Divisibility good one : KAPLAN

Is it A)?

I kno we need to consider fractions here but Still go wiht A)

Question stem: Is x even ?

Statement I

x^3 + x = x (x^2+1)

there can be 3 cases here

case I x is divisible by 4, x is even


case II x^2+1 is divisible by 4, if x is odd x^2+1 can never be divisible by 4. if x is odd x^2 - 1 will always be divisible by 4.

case III x is divisible by 2 and x^2 + 1 is divisible by 2, this cannot be possible because if x is even, x^2 +1 has to be odd, hence cannot be divisible by 2.

Sufficient.

Statement II

5x+4 is divisible by 6


if 5x + 4 = even number [ because only an even number can be divisible by 6], then x has to be an even number.

Hence D.

Here is a rule to remember:

if A is a multiple of 5 and B is also a multiple of 5, then A+B & A-B will also be multiples of 5. This is valid across all numbers not only integers.

crap, i picked D

PC explained why B is good to go, no argument there.

in A, if x^3 + x is divisible by 4, then x^3/4 and x/4. so if x is disible by 4, then it has to be divisible by 2.

niraj_a wrote:crap, i picked D

PC explained why B is good to go, no argument there.

in A, if x^3 + x is divisible by 4, then x^3/4 and x/4. so if x is disible by 4, then it has to be divisible by 2.

You are absolutely correct. I have edited my above post. Answer should be D.

OA IS D

Missed out on statement 2.

Took x=2/5 and wrongly concluded 2/5 is not divisible by 2. All fractions are divisible by 2. Someone correct me if this is a wrong generalization (pretty sure though...)


It would be D).

cramya wrote:Missed out on statement 2.

Took x=2/5 and wrongly concluded 2/5 is not divisible by 2. All fractions are divisible by 2. Someone correct me if this is a wrong generalization (pretty sure though...)


It would be D).

I bit confused myself on fractions but here is my reasoning, I hope you can add on to it

x = 2k + 0

this should be the standard equation if x is divisible by 2

Now k and r both have to be positive integers, therefore I dont think all fractions are divisible by 2.

let me know your thoughts.

let me know your thoughts

My only concern was for statement II if I took x=2/5 (positive as given in the stem and not positive integer)

5*2/5+4 is divisible by 6

So x has to be divisible by 2 if B) has to be sufficient by itself. My thought was (right or worng dont know) that any fraction can be divided in to equal parts 1/3 / 2 or 1/999 / 2 with a remainder 0 so going back to my question/assumption again, can we sagfely say are all fractions are divisible by 2??)


Correct me if I am mistaken here.

Regards,
CR

cramya wrote:
My only concern was for statement II if I took x=2/5 (positive as given in the stem and not positive integer)

5*2/5+4 is divisible by 6
So x has to be divisible by 2 if B) has to be sufficient by itself. My thought was (right or worng dont know) that any fraction can be divided in to equal parts 1/3 / 2 or 1/999 / 2 with a remainder 0 so going back to my question/assumption again, can we safely say are all fractions are divisible by 2??)

Correct me if I am mistaken here.

Regards,
CR

You are absolutely right, I agree.

niraj_a wrote:

in A, if x^3 + x is divisible by 4, then x^3/4 and x/4. so if x is disible by 4, then it has to be divisible by 2.

So according to what you say:

5x+4 is divisible by 6 MEANS

5x/6 and 4/6 are integers ?

cramya wrote:Missed out on statement 2.

Took x=2/5 and wrongly concluded 2/5 is not divisible by 2. All fractions are divisible by 2. Someone correct me if this is a wrong generalization (pretty sure though...)


It would be D).

"divisible by" means that it goes in an integer number of times. The only numbers divisible by 2 are even integers. In other words, 2/5 is NOT divisible by 2, nor is any other fraction.

Doing the math:

(2/5)/2 = (2/5) * (1/2) = 2/10 = 1/5

In fact, we could have rephrased the original question as:

If x > 0, is x an even integer?

(As an aside, the phrase "even integer" is actually redundant - only integers can be "even" or "odd".)

If x > 0, is x an even integer?

Stuart, you made it look a lot simpler! Much appreciated!

I got the right answer, but no one did it the way I did - so I'm just wondering if someone can confirm/deny my method:

St. 1)

x^3+x=4Q+0
= x*(x^2+1)=4Q+0

So, when Q={0,1,2,3...}, left side of equation={0, 4, 8, 12...}respectively.

Let's pick Q=1, which means
x*(x^2+1) (aka 'left side') = 4.
x will then = a decimal; so not an even integer and not divisible by 2. Suff.

St. 2)

5x+4 = 6Z + 0

So, when Z={0,1,2,3...}, left side of equation={0, 6, 12, 18...} respectively.

Let's pick Z=1, which means
5x+4=6
x will then = 2/5 (a decimal), so not an even integer and not divisible by 2. Suff.

Hence D.

I know it's a long explanation, but can someone criticize my approach please?