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16 Sep 2014, 01:01
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If \(n\) is a positive integer, is 9 a factor of \(n\) ?
(1) 18 is a factor of \(n^2\)
(2) 27 is a factor of \(n^3\)
(1) 18 is a factor of \(n^2\)
(2) 27 is a factor of \(n^3\)
16 Sep 2014, 01:01
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Official Solution:
If \(n\) is a positive integer, is 9 a factor of \(n\) ?
Given: \(n\) is a positive integer. The question is: can \(n\) be expressed as \(n=9k\), where \(k\) is a positive integer?
(1) 18 is a factor of \(n^2\).
From the given information, \(n^2 = 18p\), where \(p\) is a positive integer. This means \(n = 3\sqrt{2p}\). Since \(n\) must be an integer, \(\sqrt{2p}\) must also be an integer, specifically one with even values. Therefore, \(n = 3\sqrt{2p}\) can take values that are multiples of \(3*2 = 6\), such as 6, 12, 18, etc. Based on this, \(n\) might or might not be a multiple of 9. Not sufficient.
(2) 27 is a factor of \(n^3\).
From the given information, \(n^3 = 27q\), where \(q\) is a positive integer. This means \(n = 3\sqrt[3]{q}\). Since \(n\) is an integer, \(\sqrt[3]{q}\) must also be an integer. Therefore, \(n = 3\sqrt[3]{q}\) can take values that are multiples of 3, such as 3, 6, 9, 12, etc. Thus, \(n\) might or might not be a multiple of 9. Not sufficient.
(1)+(2) From (1), we know that \(n\) is a positive multiple of 6. From (2), we know that \(n\) is a positive multiple of 3. Combining these pieces of information, we only know that \(n\) is a positive multiple of 6, and thus \(n\) could be 6, 12, 18, etc. This means that \(n\) might or might not be a multiple of 9. Not sufficient.
Answer: E
If \(n\) is a positive integer, is 9 a factor of \(n\) ?
Given: \(n\) is a positive integer. The question is: can \(n\) be expressed as \(n=9k\), where \(k\) is a positive integer?
(1) 18 is a factor of \(n^2\).
From the given information, \(n^2 = 18p\), where \(p\) is a positive integer. This means \(n = 3\sqrt{2p}\). Since \(n\) must be an integer, \(\sqrt{2p}\) must also be an integer, specifically one with even values. Therefore, \(n = 3\sqrt{2p}\) can take values that are multiples of \(3*2 = 6\), such as 6, 12, 18, etc. Based on this, \(n\) might or might not be a multiple of 9. Not sufficient.
(2) 27 is a factor of \(n^3\).
From the given information, \(n^3 = 27q\), where \(q\) is a positive integer. This means \(n = 3\sqrt[3]{q}\). Since \(n\) is an integer, \(\sqrt[3]{q}\) must also be an integer. Therefore, \(n = 3\sqrt[3]{q}\) can take values that are multiples of 3, such as 3, 6, 9, 12, etc. Thus, \(n\) might or might not be a multiple of 9. Not sufficient.
(1)+(2) From (1), we know that \(n\) is a positive multiple of 6. From (2), we know that \(n\) is a positive multiple of 3. Combining these pieces of information, we only know that \(n\) is a positive multiple of 6, and thus \(n\) could be 6, 12, 18, etc. This means that \(n\) might or might not be a multiple of 9. Not sufficient.
Answer: E
General Discussion
06 Dec 2014, 17:09
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I used the method described in books from MGMAT (prime box)
for N to be divisible by 9, it must contain in its prime box two of 3's
1) 18 is a factor of N^2
prime factorization of 18 = 2 * 3^2
n^2 = n * n
which means that one n, in its prime box, will take a two, and a three, since we do not know whether n might contain another 3, statement 1 is insufficient.
2) same method used, we have that n has only one 3, it might have more than one, but we do not know, hence the statement is insufficient.
1+2 -> we know for sure that n has one three in its prime box, but it might have more than one, therefore 1+2 insufficient.
for N to be divisible by 9, it must contain in its prime box two of 3's
1) 18 is a factor of N^2
prime factorization of 18 = 2 * 3^2
n^2 = n * n
which means that one n, in its prime box, will take a two, and a three, since we do not know whether n might contain another 3, statement 1 is insufficient.
2) same method used, we have that n has only one 3, it might have more than one, but we do not know, hence the statement is insufficient.
1+2 -> we know for sure that n has one three in its prime box, but it might have more than one, therefore 1+2 insufficient.
