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31 Jan 2022, 02:30
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A
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Difficulty:
95%
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Question Stats:
43% (02:34) correct
57%
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wrong
based on 82
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What is the number of positive integers that divide k^2 but do not divide k, where k is a positive integer?
(1) k^2 has a total of 13 factors
(2) √k has a total of 4 factors
Are You Up For the Challenge: 700 Level Questions
(1) k^2 has a total of 13 factors
(2) √k has a total of 4 factors
Are You Up For the Challenge: 700 Level Questions
31 Jan 2022, 03:18
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A positive integer M that divides k^2 but does not divide k must satisfy
M = k * f where f is a positive factor (other than 1）of k.
(1) Only
A factor of 13 can only be obtained if k^2 = P^12 where P is a prime number.
So, k = P^6
The following six numbers can divide K^2 but not k: P^7, P^8, P^9, P^10, P^11, P^12.
Sufficient.
(2) Only
For √k to have a total of 4 factors, it is possible that √k = P^3 where P is a prime number, which is the same case as in (1). There should be six number.
It is, however, also possible to have a total of 4 factors if √k = P * Q where P and Q are two different prime numbers.
In this case, k=P^2*Q^2, k^2=P^4*Q^4.
The following eight numbers can divide K^2 but not k: P^2*Q^3, P^2*Q^4, P^3*Q^2, P^3*Q^3, P^3*Q^4, P^4*Q^2, P^4*Q^3, P^4*Q4.
(The eight can be calculated via 3*3-1 because we can select one from P^2, P^3, P^4 and another one from Q^2, Q^3, Q^4. Then we remove P^2*Q^2)
Not sufficient.
The answer is (A).
M = k * f where f is a positive factor (other than 1）of k.
(1) Only
A factor of 13 can only be obtained if k^2 = P^12 where P is a prime number.
So, k = P^6
The following six numbers can divide K^2 but not k: P^7, P^8, P^9, P^10, P^11, P^12.
Sufficient.
(2) Only
For √k to have a total of 4 factors, it is possible that √k = P^3 where P is a prime number, which is the same case as in (1). There should be six number.
It is, however, also possible to have a total of 4 factors if √k = P * Q where P and Q are two different prime numbers.
In this case, k=P^2*Q^2, k^2=P^4*Q^4.
The following eight numbers can divide K^2 but not k: P^2*Q^3, P^2*Q^4, P^3*Q^2, P^3*Q^3, P^3*Q^4, P^4*Q^2, P^4*Q^3, P^4*Q4.
(The eight can be calculated via 3*3-1 because we can select one from P^2, P^3, P^4 and another one from Q^2, Q^3, Q^4. Then we remove P^2*Q^2)
Not sufficient.
The answer is (A).
General Discussion
20 Feb 2022, 11:28
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zhanbo
Hey I didnt get your point can you elaborate a little bit more regarding the statement 13 factors means prime and possibility of srtk to be prime.
Thanks in advance/.
