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2 quant questions from GMATPrep

This topic has 5 member replies
BeatTheQwerty Newbie | Next Rank: 10 Posts
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2 quant questions from GMATPrep

Fri Jun 29, 2007 9:38 am
Problem 1: Data Sufficiency
If the integers a and n are greater than 1 and the product of the first eight positive integers is a multiple of a^n, what is a?
(1) a^n = 64
(2) n = 6

Problem 2: Problem Solving
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are said to be different only when the positions of people are different relative to each other. What is the total number of different possible seating arrangements for the group?

givemeanid Master | Next Rank: 500 Posts
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Fri Jun 29, 2007 11:54 am
Good luck with your test tomorrow drhomler. Go kick some GMAT butt!

drhomler Master | Next Rank: 500 Posts
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Fri Jun 29, 2007 11:42 am
Thx I didnt see that. Behold the dangers of reading too much into the 2nd clue and using that information in the first. I hope I dont do that tomorrow.

drhomler Master | Next Rank: 500 Posts
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Fri Jun 29, 2007 10:01 am
From the the question we know that A and N>1 and 8! is a multiple of a^N
8!=40,320.

Clue one tells us that a^N is 64-
40,320 divided by 64 leaves 630 which is which is equal to 2^1*5^1*7^1*3^2 so the prime factors of 40,320 are 2^7,3^2,5^1,3^2 so the only integer greater than raised to the power of another integer greater than 1 that can equal 64 and be a factor of 40,320 is 2 so the data is sufficient. Discard BCE and go with AD

Clue 2 tells us that n=6, and from the work above the only prime factor raised to a power 6 or greater is 2. so statement 2 is sufficient as well.

Please correct me if I am wrong.

givemeanid Master | Next Rank: 500 Posts
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Fri Jun 29, 2007 10:23 am
8! (40,320) is a multiple of a^n

(1) 64 = 2^6 = 4^3 = 8^2 = a^n. Not sufficient.
(2) n = 6. Factorizing, 40320 = 2^6 * 2 * 3^2 * 5^1 * 7^1. So, a=2, n=6 is the only possibility.

givemeanid Master | Next Rank: 500 Posts
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Fri Jun 29, 2007 10:46 am
Quote:
Problem 2: Problem Solving
At a dinner party, 5 people are to be seated around a circular table. Two seating arrangements are said to be different only when the positions of people are different relative to each other. What is the total number of different possible seating arrangements for the group?
If this were not a circular arrangement, then the number of different possibilities would simply be 5! = 120.

With a circular arrangement for 5, it is equivalent of choosing from 4 people without a circular arrangement which is 4! = 24.

(To imagine this, in a circular arrangement, if the first person is seated at table 1, then we have 4! = 24 different combos that other people can be seated. But when you make person 2 sit at table 1, all those arrangements are still possible with person 1 sitting at table 1 anyway).

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