Author 
Message 
TAGS:

Hide Tags

Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327

If x is an integer, then x(x  1)(x  k) must be evenly [#permalink]
Show Tags
22 May 2010, 05:15
1
This post received KUDOS
5
This post was BOOKMARKED
Question Stats:
62% (09:07) correct
38% (00:55) wrong based on 158 sessions
HideShow timer Statistics
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 Open discussion of this question is here: ifxisanintegerthenxx1xkmustbeevenlydivisible126853.html
Official Answer and Stats are available only to registered users. Register/ Login.
_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Download the Ultimate SC Flashcards
Last edited by Bunuel on 12 May 2012, 12:39, edited 1 time in total.
TOPIC LOCKED



Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
23 May 2010, 02:11
11
This post received KUDOS
i was hoping to get SOME replies on the post but anyway i'll share a very nice approach to handle such questions.
EXCEPT in the question tells us that all answer choices are alike other than one exception, the odd one out. If we can find out the odd one, we don't need to do any calculation and we can still answer this under 15secs.
now the question talks about dividing the product by 3 ... so 4/5 answers choices should be related to 3 .... let's see the difference between the answer choices
a. 4 b. 2 [2 more than A] c. 1 [3 more than A] d. 2 [6 more than A] e. 5 [9 more than A]
nice, so we do have a pattern ... 4 answers have a difference of a multiple of 3 except B ... 3, 6, 9 are all multiples of 3
so we can select B without solving much
_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Download the Ultimate SC Flashcards
Last edited by dimitri92 on 24 May 2010, 07:17, edited 2 times in total.



Manager
Joined: 17 Mar 2010
Posts: 183

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
23 May 2010, 03:20
Hi dimitri, Amazing approach. I think this is one of the best approches i have ever encountered. +1 kudo for you... But one question for you... what if one of the choice is not following the pattern?? and still that may not be the answer???



Manager
Joined: 17 Mar 2010
Posts: 183

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
23 May 2010, 03:36
7
This post received KUDOS
3
This post was BOOKMARKED
Although you method is shortest, i would prefer surest way... If x or x1 is divisible by 3 ==> x(x  1)(x  k) is divisible by 3 Assign any arbitrary number to x such that x and x1 are not disible by 3 say x=5, x1 = 4 ==> 20*(5k) assign answer choices to k.. for k=4, ==> 20*(9) ==> divisible by 3 for k=2 ==> 20(7) ==> NOT divisible by 3
Consider giving me KUDOS if you like my post



Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
23 May 2010, 13:27
1
This post received KUDOS
2
This post was BOOKMARKED
here's another approach
x(x  1)(x  k) all three are consecutive, so the product MUST be a multiple of 3
we don't know the value of k just yet ... so let's extend the series ... the extension itself reveals the answers
..(x5)..(x2)(x1)x(x+1)..(x+4)..
we can see the possible values of k too from the series
k = 2 OR 2+3n [2 & 5] k = 1 OR 1+3n [1 & 4]
B i.e. 2 does not fit in any value of k
so B it is
_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Download the Ultimate SC Flashcards



Intern
Joined: 09 May 2010
Posts: 35

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
24 May 2010, 05:14
dimitri 92
could you explain how do you come up with expansion series and how do you pick those for K? I'm still confused on your method
thank you very much.



Senior Manager
Affiliations: SPG
Joined: 15 Nov 2006
Posts: 327

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
24 May 2010, 07:14
abcd1 wrote: dimitri 92
could you explain how do you come up with expansion series and how do you pick those for K? I'm still confused on your method
thank you very much.
the given series: x(x  1)(x  k) but we don't know the value of k ... for (xk) to be a multiple of 3, it must be consecutive to x(x1) .. so it must be either (x2) or (x+1) i.e. value of k can be 2 or 1.
there are two ways of expanding the series now ..(x8)..(x5)..(x2)(x1)x.. OR ..(x1)x(x+1)..(x+4)..(x+7)..
from the above expansion you can see that the following are possible values of k 2, 5, 8, 11 .. 1, 4, 7 ..
_________________
press kudos, if you like the explanation, appreciate the effort or encourage people to respond.
Download the Ultimate SC Flashcards



Intern
Joined: 09 May 2010
Posts: 35

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
24 May 2010, 19:16
thank you very much ^_^ dimitri 92



Senior Manager
Joined: 05 Oct 2008
Posts: 273

Consecutive integers [#permalink]
Show Tags
16 Jun 2010, 00:54
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 4 2 1 2 5



Manager
Joined: 04 May 2010
Posts: 87
WE 1: 2 yrs  Oilfield Service

Re: Consecutive integers [#permalink]
Show Tags
16 Jun 2010, 01:01
The easiest way is to plug in an integer which depends on (xk) for the expression to be divisible by 3.
Example, pick 5: The expression becomes: 5* 4 * (5  k)
Now from the 5 values of k, see which value does not give you a multiple of 3.
A. 5 (4) = 9 B. 5 (2) = 7 ... Bingo, the expression would be equal to 140 which is not divisible by 3. On test day  stop.



Intern
Joined: 19 Jun 2010
Posts: 27

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
21 Jun 2010, 00:08
If your mind blanked out during the test, the easiest thing to do is to substitute each option into K. Of course, you must understand the concept that for a consecutive of three, x+1 occupies the same position as x+(1+6), x+(1+3), x+(13) or x+(16) and so on. All choices except 2 will give the consecutive numbers. Substituting 2 will give x(x1)(x2) = x(x1)(x+2) which is not a three consecutive numbers.



Intern
Joined: 20 Feb 2012
Posts: 5

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
12 May 2012, 12:00
the given series: x(x  1)(x  k) but we don't know the value of k ... for (xk) to be a multiple of 3, it must be consecutive to x(x1) .. so it must be either (x2) or (x+1) i.e. value of k can be 2 or 1.
Dimitri, for (xk) to be a multiple of 3, why MUST it be CONSECUTIVE to x(x1)?
example: if x=5 5(51)(520)= 300 which is divisible by 3 even tho 5x4x15 are not consecutive integers.
I realize that assuming they are consecutive integers helps you solve the problem but I just wanted to be clear that (xk) can be a multiple of 3 without it being consecutive to x(x1).
Please let me know if i'm missing something, thanks.



Math Expert
Joined: 02 Sep 2009
Posts: 39762

Re: consecutive integers: product divisible by three [#permalink]
Show Tags
12 May 2012, 12:38
ctiger100 wrote: the given series: x(x  1)(x  k) but we don't know the value of k ... for (xk) to be a multiple of 3, it must be consecutive to x(x1) .. so it must be either (x2) or (x+1) i.e. value of k can be 2 or 1.
Dimitri, for (xk) to be a multiple of 3, why MUST it be CONSECUTIVE to x(x1)?
example: if x=5 5(51)(520)= 300 which is divisible by 3 even tho 5x4x15 are not consecutive integers.
I realize that assuming they are consecutive integers helps you solve the problem but I just wanted to be clear that (xk) can be a multiple of 3 without it being consecutive to x(x1).
Please let me know if i'm missing something, thanks. You are right x, (x – 1), and (x – k) are not necessarily consecutive integers. Open discussion of this question is here: ifxisanintegerthenxx1xkmustbeevenlydivisible126853.html
_________________
New to the Math Forum? Please read this: All You Need for Quant  PLEASE READ AND FOLLOW: 12 Rules for Posting!!! Resources: GMAT Math Book  Triangles  Polygons  Coordinate Geometry  Factorials  Circles  Number Theory  Remainders; 8. Overlapping Sets  PDF of Math Book; 10. Remainders  GMAT Prep Software Analysis  SEVEN SAMURAI OF 2012 (BEST DISCUSSIONS)  Tricky questions from previous years.
Collection of Questions: PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.
What are GMAT Club Tests? Extrahard Quant Tests with Brilliant Analytics




Re: consecutive integers: product divisible by three
[#permalink]
12 May 2012, 12:38








Similar topics 
Author 
Replies 
Last post 
Similar Topics:




If a is an integer larger than 1, and x=60a, then x must be a multiple

Bunuel 
1 
28 Sep 2016, 13:55 

135


If x is an integer then x(x1)(xk) must be evenly divisible

enigma123 
24 
28 May 2017, 12:07 

1


If x is an even integer, is x(x+1)(x+2) divisible by 4?

napy27 
3 
11 May 2011, 21:11 

45


If x is an integer, then x(x – 1)(x – k) must be evenly divi

anilnandyala 
27 
05 Oct 2016, 16:33 

2


If x is an integer, then x(x  1)(x  k) must be evenly divisible by

shobuj 
12 
27 Apr 2015, 14:29 



