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

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If x is an integer, then x(x 1)(x k) must be evenly [#permalink]  04 Jan 2008, 20:38
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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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Re: PS - equation divisible by 3. [#permalink]  04 Jan 2008, 21:56
Expert's post
B

Obviously, N=x(x – 1)(x – 2) is product of 3 consecutive integer and N is divided by 3.
So, k should be equal 3n+2:
-7,-4,-1,2,5,8

-2 is out.
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Re: PS - equation divisible by 3. [#permalink]  04 Jan 2008, 23:20
walker wrote:
B

Obviously, N=x(x – 1)(x – 2) is product of 3 consecutive integer and N is divided by 3.
So, k should be equal 3n+2:
-7,-4,-1,2,5,8

-2 is out.

this ps works for me, but wastes time. I see yr approach helpful, but I am not clear why k should be 3n+2. Maybe this is a basis, but I cannt get it. help Pls
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 01:26
Expert's post
Take a look at the integer line:

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

we will use 3 colors to mark numbers.

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders)
we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders.
2. Now we can go back or forward from number 2 by adding or subtracting 3.

Hope this help
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 03:11
ashkrs 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

if we substitute x with 0 or 1 we always obtain 0, which is always evenly divisible by three (maybe the question should have excluded this 2 values for x).

let's try with x=2...the only value of k for which we don't have an even integer in the division is -2, therefore OA can be B
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 04:24
i was stuck between -2 and -4

lets see

if we have -4 then

if x=3 everything is divisible (no need to look at that)
if x=5 or 7, 2, 1 (i.e hint look at x=prime number)

you will quickly notice that -2 works..

B it is..
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 05:14
Any one of 3 consecutive integers is divisible by 3.
so, we have x, x-1 and x-k, for them to be consecutive,
ie we have x-1, x and x-k,
so if x-k = x+1, then they are consecutive.
so, from above k=-1.
also remember that when you add multiples of 3 to a no. divisible by 3 it again gives multiple of 3, so try adding -3, 0, 3, and 6 to -1
you get all but -2
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 18:00
walker wrote:
Take a look at the integer line:

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

we will use 3 colors to mark numbers.

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders)
we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders.
2. Now we can go back or forward from number 2 by adding or subtracting 3.

Hope this help

Hi walker,

you seems to think automatically "k should be equal to 3n+2" after you see problem, right? can you discribe that process of thinking? Thanks!
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Re: PS - equation divisible by 3. [#permalink]  05 Jan 2008, 23:31
Expert's post
sondenso wrote:
walker wrote:
Take a look at the integer line:

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

we will use 3 colors to mark numbers.

-9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15

To catch number that is divisible by 3 we should have 3 different colors (numbers with different reminders)
we can substitute any number for a number with the same color (reminder).

1. at k=2: N=x(x – 1)(x – 2) - we have 3 numbers with different reminders.
2. Now we can go back or forward from number 2 by adding or subtracting 3.

Hope this help

Hi walker,

you seems to think automatically "k should be equal to 3n+2" after you see problem, right? can you discribe that process of thinking? Thanks!

The problem took 15 sec for me and my way was exact as described above.
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Re: PS - equation divisible by 3. [#permalink]  06 Jan 2008, 11:07
ashkrs 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

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3
quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3
for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3
for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3
for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3
for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

So B.
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Re: PS - equation divisible by 3. [#permalink]  06 Jan 2008, 12:27
ashkrs 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

picked 11 for x. 11(11-)(11-k) --> 11(10)(11--2) -2 is the only one that works.

I like walker's approach b/c picking numbers can get u into trouble...

B
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Re: PS - equation divisible by 3. [#permalink]  06 Jan 2008, 20:22
ashkrs wrote:
ashkrs 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

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3
quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3
for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3
for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3
for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3
for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

So B.

ashkrs,

I like this approach,
I forgot the rule that the sum of the number that is divisible by 3 must evenly devide by 3. Thank you
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Re: PS - equation divisible by 3. [#permalink]  08 Jan 2008, 05:32
ashkrs wrote:
ashkrs 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

Well I got this question wrong because I did some calculations wrong and I thought my process was wrong . But I reworked again with same process and found B.

for x(x-1)(x-k) to be divisible by 3 , x + x-1 + x -k => 3x - ( k+1 ) should be divisible by 3
quickly substituting all values here

for -4 , 3x - ( -4 + 1 ) = 3x + 3 ->divisible by 3
for -2 , 3x - ( -2 + 1 ) = 3x - 1 not divisble by 3
for -1 , 3x - ( -1 + 1 ) = 3x ->divisible by 3
for 2 , 3x - ( 2 + 1 ) = 3x + 3 ->divisible by 3
for 5 , 3x - ( 5 + 1 ) = 3x - 6 ->divisible by 3

So B.

Hi, I went first with consecutive integers rule, which is also a good way to solve the problem..but this one is very nice. It is really: you are to have at least one factor which is divisible by three so the product of three different integers be divisible by three
Thanks a lot for the idea
Re: PS - equation divisible by 3.   [#permalink] 08 Jan 2008, 05:32
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