Composite Numbers 1-100

Introduction to composite numbers 1-100:
There are different types of numbers and here we need to fiind what are the composite numbers between 1-100, before seeing that let us learn what is composite numbers and the different types of numbers.

The following are the types of numbers:

• Natural Numbers
• Whole Numbers Integers
• Rational Numbers
• Real Numbers
• Odd numbers
• Prime numbers
• composite numbers     

Composite numbers are numbers which can be divide by more than 1 number except 1 and the number itself. The number which has more than two factors( 1 and the number itself) is called composite number

Composite Numbers 1-100
Composite numbers are numbers which can be divide by more than 1 number except 1 and the number itself

The number which has more than one factor is called composite number

Example 4 is a composite number

The factors of 4 are 1 , 2 , 4

so here except the number 1 and itself it has one more factor.The factor is 2

So it is called as composite number

In simple words it is explained that except the prime number all the other numbers are called composite numbers

1 is neither prime number nor composite number.

Prime numbers:

There are 25 prime numbers between 1 and 100

So remaining are the composite numbers

The numbers such as 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73,79,83,89,97.

are 25 prime numbers between 1 to 100

These numbers are the 25  prime numbers between 1 and 100

These numbers cannot be divide by anyother number except 1 and the number itself

And the same is called prime numbers

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Composite Numbers 1-100

Composite numbers have factors except 1and itself

Except the 25 prime numbers between 1 to 100 all the other numbers are composite numbers

These are the composite numbers list 4,6,8,9,10,12,14,15,16,18,20,21,22,24,25,26,27,28,30,32,33,34,35,36,38,39,40,42,44,45,46,48,49,50,51,52,54,55,56,57,58,60,

62,63,64,65,66,68,69,70,72,74,75,76,77,78,80,81,82,84,85,86,87,88,90,91,92,93,94,95,96,98,99,100.......

These were the composite numbers between 1 to 100

For Example:

Let us consider some examples to check whether it is composite number

Whether 45 is a composite number?

45 can be divided by 1, 3, 5 9,45

leave 1and the number 45

we have the factors of 3 ,5,9

so it is composite number.

Another example

Let us consider 50

The factors of 50 are 1,2,5,10,25,50

Living 1 and 50 we have more factors

So 50 is a composite number.


The factors of 40 are 1,2,5,10,20,40

Living 1 and 40 we have more factors

So 40 is a composite number.

Mixed Numerals Fractions

Introduction :

Mixed numerals fractions is numbers can be the mixture of a whole and a part. Mixed numerals fractions are obtained by dividing the numerator by the denominator in an improper fraction to get the quotient and the remainder. Three types of fractions such as Improper fractions, proper fractions and mixed numerals fractions. Mixed fractions used to combine two or more fraction numbers. This article we are discussing about mixed numerals fractions. The mixed numerals fractions look like as,

                                                                 Remainder

                                                  Quotient    --------------

                                                                    Divisor

Examples Problem for Mixed Numerals Fractions:

Solve mixed fraction of given two mixed numerals fractions, [`6 8/4` ] + [`3 6/2` ], using addition method.

    Solution:

    Given mixed numerals fractions is,

    The problem here is to add [6 8/4] and [3 6/2]

    These two mixed numbers do not have the same denominators or lower numbers, so we must first find a common denominator of the two mixed numbers, earlier than adding them together. For the denominators here, the 4 and 2, a common denominator for both is 8.

    With the common denominator, the [6 8/4] becomes a [6 8/4] and the [3 6/2] becomes a [3 12/4]

    So now our addition problem becomes this,

    The problem here is to add [6 8/4] and [3 6/4]

                                                       = [`6 8/4` ] + [`3 6/4` ]

    The two mixed numbers having the same denominators.

    We can add them jointly by simply adding the numerators (the 8 and 6 = 14).

    It has the same denominator (the 4).

    And, we just add the number parts, or add the quotient 6 + 3 = 9.

                                                          = (6 + 3) (8+6)/ (4)

                                                          = [`9 14/4` ]

                                         Answer is: [`9 14/4` ].

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One more Examples Problem for Mixed Numerals Fractions:

Solve mixed fraction of given two mixed numerals fractions, [`2 50/2` ] + [`6 25/1` ], using addition method.

    Solution:

    Given mixed numerals fractions is,

    The problem here is to add [2 50/2] and [6 25/1]

    These mixed numerals fractions do not have the same denominators or lower numbers, so we must first find a common denominator of the two mixed numerals fractions, earlier than adding them together. Used for the denominators here, 2 and 1, a common denominator for both is 2.

    With the common denominator, the [2 50/2] becomes a [2 50/2] and the [6 2/1] becomes a [6 50/2]

    So now our addition problem becomes this,

    The problem here is to add [2 50/2] and [6 50/2]

                                                       = [`2 50/2` ] + [`6 50/2` ]

    The two mixed numerals fractions having the same denominators.

    We can add them jointly by simply adding the numerators (the 50 and 50 = 100).

    It has the same denominator (the 2).

    And, we just add the number parts, or add the quotient 2 + 6 = 8.

                                                          = (2 + 6) (50+50)/ (2)

                                                                            = [`8 100/2` ]

                                                           Answer is: [`8 100/2` ].
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Solve square root problems

Introduction :
The Square root problems of numbers are done when the number is multiplied through itself. In the same way for polynomial the square root is obtained when the polynomials are multiplied by it.

              It is credible for solve together positive and negative numbers but when a negative number is taken out of root the value will become invented. There are two types of square problems such as ideal squares and non perfect square. It is simple to solve square root for perfect square.

Methods for Solving Square Root Problems:

The easiest method to define Square root problems in the method by using Prime Factorization:

(i) Write down the prime factorization of n from the given problems. Pair the factors such that primes in each pairs are equal.

(ii) Select one prime from every one couple and resolve multiplication of all such primes.

(iii) The multiplication obtained in (ii) is the square root of n problems.

These methods are used to solve the square root problems as easiest way.

Properties of Square Root Numbers:

If the unit’s digit of a number is 2, 3, 7 or 8, then it is not a perfect square and hence doesn’t have a square root.
If a number have square root mean it should have the numeral values of 0, 1,4,5,9.
Which are the odd numbers ends with of zeros mean it should not have a square root.
If a number ends in an odd number of zeros, then it doesn’t have a square root. If a square numeral is followed by an even number of zeros, it has a square root.
Solve Square Root Problems-example:

Perfect Squares:

Solve the square root of the following problems:

(a).v1600 = v2*v2*v5*v5*v4*v4.

According to the method, choose one prime from each pair

                 = 2*5*4.

                 = 40

So the perfect square root is 40.

(b).v900 = v3*v3*v5*v5*v2*v2.

According to the method, choose one prime from each pair,

                  = 3*5*2.

So the perfect square root is 30.

(c).v144 = v2*v2*v2*v2*v3*v3

According to the method, choose one prime from each pair,

                = v4*v4*v3

                =12.

So the solution is 12.

Non Perfect Squares:

Solve the square root of the following:

(a).v2352 = v2*v2*v2*v2*v3*v7*v7

According to the method, choose prime from each pair

                  = v4*v4*7*v3

According to the method, choose 4 as a prime from each pair

                   = 4*7*v3

                   =28*v3

So the square root of v2352 = 28*v3.

(b).v603 = v3*v3*v67

                =3*v67

So, the square root of v603 = 3*v67.

(c).v9408 =v 2*v2*v2*v2*v2*v2*v3*v7*v7

                   =2*2*2*7*v3.

                  =4*2*7*v3

                  =56*3.

So the square root of v9408= 56*3.

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Introduction for discrete random variable example

Introduction for discrete random variable example:

              In the mathematics a random variables is a variable whose value is a function of the outcome of a statistical experiment. Random variables are used in the study of probability. They were developed to get in the analysis of the games of chance, stochastic events, and the results of scientific experiments by capturing only the mathematical properties necessary to answer probabilistic questions. Further formalizations have been firmly grounded the entity in the theoretical domains of mathematics by making use of measure theory.

Examples for Discrete Random Variable:

 Discrete Random Variable If a random variable takes only a finite or a countable number of values, it is called a discrete random variable

 The number of heads obtained when two coins are the tossed is a discrete random variable as X assumes the values 0, 1 or 2 which form a countable set. Number of Aces when ten cards are drawn from a well shuffled pack of 52 cards.
The random variable X assumes 0, 1, 2, 3 or 4 which is again a countable set. X (No aces) = 0, X (one ace) = 1, X (two aces) = 2,x (three aces) = 3, X (four aces) =4
Example for Discrete Random Variable Example:

Example:  A random variables X has the following probability mass function

X             1    1     2   3    4     5     6

P(X = x)   k    3k   5k 7k   9k   11k 13k

(1) Find k.

(2) Evaluate P(X < 4), P(X = 5) and P(3< X = 6)

(3) What is the smallest value of x for which P (X = x) >1/ 2 .

Solution:

(1) Since P(X = x) is a probability mass function `sum_(n=0)^6` P(X = x) = 1

ie.,P(X=1) + P(X = 1) +P(X = 2) +P(X = 3) +P(X = 4) +P(X = 5)+P(X = 6) = 1.

? k + 3k + 5k + 7k + 9k + 11k + 13k = 1 ? 49 k = 1 ? k =1 / 49

 (2) P(X < 4) = P(X = 1) + P(X = 1 ) + P(X = 2) + P(X = 3) =1 / 49 +3 / 49 +5 / 49 +7 / 49 =16 / 49

P(X = 5) = P(X = 5) + P(X = 6) =11 / 49 +13 / 49 =24 / 49

P(3 < X = 6) = P(X = 4) + P(X = 5) + P(X = 6) =9 / 49 +11 / 49 +13 / 49 =33 / 49

(3) The minimum value of x may be determined by trial and error method.

P(X = 1) =1 / 49 <1 1="1" 2="2" 49="49" br="br" p="p">
P(X = 2) =9 / 49 <1 2="2" 3="3" 49="49" 6="6" br="br" p="p">
P(X = 4) =25 / 49 > 1 / 2

? The smallest value of x for which P(X = x) > 1 / 2 is 4.

Number Series Solver

Number series solving is needed some logical thinking
Learn now number series by solving  some examples

EX:1       1,3,5,X,9,11 THAN WHAT IS X?

Whenever you solve the number series first just absurb the pattern
Here pattern is looking like odd number series
The odd number series after 5 and before 9 we will get 7.
Here missing number from the series is 7. x=7 is the answer

EX:2    2,4,6,8,__,12,14

We observe the above series it's an  even number series
Number we have to find is followed by 8 and preceded by 12.The next even number to 8 is 10 

EX:3     1,3,5,7,9,__,13,15

We observe the above series all terms are even number

Number we have to find is followed by 9 and preceded by 13, so the next odd number to 9 is 11 

EX:4   3,6,9,4,8,12,5,10,___

Here if we first observe the first three terms are 3x1,3x2,3x3

than next three terms in the series are 4,8,12 so we can write them as 4x1,4x2,4x3.

So next three terms in the series will be 5x1,5x2,5x3 (5,10,15) , answer is 15

EX:5  1,2,4,7,11,__,22,29

The difference between second and first term is 2-1=1
Difference between third and second term is  4-2=2
Difference between fourth and third term is 7-4=3
Difference between fifth and fourth term is 11-7=4
Here we observes that the differences between two successive terms are increased by 1
let us assume next term to the 11 as x so difference between x and 11 will be x-11=4+1=5 =>x-11=5
x-11+11=5+11
x=16
Lets check  16 (the answer for this series) is satisfying the pattern or not
22-16=6
29-22=7
so pattern is satisfied with 16

EX:6 1,5,9,13,__,21

we observe the pattern here the difference between two successive terms is equal to 4 like
5-1=4 , 9-5 =4 ,13-9=4, so the numbers in the series  are in the  arthematic progression
Now assume the required number as x so x-13=4, so x-13+13=4+13
x=17

EX:7 1,2,4,8,16,__,64

We observe here  second term is equal to first term X 2,Third term is equal to second term X  2, fourth term is equal to third term  X  2 and so on .....
The required sixth term which is equal to fifth term x 2
sixth term=16 x 2= 32

Ex:8 1,4,9,16,__,36

Here the  first term is 1 which can be written as =1^2, second term 4 which can be written as 2^2
Third term is 9 which can be written as 3^2, fourth term is 16 which can be written as 4^2
The  fifth term will be 5^2 =25

EX:9: 1,8,27,64,__,216

Here first term is 1 which is equal to 1^3, second term is 8 which can be written as 2^3,third term is 27 which can be written as 3^3, fourth term is 64 which can be written as 4^3,so fifth term will be 5^3 so 125
Now check the pattern sixth term is equal to 6^3=216

EX:10: 0,1,8,27,__,125,216

Here we observe the pattern
First term is equal to (1-1)^3=0,second term is equal to (2-1)^3=1^3=1,third term is equal to (3-1)^3=2^3=8,forth term is equal to (4-1)^3=3^3=27.
Fifth term is equal to (5-1)^3=4^3 = 64
Lets check the pattern, sixth term is equal to (6-1)^3=5^3=125
seventh term is equal to (7-1)^3=6^3=216

Now Try this Example :2,5,8,11,x,17 on Number Series Solver

Examples on number series for practice:

Now after observing  the series the common difference between two successive elements is (5-2)=3=(8-5), so common difference is 3

The given series is in arithmetic progression, so the next element to the 11 is 11+3=14

so x=14 is the missing number of the series

More Examples on Number Series Solver

Ex Find 1,2,4,8,x,32 so x=?

Solution :Here common ratio between two successive terms is 2/1 =4/2=8/4=2

 the series is in geometric progression,  (x/8) = (8/4)

so x/8=2,  x=8 *2

x=16, so x=16 is the missing number of the series.

Some more examples for practice:

problem 1)   1,16,81,__,625?   tip:1^4,2^4,3^4,4^4,5^4

problem 2) 11,15,19,12,16,20,13,17,__ ?  observe first three terms than next three than next three and than compare you will get the answer

problem 3) 0,1,1,2,3,5,8,__,21............ tip third term=first term+second term

problem 4)0,2,6,12,20, __ .................. tip n^2-n for first term n=1 for second term n=2 and so on

problem 5)7,14,21,28,__  ...................tip series in multiples of 7 or in geometric progression

problem 6) abc , bcd, def , ____       no tip for this , it's a simple question