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If each of the three nonzero numbers a, b, and c is divisible by 3, th

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If each of the three nonzero numbers a, b, and c is divisible by 3, th  [#permalink]

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New post 04 Apr 2019, 00:15
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A
B
C
D
E

Difficulty:

  15% (low)

Question Stats:

93% (00:46) correct 7% (01:02) wrong based on 28 sessions

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Re: If each of the three nonzero numbers a, b, and c is divisible by 3, th  [#permalink]

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New post 04 Apr 2019, 00:32
1
As a,b and c each are divisible by 3, every number contains atleast one 3.

Let a = 3x, b = 3y, c = 3z

Product a*b*c = 27*xyz.

This product must always be divisible by 27 regardless of x,y, and z.

OPTION: B
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If each of the three nonzero numbers a, b, and c is divisible by 3, th  [#permalink]

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New post 04 Apr 2019, 02:16

Solution


Given:
    • a, b, and c non-zero numbers divisible by 3.

To find:
    • abc must be divisible by which of given number.

Approach and Working:
Each a, b, and c are divisible by 3. Hence, we can assume:
    • a = 3p, where p is a non-zero number.
    • Similarly, b = 3q and c = 3r where q and r are non-zero numbers.

Therefore, abc = 3p * 3q * 3r = 27 * p * q * r.
Hence, abc is definitely divisible by 27.

Thus, the correct answer is option B.

Answer: B

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Re: If each of the three nonzero numbers a, b, and c is divisible by 3, th  [#permalink]

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New post 04 Apr 2019, 03:21
Bunuel 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


given a,b,c are all multiples of 3
so least value of a,b,c ; 27 ; considering them all to be 3
so 27 hsa to least no a,b,c must be divisible by
IMO b
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Re: If each of the three nonzero numbers a, b, and c is divisible by 3, th  [#permalink]

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New post 07 Apr 2019, 18:45
Bunuel 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 divisible by 3, th   [#permalink] 07 Apr 2019, 18:45
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