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805+ Level|   Number Properties|         
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Bunuel
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Around the World in 80 Questions (Day 7): Is k an even integer ? 25 Jul 2023, 08:00
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Is k an even integer ?

(1) 3k is an even integer.
(2) k^3 is an even integer.


 


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C
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missionmba2025
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Around the World in 80 Questions (Day 7): Is k an even integer ? 25 Jul 2023, 08:09
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Bunuel
Is k an even integer ?

(1) 3k is an even integer.
(2) k^3 is an even integer.


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

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(1) 3k is an even integer.

If k = 2 , k is an even integer
If k = 2/3, k is not an even integer

Not sufficient.

(2) k^3 is an even integer.

If k =2 , k is an even integer
If k = cube root(2), k is not an even integer

Not sufficient.

Combined

k cannot be 1/cube root(even), hence the only possible is k is even integer.

IMO C
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Around the World in 80 Questions (Day 7): Is k an even integer ? 25 Jul 2023, 08:31
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target is k an even integer
#1
3k is an even integer
k = 2 or 2/3
insufficient
#2
k^3 is an even integer.
k= 2 or (√16)^1/3
insufficient
from 1 &2
k has to be an even integer
sufficient
option C


Bunuel
Is k an even integer ?

(1) 3k is an even integer.
(2) k^3 is an even integer.


 


This question was provided by GMAT Club
for the Around the World in 80 Questions

Win over $20,000 in prizes: Courses, Tests & more

 

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Around the World in 80 Questions (Day 7): Is k an even integer ? 25 Jul 2023, 08:48
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Is k an even integer ?

(1) 3k is an even integer. - INSUFFICIENT
Since it is not given if k is integer or not, let's take both cases.
Case 1: k = 2, therefore, 3k = 6 which is even integer. k is an even integer.
Case 2: k = 2/3, therefore, 3k = 2 which is even integer. But k is NOT an even integer.
Hence, Insufficient

(2) k^3 is an even integer. - INSUFFICIENT
k^3 = 2^a * p^b.
Case 1: if k=2^(1/3), that is k = cube root of 2, k^3 = 2, an even integer. But k is not an even integer.
Case 2: k = 2, k^3 = 8, an even integer. And k is an even integer.
Hence, Insufficient.

Taking 1 & 2, - SUFFICIENT
for 3k to be an even integer, k has to be of the form (2/3)*q, where q is any integer
Now, for k^3 to be an even integer, it cannot be fraction, so k have to be of the form 2q, where q is any integer.
Which implies k is even integer.
Hence, sufficient.

Answer is C
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Around the World in 80 Questions (Day 7): Is k an even integer ? 25 Jul 2023, 09:30
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Is k an even integer ?

(1) 3k is an even integer.
(2) k^3 is an even integer.
________________________
(1) multiplying an even integer number to any positive integer equals an even integer.
k=2 or 6 --> Yes.
but, k=2/3 --> No - Insufficient.

(2) K can be 2 power of (2/3) . In this case, we can say Yes.
but, K can be 3 power of (2/3) . In this , we can say No. Insufficient.

Combination, in this case , we cant take number in (2) option because they dont fit to requirement of (1). Moreover, we cant take the second example of (2) . Again it doesnt fit. SO, we can only take 1st example of (1). Thus, combination works well.

C is our winner.
Hope it helps.
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