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Confusing wording

This topic has 3 expert replies and 1 member reply
jenlee Newbie | Next Rank: 10 Posts Default Avatar
Joined
08 Aug 2012
Posted:
4 messages

Confusing wording

Post Tue Mar 05, 2013 3:43 pm
Elapsed Time: 00:00
  • Lap #[LAPCOUNT] ([LAPTIME])
    This question confuses me. It is from the OG 12th Ed, pg 175 #159

    How many prime numbers between 1 and 100 are
    factors of 7,150 ?
    (A) One
    (B) Two
    (C) Three
    (D) Four
    (E) Five

    It is a pretty simple question, but it asks for how many prime numbers, so I said 5 (2, 5, 5, 11, 13) and not how many DIFFERENT prime numbers (2, 5, 11, 13) which is the correct answer. Is this bad wording or should I assume next time I see a similar question that they are asking for the number of different prime numbers? Or is it something with the factors that I am missing?

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    Post Tue Mar 05, 2013 3:50 pm
    The question is properly worded.
    You're asked for the number of different primes. So, if we say that there are five different primes (2, 5, 5, 11, 13), then we're incorrectly stating that 5 and 5 are different primes, when they are not.

    Cheers,
    Brent

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    jenlee Newbie | Next Rank: 10 Posts Default Avatar
    Joined
    08 Aug 2012
    Posted:
    4 messages
    Post Tue Mar 05, 2013 3:52 pm
    Thank you for your answer! So, do they ask for the number of different primes since the word "factors" are in the question?

    Post Tue Mar 05, 2013 4:00 pm
    Sorry, I just re-read the original question and realized that the word "different" does not appear - my bad. That said, the answer is still the same.

    If you're asked to find the number of "things" you can only count each thing once.
    For example, if Al, Bob and Cam are in a room and we're asked to determine how many people are in the room, we can't answer 4: Al, Bob, Cam and Cam.
    So, even if the question doesn't include the word "different," it is implied that we can't count the same number twice.

    Cheers,
    Brent

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    Post Wed Jul 08, 2015 9:15 am
    jenlee wrote:
    This question confuses me. It is from the OG 12th Ed, pg 175 #159

    How many prime numbers between 1 and 100 are
    factors of 7,150 ?
    (A) One
    (B) Two
    (C) Three
    (D) Four
    (E) Five

    It is a pretty simple question, but it asks for how many prime numbers, so I said 5 (2, 5, 5, 11, 13) and not how many DIFFERENT prime numbers (2, 5, 11, 13) which is the correct answer. Is this bad wording or should I assume next time I see a similar question that they are asking for the number of different prime numbers? Or is it something with the factors that I am missing?
    Solution:

    We start by prime factoring 7,150.

    7,150 = 715 x 10 = 143 x 5 x 10

    At this point we must be careful. Don’t incorrectly conclude that 143 is prime. Using our divisibility rules, we can determine that 143 is divisible by 11. A number is divisible by 11 if the sum of the odd-numbered place digits minus the sum of the even-numbered place digits is divisible by 11. We can test 143 to prove this:

    1 + 3 - 4 = 4 - 4 = 0

    Since zero is divisible by 11, we know that 143 is divisible by 11. We can now finish the prime factorization.

    143 x 5 x 10 = 11 x 13 x 5 x 5 x 2

    11 x 13 x 5^2 x 2

    Thus we can see that there are 4 different prime factors of 7,150.

    Answer: D

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