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The following resources were consulted in the design of this handout:
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This is part of our GRE Math Essentials - A most comprehensive handout!!! is the best complement to our GRE Math Book. It provides a cutting-edge, in-depth overview of all the math concepts from basic to mid-upper levels. The book still remains our hallmark: from basic to the most advanced GRE math concepts tested during the exam. Moreover, the following chapters will give you many tips, tricks, and shortcuts to make your quant preparation more robust and solid.


The students who evaluate an expression must follow a precise order or nomenclature. The assumption is that only one can be the result. Turns out, if the students do not follow step by step the order the result will be different and probably wrong.

The acronym of this is the P.E. MD.AS: 

P. Parenthesis (which are a group of symbols):
E. Exponents
MD. Multiplication and/or divisions are managed simultaneously.
AS. Addition and/or Subtraction are managed simultaneously.

First: solve the expressions inside the brackets ( they could be (-) or [-] or {-})

\(25 \times 4 \div \bigl\{[25-18+3] + [7 \times 8] \times [4 \times (18 + \sqrt{16})^2]\bigl\}\)

Second: evaluation of the elements raised to the power or that have exponents

Third: perform divisions and multiplications. Usually does not matter from left to right or the reverse

Fourth: perform addition and subtractions. Usually does not matter from left to right or the reverse



 
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Note \(\sqrt{negative numbers}\) or \(\frac{number}{0-zero}\) are NOT real !!

On the GRE by default, any number mentioned in the quant section is a real number so must NOT assume that it is an integer unless stated clearly.


1) The Number Line


A real number line, or simply number line, allows us to visually display real numbers by associating them with unique points on a line. The real number associated with a point is called a coordinate. A point on the real number line that is associated with a coordinate is called its graph. To construct a number line, draw a horizontal line with arrows on both ends to indicate that it continues without bound. Next, choose any point to represent the number zero; this point is called the origin.

The scale need not always be one unit. Each tick mark might represent a unit fraction.







The opposite of any real number a is \(−a\). Opposite real numbers are the same distance from the origin on a number line, but their graphs lie on opposite sides of the origin and the numbers have opposite signs.

The opposite of a negative number is positive: \(−(−a) = a\).


2) Identify the Place Value of a Digit

The place value of a digit is simply the different value that each place has in a number.





3) Absolute Value


The absolute value of a real number a, denoted |a|, is defined as the distance between zero (the origin) and the graph of that real number on the number line. Since it is a distance, it is always positive.

\(|-4|=4\) and \(|4|=4\)





Also, it is worth noting that \(|0|=0\)


Think of a real number whose distance to the origin is 5 units. There are two solutions: the distance to the right of the origin and the distance to the left of the origin, namely, {±5}. The symbol (±) is read “plus or minus” and indicates that there are two answers, one positive and one negative.

\(|-5|=5\) and |\(5|=5\)

See more here: Absolute Numbers


Tip

If there is an even number of consecutive negative signs, then the result is positive. If there is an odd number of consecutive negative signs, then the result is negative.

KEY TAKEAWAYS

• Any real number can be associated with a point on a line.
• Create a number line by first identifying the origin and marking off a scale appropriate for the given problem.
• Negative numbers lie to the left of the origin and positive numbers lie to the right.
• Smaller numbers always lie to the left of larger numbers on the number line.
• The opposite of a positive number is negative and the opposite of a negative number is positive.
• The absolute value of any real number is always positive because it is defined to be the distance from zero (the origin) on a number line.
• The absolute value of zero is zero.




Source of the images
  1. Ryoji Iwata - Unsplash
  2. Types of Numbers at NorthAmpton Community College
  3. Beginning algebra - 2012 Book Archive under Creative Commons Attribution 3.0 International license
  4. Prealgebra by Authors: Lynn Marecek, MaryAnne Anthony-Smith on Openstax - Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License
 
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­Absolute Value Definition

The absolute value or magnitude of a real number \(x\) is denoted by |x| and is defined by 

\(|x| =\left\{ \begin{array}{cc|r} x, & \mbox{if}  & x \geq 0 \\  -x, & \mbox{if} & x <0 \end{array}\right}\)

where x is also called the "argument"


Steps for Solving Linear Absolute Value Equations: i.e. \(|ax + b| = c\)

1. Isolate the absolute value.

2. Identify what the isolated absolute value is set equal to…
  • If the absolute value is set equal to zero, remove absolute value symbols & solve the equation to get one solution.
  • If the absolute value is set equal to a negative number, there is no solution.
  • If the absolute value is set equal to a positive number, set the argument (expression within the absolute value) equal to the number and set it equal to the opposite of the number, using an ‘or’ statement in between the two equations. Then solve each equation separately to get two solutions.

Examples

\(|3x+12|+7=7\)
\(|3x+12|=0\)
Because this equal to zero, we have ONE solution

\(|3x-7|+7=2\)
\(|3x-7|=-5\)
Because this is equal to a negative number, we have NO solution

\(|3x-7|+7=9\)
\(|3x-7|=2\)
Because this is equal to a positive number, we have TWO solution


Steps for Solving Linear Absolute Value Inequalities: i.e. \(|ax+b| \leq c\)

1. Isolate the absolute value.
2. Identify what the absolute value inequality is set “equal” to…

ZERO

a. If the absolute value is less than zero, there is no solution.
b. If the absolute value is less than or equal to zero, there is one solution. Just set the argument equal to zero and solve.
c. If the absolute value is greater than or equal to zero, the solution is all real numbers.
d. If the absolute value is greater than zero, the solution is all real numbers except for the value which makes it equal to zero. This will be written as a union. 

NEGATIVE Number

e. If the absolute value is less than or less than or equal to a negative number, there is no solution. The absolute value of something will never be less than or equal to a negative number.
f. If the absolute value is greater than or greater than or equal to a negative number, the solution is all real numbers. The absolute value of something will always be greater than a negative number. 


POSITIVE Number

g. If the absolute value is less than or less than or equal to a positive number, the problem can be approached two ways. Either way, the solution will be written as an intersection.
    i. Place the argument in a 3-part inequality (compound) between the opposite of the number and the number, then solve.
    ii. Set the argument less than the number and greater than the opposite of the number using an “and” statement in between the two inequalities.


If the absolute value is greater than or greater than or equal to a positive number, set the argument less than the opposite of the number and greater than the number using an ‘or’ statement in between the two inequalities. Then solve each inequality, writing the solution as a union of the two solutions.


RELATIONSHIP BETWEEN SQUARE ROOTS AND ABSOLUTE VALUES

Recall from algebra that a number is called a square root of \(\sqrt{x}\) if its square is \(\sqrt{x}\). Recall also that every positive real number has two square roots, one positive and one negative; the positive square root is denoted by \(\sqrt{x}\) and the negative square root by \(-\sqrt{x}\). For example, the positive square root of 9 is \(\sqrt{9} = 3\), and the negative square root of 9 is \(− \sqrt{9} = −3\).

NOTE: Readers who may have been taught to write √9 as ±3 should stop doing so, since it is incorrect.


It is a common error to replace \(\sqrt{x^2}\) by x. Although this is correct when x is nonnegative, it is false for negative x. For example, if \(x = −4\), then

\(\sqrt{x^2}=\sqrt{(-4)^2}=\sqrt{16}=4 \neq x\)


For any real number \(a\), \(\sqrt{a^2}=|a|\)

Since \(a^2 = (+a)^2 = (−a)^2\), the numbers +a and −a are square roots of \(a^2\). If \(a ≥ 0\), then +a is the nonnegative square root of \(a^2\), and if \(a < 0\), then −a is the nonnegative square root of \(a^2\). Since \(\sqrt{a^2}\) denotes the nonnegative square root of \(a^2\), it follows that

\(\sqrt{a^2}=+a\) if \(a \geq 0\)
\(\sqrt{a^2}=-a\) if \(a < 0\)

That is, \(\sqrt{a^2}=|a|\)


PROPERTIES OF ABSOLUTE VALUE


If a and b are real numbers, then

(a) \(| −a|=|a|\) A number and its negative have the same absolute value.
(b) \(|ab|=|a||b|\) The absolute value of a product is the product of the absolute values.
(c) \( |\frac{a}{b}|=\frac{|a|}{|b|}\) The absolute value of a ratio is the ratio of the absolute values.


The result in part (b) can be extended to three or more factors. More precisely, for any n real numbers, \(a_1, a_2,...,a_n\), it follows that 

\(|a_1a_2 ··· a_n|=|a_1||a_2|···|a_n| \)

In the special case where \(a_1, a_2,...,a_n\) have the same value, \(a\), it follows that

\(|a^n|=|a|^n\)


GEOMETRIC INTERPRETATION OF ABSOLUTE VALUE

The notion of absolute value arises naturally in distance problems. For example, suppose that A and B are points on a number line that have coordinates \(a\) and \(b\), respectively.
Depending on the relative positions of the points, the distance \(d\) between them will be \(b − a\) or \(a − b\) (see figure below). In either case, the distance can be written as \(d = |b − a|\), so we have the following result.




(Distance Formula). If A and B are points on a number line with coordinates \(a\) and \(b\), respectively, then the distance d between A and B is \(d = |b − a|\).

This distance formula provides useful geometric interpretations of some common mathematical expressions




INEQUALITIES WITH ABSOLUTE VALUES

Inequalities of the form \(|x − a| < k\) and \(|x − a| > k\) arise so often that we have summarized the key facts about them 




NOTE: The statements above remain true if < is replaced by ≤ and > by ≥, and if the open dots are replaced by closed dots in the figure above.



source used: 
https://www.mcckc.edu/
https://science.ucalgary.ca/mathematics-statistics
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­1) Basic operations on numbers











2) Distance on a Number Line

The distance between any two real numbers a and b on a number line can be calculated using the formula \(d = |b − a|\).








KEY TAKEAWAYS

• A positive number added to a positive number is positive. A negative number added to a negative number is negative.
• The sign of a positive number added to a negative number is the same as the sign of the number with the greatest distance from the origin.
• Addition is commutative, and subtraction is not.
• When simplifying, it is a best practice first to replace sequential operations and then work the operations of addition and subtraction from left to right.
• The distance between any two numbers on a number line is the absolute value of their difference. In other words, given any real numbers a and b , use the formula \(d = |b − a|\) to calculate the distance d between them.



Source of the images
  1. Crissy Jarvis - Unsplash
  2. Trade icon vector created by rawpixel.com 
  3. A Maths Dictionary for Kids by © Jenny Eather
  4. Beginning algebra - 2012 Book Archive under Creative Commons Attribution 3.0 International license
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­1) Evens & Odds


Basic operation on Even and Odd Numbers
Math operations
Application
Examples
Addition/Subtraction Even \(\pm\) Even = Even;
Odd \(\pm\) Odd = Even; 
Even \(\pm\) Odd = Odd 
  • 8+4=6;Even
  • 9+5=14;Even
  • 7+2=9;Odd
Multiplication Even \(\times\) Even = Even;
Odd \(\times\) Odd = Odd; 
Even \(\times\) Odd = Even 

Multiply ODD \(\times\) ODD = ALWAYS ODD.
  • 8 * 2=16;Even
  • 5 * 10 =50;Even
  • 3* 7 =21;Odd
Division The division could lead us to a whole result (EVEN or ODD whole number) OR we could have a remainder

Odd / Odd = Odd;
Even / Odd = Even;
Odd / Even = Not divisible;
Even / Even = Even or Odd		 15/3=5; Odd
20/5=4; Even
13/2=? Not divisible
6/3=2; Even
25/5=5; Odd







2) Representation Evens






3) Representation Odds








Source of the images:
w3resource under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
Edge2Edge Media on Unsplash
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­1) Factors

• a whole number that divides exactly into another number.
• a whole number that multiplies with another number to make a third number.

Factors Pair

• a pair of numbers multiplied together form another number called their product.




Number of factors

A number N can be written as \( N=p^xq^yr^z\), where p,q, and r are prime factors of the number N, and x,y, and z are positive integers.

Number of factors : (including 1 and the number itself)\(=(x+1)(y+1)(z+1)\);

Sum of all the factors

\(\left[ \begin{array}{cc|r} \frac{p^{x+1}-1}{p-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{q^{y+1}-1}{q-1} \end{array} \right]\) \(\left[ \begin{array}{cc|r} \frac{r^{z+1}-1}{r-1} \end{array} \right]\)


Product all the factors

\(N^{\frac{number of factors}{2}}\)

Suppose 120 is our number and the prime factorization is \(2^3*3^1*5^1\)

1) Number of factors: simply take the exponent of each factor and add 1. \((3+1)*(1+1)*(1+1)=16\)

2) Sum of all factors: \(\frac{[(2^0+2^1+2^2+2^3)(3^0+3^1)(5^0+5^1)]}{[(2-1) (3-1)(5-1)]} = 45\)

3) Product of all factors: \(120^{\frac{16}{2}}=120^8\)



1. A number is divisible by 2 if its unit’s digit is even or zero e.g. 128, 146, 34 etc.
2. A number is divisible by 3 if the sum of its digits is divisible by 3 e.g. 102, 192, 99 etc.
3. A number is divisible by 4 when the number formed by last two right hand digits is divisible by ‘4’ e.g. 576, 328, 144 etc.
4. A number is divisible by 5 when its unit’s digit is either five or zero : e.g. 1111535, 3970, 145 etc.
5. A number is divisible by 6 when it’s divisible by 2 and 3 both. e.g. 714, 509796, 1728 etc.
6. A number is divisible by 8 when the number formed by the last three right hand digits is divisible by ‘8’. e.g. 512, 4096, 1304 etc.
7. A number is divisible by 9 when the sum of its digits is divisible by 9 e.g. 1287, 11583, 2304 etc.
8. A number is divisible by 10 when its units digit is zero. e.g. 100, 170, 10590 etc.
9. A number is divisible by 11 when the difference between the sums of digits in the odd and even places is either zero or a multiple of 11. e.g. 17259, 62468252, 12221 etc. 
For the number 17259 : Sum of digits in even places = 7 + 5 = 12, Sum of digits in the odd places = 1 + 2 + 9 =12 Hence 12 – 12 = 0.
10. A number is divisible by 12 when it is divisible by 3 & 4 both. e.g. 672, 8064 etc.
11. A number is divisible by 25 when the number formed by the last two Right hand digits is divisible by 25. e.g. 1025, 3475, 55550 etc.
12. A number is divisible by 125, when the number formed by last three right hand digits is divisible by 125. e.g. 2125, 4250, 6375 etc.

NOTE :

1. When any number with even number of digits is added to its reverse, the sum is always divisible by 11. e.g. 2341 + 1432 = 3773 which is divisible by 11.
2. If X is a prime number then for any whole number “a” \((a^X – a)\) is divisible by X e.g. Let X = 3 and a = 5. Then according to our rule \(5^3– 5\) should be divisible by 3.
Now \((5^3– 5) = 120\) which is divisible by 3



2) Primes


A number greater than 1 which has no factor other than 1 and the number itself is a Prime Number
A number with two distinct factors.
We do have infinite prime numbers and apparently, to our knowledge today, there is NO clear pattern in their unfolding toward the infinite on the number line.
Prime numbers are only positive.
Two (2) is the oNLy positive Even prime numbers
The difference between any two primes greater than 2 is always even.



Verifying the primality (checking whether the number is a prime) of a given number \(n\) can be done by trial division, that is to say dividing \(n\) by all integer numbers smaller than \(\sqrt{n}\), thereby checking whether \(n\) is a multiple of \(m\leq{\sqrt{n}}\).
Example: Verifying the primality of \(161\): \(\sqrt{161}\) is little less than \(13\), from integers from \(2\) to \(13\), \(161\) is divisible by \(7\), hence \(161\) is not prime.
Note that, it is only necessary to try dividing by prime numbers up to \(\sqrt{n}\), since if n has any divisors at all (besides 1 and n), then it must have a prime divisor.

If \(n\) is a positive integer greater than 1, then there is always a prime number \(p\) with \(n < p < 2n\).







3) Multiples





HCF (GCD / GCF) & LCM OF NUMBERS


HCF : It is the greatest factor common to two or more given numbers. It is also called GCF OR GCD (greatest common factor or greatest common divisor). e.g. HCF of 10 & 15 = 5, HCF of 55 & 200 = 5, HCF of 64 & 36 = 4 

To find the HCF of given numbers, resolve the numbers into their prime factors and then pick the common term(s) from them and multiply them if more than one. This is the required HCF.

LCM : Lowest common multiple of two or more numbers is the smallest number which is exactly divisible by all of them. e.g. LCM of 5, 7, 10 = 70, LCM of 2, 4, 5 = 20, LCM of 11, 10, 3 = 330

To find the LCM resolve all the numbers into their prime factors and then pick all the quantities (prime factors) but not more than once and multiply them. This is the LCM.

NOTE:
1. LCM x HCF = Product of two numbers (valid only for “two”)
2. HCF of fractions = HCF of numerators ÷ LCM of denominators
3. LCM of fractions = LCM of numerators ÷ HCF of denominators


Calculating LCM :

Method 1 : Factorization Method

Rule – After expressing the numbers in terms of prime factors, the LCM is the product of highest powers of all factors.

Q. Find the LCM of 40, 120, 380.

A. 4\(0 = 4 \times 10 = 2 \times 2 \times 2 \times 5 = 2^3 \times 5^1\)

\(120 = 4 \times 30 = 2 \times 2 \times 2 \times 5 \times 3 = 2^3 \times 5^1 \times 3^1\)

\(380 = 2 \times 190 = 2 \times 2 \times 95 = 2 \times 2 \times 5 \times 19 = 2^2 \times 5^1 \times 19^1\)

⇒ Required LCM = \(2^3 \times 5^1 \times 3^1 \times 19^1 = 2280\).

Method 2 : Division method

Q. Find the LCM of 6, 10, 15, 24, 39




Ans. LCM = 2 x 2 x 3 x 5 x 2 x 13 = 1560.


Calculating HCF 


Factorization
After expressing the numbers in term of the prime factors, the HCF is product of COMMON factors.

Ex. Find HCF of 88, 24, 124
 
\(88 = 2 \times 44 = 2 \times 2 \times 22 = 2 \times 2 \times 2 \times 11 = 2^3 \times 11^1\)
\(24 = 2 \times 12 = 2 \times 2 \times 6 = 2 \times 2 \times 2 \times 3 = 2^3 \times 3^1\)
\(124 = 2 \times 62 = 2 \times 2^1 \times 31^1 = 2^2 \times 31^1 ⇒ HCF = 2^2\)



Source of the images:
playworksheet.com
Claudio Schwarz on Unsplash
A Maths Dictionary for Kids by © Jenny Eather
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How to Solve: Remainders (Basics and Advanced)



Theory

• Definition: What is "Remainder"?
• Remainder Notation
• How to find Remainders : Long Division Method
• Remainder Range
• PT1: Basic Problems on Finding Remainders
• PT2: Find divisor when remainder is given
• PT3: Find dividend when one divisor and remainder are given
• PT4: Find dividend when two divisors and remainders are given
• Remainder of sum of two numbers by a number
• Remainder of difference of two numbers by a number
• Remainder of product of two numbers by a number

Video 1

• Remainder of numbers when divided by 2 , 3, 5, 9, 10
• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders

Video 2


Remainders Basics

• Definition: What is "Remainder"?

Remainder is the integer which is left over in a division, when the divisor cannot evenly divide the dividend

Dividend  = Divisor * Quotient + Remainder

Ex: 13 when divided by 3 gives 1 remainder. (13 = 3*4 + 1)
13 -> Dividend
3   -> Divisor
4   -> Quotient
1   -> Remainder

• Remainder Notation

1. A number when divided by 5 gives 3 as remainder.
Let the number be n
using Dividend  = Divisor * Quotient + Remainder
n = 5k + 3 [where k is quotient and is an integer]

2. A number when divided by 13 gives 5 as remainder.
Let the number be n
using Dividend  = Divisor * Quotient + Remainder
n = 13k + 5 [where k is quotient and is an integer]

3. A number, when divided by 16, gives 3 as remainder.
Let the number be n
using Dividend  = Divisor * Quotient + Remainder
n = 16k + 3 [where k is quotient and is an integer]


• How to find Remainders : Long Division Method -> Check the video 


• Remainder Range
"A number, when divided by a number k, can give remainder from 0 to k-1"

Ex: If we are trying to divide a number by 6 then possible values of remainder are from 0-5

Q1. x when divided by "a" gives 3 as remainder.
      y when divided by "b" gives 4 as remainder.
      Find min(a+b)

Sol: As a and b are positive numbers so min(a+b) = min(a) + min(b)
x when divided by "a" gives 3 as remainder -> Since a is giving 3 remainder that means that a >= 4. So, min a = 4
y when divided by "b" gives 4 as remainder  -> Since b is giving 4 remainder that means that b >= 5. So, min b = 5
=> min(a+b) = min(a) + min(b) = 4+ 5 =9

• PT1: Basic Problems on Finding Remainders -> Check out the video above for examples on long division method

Q1. Find the remainder when 200 is divided by 3.
Q2. Find the remainder when 80 is divided by 7.
Q3. Find the remainder when 100 is divided by 9.

Ans:
Q1. 2, Q2 3, Q3. 1

• PT2: Find divisor when remainder is given

Q1. 20 when divided by which number will give 4 as remainder?

Sol: 20 = nk + 4
=> nk= 20-4 = 16
=> n = 16/k
Now, n is giving 4 remainder that means that n>=5
Let's start putting values of k and get values of n
k=1 => n=16
k=2 => n=16/2 = 8
k=3 => n not integer
k=4 => n= 16/4 not possible as n>=5
So, possible values of the number (n) are 8 and 16

Q2. 50 when divided by which number will give 7 as remainder?

Sol: 50 = nk + 7
=> nk = 50-7 = 43 [note 43 is prime so it has only two factors 1 and 43 itself]
=> n = 43/k
k= 1 => n = 43
k = 43 not possible as n becomes 1 [ but we know that n>=8 as n is giving 7 remainder]
So, Answer is 43

• PT3: Find dividend when one divisor and remainder are given

Q1. n when divided by 7 gives 4 as remainder. Find the possible values of the number.

Sol: n when divided by 7 gives 4 as remainder. 
n = 7k + 4

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=7*0 + 4 =  4      k=5 n=7*5 + 4 = 39
k=1 n=7*1 + 4 = 11     k=6 n=7*6 + 4 = 46
k=2 n=7*2 + 4 = 18     k=7 n=7*7 + 4 = 53
k=3 n=7*3 + 4 = 25     k=8 n=7*8 + 4 = 60
k=4 n=7*4 + 4 = 32     k=9 n=7*9 + 4 = 67

Q2. n when divided by 5 gives 2 as remainder. Find the possible values of n.

n when divided by 5 gives 2 as remainder. 
n = 5k + 2

We will start taking values of k starting from k=0 and find values of n correspondingly
k=0 n=5*0 + 2 =  2      k=5 n=5*5 + 2 = 27
k=1 n=5*1 + 2 =  7      k=6 n=5*6 + 2 = 32 
k=2 n=5*2 + 2 = 12     k=7 n=5*7 + 2 = 37
k=3 n=5*3 + 2 = 17     k=8 n=5*8 + 2 = 42
k=4 n=5*4 + 2 = 22     k=9 n=5*9 + 2 = 47

• PT4: Find dividend when two divisors and remainders are given

Q1. n when divided by 7 gives 3 remainder and when divided by 5 gives 3 remainder. Find first 2 non-negative values of n

Sol: Method -1
3. n when divided by 7 gives 3 remainder
n = 7k + 3
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 ,  1,   2,   3,  4,   5,   6,   7,   8,   9
n = 3 , 10, 17, 24, 31, 38, 45, 52, 59, 66 

n when divided by 5 gives 3 remainder
n = 5t + 3
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 ,  1,   2,   3,   4,   5,   6,   7,   8,   9
n = 3 , 8,  13, 18, 23, 28, 33, 38,  43, 48

First two common values are 3 and 38

Sol: Method-2
n when divided by 7 gives 3 remainder
n = 7k + 3

n when divided by 5 gives 3 remainder
n = 5t + 3

7k+3 = 5t+3 => t = 7k/5
So, only those values of k will give us common values of n for which t is integer too.
k = 0 => t=7*0/5 = 0
k = 5 => t=7*5/5 = 7
So, k = 0 => n = 7*0 + 3 = 3
      k = 5 => n = 7*5 + 3 = 38

Q2. n when divided by 6 gives 4 remainder and when divided by 4 gives 2 remainder. Find first 2 non-negative values of n

Sol: Method-1
n when divided by 6 gives 4 remainder
n = 6k + 4
We will start taking values of k starting from k=0 and find values of n correspondingly
k = 0 ,  1,   2,   3,  4,   5,   6,   7,   8,   9
n = 4, 10, 16, 22, 28, 34, 40, 46, 52,  58 

n when divided by 4 gives 2 remainder
n = 4t + 2
We will start taking values of t starting from t=0 and find values of n correspondingly
t = 0 ,  1,   2,   3,   4,   5,   6,   7,   8,   9
n = 2,  6,  10,  14, 18, 22, 26, 30, 34, 38

First two common values are 10 and 22

Sol: Method-2
n when divided by 6 gives 4 remainder
n = 6k + 4

n when divided by 4 gives 2 remainder
n = 4t + 2

6k+4 = 4t+2 => t = (6k+2)/4 = (3k+1)/2
So, only those values of k will give us common values of n for which t is integer too.
k = 1 => t=(6*1 + 2)/4 = 2
k = 3 => t=(6*3 + 2)/4 = 5
So, k = 1 => n = 6*1 + 4 = 10
      k = 3 => n = 6*3 + 4 = 22

Note: When the numerator is smaller than the denominator then the remainder is the numerator itself
Ex: If 2 is divided by 3 then remainder is 2

• Remainder of sum of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 8 and 5 and divide 8 and 5 individually by 3 then we still get the same remainder
13/3 = (8+5)/3 = 8/3 + 5/3
    8/3 will give 2 remainder
    5/3 will give 2 remainder

Total remainder is 2+2 = 4, but remainder cant be greater than 3 so remainder will be 4-3 = 1 which is same as the remainder for 13/3

Q1. “A” when divided by 12 gives 3 as remainder. What is the remainder when A is divided by 4?

Sol: A = 12k + 3
When A is divided by 4 then 12k will give 0 remainder and 3 will give 3 remainder. Total remainder is 3
 

Q2. “B” when divided by 15 gives 6 as remainder. What is the remainder when B is divided by 5?

Sol: B = 15k + 6
When B is divided by 5 then 15k will give 0 remainder and 6 will give 1 remainder. Total remainder is 1

• Remainder of difference of two numbers by a number
13 when divided by 3 gives us 1 remainder

When we split 13 as 15 - 2 and divide 15 and 2 individually by 3 then we still get the same remainder
13/3 = (15-2)/3 = 15/3 - 2/3
    15/3 will give 0 remainder
    2/3 will give 2 remainder
Total remainder = 0-2 = -2 
But remainder cannot be negative so we are going to keep on adding 3 to -2 till the time the sum comes in the range of 0 and 2 [3-1]
=> -2 + 3 = 1
So, remainder is 1

• Remainder of product of two numbers by a number
21 when divided by 5 gives us 1 remainder. Now if we break 21 into product of two numbers and find the remainder of these individual numbers by 5 and multiply the remainders then we are going to get the same remainder as we got when we divided 21 by 5

Let's write 21 = 3*7 and divided 3 and 7 by 5
3/5 * 7/5 will give us 3 * 2 remainder respectively = 6
But we divided set of numbers by 5 so remainder cannot be more than 5, so we divide 6 again by 5 to get final remainder as 1

Q1. Find the remainder of 136 * 148 * 298 by 5.

Sol: 136/5 remainder is 1
148/5 remainder is 3
298/5 remainder is 3
Total remainder = 1*3*3 = 9 [>=5]
=> Final remainder = remainder of 9/5 = 4

Q2. Find the remainder of 1205 * 1208 * 2404 by 12.

Sol: 1205/12 remainder is 5
1208/12 remainder is 8
2404/12 remainder is 4
Total remainder = 5*8*4 = 160 [>=12]
=> Final remainder = remainder of 160/12 = 4

In next section I have just listed down the theory. For problems and Binomial theorem please refer thevideo

Check above video for examples
• Remainder of numbers when divided by 2

When we try to find remainder of a number by 2 then we just need to know if the number is odd or even.
If the number is odd  then remainder is 1
If the number is even then the remainder is 0

• Remainder of numbers when divided by 3

Remainder of a number by 3 is same as the remainder of sum of digits of the number by 3

• Remainder of numbers when divided by 5

Remainder of a number by 5 is same as the remainder of the unit's digit of the number by 5

• Remainder of numbers when divided by 9

Remainder of a number by 9 is same as the remainder of sum of digits of the number by 9

• Remainder of numbers when divided by 10

Remainder of a number by 10 is same as the unit's digit of the number

• Remainder of numbers when divided by other numbers
• Binomial Theorem
• Application of Binomial Theorem in finding remainders -> Please check Video for this

Hope it helps!
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­INTRODUCTION

1. What is a fraction? 
 
A fraction consists of a numerator (part) on top of a denominator (total) separated by a horizontal line.

For example, the fraction of the circle which is shaded is:

\(\frac{2 (parts shaded)}{4 (total parts)}\)




In the square, the fraction shaded is \(\frac{3}{8}\) and the fraction unshaded is \(\frac{5}{8}\) 




2. Equivalent Fractions – Multiplying

The three circles on the right each have equal parts shaded, yet are represented by different but equal fractions. These fractions, because they are equal, are called equivalent fractions. Any fraction can be changed into an equivalent fraction by multiplying both the numerator and denominator by the same number.





3. Equivalent Fractions – Dividing (Reducing)

Equivalent fractions can also be created if both the  numerator and denominator can be divided by the  same number (a factor) evenly.  This process is called “reducing a fraction” by dividing a common factor (a number which divides into both the numerator and denominator evenly).

\(\frac{4}{8} \div \frac{4}{4} = \frac{1}{2}\)

\(\frac{27}{81} \div \frac{9}{9} = \frac{3}{9}\)


4. Simplifying a Fraction (Reducing to its Lowest Terms)

It is usual to reduce a fraction until it can’t be reduced  any further.  A simplified fraction has no common factors which will  divide into both numerator and denominator.  Notice that, since 27 and 81 have a common factor of 9,  we find that \(\frac{3}{9}\) is an equivalent fraction. But this fraction has a factor of 3 common to both  numerator and denominator.  So, we must reduce this fraction again. It is difficult to  see, but if we had known that 27 was a factor (divides  into both parts of the fraction evenly), we could have  arrived at the answer in one step

\(\frac{8}{24} \div \frac{8}{8} = \frac{1}{3}\)



TYPES OF FRACTIONS

1. Common Fractions

A common fraction is one in which the numerator is less than the denominator (or a fraction which is less than the number 1). A common fraction can also be called a proper fraction.

\(\frac{88}{93},\frac{8}{15}\),..............are al common fractions


2. Fractions that are Whole Numbers

Some fractions, when reduced, are really whole numbers (1, 2, 3, 4… etc). Whole numbers occur if the denominator divides into the numerator evenly

\(\frac{8}{4}\) is the same as \(\frac{8}{4} \div \frac{4}{4}=\frac{2}{1}\) or \(2\)


3. Mixed Numbers

A mixed number is a combination of a whole number and a common fraction.

e.g. \(2 \frac{3}{5}\) (two and three-fifths)



4. Improper Fractions

An improper fraction is one in which the numerator is larger than the denominator. From the circles on the right, we see that \(1 \frac{3}{4}\) (mixed number) is the same as \(\frac{7}{4}\) (improper fraction). An improper fraction, like \(\frac{7}{4 }\)  can be changed to a mixed number by dividing the denominator into the numerator and expressing the remainder (3) as the numerator.

e.g. \(\frac{16}{5}=3 \frac{1}{5}\)

A mixed number can be changed to an improper fraction by changing the whole number to a fraction with the same denominator as the common fraction.

\(2 \frac{3}{5} = \frac{10}{5} and \frac{3}{5}=\frac{13}{5}\)

A simple way to do this is to multiply the whole number by the denominator,  and then add the numerator.

e.g. 4 \(\frac{5}{9} = \frac{4 \times 9 + 5 }{9} = \frac{36+5}{9 }= \frac{41}{9}\)


5. Simplifying fractions 

All types of fractions must always be simplified (reduced to lowest terms).

e.g. \(\frac{6}{9}=\frac{2}{3}\)

Note that many fractions can not be reduced since they have no common factors.


COMPARING FRACTIONS

Students can use several strategies to compare fractions. These strategies are fairly elementary. We will outline below the most common ways to confront fractions.


1. Using Visual Models  

Area models can be used to compare fractions, drawing pictures: they could be circles, stripes, boxes, and so on.



In this case , regardless of the value of the two fractions, \(\frac{3}{4} >\frac{5}{8}\) because the colored part is more significant in the first strip than the second one.


2. Using the number line 

(a) - More and less than a benchmark such as one-half or one whole.

Comparing fractions to a benchmark such as one-half or one whole is another strategy for comparing two fractions. For example, when comparing 3/8 and 5/6, we know that 3/8 is less than 1/2 and that 5/6 is more than 1/2. Thus, i is the greater fraction. Geometrically, we can illustrate this strategy by placing both fractions on the number line as shown below. 




While any two fractions that represent distinct numbers can be compared to a benchmark, this is a strategy that works particularly well when one fraction is more than a common benchmark such as 1/2 or 1 and the other fraction is less than this benchmark. Given two fractions greater than 1/2 but less than 1, finding an appropriate benchmark can be problematic. When comparing 5/6 and 13/16, it is difficult to determine a benchmark fraction. 


(b) - Distance from a benchmark such as one-ha/for one-whole.

The second strategy, comparing two fractions to a benchmark, is limited when both fractions are either greater than or less than the chosen benchmark. A strategy in these cases is to compare the distances of the fractions from the benchmark. For example, 5/6 and 7/8 are both less than the benchmark 1 and so we can compare these two fractions by considering their distance from 1. 

5/6 is 1/6 less than 1 and 7/8 is 1/8 less than 1. We can then compare 1/6 and 1/8 using our strategy for comparing two fractions that have the same number of parts, but parts of different sizes. Since 1/6 is greater than 1/8, then 1/6 is a greater distance from 1 than 1/8. Therefore, 7/8 is greater than 5/6 (or 5/6 is less than 7/8). Geometrically, we can illustrate this strategy using the number line as shown below. 







3. Common Numerator or Denominator 



Common Numerator or Denominator
Common Numerator
Common Denominator
Improper fraction > Proper fraction
\(\frac{3}{5} > \frac{2}{5}\) because \(4<5\) \(\frac{3}{5} > \frac{2}{5}\) because \(3 > 2\) \(\frac{3}{2 }> \frac{24}{25}\)
\(\frac{1}{3} < \frac{1}{2}\) because \(3>2\) \(\frac{9}{14} < \frac{11}{14}\) because \(9 > 11\) \(\frac{9}{10 }> \frac{10}{9}\)


ADDING/SUBTRACTING/MULTIPLYING/DIVIDING

Adding and Subtracting Fractions:

1. Common (like) denominators are necessary, so change all unlike fractions to equivalent fractions with like denominators. To make equivalent fractions, multiply the numerator and denominator by the same number.
2. Keep mixed numbers; DO NOT change mixed numbers into improper fractions.
3. Add (or subtract) the numerators, put the numerator answer over the common denominator. If any improper fractions arise in the answer, change the improper portion to a mixed number. (In Math 105, answers are often left in improper form)

Multiplying and Dividing Fractions:

1. Common denominators are NOT needed.
2. Always change mixed numbers to improper fractions.
3. CANCEL (reduce) between any numerator and any denominator if you can, but cancel only when a multiplication sign is present: Never cancel when you have a division sign.
4. TO MULTIPLY: Multiply numerator times numerator, denominator times denominator. Reduce answer to a mixed number in lowest terms. (In Math 105, answers are often left in improper form)
5. TO DIVIDE: Change the divide sign to a multiplication sign, then invert the second fraction and multiply as in Step 4.



Source used:
  1. Source of the image Jasmina81 Getty Images via Scientific American
  2. Vancouver Island University
  3. Math Notes: Strategies for Comparing Fractions - This work is licensed under a Creative Commons Attribution-Noncommercial-4.0 International License © 2018 Mathematics Teaching and Learning to Teach School of Education • University of Michigan • Ann Arbor, MI 48109-1259 • [email protected]
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­RATIO

The comparison between two quantities of the same kind of unit is the Ratio of one quantity to another. The ratio of a and b is usually written as a : b or a/b, where a is called the antecedent (numerator) and b the consequent (denominator).

1. \(a : b = ka : kb\) where k is a constant

2. \(a : b = \frac{a}{k} : \frac{b}{k}\)

3. \(a : b : c = X : Y : Z\) is equivalent to \(\frac{a}{X} = \frac{b}{Y} =\frac{ c}{Z}\)

4. If \(\frac{a}{b} = \frac{c}{d}\), then

(i) \(\frac{(a + b)}{b} = \frac{(c + d)}{d}\)
(ii) \(\frac{(a – b)}{b} = \frac{(c – d)}{d}\)
(iii) \(\frac{(a + b)}{(a – b)} =\frac{ (c + d)}{(c – d)}\)

5. If \(\frac{a}{b} = \frac{c}{d} = \frac{e}{f} = ...... = k\), then \(\frac{a + c + e +}{b + d + f +} .... .... = k\). 

Also note that : \(k = \frac{a}{b }= \frac{Xc }{ Xd} = \frac{–5e }{ –5f}\) ⇒ Each ratio \(= \frac{(a + Xc – 5e) }{(b + Xd – 5f)} = k\). Here, we have randomly taken X and –5. You can take any factor.

6. If \(\frac{a}{b} > 1\) or \(a > b\) then \(\frac{(a + X) }{ (b + X)} <\frac{ a}{b}\) a, b, X are natural numbers

7. If \(\frac{a}{b} < 1\) or \(a < b\) then \(\frac{(a + X) }{ (b + X )} > \frac{a}{b}\) a, b, X are natural numbers

VARIATION 

1. Direct proportion :

If two quantities X & Y are related such that any increase or decrease in ‘Y’ produces a proportionate increase or decrease in ‘X’ or vice versa, then the two quantities are said to be in direct proportion. 
In other words \(X : Y = \frac{X}{Y} = k\) (a constant) or X = KY or Y = K’X (where K and K’ are constants) 
X is directly proportional to Y is written as X ∝ Y or X = K Y

2. Inverse proportion :

Here two quantities X & Y are related such that, any increase in X would lead to a decrease in Y or any decrease in X would lead to an increase in Y. Thus the quantities X & Y are said to be inversely related and X is inversely proportional to Y is written as \(\frac{X ∝ 1}{Y}\) or \(X = \frac{k}{Y}\) or XY = k (constant) or the product of two quantities remains constant.


MIXTURES FORMULA

This rule enables us to find the proportion in which two or more ingredients at the given price must be mixed to produce a mixture at a given price. The C.P. of unit quantity of the mixture is called the MEAN PRICE. The rule says : 
If two ingredients are mixed in a ratio, then

\(\frac{Quantity of Cheaper}{Quantity of Dearer} = \frac{CP of Dearer – Mean Price}{Mean Price – CP of Cheaper}\).


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­Writing a number with a decimal is known as decimal notation. It is a way of showing parts of a whole when the whole is a power of ten. In other words, decimals are another way of writing fractions whose denominators are powers of ten. Just as the counting numbers are based on powers of ten, decimals are as well.

They are written as an extension of the place value we use to identify the whole numbers. Actually, the portion on the right of the decimal point is the portion of the number/fraction in the power of ten.

For example

\(6.004 = 6 + \frac{4}{1000}\)

Perhaps, there are no other numbers that entice people in careless calculation errors. Therefore, one of the best things a student could do is ALMOST always convert decimals to fractions or other form of numbers that are easier to deal with.





Place and Face Value


                    
Place ValueFace Value
Place value is the value represented by a digit in a number according to its position in the number.Face value is the actual value of a digit in a number.
To get the place value of a number, we multiply the digit value with its numerical value. 
For example, in the number 452, the place value of 5 is (5 × 10) = 50, since 5 is in tens place.		The face value of a digit is the number itself. For example, in the number 452, 
the face value of 4 is 4.
The place value of a number depends upon the position of the digit in the number.The face value is independent of the position of the digit in the number.
The place value of a digit in ones place is always a single digit and the place value of every next digit 
to the left increases by one more digit. 		The face value of a number is always a single digit.






Convert Decimals to Fractions or Mixed Numbers

Suppose we have \(5.03\) as decima.

5 is the ones
0 is the tenths
3 is the hundredths

\(5.03=5 \frac{3}{100}\)


Round a decimal

Step 1. Locate the given place value and mark it with an arrow.

Step 2. Underline the digit to the right of the given place value.

Step 3. Is this digit greater than or equal to 5?
Yes - add 1 to the digit in the given place value.
No - do not change the digit in the given place value.

Step 4. Rewrite the number, removing all digits to the right of the given place value.


For example 25.3679. 

We have the 7 after the six, so the number will become 25.37

Going further, we will have the 7 after the three so the number will 25.4

After the decimal point, our number is 4, which is < to 5 so the number will become 25

Helpful could be the Rounding Numbers Calculator

Source used:
  1. Source of the image Mark Turnauckas 
  2. Difference Between Place Value and Face Value
  3. Fractions and Decimals by © Tom Davis
  4. Opensatx prealgebra under a Creative Commons Attribution-NonCommercial 4.0 Unported License
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­The term percentage means parts per 100 or “for every hundred”. A fraction whose denominator is 100 is called percentage and the numerator of the fraction is called the rate percent. Thus, when we say a man made a profit of 20 percent we mean to say that he gained Rs.20 for every hundred dollars he invested in the business, i.e., 20/100 dollars for each dollar. The abbreviation of percent is p.c. and it is generally denoted by %.

Let’s start with a number X (= 1 X)

X increased by 10% would become X + 0.1 X = 1.1 X
X increased by 1% would become X + 0.01 X = 1.01 X
X increased by 0.1% would become X + 0.001 X = 1.001 X
X decreased by 10% would become X – 0.1 X = 0.9 X
X decreased by 1% would become X – 0.01 X = 0.99 X
X decreased by 0.1% would become X – 0.001 X = 0.999 X
X increased by 200% would become X + 2X = 3X
X decreased by 300% would become X – 3X = –2X

Similarly, you can work mentally with any specifically chosen number (say 500) and work out different answers.





A Percentage can be expressed as a Fraction. 10% can be expressed as \(\frac{10}{100}\) or \(\frac{1}{10}\). To express a percentage as a fraction divide it by 100, \(a\)% \(= \frac{a}{100}\).

To express a fraction as a percent multiply it by 100 ⇒ \(\frac{a}{b}= \left[ \begin{array}{cc|r} (\frac{a}{b})  \times 100 \end{array} \right]\) %


To express percentage as a decimal we remove the symbol % and shift the decimal point by two places to the left. For example, 10% can be expressed as 0.1. 6.5% = 0.065 etc.

To express decimal as a percentage we shift the decimal point by two places to the right and write the number obtained with the symbol % or simply we multiply the decimal with 100. Similarly 0.7 = 70%.


Percent INCREASE and DECREASE


Increase % \(= \frac{Increase}{Original Value} \times 100\)

Decrease % \(= \frac{Decrease }{ Original Value} \times 100\\
\)

In increase %, the denominator is smaller, whereas, in decrease %, the denominator is larger.

Change % \(= \frac{Change }{ Original Value} \times 100\)


Successive change in percentage. If a number A is increased successively by X% followed by Y%, and then by Z%, then the final value of A will be \(A(1+\frac{X}{100})(1+\frac{Y}{100})(1+\frac{Z}{100})\)

In a similar way, at any point or stage, if the value is decreased by any percentage, then we can replace the same by a negative sign. The same formula can be used for two or more successive changes. The final value of A, in this case, will be \(A(1-\frac{X}{100})(1-\frac{Y}{100})(1-\frac{Z}{100})\)


Also, let us remember that

2 = 200% (or 100% increase), 3 = 300% (or 200% increase), 3.26 = 326% (means 226% increase), fourfold (4 times) = 400 % of original = 300% increase, 10 times means 1000% means 900% increase, 0.6 means 60% of the original means 40% decrease, 0.31 times means 31% of the original means 69% decrease etc.

1/2 = 50%, 3/2 = 1 + 1/2 = 100 + 50 = 150%, 5/2 = 2 + 1/2 = 200 + 50 = 250% etc.,

2/3 = 1 - 1/3 = 100 - 33.33 = 66.66%, 4/3 = 1 + 1/3 = 100 + 33.33 = 133.33%,

5/3 = 1 + 2/3 = 100 + 66.66 % = 166.66%, 7/3 = 2 + 1/3 = 200 + 33.33 = 233.33%,

8/3 = 2 + 2/3 = 200 + 66.66 = 3 - 1/3 = 300 - 33.33 = 266.66% etc.

1/4 = 25%, 3/4 = 75%, 5/4 (1 + 1/4) = 125% (= 25 increase), 7/4 (1 + 3/4 = 2 - 1/4) = 175% (= 75% increase), 9/4 (2 + 1/4) = 225% (= 125% increase), 11/4 = 275% = (175% increase).

1/5 = 20%, 2/5 = 40%, 3/5 = 60%, 4/5 = 80%, 6/5 = 120%, 7/5 (1 + 2/5) = 140% etc.

1/6 = 16.66%, 5/6 = 83.33%, 7/6 (1 + 1/6) = 116.66%, 11/6 = 183.33%

1/8 = 12.5%, 3/8 = 37.5%, 5/8 = 62.5%, 7/8 = 87.5%, 9/8 = (1 + 1/8) = 112.5%, 11/8 = (1 + 3/8) = 137.5%, 13/8 = 162.5%, 15/8 = 187.5% etc.

1/9 = 11.11%, 2/9 = 22.22%, 4/9 = 44.44%, 5/9 = 55.55%, 7/9 = 77.77%, 8/9 = 88.88%, 10/9 = 111.11%, 11/9 = (1 + 2/9) = 122.22% etc.


If we have a problem that deals with a discount, it is useful to remember:

Marked Price - Discount = Sale Price. Also Cost Price + Profit = Sale Price
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­1) Exponents


Basic operations of indices
Math operations
Examples
\(a^m*a^n=a^{m+n}\) \(7^3*7^5=7^{3+5}=7^8\)
\(\frac{a^m}{a^n}=a^{m-n}\) \(\frac{7^3}{7^5}=7^{3-5}=2^{-2}\)
\((a^m)^n=a^{m*n}\) \((7^2)^3=7^{2*3}=7^6\)
\(x^{-n}=\frac{1}{x^n}\)\(7^{-2}=\frac{1}{7^2}\)
\((\frac{a}{b})^{-n}=(\frac{b}{a})^n\) \((\frac{1}{7})^{-2}=7^2\)  
\(\frac{1}{a^m}=a^{-m}\)\(\frac{1}{7^3}=7^{-3}\)
\(\sqrt[n]{a^m}=(a^m)^{\frac{1}{n}}=a^{\frac{m}{n}}\) \( \sqrt[3]{7^6}\)\(=(7^6)^{\frac{1}{3}}=7^{\frac{6}{3}}=7^2=49\)  
\(a^0=1\), if \(a \neq 0\) \(7^0=1\)
\((-a)^{even}=+(a)^{even}\)
\(a>0\)		 \((-7)^4=-7 \times -7 \times -7 \times -7 = 7^4=2,401\)
\((-a)^{odd}=-(a)^{odd}\)
\(a>0\)		 \(-(7)^3=-(7 \times 7 \times 7)=-343\)


Note \((a^m)^n \neq a^{m^n}\)
\((7^2)^3 = 7^6\)

However

\(7^{3^2}=7^9\). You have to elevate \(3^2\) before and then \(7^9\)

\(7^6 \neq 7^9\\
\)


2) Unit Digit Power


Unit Digit or Last Digit of a^n
a\n
1
2
3
4
cyclicity
000001
111111
224864
339714
446462
555551
666661
779314
884264
991912


Example

Find the unit digit of \(7^{99}\)?

Having a cyclicity of 4 we do have that \(7^{10}\) has a unit digit of 9. This for nine times which is \(7^{90} =\) unit digit of 9 + 9 spots the unit digit is 3



3) Unit Digit of a Square

The square of a number can never end with \(2, 3, 7, 8\) or odd number of zeros.



4) Number rearrangement


When a two-digit number is reversed then the sum of two numbers is always divisible by 11 & the difference of two numbers is always divisible by 9

Let's say we have a number 92

When reversed it will be 29
Sum = 92+29 = \(121\) ----> Divisible by \(11\)
Difference = 92-29 =  \(63\) ----> Divisible by \(9\) 



5) Square of 5

Suppose that X is a number ending in 5 or 5 itself

\((X5)^2=X(X+1),25\)

\(625^2=62 \times 63\),\(25\)\(=\)\(3906\)\(25\)

You just need to multiply \(62 \times 63\) and add \(25\)



6) Square of any number

\(n^2=(n-d)(n+d)+d^2\)

d is the distance from the nearest multiple of 10.100

\(19^2=(19-1)(19+1)+1^2=18 \times 20+1=361\)



2) Roots


1) Properties of radicals





2) Surds

• another name for an irrational number.
• a surd is a real number that can be written as a nonrepeating or nonterminating decimal but not as a fraction because the decimal goes on forever without repeating.




source: http://www.amathsdictionaryforkids.com/

Surds are irrational roots of a rational number. e.g. √6 = a surd ⇒ it can’t be exactly found. Similarly – √7, √8, \(\sqrt[3]{9}\), \(\sqrt[4]{27}\) etc. are all surds.

  • Pure Surd : The surds which are made up of only an irrational number e.g. √6, √7, √8 etc.
  • Mixed Surd : Surds which are made up of partly rational and partly irrational numbers e.g. 3√3, \(6^4√27\) etc.
  • Rationalization of Surds: In order to rationalize a given surd, multiply and divide by the conjugate of denominator [conjugate of (a + √b) is (a – √b) and vice versa].


e.g.  \(\frac{(6+\sqrt{2})}{(1-\sqrt{3})}=\frac{(6+\sqrt{2})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}=\frac{(6+6 \sqrt{3}+\sqrt{2}+\sqrt{6})}{(1-3)}=\frac{(6+6 \sqrt{3} +\sqrt{2}+\sqrt{6})}{-2}\)


Source of the images A Maths Dictionary for Kids by © Jenny Eather
Openedu - Squares, roots and powers
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­GMAT Quant Theory Master for All BrushMyQuant Units' Digit of Exponents Articles
(PDFs can be found in individual links given as Gre.MyPrepClub Article Link)



Units' Digit of Product of Numbers and Exponents





Units' Digit of Power of 2 and 3





Units' Digit of Power of 4 and 5





Units' Digit of Power of 6 and 7





Units' Digit of Power of 8 and 9




Hope it helps!­
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How To Solve: Roots


Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Hi All,

I have recently uploaded a video on YouTube to discuss Roots in Detail:




Following is covered in the video

    ¤ Square root of a number is ALWAYS positive
    ¤ Simplifying √a * √b 
    ¤ Simplifying \(\frac{√a}{√b}\)
    ¤ Simplifying \((√a)^n \)
    ¤ Simplifying √(x ± y)
    ¤ Simplifying \(a^{(x/y)}\)
    ¤ Simplifying \(√(a^2)\)
    ¤ Simplifying √n

Square root of a number is ALWAYS positive

Even root of any number will always be a positive value
Ex: √36 = + 6

Although, \(x^2\) = 36 => x = ± √36 = ±6

Simplifying √a * √b

√a * √b = √(ab)
Ex: √2 * √3 = √(2*3) = √6

Simplifying \(\frac{√a}{√b}\)

\(\frac{√a}{√b}\) = \(√(\frac{a}{b})\)
Ex: \(\frac{√4}{√2}\) = \(√(\frac{4}{2})\) = √2

Simplifying \((√a)^n \)

\((√a)^n\) = √(\(a^n\))
Ex: \((√2)^4\) = √(\(2^4\)) = \(2^2\) = 4

Simplifying √(x ± y)

√(x ± y) ≠ √x ± √y
Ex: √(2 + 3) ≠ √2 + √3
=> √5 ≠ √2 + √3
=> 2.23 ≠ 1.414 + 1.732
=> 2.23 ≠ 3.146 

Simplifying \(a^{(x/y)}\)

\(a^{(x/y)}\) = \(\sqrt[y]{a^x}\)
Ex: \(2^{(6/3)}\) = \(\sqrt[3]{2^6}\) = \(2^2\) = 4

Simplifying \(√(a^2)\)

√(\(a^2\)) = |a|
√(\(a^2\)) = -a, when a ≤ 0
√(\(a^2\)) =  a, when a ≥ 0

Ex: √(\(3^2\)) = +3 = |3|
√(\((-3)^2\)) = 3 = -(-3) = |-3|

Simplifying √n

To simplify √n, we need to express n in powers of prime numbers and then need to take out the even powers.

Example: √56 = √(4*14) = √((2^2) * 14) = 2√14

Hope it helps!
Good Luck!
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­
How To Solve: Rationalize Roots


Attached pdf of this Article as SPOILER at the top! Happy learning! :)

Hi All,

I have recently uploaded a video on YouTube to discuss Rationalize Roots in Detail:




Following is covered in the video

    ¤ How to Rationalize Roots
    ¤ Example 1 : Rationalize \(\frac{1}{√𝟑−√𝟐}\)
    ¤ Example 2 : Rationalize \(\frac{1}{√𝟒+√𝟑}\) + \(\frac{1}{√𝟑+√𝟐}\)

How to Rationalize Roots

    ¤ To Rationalize the denominator we do computations to move the root term to the numerator.
    ¤ This is usually done by multiplying the numerator and denominator with a conjugate of the denominator.
    ¤ Thus the denominator becomes a whole number.

Example 1 : Rationalize \(\frac{1}{√𝟑−√𝟐}\)

To Rationalize \(\frac{1}{√𝟑−√𝟐}\) we will multiply the numerator and the denominator with the conjugate of the denominator.
We can find the conjugate of √𝟑−√𝟐 by just inversing the sign between √𝟑 and √𝟐
=> Conjugate of √𝟑−√𝟐 will be √𝟑+√𝟐

=> \(\frac{1}{√𝟑−√𝟐}\) = \(\frac{1}{√𝟑−√𝟐}\) * \(\frac{√𝟑+√𝟐}{√𝟑+√𝟐}\)
= \(\frac{√𝟑+√𝟐}{(√𝟑−√𝟐) * (√𝟑+√𝟐)}\)

Now the denominator is of the form (a-b) * (a+b) and will be equal to \(a^2 - b^2\)

=> \(\frac{√𝟑+√𝟐}{(√𝟑−√𝟐) * (√𝟑+√𝟐)}\) = \(\frac{√𝟑+√𝟐}{(√𝟑)^2 − (√𝟐)^2}\) = \(\frac{√𝟑+√𝟐}{3 − 2}\) = √𝟑+√𝟐

Example 2 : Rationalize \(\frac{1}{√𝟒+√𝟑}\) + \(\frac{1}{√𝟑+√𝟐}\)

Following above logic we can find that \(\frac{1}{√𝟒+√𝟑}\) = √𝟒 - √𝟑 and \(\frac{1}{√𝟑+√𝟐}\) = √3 - √2

=> \(\frac{1}{√𝟒+√𝟑}\) + \(\frac{1}{√𝟑+√𝟐}\) = √𝟒 - √𝟑 + √3 - √2 = √𝟒 - √2 = 2 - √2

Hope it helps!
Good Luck!

Watch the following video to learn the Properties of Roots

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­Surds are irrational roots of a rational number.

e.g. \(\sqrt{6} =\) a surd ⇒ it can’t be exactly found. Similarly \(- \sqrt{7}, \sqrt{8}, \sqrt[3]{9}, \sqrt[4]{27}\) etc. are all surds.

Pure Surd : The surds which are made up of only an irrational number e.g. \(√6, √7, √8\) etc.
Mixed Surd : Surds which are made up of partly rational and partly irrational numbers e.g. \(3√3, 6^4√27\) etc.

Q. Convert \(√27\) to a mixed surd A. \(√27 = √9 \times 3 = 3√3\)
Q. Convert \(2√8\) to a pure surd A. \(2√8 = √8 \times 4 = √32\)

Rationalization of Surds: In order to rationalize a given surd, multiply and divide by the conjugate of denominator [conjugate of \((a + √b)\) is \((a – √b)\) and vice versa].

e.g. \(\frac{(6+\sqrt{2})}{(1-\sqrt{3})} = \frac{(6+\sqrt{2})(1+\sqrt{3})}{(1-\sqrt{3})(1+\sqrt{3})}= \frac{(6+6 \sqrt{3} +\sqrt{2}+\sqrt{6})}{(1-3)}=\frac{(6+6 \sqrt{3 }+\sqrt{2} +\sqrt{6})}{-2}\)
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­An expression is one or a group of terms and may include variables, constants, operators and grouping symbols.

Example \(-3,(x-2y),\frac{2p-3q^3}{9}, \sqrt[3]{\frac{y^6}{2-y^2}}\)




Source of the images in this post: A Maths Dictionary for Kids by © Jenny Eather
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­Any form of numeric representation representing a number is called an Expression.

The equality of two expressions is called Equation.






1. \((a ± b)^2 = a^2 ± 2ab + b^2\)

2. \((a + b)^2 – (a – b)^2 = 4ab\)

3. \((a + b)^2 + (a – b)^2 = 2 (a^2+ b^2)\)

4. \(a^2 – b^2= (a + b) (a – b)\)

5. 
(1) \((a + b)^3= a^3+ b^3+ 3ab (a + b)\); 
(2) \((a – b)^3= a^3– b^3– 3ab (a – b)\)

6. 
(1) \(a^3+ b^3= (a + b) (a^2– ab + b^2)\) ; 
(2) \(a^3– b^3= (a – b) (a^2+ ab + b^2)\)

7. \((a + b + c)^2= a^2+ b^2+ c^2+ 2ab + 2bc + 2ac\)

8. \(a^3+ b^3+ c^3– 3abc = (a+b+c)(a^2+ b^2+ c^2– ab – ac – bc) ⇒ if (a + b + c) = 0\) then \(a^3+ b + c^3= 3abc.\)


Solutions for systems of equations


1. Definition of a solution to a statement of equation. A solution of a system of equations consists of values for the variables that reduce each equation of the system to a true statement. To solve a system of equations means to find all solutions to the system.

2. Consistent systems. When a system of equations has at least one solution, it is said to be consistent.

3. Unique solutions. When a system of equations has one and only one solution, we say that the solution is unique.

4. Possible sets of solutions for a system of equations. For a given system of equations.

(a). We may have one and only one solution. In this case we say we have a unique solution.
(b). We may no solutions. In this case we say that the system of equations is inconsistent.

5. We may have more than one set of solutions but the set of solutions is not infinite. We say that the number of solutions is finite. This can occur when one of more variables appears in an equation with a power other than one and/or there is a product of variables in one of the equations. Such an equation is called a non-linear equation.

6. We have an infinite number of solutions to the system of equations. In this case we say the system is indeterminate.



Methods of finding solutions for equations


Method of substitution.

(a): Pick one of the equations and solve for one of the variables in terms of the remaining variables.
(b): Substitute the result in the remaining equations.
(c): If one equation in one variable results, solve this equation. Otherwise, repeat step a until a single equation with a single variable remains.
(d): Find the values of the remaining variables by back-substitution.
(e): Check the solution found.


Examples of the method of substitution.

\(2x + 3y = 7 \)
\(x − y = − 1\)

\(x = y − 1\)

\(2(y − 1) + 3 y = 7\)
\(⇒ 2 y − 2+3 y = 7\)
\(⇒ 5 y = 9\)
\(⇒ y =\frac{9}{5}\)

\(x − \frac{9}{5} = − 1\)
\(⇒ x = − 1 + \frac{9}{5} = \frac{4}{5}\)


Method of elimination


1. Basic idea. The idea behind the method of elimination is to keep replacing the original equations in the system with equivalent equations until an obvious solution is reached. In effect we keep replacing equations until each equation contains just one variable.

Principles for obtaining equivalent systems of equations.

(a): Changing the order in which the equations are listed produces an equivalent system.
(b): Suppose that in a system of equations, we multiply both sides of a single equation by a nonzero number (leaving the other equations) unchanged. Then the resulting system of equations is equivalent to the original system.
(c): Suppose that in a system of equations, we add a multiple of one equation to another equation (leaving the other equations) unchanged. Then the resulting system of equations is equivalent to the original system.


Method of factorization.

Suppose we wish to solve \(3x^2 = 27\). We begin by writing this in the standard form of a quadratic equation by subtracting 27 from each side to give \(3x^2 − 27 = 0\).

We now look for common factors. By observation, there is a common factor of 3 in both terms. This factor is extracted and written outside a pair of brackets. The contents of the brackets are adjusted accordingly:

\(3x^2 − 27 = 3(x^2 − 9) = 0\)

Notice here the difference between the two squares which can be factorized as

\(3(x^2 − 9) = 3(x − 3)(x + 3) = 0\)

If two quantities are multiplied together and the result is zero then either or both of the quantities must be zero. So either

\(x − 3 = 0\) or \(x + 3 = 0\)

so that

\(x = 3\) or \(x = −3\\
\)

These are the two solutions to the equation.



Method by completing the square.


Suppose we wish to solve \(x^2 − 3x − 2 = 0\).

In order to complete the square we look at the first two terms, and try to write them in the form \((?)^2\). Clearly, we need an x in the brackets:

\((x + ?)^2\) because when the term in brackets is squared this will give the term \(x^2\)

We also need the number \(− \frac{3}{2}\), which is half of the coefficient of x in the quadratic equation, \((x-\frac{3}{2})^2\) because when the term in brackets is squared this will give the term \(− 3x\)

However, removing the brackets from \((x-\frac{3}{2})^2\) we see there is also a term \((-\frac{3}{2})^2\) which we do not want, and so we subtract this again. So the quadratic equation can be written

\(x^2 − 3x − 2 = (x-\frac{3}{2})^2 - (-\frac{3}{2})^2 -2 = 0\)

after simplifying

We can write these solutions as

\(x=\frac{3+\sqrt{17}}{2}\) or \(\frac{3-\sqrt{17}}{2}\)

Again we have two answers. These are exact answers. Approximate values can be obtained using a calculator.


Method Solving quadratic equations by using graphs

In this section we will see how graphs can be used to solve quadratic equations. Whether the coefficient of \(x^2\) in the quadratic expression \(ax^2 + bx + c\) is positive then a graph of \(y = ax^2 + bx + c\) will take the form left. If the coefficient of \(x^2\) is negative the graph will take the form on the right.



What can happen when we plot a graph of \(y = ax^2 + bx + c\) for the case in which \(a\) is positive.




The horizontal line, the x axis, corresponds to points on the graph where y = 0. So points where the graph touches or crosses this axis correspond to solutions of \(ax^2 + bx + c = 0\).
The graph in (a) never cuts or touches the horizontal axis and so this corresponds to a quadratic equation \(ax^2 + bx + c = 0\) having no real roots.
The graph in (b) just touches the horizontal axis corresponding to the case in which the quadratic equation has two equal roots, also called ‘repeated roots’.
The graph in (c) cuts the horizontal axis twice, corresponding to the case in which the quadratic equation has two different roots.


What can happen when we plot a graph of \(y = ax^2 + bx + c\) for the case in which \(a\) is negative.




In case (a) there are no real roots. In case (b) there will be repeated roots. Case (c) corresponds to there being two real roots.



Method using the Quadratic formula.

Consider the general quadratic equation \(ax^2 + bx + c = 0\).

There is a formula for solving this: \(x =\frac{−b ± \sqrt{b^2-4ac}}{2a}\)



Nature of Roots of Quadratic Equation






The following resources were consulted in the design of this handout:
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­Let a and b be real numbers. If a – b is negative we say that a is less than b and write a < b. If a – b is positive then a is greater than b, i.e., a > b.

1. For any two real numbers a and b, we have a > b or a = b or a < b.

2. If a > b and b > c, then a > c. If a > b then (a + c) > (b + c) and (a - c) > (b - c), however, ac > bc and (a/c) > (b/c) (not sure) (is true only when c is positive)

3. If a > b, then a + m > b + m, for any real number m.

4. If a ≠ 0, b ≠ 0 and a > b, then \(\frac{1}{a} < \frac{1}{b}\).

5. If a > b, then am > bm for m > 0 and am < bm for m < 0, that is, when we multiply both sides of inequality by a negative quantity, the sign of inequality is reversed.

6. If a > X, b > Y, c > Z then (1) a + b + c + .... > X + Y + Z + .... (2) abc .... > XYZ .... (Provided none is negative)

7. If x > 0 and a > b > 0, then \(a^x > b^x\)

8. If a > 1 and x > y > 0, then \(a^x > a^y\)

9. If 0 < a < 1 and x > y > 0, then \(a^x < a^y\)

10. Do not cancel anything from both sides of inequality unless you are sure that the canceled quantity is positive, so ax > ay does not necessarily mean x > y, etc.

11. The concept of number line is very useful in checking inequalities. The common values to check are x = 0, 1, -1, >1 (preferred value = 2), between 0 and 1 (preferred value = 1/2), between - 1 and 0 (preferred value = -1/2), and less than -1 (preferred value = -2). So in short, there are 7 points: -2, -1, -1/2, 0, 1/2, 1, 2.

12. |x| is defined as the non-negative value of x and hence is never negative. On the GRE, \(\sqrt{x^2} = |x|\), that means, the square root of any quantity is defined to be non-negative, so \(\sqrt{36} = 6\) and not − 6 on the GRE. BUT if \(x^2=36\) ⇒ \(x=6\) or \(-6\) both. So \(\sqrt{x^2}= x\) or \(-x\) both are possible. If, x is negative, then \(\sqrt{x^2}=-x\) as it has to be +ve eventually. In this case x is negative and -x is positive.

13. |5| = 5, |-5| = 5, so |x| = x, if x is positive or 0 and |x| = -x if x is negative.

14. If |x| > x, then x is negative.

15. If |x| = a, then x = a or x = -a.

16. If |x| > a, then x > a or x < -a.

17. If |x| < a, then x < a or x > -a.

18. If |x - a| > b, then either x - a > b or x - a < -b

19. If |x - a| < b, then either x - a < b or x - a > -b.

20. If |x| = x, then x is either positive or 0.

21. |a + b| ≤ |a| + |b|, |a – b| ≥ ||a| – |b||, |ab| = |a| |b|, \(|\frac{a}{b}|=|\frac{a}{b}|\), \(b \neq 0\), \(|a^2|=a^2\)

22. If (x - a) (x - b) < 0, then x lies between a and b. OR a < x < b.

23. If (x - a) (x - b) > 0, then x lies outside a and b. OR x < a, x > b.

24. If \(x^2> x\), then either x > 1 or x is negative (x < 0)

25. If \(x^2< x\), then x lies between 0 and 1. (0 < x < 1)

26. If \(x^2= x\), then x = 0 or x = 1.

27. If \(x^3> x\), then either x > 1 or x is between -1 and 0(either x > 1 or -1 < x < 0).

28. If \(x^3< x\), then either x lies between 0 and 1 or x is less than -1. (either 0 < x < 1 or x < -1)

29. If \(x^3= x\), then x = 0 or x = 1 or x = -1.

30. If \(x^3= x\), then x = 0 or x = 1 or x = -1.

31. If x > y, it is not necessary that \(x^2> y^2\) or \(\sqrt{x} > \sqrt{y}\) etc. So even powers can’t be predicted.

32. If x > y, it is necessarily true that \(x^3 > y^3\) or \(\sqrt[3]{x} >  \sqrt[3]{y}\) etc. So odd powers and roots dont change sign.

33. ab > 0 means \(\frac{a}{b} > 0\) and vice versa. The two are of the same sign.

34. ab < 0 means \(\frac{a}{b} < 0\) and vice versa. The two are of the opposite sign.

35. If x is positive, \(x + \frac{1}{x} ≥ 2\) .

36. If X is positive, then

(1) \(\frac{(a + X) }{ (b + X)} > \frac{a}{b}\) if a < b 

(2) \(\frac{(a + X) }{ (b + X)} < \frac{a}{b}\) if a > b

37. If X is negative, then

(1) \(\frac{(a + X) }{ (b + X)} > \frac{a}{b}\) if a > b 

(2) \(\frac{(a + X) }{ (b + X)} < \frac{a}{b}\) if a < b

38. \(\frac{(a + c + e + ....) }{ (b + d + f + ....)}\) is less than the greatest and greater than the least of the fractions \(\frac{a}{b}, \frac{c}{d}, \frac{e}{f}, .....\\
\)
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