School, Cambridge, and Oxford, top Canadian schools like Rotman and top Asian schools like ISB and NUS. Our students have won over USD 2.1 Million in scholarships over the last 3 admissions years.
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.
_________________
Quote:
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.
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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?
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?
Game
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.
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enigma123
Responding to a pm:
Quote:
You are comparing factors of n with factors of (n+1). You actually have to compare factors of n! with factors of n! + 1.
Say n = 4.
n! = 1*2*3*4 = 24
n! + 1 = 25
Will 25 have any factors common with n!? No, because n! has all factors from 1 to n.
n! + 1 has 5 as a factor which is larger than n.
Similarly, say n = 5
n! = 1*2*3*4*5 = 120
n! + 1 = 121
Will 121 have any factors from 1 to 5? No. All these numbers are factors of n! so they cannot be factors of n!+1.
Does this make sense?
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