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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
5-Day Free Trial
5-day free, full-access trial TTP
Available with Beat the GMAT members only code
MORE DETAILS
Consecutive Integers
This topic has expert replies
-
diebeatsthegmat
- Legendary Member
- Posts: 1119
- Joined: Fri May 07, 2010 8:50 am
- Thanked: 29 times
- Followed by:3 members
the best solution is choose a number...tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
x=2 so k=-1 or 5 -4
x=8 so k=5,2,-1,-4
the answer should be B
thus the answer is
-
limestone
- Master | Next Rank: 500 Posts
- Posts: 307
- Joined: Sun Jul 11, 2010 7:52 pm
- Thanked: 36 times
- Followed by:1 members
- GMAT Score:640
Trick for this question:
5 and 2 is the same to 3:
5=3+2
Both 5 and 2 if be divided to 3, will give out the remainder 2. Then if 5 is the answer choice, 2 must be too.
-4 and -1 is the same to 3:
-4 = -3 + (-1)
B is odd, so B must be the correct choice.
5 and 2 is the same to 3:
5=3+2
Both 5 and 2 if be divided to 3, will give out the remainder 2. Then if 5 is the answer choice, 2 must be too.
-4 and -1 is the same to 3:
-4 = -3 + (-1)
B is odd, so B must be the correct choice.
"There is nothing either good or bad - but thinking makes it so" - Shakespeare.
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
B[/spoiler]
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
ankur.agrawal
- Master | Next Rank: 500 Posts
- Posts: 261
- Joined: Wed Mar 31, 2010 8:37 pm
- Location: Varanasi
- Thanked: 11 times
- Followed by:3 members
sanju09 wrote:The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
Hey Sanju Can u explain Why -4 & 5 is allowed?
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
ankur.agrawal wrote:sanju09 wrote:The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
Hey Sanju Can u explain Why -4 & 5 is allowed?
If x is one of the three numbers whose product is divisible by 3, then using x - 3 or x + 3 in place of x, would still keep the product divisible by 3, no matter is x is divisible by 3 or not.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
ankur.agrawal
- Master | Next Rank: 500 Posts
- Posts: 261
- Joined: Wed Mar 31, 2010 8:37 pm
- Location: Varanasi
- Thanked: 11 times
- Followed by:3 members
sanju09 wrote:ankur.agrawal wrote:sanju09 wrote:The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
Hey Sanju Can u explain Why -4 & 5 is allowed?
If x is one of the three numbers whose product is divisible by 3, then using x - 3 or x + 3 in place of x, would still keep the product divisible by 3, no matter is x is divisible by 3 or not.
Sorry but i couldnt grasp this concept. If we take k=-4 then
(x) (x-1) (x+4) ---- is this a series of consecutive integers.? What am i missing. Plz explain
-
sanju09
- GMAT Instructor
- Posts: 3650
- Joined: Wed Jan 21, 2009 4:27 am
- Location: India
- Thanked: 267 times
- Followed by:80 members
- GMAT Score:760
ankur.agrawal wrote:sanju09 wrote:ankur.agrawal wrote:sanju09 wrote:The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
Hey Sanju Can u explain Why -4 & 5 is allowed?
If x is one of the three numbers whose product is divisible by 3, then using x - 3 or x + 3 in place of x, would still keep the product divisible by 3, no matter if x is divisible by 3 or not.
Sorry but i couldnt grasp this concept. If we take k=-4 then
(x) (x-1) (x+4) ---- is this a series of consecutive integers.? What am i missing. Plz explain
x and x - 1 are two consecutive integers, one of which may or may not be a multiple of 3, and since the product of three consecutive integers is always divisible by 3, hence the nearest third integer to make sure the product is divisible by 3, could be either x + 1 or x - 2, that's why k can initially be taken as -1 or 2.
Now, if x and x - 1 are two consecutive integers, one of which is a multiple of 3, then x (x - 1) (x - k) will always be divisible by 3 for any value of k; but if it's not so, then it will be divisible by 3 for some specific values of k only. Because, then x - k has to be divisible by 3, and if so, then both x + 1 and x - 2 are divisible by 3. And if so, then 3 more than x + 1 and 3 less than x - 2 will also be divisible by 3. That is, x + 4 and x - 5 will also be divisible by 3, and hence all of the following products will also be divisible by 3
A. x (x - 1) (x + 4)
C. x (x - 1) (x + 1)
D. x (x - 1) (x - 2)
E. x (x - 1) (x - 5)
EXCEPT
B. x (x - 1) (x + 2)
It's not necessary for the three integers to be always consecutive for their product to be divisible by 3.
The mind is everything. What you think you become. -Lord Buddha
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
Sanjeev K Saxena
Quantitative Instructor
The Princeton Review - Manya Abroad
Lucknow-226001
www.manyagroup.com
-
ankur.agrawal
- Master | Next Rank: 500 Posts
- Posts: 261
- Joined: Wed Mar 31, 2010 8:37 pm
- Location: Varanasi
- Thanked: 11 times
- Followed by:3 members
sanju09 wrote:ankur.agrawal wrote:sanju09 wrote:ankur.agrawal wrote:sanju09 wrote:The product of three consecutive integers is always divisible by 3. Hence, k can initially take any of the values -1 and 2, and 3 thereafter in each direction, which allows -4 and 5, what is not allowed is [spoiler]-2, from the choices.tlt2372 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
OA: B
Does anyone know a quick way to solve this problem?
THanks!
Hey Sanju Can u explain Why -4 & 5 is allowed?
If x is one of the three numbers whose product is divisible by 3, then using x - 3 or x + 3 in place of x, would still keep the product divisible by 3, no matter if x is divisible by 3 or not.
Sorry but i couldnt grasp this concept. If we take k=-4 then
(x) (x-1) (x+4) ---- is this a series of consecutive integers.? What am i missing. Plz explain
x and x - 1 are two consecutive integers, one of which may or may not be a multiple of 3, and since the product of three consecutive integers is always divisible by 3, hence the nearest third integer to make sure the product is divisible by 3, could be either x + 1 or x - 2, that's why k can initially be taken as -1 or 2.
Now, if x and x - 1 are two consecutive integers, one of which is a multiple of 3, then x (x - 1) (x - k) will always be divisible by 3 for any value of k; but if it's not so, then it will be divisible by 3 for some specific values of k only. Because, then x - k has to be divisible by 3, and if so, then both x + 1 and x - 2 are divisible by 3. And if so, then 3 more than x + 1 and 3 less than x - 2 will also be divisible by 3. That is, x + 4 and x - 5 will also be divisible by 3, and hence all of the following products will also be divisible by 3
A. x (x - 1) (x + 4)
C. x (x - 1) (x + 1)
D. x (x - 1) (x - 2)
E. x (x - 1) (x - 5)
EXCEPT
B. x (x - 1) (x + 2)
It's not necessary for the three integers to be always consecutive for their product to be divisible by 3.
Thanks for the detailed explaination man. U rock.
-
samirnajeeb
- Junior | Next Rank: 30 Posts
- Posts: 12
- Joined: Sat May 19, 2007 8:59 am
- Thanked: 1 times
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
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
