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is K divisible by 4

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by MartyMurray » Sun Jun 26, 2016 10:38 pm
750+ wrote:Is the integer k divisible by 4?

(1) 8k is divisible by 16

(2) 9k is divisible by 12
Statement 1:

You know that 8 x 2 = 16. So k could be 2, in which case the answer to the question is no.

Alternatively you could consider that for 8k to be divisible by 16, you just need one 2 among the prime factors of k. If k has just one 2 among its prime factors, then k is not divisible by 4.

Meanwhile, any k such at there is a 2 in its prime factors will create a number divisible by 16 when k is multiplied by 8. So k could be any even number, including 4 or other numbers divisible by 4. So the answer to the question could be yes.

Insufficient.

Statement 2:

In order for a number to be divisible by 12, it has to be divisible by all of the factors of 12, which factors include 4.

There are no 4's in the factors of 9. So for 9k to be divisible by 12, there must be a 4 in the factors of k.

Sufficient.

The correct answer is B.
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by GMATGuruNY » Mon Jun 27, 2016 5:21 am
Is the integer k divisible by 4?

(1) 8k is divisible by 16

(2) 9k is divisible by 12
Statement 1:
8k = 16, 32, 48

Dividing all of these values by 8, we get:
k = 2, 4, 6...

If k=2, then k is NOT divisible by 4.
If k=4, then k IS divisible by 4.
INSUFFICIENT.

Statement 2:
9k = 12, 24, 36, 48, 64, 72...

Dividing all of these values by 9, we get:
k = 12/9, 24/9, 4, 48/9, 64/9, 8...

Since k must be an integer, only the values in blue are viable:
k = 4, 8...
Both of these options for k are divisible by 4.

If we extend our list of options for 9k, we get:
9k = 84, 96, 108...
k = 84/9, 96/9, 12...

Since the next viable option for k -- 12 -- is also a multiple of 4, k must be divisible by 4.
SUFFICIENT.

The correct answer is B.
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by 800_or_bust » Mon Jun 27, 2016 8:08 am
750+ wrote:Can someone please provide a solution
K will always be divisible by 4 if we could prove either that K either equals zero or is an integer that contains at least 2 factors of 2.

(1) Not sufficient. If 8k is divisible by 16, then 8k is either zero or contains at least four factors of 2. However, 8 alone contains three of those factors of two. Thus, k could contain only one factor of two (or it could contain more). So, this is not sufficient.

(2) Sufficient. If 9k is divisible by 12, then 9k is either zero or contains at least two factors of two and one factor of three. Since 9 does not contain any factors of two, k must contain both factors of two. Thus, k is either 0 or an integer containing two factors of two. In either case, k is a multiple of 4, and thus must be divisible by 4.

Note - the case where k = 0 is somewhat of a trivial solution here, since 0 is divisible by any number. But if you think strictly that k must have two factors of 2, you could miss 0. They could have given two facts that when combined implied k = 0, for instance, which also would have been sufficient (e.g. (1) kx = 0, (2) x does not equal zero). Because I sometimes overlook zero, I like to expressly include it unless the prompt states that the number cannot be zero.

Answer is B.
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by GMATGuruNY » Mon Jun 27, 2016 8:32 am
Is the integer k divisible by 4?

(1) 8k is divisible by 16

(2) 9k is divisible by 12
Another approach:

Statement 1:
8k = 16a, where a is an integer.

Dividing both sides by 8, we get:
k = 2a.

If a=1, then k = 2*1 = 2, in which case k is NOT divisible by 4.
If a=2, then k = 2*2 = 4, in which case k IS divisible by 4.
INSUFFICIENT.

Statement 2:
9k = 12b, where b is an integer.

Dividing both sides by 9, we get:
k = (12/9)b
k = (4/3)b.

Since k must be an integer, b must be a multiple of 3.
If b=3, then k = (4/3)(3) = 4.
If b=6, then k = (4/3)(6) = 8.
If b=9, then k = (4/3)(9) = 12.
In every case, k is divisible by 4.
SUFFICIENT.

The correct answer is B.
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by Matt@VeritasPrep » Fri Jul 22, 2016 2:35 am
Plugging in numbers is helpful too, so here's an approach that would help you make a confident guess without having to do any real theoretical puzzling.

S1:

k = 2 makes 8k = 16, which is divisible by 16

k = 4 makes 8k = 32, which is also divisible by 32

So we could have k = a multiple of 4 but we could also have k = some even number that ISN'T a multiple of 4; NOT SUFFICIENT

S2:

If 9k / 12 = an integer, we must have 9k = even. (Odd divided by Even never gives an integer.)

Now let's try our approach from above.

If k = 2, 9k ISN'T divisible by 12.

If k = 4, 9k IS divisible by 12.

k = 6 doesn't work, k = 8 does, k = 10 doesn't work, k = 12 does, ...

So without doing much algebra (or any other mental heavy lifting), we can guess with reasonable confidence that k is divisible by 4.
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