Even and Odd
Definition: Suppose k is an integer. If there exists an integer r such that k=2r+1, then k is an odd number. If there exists an integer r such that k=2r, then k is an even number.
Explanation: as long as an integer can be divided by 2, it is an even number.
Zero is an even number.

Positive and Negative
Definition: A positive number is a real number that is greater than zero. A negative number is a real number that is smaller than zero.
Zero is not positive, nor negative.

very useful thread

do you, guys, know what the "line test method" is, which is discussed here

http://www.gmatclub.com/phpbb/viewtopic ... equalities

Did those guys mean just picking numbers or ..?

Cancelling out "common terms" on both sides of an equation

You need to be very careful when you do algebra derivations. One of the common mistake is to divide both side by "a common term". Remember you can only do this safely if the "common term" is a constant. However you CAN't do it if it contains a variable.

Example:

x(x-2)=x
You can't cancel out the x on both side and say x=3 is the solution. You must move the x on the right side to the left side.
x(x-2)-x=0
x(x-2-1)=0
The solutions are: x=0 and x=3
The reason why you can't divided both sides by x is that when x is zero, you can't divide anything by zero.

Equally important if not more, is that you CAN'T multiple or divide a "common term" that includes a variable from both side of an inequality. Not only it could be zero, but it could also be negative in which case you would need to flip the sign.

Example:

x^2>x
You CAN'T divided both sides by x and say x>1. What you have to do is to move the right side to the left:
x^2-x>0
x(x-1)>0
Solution would be either both x and x-1 are greater than zero, or both x and x-1 are smaller than zero. So your solution is: x>1 or x<0

Example:

x>1/x
Again you CAN'T multiply both sides by x because you don't know if x is positive or negative. What you have to do is to move the right side to the left:
x-1/x>0
(x^2-1)/x>0
If x>0 then x^2-1>0 =>x>1
If x<0 then x^2-1<0 =>x>-1
Therefore your solution is x>1 or 0>x>-1.
You could also break the original question to two branches from the beginning:
x>1/x
if x>0 then x^2>1 =>x>1
if x<0 then x^2<1 => x>-1
Therefore your solution is x>1 or 0>x>-1.

I really struggle with the number property stuff. i guess i didn't pay enough attention back in 9th grade. do you have any broader suggestions if you will. i always try plugging but a lot of times, you lead oyuirself down the wrong path.

Hong thanks for this thread, has been a big help!

and for 11 which i got on an OG problem recently

Start with the units digit, add every other digit and remember this number. Form a new number by adding the digits that remain. If the difference between these two numbers is divisible by 11, then the original number is divisible by 11.

Examples:
Is the number 824472 divisible by 11? Starting with the units digit, add every other number:2 + 4 + 2 = 8. Then add the remaining numbers: 7 + 4 + 8 = 19. Since the difference between these two sums is 11, which is divisible by 11, 824472 is divisible by 11.

Is the number 49137 divisible by 11? Starting with the units digit, add every other number:7 + 1 + 4 = 12. Then add the remaining numbers: 3 + 9 = 12. Since the difference between these two sums is 0, which is divisible by 11, 49137 is divisible by 11.

Sometime back, someone asked me how to find the GCF and LCM. Thought it'll be useful to throw it in this thread so members preparing for gmat and need a quick refresh can go through this.

GCF -> Greatest Common Factor
-> Largest possible common factor between numbers

LCM -> Lowest Common Multiple
-> Largest possible common multiple between numbers

To find the GCF/LCM, you will need to do prime-factorization. This means reducing a number to its prime-factor form.

E.g. 1

GCF/LCM of 4,18

4 = 2*2
18= 2*3*3

To find the GCF, take the multiplication of the common factors (pick the lowest power of the common factors) In this case, GCF = 2.

To find the LCM, take the multiplication of all the factors (pick the higest power of the common factors). In this case, LCM=2*2*3*3=36

E.g. 2

GCF/LCM of 4,24

4 = 2*2
24 = 2*2*2*3

GCF = 2*2 = 4
LCM = 2*2*2*3 = 24

Good point. Negative numbers need to flip the signs again. I'll edit the post. Thanks!

alright, some more as I've promised....

Working with absolute numbers:
When solving for absolute equations, be sure to test both positive and negative values since the absolute number only represent the distance from 0 on the number line.

For instance, |x| = 2 could be x=-2 or x=2

Here's one way to solve absolue inequalities:

|x-4| < 9
We need to solve both positive and negative x

Solving for positive: x-4 < 9 --> x <13
Solving for negative:
-x+4 < 9 --> - x < 5 --> x>-5 (dividing by negative number, switch the sign)

So now we have -5<x<13

Yes, that's a good one. I've been long thinking of doing one about the absolute value questions but have forgot about it.

Absolute values
The way to solve this kind of questions is to break the equation (inequality) into two parts, one is when the value is non negative, the other is when the value is negative.

For example:

|x-4|<9

You break it into two parts:
If x-4>=0, then x-4<9, solve for both you get x>=4, x<13. So your solution is 4<=x<13.
If x-4<0, then -(x-4)<9, ie x-4>-9. Solve for both you get x<4, x>-5. So your solution for this part is -5<x<4.

Combine the two solutions, you get -5<x<13 as your final solution.

Another example:
|x+4|>4
If x+4>=0, then x+4>4. Solve for both you get x>=-4, x>0. So your solution is x>0.
If x+4<0, then -(x+4)>4, ie. x+4<-4. Solve for both you get x<-4, x<-8. So your solution is x<-8.
You final solution is x>0 or x<-8.

The same strategy can apply to square questions.
For example: (x+4)^2>4
You could solve it this way:
x^2+8x+12>0
(x+2)(x+6)>0
x>-2 or x<-6
Or you can solve it this way:
If x+4>=0 then x+4>2. Solve for them you get x>-2.
If x+4<0 then x+4<-2. Solve for them you get x<-6.

|y|>|y+1|
if y>=0, y+1>=0, y>y+1, no solution.
if y<0, y+1<0, -y>-(y+1), solution is y<-1
if y>=0, y+1<0, y>-(y+1), no solution.
if y<0, y+1>=0, -y>y+1, solution is -1<=y<-1/2
So your final solution is y<-1/2
You could also solve this question by going the square route.
y^2>(y+1)^2
y^2>y^2+2y+1
2y+1<0
y<-1/2