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13 Dec 2016, 10:55
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
35%
(medium)
Question Stats:
68% (01:46) correct
32%
(01:42)
wrong
based on 273
sessions
History
Date
Time
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Not Attempted Yet
If a is a multiple of 3 and a = y*z^2, where y and z are prime numbers, which of the following must be a multiple of 9 ?
A. y^2
B. z^2
C. yz
D. y^2*z^2
E. y^3*z
A. y^2
B. z^2
C. yz
D. y^2*z^2
E. y^3*z
13 Dec 2016, 12:16
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Bunuel
Given a = y*z^2 and a is a multiple of 3
There can be 3 possibilities -
1. Either y = 3k
2. Or z = 3k
3. Both y & z are multiple of 3k
(A) Not possible if z is a multiple of 3 and y is not a multiple of 3
(B) Not possible if y is a multiple of 3 and z is not a multiple of 3
(C) Not possible if either y or z is a multiple of 3 alone
(D) Possible, if y = miltiple y^2 = Miltiple of 9 & if if z = miltiple z^2 = Miltiple of 9
(E) Not possible if y is a multiple of 3 and z is not a multiple of 3
Hence, correct answer will definitely be (D)
15 Dec 2016, 05:38
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"a" is a multiple of 3, so 3 is a factor of "a". Given "a" = yz^2 then it's a prime factorization of "a". Taken that 3 is a prime number and it's a factor of "a" then 3 should be either y or z.Taken that prime factorization of 9 = 3^2, so to be the multiple of 9 "a" should contain at least 3^2, therefore it should be either y^2*z or y*z^2 or y^2*z^2, only answer "D" conforms to the basic terms.
