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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!
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!
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?
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.
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
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.
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.
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