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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
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 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

_________________
Jeffrey Miller Head of GMAT Instruction

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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

_________________
Brent Hanneson – Founder of GMATPrepNow.com
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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

_________________
Jeffrey Miller Head of GMAT Instruction

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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

_________________
Brent Hanneson – Founder of GMATPrepNow.com
Use our video course along with Beat The GMAT's free 60-Day Study Guide

Check out the online reviews of our course
Come see all of our free resources

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