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Divisibility&Primes DS Qstn

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Divisibility&Primes DS Qstn

by Suz » Fri Aug 10, 2012 8:47 pm
Q. K is a set of numbers such that

I) if x is in K, then - x is in K, and
II) if each of x and y is in K, then xy is in K

Is 12 in K?

1) 2 is in K
2) 3 is in K


Thanks!
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Source: — Data Sufficiency |
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by niketdoshi123 » Fri Aug 10, 2012 9:08 pm
Suz wrote:Q. K is a set of numbers such that

I) if x is in K, then - x is in K, and
II) if each of x and y is in K, then xy is in K

Is 12 in K?

1) 2 is in K
2) 3 is in K
Question : Is 12 in K?

if each of x and y is in K, then xy is in K
we can check for the factors of 12
12 = 2*6 = 2*(2*3)
12 = 4*3 = (2*2)*3
12 = 12*1 = (2*2*3)*1

rephrased question: Are 2 & 3 in K

Since if 2 and 3 are in K , then 2*3 =6 is also in K
and if 2 and 6 are in K, then 2*6 = 12 is also in K

Hence combining both the statements will answer our question.

the correct answer is C
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by ncavaggn » Fri Aug 10, 2012 9:24 pm
Given that,
(I) K: { ...,-x^2, x, -x, x^3,... }
(II) K: { ...,-x^2, -x ,x, y, xy, x^2y,...}

1)Now, 2 is in K
If only 2 is in K, K: { ...,-4, -2, 2, 8,...} => 12 is not in K
Considering other numbers, say 6 is also in K, K: { ...,-12, -6, -2, 2,6, 12...} => 12 is in K
K may or may not contain 12.
Hence, NOT SUFFICIENT

2)3 is in K
If only 3 is in K, K: {...-9,-3,3,9,...} => 12 is not in K
Considering other numbers say, 4 is also in K, K: {...,-12, -4,-3,3,4,12,...} => 12 is in K
K may or may not contain 12.
Hence, NOT SUFFICIENT

Combining (1) and (2),
2 is in K and 3 is in K,
K: {...,-12, -9, -3, -4, -2, 2, 3,4, 8, 12...} => Irrespective of any other numbers, K definitely contains 12.

Hence, C.
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