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

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.

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.

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Hi amma,

I am sorry but I think your answer is wrong. Please read the above explaination by Bunuel. Correct answer is D.

Regards,
Ahirjoy

Could you explain with a little more of detail the step above?
Many thanks,
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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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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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Consider kudos, they are good for health

thanks for discussion

very helpful