Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 350,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

At a dinner party, 5 people are to be seated around a [#permalink]
17 Mar 2010, 06:01

00:00

A

B

C

D

E

Difficulty:

(N/A)

Question Stats:

50% (01:16) correct
50% (00:00) wrong based on 1 sessions

1) At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the position of the people are different relative to each other. What is the total number of different possible arrangements possible?

A: - 5/10/24/32/120

2) For every positive 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 lies between

A:- 2 and 10/ 10 and 20/ 20 and 30/ 30 and 40/ > 40

3) If n is a multiple of 5 and n=p*p*q where p,q prime, which of the following is multiple of 25,

Re: GMATPrep Problem Solving Questions [#permalink]
17 Mar 2010, 07:54

5

This post received KUDOS

Expert's post

ahirjoy wrote:

1) At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the position of the people are different relative to each other. What is the total number of different possible arrangements possible?

A: - 5/10/24/32/120

2) For every positive 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 lies between

A:- 2 and 10/ 10 and 20/ 20 and 30/ 30 and 40/ > 40

3) If n is a multiple of 5 and n=p*p*q where p,q prime, which of the following is multiple of 25,

A:- p*p/ q*q/ p*q/ (p*p)*(q*q)/ (p*p*p)*q

1. At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the position of the people are different relative to each other. What is the total number of different possible arrangements possible?

This is the case of circular arrangement. The number of arrangements of n distinct objects in a row is given by \(n!\). The number of arrangements of n distinct objects in a circle is given by \((n-1)!\).

From Gmat Club Math Book (combinatorics chapter): "The difference between placement in a row and that in a circle is following: if we shift all object by one position, we will get different arrangement in a row but the same relative arrangement in a circle. So, for the number of circular arrangements of n objects we have:

\(R = \frac{n!}{n} = (n-1)!\)"

\((n-1)!=(5-1)!=24\)

Answer: C.

2. For every positive 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 lies between

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 numbers from 1 to 50 as its factors, according to above \(h(100)+1=2^{50}*50!+1\) won't have ANY factor from 1 to 50. Hence \(p\) (\(>1\)), the smallest factor of \(h(100)+1\) will be more than 50.

Answer: E.

3. If \(n\) is multiple of \(5\), and \(n = p^2q\) where \(p\) and \(q\) are prime, which of the following must be a multiple of \(25\)?

A \(p^2\) B. \(q^2\) C. \(pq\) D. \(p^2q^2\) E. \(p^3q\)

\(n=5k\) and \(n=p^2p\), (\(p\) and \(q\) are primes). Q: \(25m=?\)

Well obviously either \(p\) or \(q\) is \(5\). As we are asked to determine which choice MUST be multiple of \(25\), right answer choice must have BOTH, \(p\) and \(q\) in power of 2 or higher to guarantee the divisibility by \(25\). Only D offers this.

Answer: D.

Hope it helps.

P.S. Please post one question per post. _________________

Re: GMATPrep Problem Solving Questions [#permalink]
19 Mar 2010, 11:57

ahirjoy wrote:

1) At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are considered different only when the position of the people are different relative to each other. What is the total number of different possible arrangements possible?

A: - 5/10/24/32/120

2) For every positive 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 lies between

A:- 2 and 10/ 10 and 20/ 20 and 30/ 30 and 40/ > 40

3) If n is a multiple of 5 and n=p*p*q where p,q prime, which of the following is multiple of 25,

A:- p*p/ q*q/ p*q/ (p*p)*(q*q)/ (p*p*p)*q

Q1. no. of arrangements = (5-1)! = 24 Q2. h(100)+1 = 2^50 * 50! +1 due to coprime rule factor will be >50 or >40 q3. n = p^2*q => q is the multiple of 5 or p is the multiple of 5 or both. only P8P*q*q satisy the condition since both p and q are are sqaured here so, in any case it will be multiple of 25. _________________

Re: GMATPrep Problem Solving Questions [#permalink]
14 Aug 2010, 09:00

2. For every positive 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 lies between

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 numbers from 1 to 50 as its factors, according to above \(h(100)+1=2^{50}*50!+1\) won't have ANY factor from 1 to 50. Hence \(p\) (\(>1\)), the smallest factor of \(h(100)+1\) will be more than 50.

Bunuel, As I read this I wanted to ask you this. If a number is composite as in this h(100) and you correctly point out that h(100)+1 is coprime with h(100), can one however then deduce that h(100)+1 is prime or composite itself? I guess we do not have consecutive prime numbers, so if X is composite, then X+1 can be prime (10, 11 for example). For coprimes both the numbers can be primes or composites or 1 prime 1 composite 3 5 - both primes 3 77 - 1 prime 1 composite 15 77 - both composites So my question then is how do we know for sure that h(100)+1 is composite and hence has a factor greater than 1? I see that the question is asking for the smallest prime factor of h(100)+1, how do we know it is not itself? Perhaps a dumb question, would appreciate your thoughts. _________________

Re: GMATPrep Problem Solving Questions [#permalink]
14 Aug 2010, 09:32

Expert's post

mainhoon wrote:

So my question then is how do we know for sure that h(100)+1 is composite and hence has a factor greater than 1? I see that the question is asking for the smallest prime factor of h(100)+1, how do we know it is not itself? Perhaps a dumb question, would appreciate your thoughts.

Actually we don't and don't even care. \(h(100)+1\) might be a prime number and then \(p\) will be equal to \(h(100)+1\) itself (number is factor of itself). But in any case we know that \(50<p\leq{h(100)+1}\) (the smallest factor of \(h(100)+1\) will be more than 50 and less than or equal to \(h(100)+1\)).

Re: GMATPrep Problem Solving Questions [#permalink]
14 Aug 2010, 09:42

Yes. Got your point. Just looking at the question, my first reaction was that well it has a factor > 50 so this is most likely a composite number. Same solution holds even if it were prime. If GMAT had perhaps if they had thrown in a twist as to whether this number is prime or not, it might have confused me! Is that beyong GMAT's scope then to infer as to h(100)+1 primality? Thanks _________________

Type of Visa: You will be applying for a Non-Immigrant F-1 (Student) US Visa. Applying for a Visa: Create an account on: https://cgifederal.secure.force.com/?language=Englishcountry=India Complete...

I started running back in 2005. I finally conquered what seemed impossible. Not sure when I would be able to do full marathon, but this will do for now...