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You might find this useful . I got this from other sources and compiled.

Number property theory Divisibility rules: - Know them well

Integer is divisible by: 2 - Even integer 3 - Sum of digits are divisible by 3 4 - Integer is divisible by 2 twice or Last 2 digits are divisible by 4 5 - Last digit is 0 or 5 6 - Integer is divisbile by 2 AND 3 8 - Integer is divisible by 2 three times 9 - Sum of digits is divisible by 9 10 - Last digit is 0

Some other things to note: - If 2 numbers have the same factor, then the sum or difference of the two numbers will have the same facor.

(e.g. 4 is a factor of 20, 4 is also a factor of 80, then 4 will be a factor of 60 (difference) and also 120 (sum))

- Remember to include '1' if you're asked to count the number of factors a number has ----------------------------------------------------------------------------------------------------------- Adding/subtracting two odds or two evens --> even Add/ Subtract an odd and an even --> od

Multiplication with at no even number --> odd ** Even number in a multiplication will always ensure an even product

Interesting properties: - Adding n consecutive integers will yield a sum that is divisible by n (i.e. n will be a factor of the sum)

E.g. Adding 3,4,5,6 will give a sum with 4 as a factor Adding 2,3,4 will give a sum with 3 as a factor 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

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 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 Working with Ratios

Ratio questions are very easy to solve if you have mastered the way of thinking.

Basically if you have a/b=c/d (or a:b=c:d) then you can immediately derive a variaty of correlated ratios, such as: a/(a+b)=c/(c+d) a/(a-b)=c/(c-d) (a+b)/(a-b)=(c+d)/(c-d) (a+c)/(b+d)=c/d (a-c)/(b-d)=c/d etc

Basically, you can do all kinds of additions and subtractions.

Example:

a/b=3/5 (1) 2a-b=4 (2) What is a? From (1) we get a/(2a-b)=3/1, so a=3*4=12 Explanation: a is 3 share, b is 5 share. Two a is 6 share, 2a-b is one share. If one share is 4, then 3 share is 12.

Of course this question can be solved using the more traditional algebra approach: b=5/3a substitute in (2) 2a-5/3a=4 1/3a=4 a=12

You can see the two approaches are really the same in nature. However the first approach is very straight forward and does not involve calculation in fractions. Sometimes it can save you lots of time, especially when using this method with word problems such as mixture problems. -=-------------------------------------------------------------------------------------------------------- Square root

A square root, also called a radical or surd, of x is a number r such that r^2=x. The function r=sqrt(n) is therefore the inverse function of f(x)=x^2 for x>=0.

Eg. if x<0, sqrt (x^2)=-x

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.

Basic rules for inequalities: (in the example: a>b>0, c>d>0)

You need to flip signs when both side are multiplied by a negative number: -a<-b, -c<-d

You need to flip signs when 1 is divided by both side: 1/a<1/b, 1/c<1/d

You can only add or multiply them when their signs are in the same direction: a+c>b+d ac>bd

You can only apply substractions and divisions when their signs are in the opposite directions: a>b, d<c a-d>b-c a/d>b/c (You can't say a/c>b/c. It is WRONG)

Deal with negative numbers: -a<-b<0, -c<-d<0 Then -a-c<-b-d<0 -a-(-d)<-b-(-c) However the sign needs to be flipped one more time if you are doing multiplication or division (because you are multiplying/dividing a negative number): (-a)*(-c)>(-b)*(-d) (-a)/(-d)>(-b)/(-c)

For example: If x<-4, y<-2, we know that xy>8, but we don't know how x/y compare to (-4)/(-2)=2 since you can only do division when their signs are in different directions If x>-4 and y<-2 then x/y<2 but we don't know how xy is compared to 8 since we can only do multiplication when their signs are the same direction.

It is easier to do the derivation, though, if you first change them to postive. For example: If x<-4, y<-2, then -x>4, -y>2, xy>8 If x<-4, y<2, then -x>4, y<2, -x/y>2, x/y<-2

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.

Moved to the proper forum. Also have a quick suggestion - you may want to add some formatting/style to make it easier to read/use. Something similar to this: math-number-theory-88376.html _________________

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Thanks to another GMAT Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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