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If x is an integer, then x(x 1)(x k) must be evenly [#permalink]
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04 Jan 2008, 21:38
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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
Show work please.



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Re: PS  equation divisible by 3. [#permalink]
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04 Jan 2008, 22:56
BObviously, 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]
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05 Jan 2008, 00:20
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]
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05 Jan 2008, 02:26
Take a look at the integer line: 9,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15we will use 3 colors to mark numbers. 9,8,7,6,5,4,3,2,1,0,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15To 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]
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05 Jan 2008, 04:11
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
Show work please. 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]
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05 Jan 2008, 05:24
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]
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05 Jan 2008, 06:14
ANSWER IS 2. Any one of 3 consecutive integers is divisible by 3. so, we have x, x1 and xk, for them to be consecutive, ie we have x1, x and xk, so if xk = 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]
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05 Jan 2008, 19:00
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]
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06 Jan 2008, 00:31
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]
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06 Jan 2008, 12:07
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
Show work please. 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(x1)(xk) to be divisible by 3 , x + x1 + 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]
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06 Jan 2008, 13:27
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
Show work please. picked 11 for x. 11(11)(11k) > 11(10)(112) 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]
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06 Jan 2008, 21:22
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
Show work please. 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(x1)(xk) to be divisible by 3 , x + x1 + 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]
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08 Jan 2008, 06:32
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
Show work please. 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(x1)(xk) to be divisible by 3 , x + x1 + 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.
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