guide 1, pg 98, the data sufficiency example...

[mdobejr]

If x is a positive integer, is x³-3x²+2x divisible by 4?

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

...

My problem is with (2). In the example the rephrased the problem to be "Is x even?" and because z is an integer, 2z must be a multiple of 2. The result will always be a multiple of 2, aka even. so x is even for (2)....

but take z = 0, 0 is an integer... x=0+2= 2... 2 is even BUT
plug it in and (2)³-3(2)²+2(2) = 8-12+4= 0... 0 isn't divisible by 4...so seems to me (2) is not sufficient to answer the question.

discuss:)

[adiagr]

If x is a positive integer, is x³-3x²+2x divisible by 4?

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

...

My problem is with (2). In the example the rephrased the problem to be "Is x even?" and because z is an integer, 2z must be a multiple of 2. The result will always be a multiple of 2, aka even. so x is even for (2)....

but take z = 0, 0 is an integer... x=0+2= 2... 2 is even BUT
plug it in and (2)³-3(2)²+2(2) = 8-12+4= 0... 0 isn't divisible by 4...so seems to me (2) is not sufficient to answer the question.

discuss:)

We have to check whether x³-3x²+2x divisible by 4?


Taking statement (2)

x=2z+2, where z is an integer,

so x is divisible by 2

x³ will be divisible by 4

3x² will be divisible by 4

also 2x will be divisible by 4

So, x³-3x²+2x will be divisible by 4, by remainder theorem

For sake of completion,

similar approach has to be taken for statement (1).

Statement (1) is also sufficient.

Answer to be D

[dinesh19aug]

Hi,
As aditya stated, the answer is D.
An alternative approach that I use in this question is break the problem in simple terms(Primary rule for DS).

Given:
x>0
Eq: can be simplified to x * (x-1) * (x-2)

Stmt1: x= 4y+4 ==> Clearly if we replace the terms in eqn above, the first term x is divisible by 4 . Hence SUFFICIENT.

STMT2: x= 2z+2, put the values in eqn above ==>
(2z + 2) * (2z-1) * (2z). Now any value of z you pick up will either divide (2z+2) or 2z. Because an even number will either be divided by 0 or will leave remaider as 2 or in short will require an extra 2 to be divisible. For example if :

z = 4 => 2z is divisble.
z= 3 > 2z = 6 which is short by 2 hence 2z+2 will be divided by 4.

[mdobejr]

how do you explain z = 0

"but take z = 0, 0 is an integer... x=0+2= 2... 2 is even BUT
plug it in and (2)³-3(2)²+2(2) = 8-12+4= 0... 0 isn't divisible by 4...so seems to me (2) is not sufficient to answer the question."

[dinesh19aug]

:-) The question says that x is a positive integer.
Hence z cannot be 0. z will always be greater than 0. z>0 because 0 is neither positive nor negative.

[debmalya.dutta]

factorising x³-3x²+2x
x(x-1)(x-2)

from statement 1
x=4y+4. so from here since x is divisible by 4 , so is x(x-1)(x-2) . you can explicitly substitute the value of x and also check

from statement x=2z+2 means that x is even
that mean x-1 is odd and x-2 is again even
So x(x-1)(x-2) is basically a product of 2 even and an odd number and hence has be divisible by 4

[dinesh19aug]

factorising x³-3x²+2x
x(x-1)(x-2)

from statement 1
x=4y+4. so from here since x is divisible by 4 , so is x(x-1)(x-2) . you can explicitly substitute the value of x and also check

from statement x=2z+2 means that x is even
that mean x-1 is odd and x-2 is again even
So x(x-1)(x-2) is basically a product of 2 even and an odd number and hence has be divisible by 4

Deb,
Your solution is correct but the reasoning is a little doubtful??product of tow even number and an odd is divisible by 4 -- 2*5*3 not divisible by 4. :-)

It would be true ONLY when the 3 numbers are consecutive and begin with an even number . I read about this rule in in some problem solution in MGMAT guide.

Ex - 2,34
4,5,6
5,6,7 -- No because starts with odd number
18,19,20 etc.

[debmalya.dutta]

Ok.. So let me elaborate....u agree that it's a product of 2 even numbers and an odd number... Let's start from there.. so an even number is always divisible by 2.. Product of 2 even numbers is divisible by 4... You can try it out.... Here on you can follow my reasoning in the above post

[dinesh19aug]

Ok.. So let me elaborate....u agree that it's a product of 2 even numbers and an odd number... Let's start from there.. so an even number is always divisible by 2.. Product of 2 even numbers is divisible by 4... You can try it out.... Here on you can follow my reasoning in the above post

I think we both are saying same thing but using different style. :-) I understand your reasoning now.

[tim]

Okay, everyone seems to be overlooking the original question. Question says x>0, not z>0, so if z=0 and x=2, then we still have the question of what to do when the polynomial is 0..

The answer is super simple: ZERO IS DIVISIBLE BY FOUR! A lot of people miss this, just as they often make the mistake of thinking zero is not an integer. But the best way to understand this is to go back to the definition of divisibility: a is divisible by b if a/b is an integer:

8 is divisible by 4 because 8/4=2 (integer)
5 is not divisible by 4 because 5/4=1.25 (non-integer)
0 is divisible by 4 because 0/4=0 (integer)

But wait, doesn't this mean that 0 is divisible by EVERY integer? Absolutely. The bottom line here is don't make up math rules that aren't true, because they will only frustrate you.. :)