Easy Way to Multiply Big Numbers

Multiplication is one of the basic operation in mathematics and its symbolic representation is ‘’×’’. It is difficult while multiplying big numbers. So, we are going to use the easy for multiplying big numbers. Easy way is to split the numbers into smaller parts and then multiply with 5, 10 and then add the result to the multiplication of remaining parts, we get the answer. Let us see about easy way to multiply big numbers in this article.

Worked Examples for Easy Way to Multiply Big Numbers

Example 1 for Easy Way to Multiply Big Numbers:

Multiply the Number which Ends with 5:

How to square the number 85?

Step 1:

Take the number which is on the left of the given number. The left of the given number is 8.

Step 2:

Multiply the left of the given number 8 with its consecutive number 9.

Multiply 8 × (8 + 1) = 8 × 9 = 72

Step 3:

Then multiply the ending number with the same ending number 5 is 5×5 = 25.

Step 4:

Grouping the step 2 and step 3 results, we get 7225.

Step 5:

Hence, squaring the given number 85 is 7225.

Example 2 for Easy Way to Multiply Big Numbers:

Multiplying Big Numbers by 11:

How to multiply 986 by 11?

Step 1:

Multiply 986 × 11

Step 2:

The number 11 can be written as (10 + 1).

986 × (10 + 1)

Step 3:

Now multiply 986 by 10, we get 9860.

Step 4:

Multiply 986 by 1, we get 986.

Step 5:

Add the result of above two steps.

9860 + 986 = 10846

Step 6:

Hence, multiplying 986 by 11 is 10846.

Example 3 for Easy Way to Multiply Big Numbers:

Multiplying two Big Digit Numbers:

55 × 57

13 × 14

Solution:

a) 55 × 57

Step 1:

Split the multiplier into two parts.

(55) × (55 + 2)

Now, multiply 55 with 55 and 2 respectively and add both the results, we get the answer.

Step 2:

Multiply the first part 55 × 55, by using the multiplication concept of numbers end with 5.

Step 3:

Take the number which is on the left of the given number 55. The left of the given number is 5.

Step 4:

Multiply the left of the given number 5 with its consecutive number 6.

Multiply 5 × (5 + 1) = 5 × 6 = 30

Step 5:

Then multiply the ending number with the same ending number 5 is 5×5 = 25.

Step 6:

Grouping the step 4 and step 5 results, we get 3025.

Step 7:

Multiply the second part 55 ×2, by using the multiplication concept of numbers end with 5.

55 × 2 = 110

Step 8:

Hence, multiplying 55 × 57, we have to add the step 6 and step 7 results, we get the answer.

3025 + 110 = 3135

Step 9:

Hence, the solution for multiplying 55 × 57 is 3135.

b) 13 × 15

Step 1:

13 × (10 + 5)

Step 2:

13 × 10 = 130

Step 3:

13 × 5 = 65

Step 4:

130 + 65 = 195

Step 5:

Hence, the solution for multiplying 13 × 15 is 195.

Practice Problems for Easy Way to Multiply Big Numbers

1. Multiply the 788 by 11.

2. What is the value of square of 75?

Solutions:

1. 8668

2. 5625

Adding Sine and Cosine

Here we are going to see the article as adding sine and cosine, generally sine and cosine are trigonometric functions it is used for measuring the side length and angle of the right angle triangle .Sine is defined as the ration of the opposite side to the hypotenuse and cosine is defined as the ratio of the adjacent side to the hypotenuse. So here the right angle triangle will be given with the corresponding sides name like adjacent, opposite and hypotenuse.

Example 1-adding Sine and Cosine:

Sin 45 +cos 45

Solution:

We know that the value of sin 45 and cos 45 (both are equal) =1/`sqrt 2` , plug these values into above expression.

`1/sqrt 2` +`1/sqrt2`

If the denominators of both fractions are same so we add the numerator part only we get the answer.

`(1+1)/sqrt 2 =2/sqrt 2`

Example 2-Adding sine and cosine:

Sin260+cos260

Solution:

Already we know that the formula for sin2 A+cos 2A=1

Here the value of A=60 so we use the formula as Sin260+cos260=1.

It will be check with the another method,

We know that the value of sin 60 is sqrt `3/2 ` and cos 60 is `1/2` plug these values into above expression

`(sqrt(3)/2)^2 + (1/2)^2` =`3/4` +`1/4` =1

We get the same answer in both the method.

Example 3-adding Sine and Cosine:

Sin 30+cos 90

Solution:

We know that the value of sin 30 and cos 90 the value of sin 30 is `1/2` and cos 90 is 0plug these values into above expression.

`1/ 2` +0= `1/2`



Example 4-Adding sine and cosine:

Sin 0+cos (-30)

Solution:

We know that the value of sin 0 is 0 and cos (-30) =cos 30 value of cos 30 is sqrt 3/2

0+`(sqrt3)/2` =`(sqrt 3)/2.`

Practice problem -Adding sine and cosine:

1)    Add sin2 25+cos2 25

Answer =1

2)    Add sin 67+cos 13

Answer =1.8949

Statistical Techniques

Let fp( .| `theta` )g be a family of probability densities indexed by a parameter `theta` .Let y = (y1,....., yn) be a sample from p(.|`theta` ) for some unknown `theta` . Let T(y) be a statistic such that the joint distribution factors as follows

`prod` p(yi | `theta` ) = g(T(y); `theta` )h(y):

for some functions g and h. Then T is called a sufficient statistic for `Theta``theta` .

The  statistical techniques involve to study the two important parts namely mean deviation and standard deviation.

Mean Deviation for un grouped data:

Let x1,x2,x3,………xn be the given n observation. Let  `barx`   be the AM and M be the median .then

i.) MD(x¯) = `(sum_(i= 1)^n)/(n)` | xi –x|           where `barx`

Statistical Techniques in Mean, Mean Deviation

1.) Find the mean deviation about the mean for the given data.

15,17,10,13,7,18,9,6,14,11.

Sol: Let the mean of the given data be `barx` Then,

`barx` = `(sum_(i= 1)^n)/(n)` | xi –x|
= `(120)/(10)`  = 12    [n = 10]

so, the values of (xi – x) are:

3, 5, 2, 1,-5,6,-3,-6,2,-1

So, the values of | xi –x| are:

3 , 5 , 2 , 1 , 5 , 6 , 3 , 6 , 2 , 1

MD(`barx` ) =   `(sum_(i= 1)^n)/(n)` | xi –x|


=  `(34)/(10)`  = 3.4

Hence the MD (`barx` ) = 3.4

Statistical Techniques in Variance with Example:

Variance and Standard Deviation for statistical techniques:

Definition of variance :

Mean of the squares of the deviations from the mean is called the variance , to be denoted by σ2. The variance of n observations x1 , x2 ,.....xn is given by σ2 =   `(1)/(n)` `sum_(i=1)^n` (xi - x¯)2 .

Definition  of Standard deviation :

The  positive square root of variance is called standard deviation and it is denoted by σ  .

σ =`sqrt((sum_(i=1)^n( x i-x)^2)/(n))`

Ex: Find the mean, standard deviation and variance of first n natural numbers

Sol:

Mean,x = ( 1 + 2 + 3 +........+ n)    = 1  .1   n( n + 1 ) =  1  ( n + 1 )
n               n   2                      2

Variance, σ2 = `(sum)/(n)` x-2
=  x¯2



Variance, σ2 = `(sum)/(n)`  n2  =  { 1/2 (n + 1) }2



=          `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=     `(1)/(n)` `(n(n+1)(2n+1))/(6)`  -  `((n+1)^2)/(4)`

=  `((n+1))/(n)` ×   `(n(2n+1))/(6)`  -  `((n+1))/(4)`

= `((n+1))/(1)`   × `((2n+1))/(6)`  -  `((n+1))/(4)`

=`((n+1)(n-1))/(12)`

Variance,σ2  = `((n^2 -1))/(12)`

Standard deviation ,σ2 = `((sqrt(n^2 -1)))/(12)`

= `((1/2) (sqrt(n^2 -1)))/(3)`

Change in Percentage

Change is something which never dies and never changes. Change can be termed in the mathematical form as Variable, which has no constant value and it can take any value. It is usually denoted using some alphabet like the usual one ‘x’, the value of x is always unknown. There is a change in lot of things around us like every instant that is every second, and even that every second is a change in terms of time, distance and anything possible. The word possible is very challenging and there is nothing impossible in the world, impossible can be changed into possible through hard work and perseverance.

Coming to the concept of numbers, in which the topic concerned is about percentage  This concept when thought of, the general perception that every single has is that it is tough and not an easy concept to understand. The Percentage Change can always be an increase or a decrease in per-cent-age, it can be either way round. So it is necessary for us to learn both the method of calculations, actually both require the same Percentage Change Formula, the Calculating Percentage Change here will be only difference in the final answer, that too in the interpretation part, since obviously without understanding the number one can never say if it is an increase or a decrease in the change occurred.

Let us now try to answer the question of  How to Calculate Percentage Change. Answering this question is never a great task, it is all about the numbers we get for solving, and one can easily solve such type of questions, if one understands the problem and interprets the Percentage of Change. Let us consider an example to get a better understanding Calculate Percentage Change of the numbers whose initial value that is the original value is 50 and the final value is 100; the answer for this question is 50 – 100 / 50 * 100 = 100%, the answer shows us that there is an 100 % increase in the final value from the original value and that is the process of Calculating Percentage Change.

The fact is that anyone can solve such problems, but the interpretation part can only be done by the person who completely understood the concept. The final interpretation shows us the direction of change that is it shows whether the change is positive or negative and that makes the concept complete.

First Decimal Place

Binary numerals are usually read digit-by-digit, in order to distinguish them from decimal numbers. For example, the binary numeral 100 is pronounced one zero zero, rather than one hundred, to make its binary nature explicit, Since the binary numeral 100 is equal to the decimal value four, it would be confusing to refer to the numeral as one hundred. (Source: Wikipedia)

First Decimal Place:

Let us take a number, 19.7.

1 indicates place value ten.

9 indicates place value one.

7 indicate 7/10(tenths decimal place)

Example problems for first decimal place:

Example 1:

Calculate the following 15.9+15.7-0.15+0.1

Solution:

Given that, 15.9+15.7-0.15+0.1

Numbers of digits behind decimal point are equal.

Step 1: Adding positive numbers,

= 15.9+15.7+0.1

= 31.7.

Step 2: Negative integer add to the primary step

= 31.7 – 0.15

= 31.55

The solution is 31.55

Example 2:

Calculate the following 18.2+0.1-18.7+1.8

Solution:

= 18.2+0.1-18.7+1.8

Numbers of digits behind decimal points are equal.

Step 1: Adding positive numbers,

= 18.2+0.1+1.8

= 20.1

Step 2: Negative number add to the first step

= 20.1– 18.7

= 1.4

The solution is 1.4

Addition for first decimal place:

Problem 1:

Add the following 1.3+1.5+1.7+1.9

Add the positive value,

=1.3+1.5+1.7+1.9

= 6.4

The solution is 6.4

Problem 2:

Add the following 1.6+1.5+1.7+1.7

Add the positive value,

=1.6+1.5+1.7+1.7

=6.5

The solution is 6.5

Example problem for first decimal place:

Subtraction for first decimal place:

Problem 1:

Find 9.3-9.5-9.8-9.7

= 9.3-9.5-9.8-9.7

Subtract the negative value,

= -19.7

The solution is -19.7

Problem 2:

Find 8.6-8.5-8.7-8.9

Subtract the negative value,

= 8.6-8.5-8.7-8.9

= -17.5

The solution is -17.5

Multiplication for first decimal place:

Problem 1:

3.5 * 3.5 * 3.5

Multiply the positive value,

= 3.5 * 3.5 * 3.5

= 12.25* 3.5

= 42.8

The solution is 42.8

Problem 2:

1.8 * 1.8 * 1.8

Multiply the positive terms,

= 1.8 * 1.8 * 1.8

= 3.24 * 1.8

= 5.8

The solution is 5.8

Division for first decimal place:

Problem 1:

`7.8/0.6`

Divide the positive numerator value,

= `7.8/0.6`

= 13.0

The solution is 13.0

Problem 2:

`9.7/8.8`

Divide the numerator value,

= `9.7/8.8`

=1.1

The solution is 1.1

Practice Problem for first Decimal Place:

Problem1: 7.2+8.3+8.2

Answer: 23.7

Problem2: 9.2+9.2+9.2

Answer: 27.6

Problem3: 15.3+15.2+15.2

Answer: 45.7

Practice Model Definition

If any student wants to know about the definition for math practice model with examples, they can be referring the below examples for them. In this we can see some definitions for practice model in statistics with example problems. Let us see some of the definition in statistics such as mean, median and mode with example and practice problems with solve as follows.

Practice Model Definition - Definitions:

In this following we can see some definition for math practice model in statistics.

Practice Model Definition 1:

Mean:

It is a set of data which can find the average in dividing by sum of numbers with total numbers in the given data.

Practice Model Definition 2:

Median:

It is a set of numbers which can find the middle numbers and arrange the numbers in order to select the middle number.

Mode:

It is a set of numbers that can occur frequently in a set of data and no number can occur more than once.
Practice Model Definition - Examples and Practice Problems:

In this following we can see some definition for math practice model in statistics with example and practice problems.

Find the mean of weights for 6 peoples in kilograms are 15, 9, 10, 7, 16, and 23.

Solution:

                Sum of numbers
Mean = ------------------------
                  Total number

= `(15+9+10+7+16+23)/6`

= `80/6`

= 13.33

Find the median of 10, 24, 12, 21, 7 and 18.

Solution:

Arrange the data in ascending order as 7, 10, 12, 18, 21 and 24.

N = 6

Since n is even, median = `1 / 2` [`(nth)/2` item value + (`n / 2` + 1)th item value]

= `1 / 2` [6th item value + (`6/2` + 1)th item value]

= `1 / 2` [3rd item value + 4th item value]

= `1 / 2` [12 + 18]

= `1 / 2 xx 30`

= 15

Find the mode of 22, 17, 19, 15, and 22.

Solution:

22 are repeated twice.

Mode = 22

Practice Problems:

Find the mean of weights for 6 peoples in kilograms are 5, 9, 10, 7, 16, and 23.

Solution:

= 11.66

Find the median of 8, 25, 11, 21, 6 and 17.

Solution:

= 14

Find the mode of 20, 17, 11, 13, and 20.

Solution:

= 20

Integral of X Sinx

The term integral may also refer to the notion of anti-derivative, a function F whose derivative is the given function ƒ. In this case it is called an indefinite integral, while the integrals discussed in this article are termed definite integrals. Integrate the function f(x) can be written as`int ` f(x) dx. The integral is used to find the surface area, volume geometric solids by using stokes theorem, divergence theorem (like double integral, triple integral).
Source Wikipedia.

Integral Formulas:

1. `int` x n dx = `(x^n+1) / (n+1) ` + c

2. `int`sin x. dx = - cos x + c

3.` int ` cos x .dx = sin x. + c

4.  ` int` sec2x. dx = tan x + c

5. `int` cosec x.cot x. dx = -cosec x + c

6.`int dx / sqrt(a^2 - x^2) = sin^-1(x / a) + c`

Integral Problems:

Integral problem 1:

Find the integration of  x sin x with respect to x.

Solution:

Given term is x sin x

Integral of x sin x can be expressed as `int` x sin x dx

let u = x                         dv = sin x dx

`(du)/(dx)` = 1            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

= x . (-cos x ) - ` int` (-cos x) 1.dx

= - x cos x +` int` cos x dx

= - x cos x + sin x + c

Answer:  `int` x sin x dx  = - x cos x + sin x + c

Integral problem 2:

Find the integration of  (2x + sin x) with respect to x.

Solution:

Given term is 2x + sin x

Integral of (2x + sin x )can be expressed as `int`(2x + sin x )  dx

`int`(2x + sin x )  dx =  `int` 2x dx +` int` (sin x).dx

= 2 `int` x dx +` int`sin x dx

= 2` (x^2/2)`+ (-cos x) + c

= x2 - cos x + c

Answer:  `int` (2x + sin x) dx  =   x2 - cos x + c

Integral problem 3:

Find the integration of  (sin x  + 3cosec2x)  with respect to x

Solution:

Given term is  (sin x  + 3cosec2x)

Integral of (sin x  + 3cosec2x) can be expressed as  `int ` (sin x + 3cosec2x) dx

`int ` (sin x + 3cosec2x) dx  =` int` (sin x dx) +` int` (3cosec2x) dx

= `int` (sin x dx) + 3` int ` (cosec2x) dx

=  (- cos x) + 3 (-cot x) + c

= - cos x - 3 cot x + c

Answer: `int `(sin x  + 3cosec2x) dx  = - (cos x + 3 cot x) + c

Integral problem 4:

Find the integration of  x2 sin x with respect to x.

Solution:

Given term is x2 sin x

Integral of x2 sin x can be expressed as `int`x2 sin x  dx

let u = x2                          dv = sin x dx

`(du)/(dx)` = 2x            and        v = - cos x

We know `int ` u dv  = uv - `int` v du

`int`x2 sin x  dx   = x2 . (-cos x ) - ` int` (-cos x) 2x.dx

= - x2 cos x +` int` 2x cos x dx

=  - x2 cos x + 2` int` x cos x dx + c

= - x2 cos x + 2 I1+c

Take I1 = ` int` x cos x dx

let u = x                       and     dv = cos x dx

`(du)/(dx)` = 1                                   v =  sin x

We know `int ` u dv  = uv - `int` v du

I1 =` int` x cos x dx = x sin x - `int` sin x dx

= x sin x - (-cos x) + c

I1 = x sin x + cos x + c

`int`x2 sin x  dx  =  - x2 cos x + 2 I1+ c

=  - x2 cos x + 2 (x sin x + cos x )+ c

= - x2 cos x + 2x sin x + 2cos x + c

= 2x sin x - (x2 - 2 )cos x + c

Answer:  `int`x2 sin x  dx   =  2x sin x - (x2 - 2 )cos x + c