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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
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
04 Jan 2011, 02:22
Expert's post
vjsharma25 wrote:
Hi bunuel, I don't understand the problem language, it says
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
how does it matter whats the value of K, i can choose x = 3 and the expression will always be divisible by 3.
Am i missing any minor yet important point?
Stem says: "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"
The important word in the stem is "MUST", which means that we should guarantee the divisibility by 3 no matter the value of x (for ANY integer value of x), so you cannot arbitrary pick its value.
To be divisible by 3, one of these sequences must be divisible by 3.
X(X-1) (X-k)
Any 3 sequence number will always be divisible by 3. So X(X-1) (x-2) is divisible by 3.
K = 2, divisible by 3 K= 5, also a sequence ( parallel ) divisible by 3 K= -1, sequence is (X-1) X (X+1) so divisible by 3 K= -4, also a sequence ( parallel ) divisible by 3 K=-2, not a sequence, may not be divisible by 3
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
I am providing the theoretical explanation below. Once you get it, you can solve such questions in a few seconds in future!
Notice a few things about integers: -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16......
Every number is a multiple of 1 Every second number is a multiple of 2 Every third number is a multiple of 3 Every fourth number is a multiple of 4 and so on...
So if I pick any 3 consecutive integers, one and only one of them will be a multiple of 3: e.g. I pick 4, 5, 6 (6 is a multiple of 3) or I pick 11, 12, 13 (12 is a multiple of 3) etc..
x(x - 1)(x - k) will be evenly divisible by 3 if at least one of x, x - 1 and x - k is a multiple of 3. We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it. Also, (x-1)x(x + 1) will have a multiple of 3 in it because they both are products of 3 consecutive integers. So k can be 2 or -1. Eliminate these options. Now let me write down consecutive integers around x:
(x - 2)(x - 1)x will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x - 2) could be the multiple of 3, in which case (x - 5) will also be a multiple of 3. So in any case, (x - 5)(x - 1)x will have a multiple of 3 in it. So k can be 5.
Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3. So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.
We cannot say whether (x-1)x(x + 2) will have a multiple of 3 in it and hence if k = -2, we cannot say whether the product is evenly divisible by 3.
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
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
Bunnel, The second approach is too good... Very helpful,...
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
29 Nov 2013, 20:22
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
02 Aug 2014, 00:42
anilnandyala 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
Since x(x-1)(x-k) is divisible by 3, take a case when x(x-1) is not divisible by 3 and so (x-k) has to be divisible by 3. Let us take x=8 and x-1=7. Only for the second option we do not get x-k divisible by 3. _________________
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
01 Sep 2014, 23:21
Expert's post
VeritasPrepKarishma wrote:
I am providing the theoretical explanation below. Once you get it, you can solve such questions in a few seconds in future!
Notice a few things about integers: -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16......
Every number is a multiple of 1 Every second number is a multiple of 2 Every third number is a multiple of 3 Every fourth number is a multiple of 4 and so on...
So if I pick any 3 consecutive integers, one and only one of them will be a multiple of 3: e.g. I pick 4, 5, 6 (6 is a multiple of 3) or I pick 11, 12, 13 (12 is a multiple of 3) etc..
x(x - 1)(x - k) will be evenly divisible by 3 if at least one of x, x - 1 and x - k is a multiple of 3. We know from above, (x - 2)(x - 1)x will have a multiple of 3 in it. Also, (x-1)x(x + 1) will have a multiple of 3 in it because they both are products of 3 consecutive integers. So k can be 2 or -1. Eliminate these options. Now let me write down consecutive integers around x:
(x - 2)(x - 1)x will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x - 2) could be the multiple of 3, in which case (x - 5) will also be a multiple of 3. So in any case, (x - 5)(x - 1)x will have a multiple of 3 in it. So k can be 5.
Similarly, (x-1)x(x + 1) will have a multiple of 3 in it. x could be the multiple of 3, (x - 1) could be the multiple of 3 or (x + 1) could be the multiple of 3, in which case (x + 4) will also be a multiple of 3. So in any case, (x - 1)x(x + 4) will have a multiple of 3 in it. So k can be -4.
We cannot say whether (x-1)x(x + 2) will have a multiple of 3 in it and hence if k = -2, we cannot say whether the product is evenly divisible by 3.
Answer (B).
Quote:
Plz Could you please explain how x-5 will also be a multiple of 3. I couldnot understand that part.
If (x - 2) is a multiple of 3, (x - 5), a number 3 places away from (x - 5) will also be divisible by 3.
Say (x - 2) = 9 (a multiple of 3) then (x - 5) = 6 (previous multiple of 3) _________________
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
03 Oct 2014, 07:45
Bunuel wrote:
anilnandyala 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
-4 -2 -1 2 5
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
Hi Bunuel, does "evenly divisible" mean that the dividend on being divided by 3, leave a quotient that is even?? please correct me if i am wrong.
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
03 Oct 2014, 07:49
Expert's post
arnabs wrote:
Bunuel wrote:
anilnandyala 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
-4 -2 -1 2 5
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
Hi Bunuel, does "evenly divisible" mean that the dividend on being divided by 3, leave a quotient that is even?? please correct me if i am wrong.
No, evenly divisible means divisible without remainder, so simply divisible. _________________
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
03 Oct 2014, 07:56
Bunuel wrote:
anilnandyala 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
-4 -2 -1 2 5
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
I am sorry but i did not really get the solution. A little more elaboration would help Bunuel. My main concern here is, if x(x-1)(x-k) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3.
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
03 Oct 2014, 08:59
Expert's post
arnabs wrote:
Bunuel wrote:
anilnandyala 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
-4 -2 -1 2 5
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
Answer: B.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
I am sorry but i did not really get the solution. A little more elaboration would help Bunuel. My main concern here is, if x(x-1)(x-k) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3.
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
03 Oct 2014, 10:17
Bunuel wrote:
vjsharma25 wrote:
Hi bunuel, I don't understand the problem language, it says
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
how does it matter whats the value of K, i can choose x = 3 and the expression will always be divisible by 3.
Am i missing any minor yet important point?
Stem says: "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"
The important word in the stem is "MUST", which means that we should guarantee the divisibility by 3 no matter the value of x (for ANY integer value of x), so you can not arbitrary pick its value.
Hope it's clear.
that was so helpful bunuel ,thank you so much!!!
gmatclubot
Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi
[#permalink]
03 Oct 2014, 10:17
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