NUmber Prop

[nycknicks11]

Is there a quick way to do this problem using number prop? (oops wrong forum. i can't move this. sorry)

[Brian@VeritasPrep]

Hey Knicks,

Definitely! And I love that you're thinking that way...the more you're thinking through the Number Properties lens, the easier it is to save time and energy on quant problems.

When they ask "is x divisible by 2", you should really start thinking "what does it mean to be divisible by 2?". Well, that's an even number - the definition of "even" is essentially "divisible by 2". So with that in mind, your only options for the two statements (since we know that x is an integer) are "even", "odd", or "could be either". And we have some really defined rules for addition/subtraction and for multiplication when it comes to evens and odds:

Add/Subtract
Even + Even = Even
Odd + Odd = Even
Odd + Even = Odd

Multiply
Even * Even = Even
Even * Odd = Even
Odd * Odd = Odd

***Note: it's really important to know that these rules exist, but you probably don't need to memorize them...as long as you know they exist you can prove them to yourself with 1s and 2s really quickly.

So for this problem, statement 1 tells us that x^3 + x is divisible by 4. Well, that means that it's even. And we know that the two ways to add to an even number are:

Odd + Odd
or
Even + Even

And since x^3 and x will both be the same, as far as even/odd, we can have either of the above - if x is odd, x^3 will be odd, and we have the Odd + Odd = Even version. If x is even, we have the Even + Even version. So x could be either even or odd, and statement 1 is not sufficient.

Statement 2 says that 5x + 4 is divisible by 6, which means that it's also even. And, again, our only options to add to even are:

Odd + Odd ---> But we can't have that, since we're already stuck with an even + 4 by the statement, so we can't use this one.

Even + Even

This one we can use, but if x were odd, 5x would be odd (Odd*Odd=Odd), so in order for us to use this, x MUST BE even. Therefore, we can prove that x must be even, as that's the only type of number that allows us to satisfy the statement.



I think the biggest key here is to recognize up front that the question is asking "is x EVEN?". From there, "...divisible by 4"; "...divisible by 6" and statements like that are really only worthwhile to you as indications of even or odd, so you don't have to do the algebra...you can just test the number properties.

[cpanagio]

Hi all,

I been working down the list of problems and I cant seem to figure out how the first statement is NOT sufficient.

If (x^3+x)/4 is an integer, then it would follow that x^3/4 + x/4 is an integer. If X is odd, the previous statement cant be an integer therefore forcing x to be even and statement 1 sufficient, right?

[boymeetsworld]

Hi all,

I been working down the list of problems and I cant seem to figure out how the first statement is NOT sufficient.

If (x^3+x)/4 is an integer, then it would follow that x^3/4 + x/4 is an integer. If X is odd, the previous statement cant be an integer therefore forcing x to be even and statement 1 sufficient, right?

But couldn't (x^3)/4 and X/4 both be mixed numbers that possibly add up to an integer?

[Brent@GMATPrepNow]

I noticed that users not logged in cannot view attachments, so I posted my solution in an identical thread in the DS section where everyone can see the original question: https://www.beatthegmat.com/algebra-t83990.html

Cheers,
Brent

[cans]

x is +ve integer. is x even??
a)x(x^2+1) is divisible by 4.
if x=odd, x^2 is odd and thus x^2+1 is even.
x=2m+1; x^2 = 4m^2 + 4m +1. Thus x^2 + 1 is only divisible by 2 and not 4.
and thus x can't be odd.
and thus x is even
Sufficient.
b)5x + 4 is divisible by 6.
thus 5x has to be even. thus x is even.
Sufficient
IMO D

[Gurpinder]

the simplest answer is to remember the factor foundation rule:

if A is a factor of B and B is a factor of C, then A is also a factor of C.


Now. the question is asking if x is divisible by 2.

(1) since W.E is divisible by 4 and 4 is divisible by 2, W.E is also divisible by 2.

Same applies to (2).

Hence (D).

[Ian Stewart]

So for this problem, statement 1 tells us that x^3 + x is divisible by 4. Well, that means that it's even.

But that's not *all* it means. If you rephrase the statement "x^3 + x is divisible by 4" as "x^3 + x is even", you're losing a lot of information, so much information that you'll get the wrong answer here.

I think the biggest key here is to recognize up front that the question is asking "is x EVEN?". From there, "...divisible by 4"; "...divisible by 6" and statements like that are really only worthwhile to you as indications of even or odd, so you don't have to do the algebra...you can just test the number properties.

No, this is not the case. You can see this easily enough by taking a very simple example. If I ask if x is even, and tell you that x + x is divisible by 4, then 2x must be divisible by 4, so x must be divisible by 2. You would get the wrong answer if you rephrased the statement 'x + x is divisible by 4' as 'x + x is even', since then you would think x might be odd.

The same thing happens in the question above. We know from Statement 1 that x(x^2 + 1) is not only even, but is divisible by 4. Since one of x or x^2+1 is odd, the other even, we know that either x is divisible by 4, or that x^2 + 1 is divisible by 4. It's certainly possible for x to be divisible by 4, in which case x is even. It is absolutely impossible, however, for x^2 + 1 to be divisible by 4. For that to be true, x would need to be an odd number. If x is an odd number, then x^2 - 1 = (x+1)(x-1) is always equal to the product of two even numbers, so is always divisible by 4 (in fact, it is always divisible by 8, since it's the product of two consecutive even numbers, and among two consecutive even numbers you always have one multiple of 4). Since when x is odd, x^2 - 1 and x^2 + 1 are consecutive even numbers, only one of them can be divisible by 4. Since x^2 - 1 is divisible by 4, x^2 + 1 is never divisible by 4. So from Statement 1, it's impossible for x to be odd, and Statement 1 is sufficient.