Hello,
I can't seem to wrap my head around the explanation to this problem. I understand that you can pick numbers for this but it seems to be the most time consuming solution. Can someone explain the best way to get to the right answer to this in >2 minutes?
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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- alex.baboulas
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- RonPurewal
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Re: CAT #2- Question 25 (Reject a Value for K)
hey,
* first, i'll suppose you meant to write "<2 minutes" and not ">2 minutes".
by the way, this is a bad assumption to go into every problem with. remember, two minutes is the average time allocated to a quant problem, NOT an upper limit.
because an average time is ... well, an average time, you should expect to run over 2 minutes on just about half of all the problems on the test. (these will be counterbalanced by problems that you finish in under 2 minutes, especially short and sweet things like odd/even, pos/neg, and the like.)
* if you are organized when you plug in numbers, you shouldn't take anywhere close to the two-minute limit here anyway.
viz.:
let's just start from choice (a) and go through the choices.
choice (a):
this is (x)(x - 1)(x + 4).
if x is 1, this is (1)(0)(5) = 0, which is a multiple of every number in the world, including 3.
if x is 2, this is (2)(1)(6), which is divisible by 3.
if x is 3, this is (3)(2)(7), which is divisible by 3.
if x is 4, this is (4)(3)(8), which is divisible by 3.
... and then you'll notice that the red things (the multiples of 3) repeat in cycles of three.
done; that's no more than 15-20 seconds of work if you are organized about it.
choice (b):
this is (x)(x - 1)(x + 2).
if x is 1, this is (1)(0)(3) = 0 again.
if x is 2, this is (2)(1)(4) = 8. hey! that's not a multiple of 3.
done.
ask yourself whether that would take two minutes. or even one minute.
(unless you sit there staring at the problem.)
--
you can also picture all the numbers on the number line, with the multiples of 3 in red:
... 0 1 2 3 4 5 6 7 8 9 ...
once you have this diagram, you can visualize each of the choices, and see whether it's possible to place any of the choices without hitting the red things.
choice (a), for instance, is (x)(x - 1)(x + 4); that's one spot (x), the spot to the left of that (x - 1), and the spot four to the right of it (x + 4). if you try that in different places, you'll notice pretty quickly that you can't avoid the red things.
choice (b) is (x)(x -Â 1)(x + 2). if x is one of the numbers directly before a red one, then this product won't contain anything red. so it's the answer you want.
* first, i'll suppose you meant to write "<2 minutes" and not ">2 minutes".
by the way, this is a bad assumption to go into every problem with. remember, two minutes is the average time allocated to a quant problem, NOT an upper limit.
because an average time is ... well, an average time, you should expect to run over 2 minutes on just about half of all the problems on the test. (these will be counterbalanced by problems that you finish in under 2 minutes, especially short and sweet things like odd/even, pos/neg, and the like.)
* if you are organized when you plug in numbers, you shouldn't take anywhere close to the two-minute limit here anyway.
viz.:
let's just start from choice (a) and go through the choices.
choice (a):
this is (x)(x - 1)(x + 4).
if x is 1, this is (1)(0)(5) = 0, which is a multiple of every number in the world, including 3.
if x is 2, this is (2)(1)(6), which is divisible by 3.
if x is 3, this is (3)(2)(7), which is divisible by 3.
if x is 4, this is (4)(3)(8), which is divisible by 3.
... and then you'll notice that the red things (the multiples of 3) repeat in cycles of three.
done; that's no more than 15-20 seconds of work if you are organized about it.
choice (b):
this is (x)(x - 1)(x + 2).
if x is 1, this is (1)(0)(3) = 0 again.
if x is 2, this is (2)(1)(4) = 8. hey! that's not a multiple of 3.
done.
ask yourself whether that would take two minutes. or even one minute.
(unless you sit there staring at the problem.)
--
you can also picture all the numbers on the number line, with the multiples of 3 in red:
... 0 1 2 3 4 5 6 7 8 9 ...
once you have this diagram, you can visualize each of the choices, and see whether it's possible to place any of the choices without hitting the red things.
choice (a), for instance, is (x)(x - 1)(x + 4); that's one spot (x), the spot to the left of that (x - 1), and the spot four to the right of it (x + 4). if you try that in different places, you'll notice pretty quickly that you can't avoid the red things.
choice (b) is (x)(x -Â 1)(x + 2). if x is one of the numbers directly before a red one, then this product won't contain anything red. so it's the answer you want.
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