integer k ...CAT 6

[kishankolli]

For positive integer k, is the expression (k + 2)(k2 + 4k + 3) divisible by 4?

(1) k is divisible by 8.
(2) k + 1/3 is an odd integer

NOTE :In the expression K2 is k square.

Solution: The answer posted is A. But, as per my analysis it was D. Can you please share your views about this ??

My Analysis :
B is also right as K+1 is odd, so K is even. The number that leads K to be divisible by 3(K+1) are 2,8 & 14. Which are all divisible by 4 when we input these even integers into the expression.

Waiting for your reply.

[anoo.anand]

(k + 2)(k2 + 4k + 3)

=> k^3 + 4k^2 + 3K + 2K^2 + 8K + 6

A) using A if K is divisible by 8 thn it is alo divisible by 4

however ....+6 is never divisible.... thus answer is always NO >> A / C / D left.


B) the option says K+1/3 is a integer, but i am unable to find any such number.

k = 3 + 1/3
k = 4 + 1/3
....

Thus ...A is the only option left.

[gorav.s]

I agree with Answer A ) -->> never divisible by 4.

B) there is no real value of K it seems as K is a positive integer - as per question statement- then you can never have K+1/3 = odd integer --> implies K is a fraction - like 1/3

[kishankolli]

For the second point I am sorry that my question was not specific..It goes like this:

K+1
2) ---- is an odd integer
3

K+1 to be divisible by 3 has to be an odd number, so K is even !!

The numbers that leads to K+1 to be divisible by 3 to be odd are 2,8 & 14.

All went well till I tried 8, by inserting it into the expression..

So, only A is right..

Tx guys.

[Ben Ku]

Let's continue to factor the given expression:
(k+2)(k^2 + 4k + 3)
= (k+2)(k+1)(k+3)
= (k+1)(k+2)(k+3).

In other words, this is the product of three consecutive integers. We can apply consecutive integer, even/odd, and divisibility rules.

There are two ways the answer is "Yes." Either:
- (K+1) and (k+3) are even, OR
- (k+2) is divisible by 4.
The answer is "No" IF
- (k+2) is an even number that is NOT divisible by 4.

Statement (1) says that k is divisible by 8. This means:
(k+1) is odd
(k+2) is even, but is NOT divisible by 4, since k is already divisible by 4
(k+3) is odd.
From statement (1), we know the answer to the question is NO, so statement (1) is sufficient.

From statement (2), (k+1)/3 is an odd integer. Since 3 * (Odd) is still odd, then we know that k + 1 is odd, which makes k even. We know that:
k + 1 is odd
k + 2 is even, but we don't know if it's divisible by 4
k + 3 is odd
So because we don't know if k+2 is divisible by 4, this statement is INSUFFICIENT.


Hope that helps.

[jigar24]

I think there is something wrong with question.. statements contradict each other.. here's how..

When k=2

(K+1)/3 = 3/3 = 1 which is an odd integer so satisfies stat 2

however, statement 1 says k divisible by 8 BUT 2 is NOT divisible by 8.. there s got to be some mistake.. please some prof. check my logic and correct the question or my reasoning, whichever is flawed

[jnelson0612]

jigar,
The statements actually don't contradict each other. They just give you different sets of numbers.

Statement 1 says k is divisible by 8, so k could be 8, 16, 24, 32, etc.

Statement 2 says that the result of (k+1)/3 is odd, thus k+1 is odd and k is always even, with possible numbers such as 2, 4, 6, 8, 10, and so on.

There is no problem with the statements indicating different sets of numbers as possible values for k. That is completely allowed on the GMAT.

Also, look at it this way--statement 1 indicates values that are a subset of the values of statement 2. If statement two is all positive even integers, then multiples of 8 are contained in that group of numbers. Statement 1 is only indicate a subset of the values in statement 2.

I hope this helps clear things up.

Thank you,

[wallstreetcat]

Originally did this one with abstract math, as mentioned above but got it wrong even during review. I realized there are a few things here that might make this problem easier upon review and for future problems.

I classified this one as a divisibility problem, Y/N, single variable Data Sufficiency. I remember from the recently watched Y/N testing lab that these one variable, Y/N questions are best for testing #s - CLA. I realized this works a lot better for this question than trying to think through the odd and even calculations.

After the rephrase of the question, as mentioned above:
1) k/8, K>0 (from question stem). Since K is even, is k+2 divisible by 4 (from rephrasing)?
k = 8, 16, 24, 32, 40
K+2 = 10, 18, 26, 34, 42
K+2 / 4? N, N, N, N, N
SUFFICIENT

2) K+1/3 = odd int, K>0
K+1 = 3, 9, 15, 21, 27
K= 2, 8, 14, 20, 26
equation / 4? = Y, N --> Stop here
Insufficient

Kind of forced this solution upon myself since the abstract nature of the problem, with odd/even number theory + consecutive integers made it difficult to follow.

Do the mgmat instructors agree with the above approach or is my method off?

thanks

[tim]

You haven't addressed the correct question here. Please go back and proofread your work and let us know if we can help you further. Your approach makes sense, you just need to implement it properly.