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04 Apr 2019, 00:15
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B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
5%
(low)
Question Stats:
90% (00:53) correct
10%
(00:57)
wrong
based on 73
sessions
History
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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
(A) 8
(B) 27
(C) 81
(D) 121
(E) 159
eswarchethu135
GMAT 2: 640 Q49 V27
Posts: 276
04 Apr 2019, 00:32
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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
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
04 Apr 2019, 02:16
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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
