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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

Originally posted on MIT Sloan School of Management : We are busy putting the final touches on our application. We plan to have it go live by July 15...