If x is a positive integer, is x^3-3x^2+2x divisible by 4?

Yes, I'd say the difficulty level is ~650.

If x is a positive integer, is x^3-3x^2+2x divisible by 4?

(1) x=4y+4, where y is an integer
(2) x=2z+2, where z is an integer

(1): [4(y+1)]^3 - 3 [4(y+1)]^2 + 2 [4(y+1)] = 4^3(y+1)^3 - 3 (4^2) (y+1)^2 + 2 (4) (y+1) --> divisible by 4: sufficient
(2): 2^3(z+1)^3 - 3 (2^2) (z+1)^2 + 2 (2) (z+1) --> divisible by 4: sufficient
==> Answer is D

For the second statement could one only say that since x is a multiple of 2 and thus even then x-2 will also be even and that means that E*E = E and multiple of 4?

Thanks

Cheers
J

Bunuel - could we have solved/approached the question this way?

Given the question stem - x^3-3x^2+2x - factor out an "x" and apply the FOIL method --> therefore we are left with
x(x^2 - 3x +2x) = 0 --> x(x - 2)(x - 1)
Therefore, x must equal 0, 1, or 2

With that said, plug in the values of 0, 1, and 2 into each of statement one and two to determine if they are sufficient
Statement 1 --> only 0 works
Statement 2 --> both 0 and 2 work (due to the value being an integer)

Therefore, OA is D because we can determine what exact values of of the question stems can be valid.

Let me know what you think, just trying to help out with different ways to approach this question.

Your approach is not correct. Notice that we are NOT told that x^3-3x^2+2x is 0, thus your derivation that x is 0, 1, or 2 is not right.

Is it enough to note that we have (x)(x-2)(x-1)

So then if x is even so will (x-2), so basically just knowing that x is even for each statement separately?

Or do we need to go through all the replacing and factorization for each statement?

Please advice
Thanks!
Cheers!
J

Didn't you answer your own question?

\(x^3 - 3x^2+2x=x (x-1) (x-2)\) --> if x is even, then x-2 is also even, thus x(x-1)(x-2) is divisible by 4.

This is how I did it,

I think that if we have 3 consecutive integers in the format x(x-1)(x-2) and x is even, then it will be divisible by 4 (2)(3)(4)

If x = Odd it may not be divisible by 4 (3)(2)(1)

Forget conventional ways of solving math questions. In DS, Variable approach is the easiest and quickest way to find the answer without actually solving the problem. Remember equal number of variables and independent equations ensures a solution.

If x is a positive integer, is x^3-3x^2+2x divisible by 4?

(1) x=4y+4, where y is an integer
(2) x=2z+2, where z is an integer

If we modify the question, x^3-3x^2+2x=4t? (t is an integer), --> x(x^2-3x+2)=4t? and we ultimately want to know whether
x(x-1)(x-2)=4t, which is the same as whether x-2 is even
For condition 1, from x=4y+4, x-2=4y+2=2(2y+1). This is even and the condition is sufficient
For condition 2, from x=2z+2, x-2=2z. This is also even and the condition sufficiently answers the question 'yes'
The answer becomes (D).

Once we modify the original condition and the question according to the variable approach method 1, we can solve approximately 30% of DS questions.

x^3 - 3x^2+2x divisible by 4
x(X^2-3x+2)
x(x-1)(x-2)

so it is consecutive number.
any three positive consecutive numbers are divisible by 4..
Am i thinking something wrong here?

I simplified the prompt to is x(x-2)(x-1)/4.?

In S1: Any value of y would satisfy the equation, hence it is sufficient
In S2: All the values of z above integer 0 would satisfy the condition, and if I consider z=0, then the numerator becomes 0, and 0 divided by any integer will yield 0. Hence it should be sufficient.

I always pick numbers for testing cases in DS problems. Is that a good practice.?

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