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# If x is an integer, then x(x – 1)(x – k) must be evenly divi

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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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26 Oct 2015, 18:54
VeritasPrepKarishma wrote:
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-5), (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, (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.

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)

Hi Karishma, Initially I was trying the same method only, though I could not figure out the following scenario (x-1) x (x+1) (x+2), now say (x-1) is divisible by 3, in that case (x+2) i.e. k = -2 will also be divisible by 3 and hence in that case (x-1) x (x+2) would be evenly divisible by 3. Could you please tell me what am I missing here?
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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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06 Mar 2016, 19:39
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

to be evenly divisible by 3, we either have to have 3 consecutive integers, x(x-1)(x-2), or 2+/-3 or 2+/-6
D=2 - can be so out
2+3 = 5 - E can be so out
2-3 = -1 - C can be so out
-1-3=-4 - A can be so out
only B remains
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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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14 Mar 2016, 06:04
Here Take 5 for x hence 5 is divisible by all except when k=-2
so 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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15 Mar 2016, 08:35
1
shameekv wrote:

Hi Karishma, Initially I was trying the same method only, though I could not figure out the following scenario (x-1) x (x+1) (x+2), now say (x-1) is divisible by 3, in that case (x+2) i.e. k = -2 will also be divisible by 3 and hence in that case (x-1) x (x+2) would be evenly divisible by 3. Could you please tell me what am I missing here?

So the question is "... MUST BE divisible by ..."

Note that (x - 1)x(x + 2) may or may not be divisible by 3. This happens in case (x +1) is a multiple of 3.

e.g. say x = 5
(x - 1)x(x + 2) = 4*5*7 --> Not divisible by 3

Say x = 6
(x - 1)x(x + 2) = 5*6*8 --> Divisible by 3

So it is not necessary that (x - 1)x(x + 2) will be divisible by 3.

Why must the others be divisible by 3?
Say, for example take k = -4
(x - 1)x(x + 4)
This is equivalent to (x - 1)x(x + 1) in case we are talking about divisibility by 3. We know that (x - 1)x(x + 1) MUST have a multiple of 3.
So (x - 1)x(x + 4) MUST have a 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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16 Mar 2016, 00:50
Take X as % and and we can easily say that for k=-2 => the equation will not be divisible by 3=> 5*4*7 => not a multiple of 3
hence 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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05 Oct 2016, 16:33
tfincham86 wrote:
There is probably an easier way, but I just used the picking numbers option for this.

I chose x=2
2(1)(2-k) 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.

that is exactly what I did and got lucky I tested B: -2 second and just 8/3 was the only incorrect result! But since we both chose x=2 to start and that followed the prompt's "rules," we should be able to trust the answer without plugging in all 5. Bunuel had a nice explanation, I didn't think that route
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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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23 Feb 2019, 05:54
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Re: If x is an integer, then x(x – 1)(x – k) must be evenly divi   [#permalink] 23 Feb 2019, 05:54

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