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If x and y are integers and xy = 660, x or y must be divisible by whic

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If x and y are integers and xy = 660, x or y must be divisible by whic  [#permalink]

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New post 19 Apr 2016, 03:04
00:00
A
B
C
D
E

Difficulty:

  25% (medium)

Question Stats:

78% (01:28) correct 22% (01:17) wrong based on 126 sessions

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Re: If x and y are integers and xy = 660, x or y must be divisible by whic  [#permalink]

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New post 19 Apr 2016, 05:59
1
xy = 660 = 2 * 2 * 3 * 5 * 11

660 is even and a multiple of 3 --> So x or y must be a multiple of 3.

Let's check it out:
Consider 660 = 2 * 330 --> x or y cannot be a multiple of 4, 20
Consider 660 = 4 * 165 --> x or y cannot be a multiple of 6, 30

Answer: A
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Re: If x and y are integers and xy = 660, x or y must be divisible by whic  [#permalink]

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New post 19 Apr 2016, 06:20
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1
Bunuel wrote:
If x and y are integers and xy = 660, x or y must be divisible by which of the following?

A. 3
B. 4
C. 6
D. 20
E. 30


Vyshak a good solution..
The reasoning is that the solution has to be a PRIME number because any other composite number can be expressed in terms which may not necessarily contain that composite number..
example 36 may be 6*6 BUT 36 can still be expressed in PRODUCT of factors that may not contain..
36 = 4*9..

so a solution would bea prime factor of 660..
here 3 is given
A
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Re: If x and y are integers and xy = 660, x or y must be divisible by whic  [#permalink]

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New post 19 Apr 2016, 10:38
Just using prime factorization or normal factorization will easily tell you its going to be a multiple of 3 :)

330 times 2
55 times 12
etc
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Re: If x and y are integers and xy = 660, x or y must be divisible by whic  [#permalink]

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New post 31 Jul 2018, 23:18
I don't completely understand why it needs to be divisible by 3 and why it can't be divisible by 4 i.e. 2^2, which is also contained in the prime factorisation.

Anyone able to clarify?
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Re: If x and y are integers and xy = 660, x or y must be divisible by whic &nbs [#permalink] 31 Jul 2018, 23:18
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