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If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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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 A. 4 B. 2 C. 1 D. 2 E. 5
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Last edited by Bunuel on 09 Jul 2013, 09:59, edited 1 time in total.
Renamed the topic and edited the question.



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Re: x(x – 1)(x – k) [#permalink]
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15 Dec 2010, 07:41
There is probably an easier way, but I just used the picking numbers option for this. I chose x=2 2(1)(2k) then just plugged in the answer choices for K until one wasn't evenly divisible by 3. B gives you 8. 8/3 is not an integer. B is the answer.
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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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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.
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Re: x(x – 1)(x – k) [#permalink]
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03 Jan 2011, 22:00
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?



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If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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04 Jan 2011, 03:22
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. Hope it's clear.
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Re: x(x – 1)(x – k) [#permalink]
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04 Jan 2011, 08:46
OK. Now i get it.
Thanks Bunuel.



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Re: x(x – 1)(x – k) [#permalink]
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04 Jan 2011, 09:41
To be divisible by 3, one of these sequences must be divisible by 3.
X(X1) (Xk)
Any 3 sequence number will always be divisible by 3. So X(X1) (x2) is divisible by 3.
K = 2, divisible by 3 K= 5, also a sequence ( parallel ) divisible by 3 K= 1, sequence is (X1) 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
So Answer is (B)



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Re: x(x – 1)(x – k) [#permalink]
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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 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, (x1)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: (x5), (x  4), (x  3), (x  2), (x  1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) etc (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, (x1)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 (x1)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).
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Re: x(x – 1)(x – k) [#permalink]
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05 Jan 2011, 01:29
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
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Re: x(x – 1)(x – k) [#permalink]
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07 Jan 2011, 23:49
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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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,...



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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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07 Mar 2014, 04:52
Plugged in value of x = 8 It comes up 8 x 7 x (8k) Checking for each option available 4 >>> 8+4 = 12.. Divisible by 3 2 >>> 8+2 = 10.. Not divisible by 3 1 >>> 8+1 = 9 .. Divisible by 3 2 >>> 82 = 6 .. Divisible by 3 5 >>> 85 = 3 .. Divisible by 3 Answer = B
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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08 Mar 2014, 05:51
Hats off to Bunuel for the 30 sec. Approach!Couldn't visualize that solution!
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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02 Aug 2014, 01:42
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(x1)(xk) is divisible by 3, take a case when x(x1) is not divisible by 3 and so (xk) has to be divisible by 3. Let us take x=8 and x1=7. Only for the second option we do not get xk divisible by 3.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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02 Sep 2014, 00:21
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, (x1)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:
(x5), (x  4), (x  3), (x  2), (x  1), x, (x + 1), (x + 2), (x + 3), (x + 4), (x + 5) etc
(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, (x1)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 (x1)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 x5 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)
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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03 Oct 2014, 08:45
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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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.



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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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03 Oct 2014, 08:49
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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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.
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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03 Oct 2014, 08:56
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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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(x1)(xk) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3.



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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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03 Oct 2014, 09:59
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: (x1)x(xk). 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 (x1)x(x+2)=(x1)x(x1+3) > which means that (x1) and (x+2) will have the same remainder upon division by 3. Thus for k=2 we won't be sure whether (x1)x(xk) 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(x1)(xk) were to be evenly divisible, then plugging any value for x(lets say 3) should make it evenly divisible by 3. Have you checked this: ifxisanintegerthenxx1xkmustbeevenlydivi106310.html#p846137 ?
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi [#permalink]
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03 Oct 2014, 11:17
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!!!




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