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Set B contains all factors and multiples of 20. If x is divisible by

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Set B contains all factors and multiples of 20. If x is divisible by  [#permalink]

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New post 08 Jan 2017, 07:12
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Question Stats:

62% (01:10) correct 38% (01:39) wrong based on 65 sessions

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Set B contains all factors and multiples of 20. If x is divisible by 20, then x must also be divisible by how many members of set B?

A. 4
B. 5
C. 6
D. 10
E. 20

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Re: Set B contains all factors and multiples of 20. If x is divisible by  [#permalink]

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New post 08 Jan 2017, 08:51
Bunuel wrote:
Set B contains all factors and multiples of 20. If x is divisible by 20, then x must also be divisible by how many members of set B?

A. 4
B. 5
C. 6
D. 10
E. 20


x is divisible by 20 means that x is definitely divisible by all factors of 20. But we can’t be sure about divisibility of x by other multiples of 20: 20*2, 20*3 …. Our B is infinite set {0; +∞}, but we are definite only with part of it’s elements, which represent factors of 20.

\(20 = 2^2*5\)

# of factors =\(3*2 = 6\)

Answer C.
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Re: Set B contains all factors and multiples of 20. If x is divisible by  [#permalink]

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New post 08 Jan 2017, 08:51
Bunuel wrote:
Set B contains all factors and multiples of 20. If x is divisible by 20, then x must also be divisible by how many members of set B?

A. 4
B. 5
C. 6
D. 10
E. 20


Factors of 20 = 1, 2, 4, 5, 10, 20
Multiples of 20 = 20 , 40 , 60, 80.....................

So, B = { 1, 2, 4, 5, 10, 20 40 , 60, 80..................... }

x/20 = Remainder 0

Let the minimum value of x = 20

Thus , x must be divisible by 1, 2, 4, 5, 10, 20

So, x must also be divisible by 6 members of set B

Answer will be (C) 6
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Re: Set B contains all factors and multiples of 20. If x is divisible by  [#permalink]

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New post 08 Jan 2017, 14:42
Set B = ( 1,2,4,5,10,20,40... )
x can be divisible only by 20 and factors of 20.
6
C
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Re: Set B contains all factors and multiples of 20. If x is divisible by  [#permalink]

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Re: Set B contains all factors and multiples of 20. If x is divisible by   [#permalink] 23 Jan 2019, 21:36
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