Data Sufficiency

If x is any positive integer, is x divisible by 2?

1) x3(x cubed) + x is divisible by 4.
2) 5x + 4 is divisible by 6.

I would go with D)

Stmt II

5x+4/6 = k (some integer)

x = 6k-4/5

6k-4 is always even so even/odd is even or decimal but since x is given to be a positive integer x is always divisible by 2 since its even

SUFF

Stmt I

x^3+x is divisible by 4

This means x^3 is divisible by 4 and x is diviisble by 4

So x is defnitely divisible by 2

SUFF

cramya wrote:I would go with D)

Stmt II

5x+4/6 = k (some integer)

x = 6k-4/5

6k-4 is always even so even/odd is even or decimal but since x is given to be a positive integer x is always divisible by 2 since its even

SUFF

Stmt I

x^3+x is divisible by 4

This means x^3 is divisible by 4 and x is diviisble by 4

So x is defnitely divisible by 2

SUFF

cramya can you explain where all we can apply similar rules
x^3+x is divisible by 4

"This means x^3 is divisible by 4 and x is diviisble by 4"

Amit,
Here it goes(hope I dint add too much confusion here):

Two cases to consider here:

Case 1: x+x^3 will be divisible by 4 if x and x^3 individually are divisible by 4. If 2 components individiaully are divisible by a number then their sum definitely is

This means x is divisible by 2

Case 2: x and x^3 are both not divisible by 4 then x+x^3 will be divisible by 4 if x and x^3 remainders sum is divisible by 4

Whats unique about x^3 if I am not mistaken is that x will give the same remainder as x^3 when x^3 is not divisible by that number. Therefore their sum will never be diviisble by 4

Eg: x=3 x^3=27 when divided by 4 Both give a remainder of 3 so the sum of the remainders 6 is not divisble by 4 so x+x^3 is not divisible by 4

The 2nd CASE is not possible since x^3+x is given to be divisible by 4 so it has to be case 1

Hope this helps!

Someone feel free to correct me if I am mistaken in any of the statements made above.

I have tried to adapt the explanation to this problem (since we are dealing wiht the same number's different powers) but please see Ian's explanation to my question here:

https://www.beatthegmat.com/is-x-16-y-8- ... 26911.html


Regards,
Cramya

Thanks cramya...

If x is any positive integer, is x divisible by 2?

1) x3(x cubed) + x is divisible by 4.

X(x^2+1)

Either X or X^2+1 is divisible with 4

or they both are divisible with 2


X=even , x^2+1 can not be even so we rule out that option
X=odd NOPE

so either X or X^2+1 needs to be divided by 4

if X is divided by 4 it is divisible by 2

X^2 needs to be ODD to be divisible with 4 because there is + 1

SO SUF!

try 4,8,12, 16, and etc...



2) 5x + 4 is divisible by 6.[/quote]


means it is divisible by 3 and 2 at the same time

well if it is divisible by 2 it is an even number

5x+4 = EVEN

5x = EVEN-4

5x = ANOTHER EVEN

x can not be ODD

X can be EVEN

so X can be divided by two

SUF

Choose D

Thanks cramya...

No probs buddy; anytime! Good luck.

Logitech, camya

appreciate your inputs!

x * ( x^2 + 1)

now if x is multiple of 4 ... case suff
now if x is odd

odd * ( odd * odd + 1)

now this has to be the odd*odd + 1 has to be the multiple of 4


now is there any rule that

odd * odd + 1 cannot be multiple of 4


PLs help

ronniecoleman wrote:Logitech, camya
appreciate your inputs!

x * ( x^2 + 1)
now if x is multiple of 4 ... case suff
now if x is odd
odd * ( odd * odd + 1)
now this has to be the odd*odd + 1 has to be the multiple of 4

now is there any rule that
odd * odd + 1 cannot be multiple of 4

PLs help

odd*odd + 1 may or may not be divisible by 4

but here we are considering the square of a number
lets take a odd number x = 2b+1
where b is any non-negetive integer

x^2 + 1 = (2b+1)^2 + 1
(4b^2 + 4b + 1) + 1
(4b^2 + 4b + 2)
this number will always leave a rem. = 2.

hope that clears your doubt