For positive integer k, divisible by 4?

[konkona]

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

(1) k is divisible by 8.

(2) (k + 1)/3 is an odd integer


The above problem was one of the manhattan exam questions the answer provided was A but I think it is D.
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(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer. However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.

The question might therefore be rephrased "Is k odd, OR is k + 2 divisible by 4?" Note that a "˜yes’ to either of the conditions would suffice, but to answer 'no' to the question would require a "˜no’ to both conditions.

(1) SUFFICIENT: If k is divisible by 8, it must be both even and divisible by 4. If k is divisible by 4, k + 2 cannot be divisible by 4. Therefore, statement (1) yields a definitive "˜no’ to both conditions in our rephrased question; k is not odd, and k + 2 is not divisible by 4.

(2) INSUFFICIENT: If k + 1 is divisible by 3, k + 1 must be an odd integer, and k an even integer.
However, we do not have sufficient information to determine whether k or k + 2 is divisible by 4.

The correct answer is A.
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But I think we have sufficient information for (2). My logic is as follows:

if (k + 1)/3 is odd then we can say (k + 1)/3 = 2n + 1 so k = 6n + 2
the original equation can be expressed as (k+2)(k+3)(k+1) so replacing k by 6n + 2 we get
(6n+4)(6n+5)(6n+3) = (2)(3n+2)(6n+5)(6n +3)

[cyapt81]

you said "if (k + 1)/3 is odd then we can say (k + 1)/3 = 2n + 1 so k = 6n + 2
the original equation can be expressed as (k+2)(k+3)(k+1) so replacing k by 6n + 2 we get
(6n+4)(6n+5)(6n+3) = (2)(3n+2)(6n+5)(6n +3)"

I think everything you said is true except that (2)(3n+2)(6n+5)(6n +3) is not necessarily a multiple of 4. If n is 1, then (6n+4)(6n+5)(6n+3) would equal 9x10x11 which is not a multiple of 4. If n is 2 than (6n+4)(6n+5)(6n+3) would equal 16x17x15 which is a multiple of 4.

Hence A.

[Ben Ku]

But I think we have sufficient information for (2). My logic is as follows:

if (k + 1)/3 is odd then we can say (k + 1)/3 = 2n + 1 so k = 6n + 2
the original equation can be expressed as (k+2)(k+3)(k+1) so replacing k by 6n + 2 we get
(6n+4)(6n+5)(6n+3) = (2)(3n+2)(6n+5)(6n +3)

This conclusion tells us that it's divisible by 2, but not divisible by 4.

[sylvester.marvyn]

On this question, I do not understand why statement 1 is not insuff. The numbers 1 and 8 are both divisble by 8. if you plug in 1 for k you get 2, 3, 4 which answers question w 'yes'. Then if you test 4, another factor of 8, you get 5, 6, 7 which results in answer 'no'. Shouldn't the answer by E? What am I missing?

-Marvyn

[jnelson0612]

On this question, I do not understand why statement 1 is not insuff. The numbers 1 and 8 are both divisble by 8. if you plug in 1 for k you get 2, 3, 4 which answers question w 'yes'. Then if you test 4, another factor of 8, you get 5, 6, 7 which results in answer 'no'. Shouldn't the answer by E? What am I missing?

-Marvyn

Hi Marvyn,
I agree with you that 8 is divisible by 8, but 1 is not divisible by 8. Given statement 1, I can only use k values that are divisible by 8, so I could say k is 8, 16, 24, 32, etc. I think that you are thinking of *factors* of 8, rather than numbers that are divisible by 8 (in other words, multiples of 8).

[MarioN389]

I just have a potential correction for the explanation of the solution.

It states that "If K+1 is divisible by 3, k+1 MUST be an odd integer..."

This isn't true in all cases because k+1 could be 30 in which case it is divisible by 3 and is an EVEN integer.

I realize that (2) mentions that (k+1)/3 yields an odd integer meaning that K+1 cannot be 30 (or any multiple of 30) since dividing it by 3 yields an even integer. However, this explanation is still misleading and could cause people to incorrectly assume that if (k+1) is divisible by 3 that it'll always be odd in other problems.

It would be better if it read something like "If k+1 is divisible by 3 AND yields an odd integer, k+1 MUST be odd..."

[RonPurewal]

the original problem says that (k + 1)/3 is an odd integer. (the top post actually left out the words completely; i've edited it)

so, in that case, (k + 1) is 3 times whatever odd integer, and is thus odd as well.