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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 -4 -2 -1 2 5

Any short cut, the OE is not helpful

make or break the expression into a form such that it is either in a consecutive form or a multiple of 3 as under: 1. x(x – 1)(x + 4) = x(x – 1)(x + 1 +3) = x(x – 1)[(x + 1) +3] 2. x(x – 1)(x + 2) 3. x(x – 1)(x + 1) 4. x(x – 1)(x - 2) = x(x – 1)(x + 1 - 3) 5. x(x – 1)(x - 5) = x(x – 1)(x +1 - 6)

B. except -2, all could be k such that x(x – 1)(x - k) is divisible by 3. in all options except second, the integers are in eith consecutive order or in at least one of them is divisible by 3 but B is not at any case. _________________

I say shortcut for such problems is always trying a couple of numbers and see which one satisfies. for eg Take x=5 Just plug in the options. B doesnt satisfy

1. x(x – 1)(x – k) is divisible by 3 if x,x-1,x-k is consecutive integers: x(x – 1)(x – 2) for k=2 2. Now, subtraction or addition of 3 to any of members does not change divisibility: -7,-4,-1,2,5,8. -2 is out of the set. _________________

i dont fully get it ... what if x=3 ? thats what i used, and it turned out that if you use that, then for all k values in the question, the product is divisible by 3 :S

I guess for these types of questions, you shouldnt use the value youre dividing by ?

i dont fully get it ... what if x=3 ? thats what i used, and it turned out that if you use that, then for all k values in the question, the product is divisible by 3 :S

I guess for these types of questions, you shouldnt use the value youre dividing by ?

Reread the question - must be evenly divisible by three [at any x]

x=3 and k=any - works x=5 and k=0 - does not work _________________

i dont fully get it ... what if x=3 ? thats what i used, and it turned out that if you use that, then for all k values in the question, the product is divisible by 3 :S

I guess for these types of questions, you shouldnt use the value youre dividing by ?

Reread the question - must be evenly divisible by three [at any x]

x=3 and k=any - works x=5 and k=0 - does not work

thanks walker, but im still confused. i get the general concept, but having problems applying it. i assume evenly divisible means that the quotient is an even number ? how does this change things ? gah im so confused !

thanks walker, but im still confused. i get the general concept, but having problems applying it. i assume evenly divisible means that the quotient is an even number ? how does this change things ? gah im so confused !

the quotient x(x-1)(x-k) div 3 is always even number, because one of the number, x or (x-1), is always even. _________________