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

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If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 31 Jan 2012, 14:49
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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

[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
[Reveal] Spoiler: OA

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Re: Divisible by 3 [#permalink] New post 31 Jan 2012, 15:17
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Hi, there. I'm happy to help with this. :)

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

Think about it this way:

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question:
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


All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike :)
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Re: Divisible by 3 [#permalink] New post 31 Jan 2012, 15:06
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enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.


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.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 04 Oct 2012, 22:10
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enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.


Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.
In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.
(A) 2 - (-4) = 6 NO
(B) 2 - (-2) = 4 BINGO!

Answer B.
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If x is an integer.then x(x-1)(x-k) [#permalink] New post 24 Feb 2013, 09:45
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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


Need help..............
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Re: If x is an integer.then x(x-1)(x-k) [#permalink] New post 24 Feb 2013, 09:54
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mun23 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


Need help..............


Merging similar topics. Please refer to the solutions above.
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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.


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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 31 Jan 2012, 15:09
Thanks very much for a thorough explanation.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 31 Jan 2012, 15:18
Thanks Mike. Really appreciate your solution too.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 31 Jan 2012, 15:34
This is my 1st post :) finally thought of jumping in instead of just being an observer :-D
I attacked this problem in a simple way. As it states it is divisible by 3
that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3
so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 31 Jan 2012, 15:37
Thanks and welcome to GMAT Club Azim. I am sure you will have a great experience from a very helpful community.
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Re: Divisible by 3 [#permalink] New post 04 Oct 2012, 17:59
Bunuel wrote:
enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.


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.



Bunuel,

Would this approach work for all integer divisors (NOT the 30 sec approach)?
Say the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 05 Oct 2012, 02:37
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 05 Oct 2012, 02:56
Avantika5 wrote:
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.


On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.
It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 06 Oct 2012, 00:10
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot... :)
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 06 Oct 2012, 01:10
Avantika5 wrote:
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot... :)


Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.
But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.
Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...
Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)

Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?
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Last edited by EvaJager on 06 Oct 2012, 01:46, edited 1 time in total.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 06 Oct 2012, 01:25
1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases.
2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7
3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3
4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3
5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3
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Re: Divisible by 3 [#permalink] New post 16 Feb 2013, 11:34
mikemcgarry wrote:
Hi, there. I'm happy to help with this. :)

The rule that the product of three consecutive integers is a good start, but not the be all and end all.

Think about it this way:

number = x(x – 1)(x – k)

So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --

k = 2 ----> x(x – 1)(x – 2)

k = -1 ----> x(x – 1)(x + 1)

Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.

(x - 2) - 3 = (x - 5)

(x - 5) - 3 = (x - 8)

(x - 2) + 3 = (x + 1), which we have already

(x + 1) + 3 = (x + 4)

(x + 3) + 3 = (x + 7)

So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.

Back to the question:
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


All of those choices give us a term on our list except for (B) -2.

BTW, notice all the answer choices are spaced apart by three except for (B).

Does that make sense? Please do not hesitate to ask if you have any questions.

Mike :)


Great posts everyone, including you Mike, who I am now quoting, but I would like to add that..

The first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...

It appears that was left out :beer
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink] New post 17 Oct 2013, 14:25
enigma123 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

The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?

Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.


Hi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks
Re: If x is an integer then x(x-1)(x-k) must be evenly divisible   [#permalink] 17 Oct 2013, 14:25
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