• e-gmat Exclusive Offer
    Free Resources for GMAT Preparation
    Practice Questions, Videos & webinars

    Available with Beat the GMAT members only code

    MORE DETAILS
    e-gmat Exclusive Offer
  • Veritas Prep
    Free Veritas GMAT Class
    Experience Lesson 1 Live Free

    Available with Beat the GMAT members only code

    MORE DETAILS
    Veritas Prep
  • Magoosh
    Magoosh
    Study with Magoosh GMAT prep

    Available with Beat the GMAT members only code

    MORE DETAILS
    Magoosh
  • Kaplan Test Prep
    Free Practice Test & Review
    How would you score if you took the GMAT

    Available with Beat the GMAT members only code

    MORE DETAILS
    Kaplan Test Prep
  • PrepScholar GMAT
    5 Day FREE Trial
    Study Smarter, Not Harder

    Available with Beat the GMAT members only code

    MORE DETAILS
    PrepScholar GMAT
  • EMPOWERgmat Slider
    1 Hour Free
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    EMPOWERgmat Slider
  • Target Test Prep
    5-Day Free Trial
    5-day free, full-access trial TTP Quant

    Available with Beat the GMAT members only code

    MORE DETAILS
    Target Test Prep
  • Economist Test Prep
    Free Trial & Practice Exam
    BEAT THE GMAT EXCLUSIVE

    Available with Beat the GMAT members only code

    MORE DETAILS
    Economist Test Prep

2 quant questions from GMATPrep

This topic has 5 member replies
BeatTheQwerty Newbie | Next Rank: 10 Posts Default Avatar
Joined
06 May 2007
Posted:
4 messages

2 quant questions from GMATPrep

Post Fri Jun 29, 2007 9:38 am
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    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?

    Need free GMAT or MBA advice from an expert? Register for Beat The GMAT now and post your question in these forums!
    drhomler Master | Next Rank: 500 Posts Default Avatar
    Joined
    03 May 2007
    Posted:
    141 messages
    Post 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.
    Answer D

    Please correct me if I am wrong.

    givemeanid Master | Next Rank: 500 Posts
    Joined
    17 Jun 2007
    Posted:
    277 messages
    Followed by:
    1 members
    Thanked:
    6 times
    Post 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.

    Answer is (B)

    givemeanid Master | Next Rank: 500 Posts
    Joined
    17 Jun 2007
    Posted:
    277 messages
    Followed by:
    1 members
    Thanked:
    6 times
    Post 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).

    drhomler Master | Next Rank: 500 Posts Default Avatar
    Joined
    03 May 2007
    Posted:
    141 messages
    Post 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.

    givemeanid Master | Next Rank: 500 Posts
    Joined
    17 Jun 2007
    Posted:
    277 messages
    Followed by:
    1 members
    Thanked:
    6 times
    Post Fri Jun 29, 2007 11:54 am
    Good luck with your test tomorrow drhomler. Go kick some GMAT butt!

    Best Conversation Starters

    1 Vincen 181 topics
    2 lionsshare 46 topics
    3 Roland2rule 45 topics
    4 lheiannie07 44 topics
    5 ardz24 32 topics
    See More Top Beat The GMAT Members...

    Most Active Experts

    1 image description Brent@GMATPrepNow

    GMAT Prep Now Teacher

    128 posts
    2 image description Rich.C@EMPOWERgma...

    EMPOWERgmat

    116 posts
    3 image description Jay@ManhattanReview

    Manhattan Review

    105 posts
    4 image description GMATGuruNY

    The Princeton Review Teacher

    99 posts
    5 image description DavidG@VeritasPrep

    Veritas Prep

    86 posts
    See More Top Beat The GMAT Experts