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.
It appears that you are browsing the GMAT Club forum unregistered!
Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club
Registration gives you:
Tests
Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.
Applicant Stats
View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more
Books/Downloads
Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!
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