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The number of distinct prime factors of 100,000
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12 Aug 2018, 16:01
12 Aug 2018, 16:01
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Expert Reply
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Question Stats:
76% (00:45) correct
23% (00:35) wrong based on 150 sessions
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Quantity A |
Quantity B |
The number of distinct prime factors of 100,000 |
The number of distinct prime factors of 99,000 |
A)The quantity in Column A is greater.
B)The quantity in Column B is greater.
C)The two quantities are equal.
D)The relationship cannot be determined from the information given.
Re: The number of distinct prime factors of 100,000
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15 Aug 2018, 23:22
15 Aug 2018, 23:22
Quantity A gives 2 distinct factors 2 and 5.
Quantity B gives 4 distinct factors 2, 3, 5, 11.
Hence B is greater.
Option B is the answer.
Quantity B gives 4 distinct factors 2, 3, 5, 11.
Hence B is greater.
Option B is the answer.
Re: The number of distinct prime factors of 100,000
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17 Aug 2018, 16:10
17 Aug 2018, 16:10
2
Expert Reply
Explanation
Since only the number of distinct prime factors matter, not what they are or how many times they are present, it is possible to tell on sight that Quantity A has only 2 distinct prime factors, because 100,000 is a power of 10. (Any prime tree for 10, 100, or 1,000, etc. will contain only the prime factors 2 and 5, occurring in pairs.)
In Quantity B, 99,000 breaks down as 99 × 1,000. Since 1,000 also contains 2’s and 5’s, and 99 contains even more factors (specifically 3, 3, and 11), Quantity B is greater. It is not necessary to make prime factor trees for each number.
Since only the number of distinct prime factors matter, not what they are or how many times they are present, it is possible to tell on sight that Quantity A has only 2 distinct prime factors, because 100,000 is a power of 10. (Any prime tree for 10, 100, or 1,000, etc. will contain only the prime factors 2 and 5, occurring in pairs.)
In Quantity B, 99,000 breaks down as 99 × 1,000. Since 1,000 also contains 2’s and 5’s, and 99 contains even more factors (specifically 3, 3, and 11), Quantity B is greater. It is not necessary to make prime factor trees for each number.
