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MGMAT test question

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by sherlock » Wed Mar 14, 2012 11:24 am
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.

i chose D but the answer says A .. i think the problem could be solved without the options!! help

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Source: — Data Sufficiency |
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by GMATGuruNY » Wed Mar 14, 2012 11:50 am
sherlock wrote: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.

i chose D but the answer says A .. i think the problem could be solved without the options!! help

MGMAT test question
(k+2)(k² + 4k + 3) = (k+2)(k+1)(k+3) = (k+1)(k+2)(k+3).

Since k is a positive integer, k+1, k+2, and k+3 are 3 consecutive positive integers.

Statement 1: k is divisible by 8.
Thus, k is a multiple of 4.
Since EVERY OTHER even integer is a multiple of 4, k+2 is NOT a multiple of 4.
Thus, (k+1)(k+2)(k+3) = (odd)(even non-multiple of 4)(odd).
Thus, (k+1)(k+2)(k+3) is not a multiple of 4.
SUFFICIENT.

Statement 2: (k + 1)/3 is an odd integer.
It's possible that k=2, since (2 + 1)/3 = 1.
In this case, (k+1)(k+2)(k+3) = 3*4*5, which is a multiple of 4.
It's possible that k=8, since (8 + 1)/3 = 3.
In this case, (k+1)(k+2)(k+3) = 9*10*11, which is a not a multiple of 4.
INSUFFICIENT.

The correct answer is A.
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by mevicks » Tue Oct 15, 2013 4:12 am
sherlock wrote:i think the problem could be solved without the options!! help
Sorry for bumping into an old post but would just like to point out the following:
If k = 1; (k+2)(k² + 4k + 3) becomes 24 which is divisible by 4
If k = 4; (k+2)(k² + 4k + 3) becomes 210 which is not divisible by 4

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Vivek
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by lunarpower » Wed Oct 16, 2013 2:22 am
mevicks wrote:
sherlock wrote:i think the problem could be solved without the options!! help
Sorry for bumping into an old post but would just like to point out the following:
If k = 1; (k+2)(k² + 4k + 3) becomes 24 which is divisible by 4
If k = 4; (k+2)(k² + 4k + 3) becomes 210 which is not divisible by 4

Regards,
Vivek
^^ That doesn't work. k = 4 doesn't satisfy either statement, so it's irrelevant.
Ron has been teaching various standardized tests for 20 years.

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by mevicks » Wed Oct 16, 2013 2:50 am
lunarpower wrote:
mevicks wrote:
sherlock wrote:i think the problem could be solved without the options!! help

^^ That doesn't work. k = 4 doesn't satisfy either statement, so it's irrelevant.
Hi Ron,

Yes 4 doesnt work for either of the statements; the OP had a doubt about the integrity of the question itself (ie we dont need the statements (1)&(2)) so used the above examples.

Regards,
Vivek
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by lunarpower » Wed Oct 16, 2013 2:54 am
mevicks wrote:
lunarpower wrote:
mevicks wrote:
sherlock wrote:i think the problem could be solved without the options!! help

^^ That doesn't work. k = 4 doesn't satisfy either statement, so it's irrelevant.
Hi Ron,

Yes 4 doesnt work for either of the statements; the OP had a doubt about the integrity of the question itself (ie we dont need the statements (1)&(2)) so used the above examples.

Regards,
Vivek
Ok.
But Mitch's choices (k = 2 and k = 10) already established the integrity of the question. And he posted those before the whole k = 4 thing. So, I couldn't see what value would be added by that post, beyond what Mitch had already established.
Ron has been teaching various standardized tests for 20 years.

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