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# How many divisors of 1728 are also mutiples of 6

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Joined: 21 Aug 2003
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How many divisors of 1728 are also mutiples of 6 [#permalink]  27 Sep 2003, 07:24
How many divisors of 1728 are also mutiples of 6.
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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
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