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Is k an odd integer? (1) k is an integer divisible by 3. (2) The squa

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Bunuel
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Is k an odd integer? (1) k is an integer divisible by 3. (2) The squa [#permalink] New post   05 Nov 2017, 02:05
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E

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74% (01:12) correct 26% (01:05) wrong based on 89 sessions
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Is k an odd integer?

(1) k is an integer divisible by 3.
(2) The square root of k is an integer divisible by 3.
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E
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Is k an odd integer? (1) k is an integer divisible by 3. (2) The squa [#permalink] New post   05 Nov 2017, 05:08
Bunuel wrote:
Is k an odd integer?

(1) k is an integer divisible by 3.
(2) The square root of k is an integer divisible by 3.


E is the answer as follows

(1) k is an integer divisible by 3
Let k=3m => k=0,3,6,9,12,15,18,21,24,27,30, ...........................
The numbers marked in RED are odd and the numbers marked in BLUE are even
NOT SUFFICIENT

(2) The square root of k is an integer divisible by 3
Let k=9\(n^2\) = 0,9,36,81,144,..........
\(\sqrt{k}\) = 0,3,6,9,12
NOT SUFFICIENT

Combining both the conditions we get k = 0,9,36,81,144,..........
Clearly, some numbers are odd and some are even.
Hence, NOT SUFFICIENT

E is the answer
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Is k an odd integer? (1) k is an integer divisible by 3. (2) The squa [#permalink] New post   05 Nov 2017, 12:17
Bunuel wrote:
Is k an odd integer?

(1) k is an integer divisible by 3.
(2) The square root of k is an integer divisible by 3.


1) Clearly NS, as both 3 and 6 are divisible by 3 but are odd and even respectively.
2) NS: If the square root of k can be divided by 3, then k has 9 as a factor. \(\sqrt{k} = \sqrt{3*3*(other int)^2} = 3*(other int)\)

1+2) If K has a factor of 9, as per 2, then we already know that k is an integer divisible by 3. No information added, 1+2 are not sufficient.

The answer is E.
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Is k an odd integer? (1) k is an integer divisible by 3. (2) The squa [#permalink]
05 Nov 2017, 12:17
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Bunuel
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