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If each of the three nonzero numbers a, b, and c is divisibl

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If each of the three nonzero numbers a, b, and c is divisibl [#permalink] New post 21 Apr 2018, 04:58
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If each of the three nonzero numbers a, b, and c is divisible by 3, then abc must be divisible by which one of the following the numbers?

(A) 8
(B) 27
(C) 81
(D) 121
(E) 159
[Reveal] Spoiler: OA

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Re: If each of the three nonzero numbers a, b, and c is divisibl [#permalink] New post 25 Apr 2018, 01:53
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\(A,B, and C\) are non zero integer divisible by \(3\)so at minimum \(a=3,b=3,c=3\)
therefore abc at minimum \(= 27\) and abc should be a multiple of \(27\)
option B
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Re: If each of the three nonzero numbers a, b, and c is divisibl [#permalink] New post 07 Apr 2019, 17:45
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Carcass wrote:
If each of the three nonzero numbers a, b, and c is divisible by 3, then abc must be divisible by which one of the following the numbers?

(A) 8
(B) 27
(C) 81
(D) 121
(E) 159



We see that a = 3r, b = 3s and c = 3t for some nonzero integers r, s, and t. Therefore, abc = (3r)(3s)(3t) = 27rst. We see that abc must be divisible by 27.

Answer: B
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Re: If each of the three nonzero numbers a, b, and c is divisibl   [#permalink] 07 Apr 2019, 17:45
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If each of the three nonzero numbers a, b, and c is divisibl

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