Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

Re: PS - equation divisible by 3. [#permalink]
04 Jan 2008, 23:20

walker wrote:

B

Obviously, N=x(x – 1)(x – 2) is product of 3 consecutive integer and N is divided by 3. So, k should be equal 3n+2: -7,-4,-1,2,5,8

-2 is out.

this ps works for me, but wastes time. I see yr approach helpful, but I am not clear why k should be 3n+2. Maybe this is a basis, but I cannt get it. help Pls

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders) we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders. 2. Now we can go back or forward from number 2 by adding or subtracting 3.

Re: PS - equation divisible by 3. [#permalink]
05 Jan 2008, 03:11

ashkrs wrote:

If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT A -4 B -2 C -1 D 2 E 5

Show work please.

if we substitute x with 0 or 1 we always obtain 0, which is always evenly divisible by three (maybe the question should have excluded this 2 values for x).

let's try with x=2...the only value of k for which we don't have an even integer in the division is -2, therefore OA can be B

Re: PS - equation divisible by 3. [#permalink]
05 Jan 2008, 05:14

ANSWER IS -2. Any one of 3 consecutive integers is divisible by 3. so, we have x, x-1 and x-k, for them to be consecutive, ie we have x-1, x and x-k, so if x-k = x+1, then they are consecutive. so, from above k=-1. also remember that when you add multiples of 3 to a no. divisible by 3 it again gives multiple of 3, so try adding -3, 0, 3, and 6 to -1 you get all but -2

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders) we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders. 2. Now we can go back or forward from number 2 by adding or subtracting 3.

Hope this help

Hi walker,

you seems to think automatically "k should be equal to 3n+2" after you see problem, right? can you discribe that process of thinking? Thanks!

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders) we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders. 2. Now we can go back or forward from number 2 by adding or subtracting 3.

Hope this help

Hi walker,

you seems to think automatically "k should be equal to 3n+2" after you see problem, right? can you discribe that process of thinking? Thanks!

The problem took 15 sec for me and my way was exact as described above.
_________________

Re: PS - equation divisible by 3. [#permalink]
06 Jan 2008, 11:07

ashkrs wrote:

If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT A -4 B -2 C -1 D 2 E 5

Show work please.

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3 quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3 for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3 for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3 for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3 for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

Re: PS - equation divisible by 3. [#permalink]
06 Jan 2008, 20:22

ashkrs wrote:

ashkrs wrote:

If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT A -4 B -2 C -1 D 2 E 5

Show work please.

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3 quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3 for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3 for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3 for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3 for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

So B.

ashkrs,

I like this approach, I forgot the rule that the sum of the number that is divisible by 3 must evenly devide by 3. Thank you

Re: PS - equation divisible by 3. [#permalink]
08 Jan 2008, 05:32

ashkrs wrote:

ashkrs wrote:

If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT A -4 B -2 C -1 D 2 E 5

Show work please.

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3 quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3 for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3 for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3 for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3 for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

So B.

Hi, I went first with consecutive integers rule, which is also a good way to solve the problem..but this one is very nice. It is really: you are to have at least one factor which is divisible by three so the product of three different integers be divisible by three Thanks a lot for the idea

gmatclubot

Re: PS - equation divisible by 3.
[#permalink]
08 Jan 2008, 05:32