Algebra

[voodoo_child]

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

(1) x^3 + x is divisible by 4.

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

Is there any algebraic/shortcut method for solving 1) and 2) ?
Here's what I tried for 2):-

m=quotient;
5x+4 = 6m + 0 (0=remainder)
=> x= = 6m/5 - 4/5 => x=2(3m/5 - 2/5) => divisible by 2. Is my
understanding correct ?

-Voodoo

[Frankenstein]

Hi,
For(2): 5x+4 = 6m. RHS is 6m which is even. So LHS should be even too. As 4 is even 5x should be even too. That means x has to be even.

Cheers!

[Frankenstein]

Here's what I tried for 2):-

m=quotient;
5x+4 = 6m + 0 (0=remainder)
=> x= = 6m/5 - 4/5 => x=2(3m/5 - 2/5) => divisible by 2. Is my
understanding correct ?

-Voodoo

Hi,
Lets say m=1, u will get x=2(3/5-1/5) => not divisible by 2. So, that is not how you get x even.

Cheers!

[Frankenstein]

Hi again,
For(1): x can be either odd(2k+1) or even(2k)
Case 1: x = 2k+1, x^3+x = x(x^2+1) = (2k+1)[(2k+1)^2+1] = (2k+1)[4k^2+4k+2] = (2k+1)[4k(k+1)+2] = 4k(k+1)(2k+1)+4k+2 = 4(integer)+2. This is never divisible by 4.
So, x has to be even.

Cheers!

[manpsingh87]

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

(1) x^3 + x is divisible by 4.

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

Is there any algebraic/shortcut method for solving 1) and 2) ?
Here's what I tried for 2):-

m=quotient;
5x+4 = 6m + 0 (0=remainder)
=> x= = 6m/5 - 4/5 => x=2(3m/5 - 2/5) => divisible by 2. Is my
understanding correct ?

-Voodoo

interesting thing to note here is that x is a positive integer,,, x=(6m-4)/5; now for m=2; 12-8/5 is not an integer; hence m should be of the form that 6m-4/5 becomes an integer which is possible for m=4; m=9; which will result in the number of the form 10k/5=2k; hence x will always be even...!!!

[Brent@GMATPrepNow]

If x is a positive integer, is x divisible by 2?
(1) x^3 + x is divisible by 4.
(2) 5x + 4 is divisible by 6.
-Voodoo

Statement 1: x^3 + x is divisible by 4
When we factor the expression we see that x(x^2 + 1) is divisible by 4

For x(x^2 + 1) to be divisible by 4, there are essentially 3 possible cases to consider:

case a: x is divisible by 4 (in which case x is divisible by 2)
case b: x is divisible by 2, and x^2 + 1 is divisible by 2 (in which case x is divisible by 2)
case c: x^2 + 1 is divisible by 4, and x is not divisible by 2

case b is impossible, for the following reason: If x is even (i.e., divisible by 2) then x^2 is also even, in which case x^2 + 1 is odd. If x^2 + 1 is odd, it cannot be divisible by 2

case c is impossible for the following reason: First, if x^2 + 1 is divisible by 4, then x^2 + 1 is even, in which case x^2 is odd. If x^2 is odd, then x must be odd.
Now, if x must be odd, it is not possible for x^2 + 1 to be divisible by 4
We know this because, if x is odd, we can say that x = 2k+1 for some integer k (all odd numbers can be written as 2k+1)
This means that:
x^2 + 1 = (2k+1)^2 + 1
= 4k^2 + 4k + 2
= 4(k^2 + k) + 2
Since 4(k^2 + k) is definitely divisible by 4, we can see that 4(k^2 + k) + 2 is NOT divisible by 4, which means x^2 + 1 is NOT divisible by 4.
If x^2 + 1 is NOT divisible by 4, we can eliminate case c, which means case a MUST be true, which means x MUST be divisible by 2.

So, statement 1 is sufficient.

Statement 2: 5x + 4 is divisible by 6
If 5x + 4 is divisible by 6, we know that 5x+4 must be divisible by 2, which means 5x+4 is even
If 5x+4 is even, 5x must be even (since EVEN + EVEN = EVEN)
If 5x is even, x must be even (since ODD x EVEN = EVEN)
If x is even, then x is divisible by 2

So, statement 2 is sufficient, and the answer is D