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

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible  [#permalink]

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New post 25 Jan 2017, 08:02
1) Three consecutive integers or their multiples will always be divisible by 3.
2) We need to check for the multiples of (x-2) in the answer choices.
a)(x-(-4))=x+4. If we deduct (-6) from (x+4) we get (x-2)
b)(x-(-2))=x+2. The closest we can get to (x-2) is x+2-3=x-1. This must be the odd answer.
c)(x-(-1))=x+1. if we deduct (-3) from (x+1) we get (x-2)
d)(x-2) is already in the right form
e)(x-5). If we add 3 to (x-5), we get the desired (x-2)

As predicted, the correct answer is B.

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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible  [#permalink]

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New post 28 May 2017, 11:07
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.


This problem is best done via testing cases.

If x = 0, then all values are equal to zero and thus divisible by 3 evenly.

If x = 1, then all values are divisible by 3.

If x = 2, then we have the following situation:

A. K = -4

2*1*6 = 12, divisible by 3.

B. K = -2
2*(1)*2 = 4, not divisible by 3.

C. K = -1

2*1*3 = 6, divisible by 3.

D. K = 2

2*(1)*0 = 0, divisible by 3.

E. K = 5

2*(1)(-3) = -6, divisible by 3.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible  [#permalink]

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New post 04 Jul 2018, 06:42
santorasantu wrote:
I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.

Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)


hi pushpitkc in the solution above the product of x*(x-1)*(x-k) is converted into SUM x+x-1+x-k how can that be possible, that we convert PRODUCT of integers into the SUM of integers ? :? can you please explain.

thank you and have a good day :)
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If x is an integer then x(x-1)(x-k) must be evenly divisible  [#permalink]

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New post 04 Jul 2018, 11:14
1
dave13 wrote:
santorasantu wrote:
I tried the problem in a different way:

for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.
Taking the sum of x, (x-1) and (x-k) we have:

x+x-1+x-k = 3x-1-k

now looking at the choices

k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3

so answer choice is (b)


hi pushpitkc in the solution above the product of x*(x-1)*(x-k) is converted into SUM x+x-1+x-k how can that be possible, that we convert PRODUCT of integers into the SUM of integers ? :? can you please explain.

thank you and have a good day :)


Hey dave13

Look at the highlighted portion!

This says - If a number is divisible by 3, the sum of the integers must be a factor of 3

Eg. If the number be 123, the sum of the digits is a factor of 3.
Similarly, if the number is 241 - the sum of the digits is not a factor of 3.

This problem has been explained properly here.

For clarification regarding the approach, you can see this link
https://gmatclub.com/forum/if-x-is-an-i ... l#p1539020

Hope this helps you!
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If x is an integer then x(x-1)(x-k) must be evenly divisible &nbs [#permalink] 04 Jul 2018, 11:14

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