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At a dinner party, 5 people are to be seated around a

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17 Mar 2010, 06:01
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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
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17 Mar 2010, 07:54
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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:

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

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

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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.

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.

Hope it helps.

P.S. Please post one question per post.
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17 Mar 2010, 12:20
I am only trying 3) because I've already worked on 1 and 2:

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

by plugging numbers:
if n=50=p*p*q=5*5*2. therefore p=5 & q=2, so a) is the only one which will be divisible by 25
n= 75 (same method).
IMO: A
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17 Mar 2010, 19:22
Hi amma,

Regards,
Ahirjoy
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Re: GMATPrep Problem Solving Questions [#permalink]

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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.
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Re: GMATPrep Problem Solving Questions [#permalink]

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31 Jul 2010, 10:39
Bunuel wrote:
[m]h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1

Could you explain with a little more of detail the step above?
Many thanks,
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31 Jul 2010, 18:45
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noboru wrote:
Bunuel wrote:
h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1

Could you explain with a little more of detail the step above?
Many thanks,

$$h(100)+1=2*4*6*...*100+1=(2*1)*(2*2)*(2*3)*(2*4)*...*(2*50)+1=2^{50}*(1*2*3*..*50)+1$$.

Hope it's clear.
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Re: GMATPrep Problem Solving Questions [#permalink]

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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

$$h(100)+1=2*4*6*...*100+1=2^{50}*(1*2*3*..*50)+1=2^{50}*50!+1$$

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 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.
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14 Aug 2010, 09:32
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 $$h(100)+1$$).

Hope it's clear.
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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
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14 Aug 2010, 10:01
No way you can factor such a huge number as $$h(100)+1$$ without a computer. So don't worry about it.
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Re: GMATPrep Problem Solving Questions [#permalink]

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22 Jul 2011, 13:10
thanks for discussion

Re: GMATPrep Problem Solving Questions   [#permalink] 22 Jul 2011, 13:10
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