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27 Sep 2003, 08:24
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How many divisors of 1728 are also mutiples of 6.
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27 Sep 2003, 08:49
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1728=[2^6][3^3]
(I) they should have at least one 2 and at least one 3.
therefore, 6*3=18, since the question does not say whether divisors are positive, we have to allow for negative divisors. 18*2=36
(II) or another way
(1) [3^1][2, or 2^2, or 2^3 ... 2^6] 6 numbers
(2) [3^2][2, or 2^2, or 2^3 ... 2^6] 6 numbers
(3) [3^3][2, or 2^2, or 2^3 ... 2^6] 6 numbers
total 18 numbers. With negatives:36
(III) or yet another way
1728=[2^6][3^3], there are 7*4=28 positive divisors overall. The divisors that are not multiples of 6 are: 1, 2, 4, 8, 16, 32, 64, 3, 9, and 27. 28-10=18. 18*2=36
(I) they should have at least one 2 and at least one 3.
therefore, 6*3=18, since the question does not say whether divisors are positive, we have to allow for negative divisors. 18*2=36
(II) or another way
(1) [3^1][2, or 2^2, or 2^3 ... 2^6] 6 numbers
(2) [3^2][2, or 2^2, or 2^3 ... 2^6] 6 numbers
(3) [3^3][2, or 2^2, or 2^3 ... 2^6] 6 numbers
total 18 numbers. With negatives:36
(III) or yet another way
1728=[2^6][3^3], there are 7*4=28 positive divisors overall. The divisors that are not multiples of 6 are: 1, 2, 4, 8, 16, 32, 64, 3, 9, and 27. 28-10=18. 18*2=36
Archived Topic
Hi there,
This topic has been closed and archived due to inactivity or violation of community quality standards. No more replies are possible here.
Where to now? Join ongoing discussions on thousands of quality questions in our Quantitative Questions Forum
Still interested in this question? Check out the "Best Topics" block above for a better discussion on this exact question, as well as several more related questions.
Thank you for understanding, and happy exploring!
