Hello Everyone,
Below is the summary of my journey in ECI831. I gained lot of knowledge in this course and understood the part that social media can play in education. Thank you Dr.Alec Couros and my fellow classmates, you made this journey memorable. Hope you enjoy my video.

This blog post marks the completion of my learning project. Finally, I have successfullylearned and implementedVedic Techniquesin solving problems onArithmetic

operations . It was indeed fun and my journey was mesmerizing.Vedic Mathematicsis a great way to make Mathematics simpler and joyful for everyone and yes, it is Magic math.

In my first blog post,we learned what is Vedic Mathematics and how it can make Mathematics simpler and joyful. I started this journey of learning and knowledge sharing with a great interest in the area of Vedic mathematics. My goal was to implemented Vedic Technic’s in solving problems on Arithmetic operations. Mysecond blog postconcentrated on implementingVedic tricks in multiplication, we witnessed how easily one can multiply complex numbers in a minute and that too without using acalculator. Mythirdandfourth blog post focused on findingSquareandSquare rootof any numbers quickly and am sure most of us start looking for calculators if we want tofind square or square root of numbers more than three digits and now we can do it mentally and that too within few seconds.

In myFifth and Sixth blog postwe learned how cool Vedic math is when it comes tofinding cubeandcube rootof any numbers.This awe-inspiring technique helps us to easily find cube and cube root of any given digits mentally. Further Vedic Techniquesmade myseventh blogpost more interesting by simplifying thedivision process, now we can do more than just painlessly split a check.In myNinth blog postVedic mathematics offered a neat solution to calculatepercentagein our mind to overcome our day to day challenges.

In this blog post we are going to learn how to convert any fraction ending with 9 and denominator into its decimal equivalent. Vedic mathematics helps us to shorten the complex fraction process into simpler way ,In Sanskrit this technic is called as “Ekadhikena Purvena“ To solve fractions ending with 9 there is a Sutra or formula i.e. BY ONE MORE THAN THE ONE BEFORE ,by using this formula let me solve two examples

Below image is a simple illustration of the difference between usual method and Vedic method

1)Find the fraction of1/39 STEPS: i) In 1st step, i will find the divisor ,here the number 9 is the one before, previous . number of 9 is 3 and the number 3 will be increase by 1, so we will get . 3+1=4 ii) In 2nd step, on every example, i will start with zero point, and i will divide 1 by 4i.e.1/4=0.25 ,i will write 0.25 as 025 i.e 1/39=0.025 iii) In 3rd step, i will divide 25/4, i will get remainder as 1 and quotient as 6 i.e 1/39=0.0256 . 1 (REMAINDER) iv) In 4th step, i will combine remainder and quotient, it will become 16, then i will . divide again 16 by 4 which gives me 4 i.e.1/39=0.02564 . 1 (REMAINDER) v) In 5th step, i am going to divide 4 by 4 which gives me the value as 1 i.e.1/39=0.025641 . 1 (REMAINDER) vi) In 6th step, last digit is 1,if we divide 1 by 4, i will get as 0.25again i will write it . as 025,it became same as in 1st step 25 divide it by 4 gives me 1 as remainder and 6 . as quotient i.e. 1/39=0.0256410256 . 1 1 (REMAINDER) vii) In next step again we are going to combine remainder and quotient i.e.16, and i will . divide it by 4which gives me4 and again 4 by 4 gives me 1,i will stop up-to last . digit 1, if i go on solving the digit i will get recurring values ,we can stop up-to 5 . decimal values. i.e.1/39=0.025641025641 . 1 1 (REMAINDER) viii) Final answer : 1/39=0.025641025641

2) Let us solve one more example. . 2/39 i) In 1st step,i will find the divisor ,here the number 9 is the one before, previous . number 9 is 3 and the number 3 we will be increase by 1,we will get. . 3+1=4 ii) In 2nd step,on every example, i will start with zero point and we will divide 2 by 4i.e.2/4 = 0.5, i will write 0.5 as 05. i.e 2/39=0.05 iii) In 3rd step, i will divide 5/4, we will get remainder as 1 and quotient as 1. i.e2/39=0.051 . 1 (REMAINDER) iv) In 4th step, i will combine remainder and quotient, it will become 11, then i will . divide 11 by 4, which gives me remainder as 3 and quotient as 2 . i.e. 2/39=0.0512 . 1 3 (REMAINDER) v) In 5th step, if i combine remainder and quotient, it will become 32 .then i will divide .32 by 4 which gives me 8 . i.e.2/39=0.05128 . 1 3 (REMAINDER) vi) In 6th step last digit is 8 ,i will divide 8by 4, i will get the value as 2. i.e. 2/39=0.051282 . 1 3 (REMAINDER)
vii) In 7th step last digit is 2, if i divide 2 by 4, i will get the values as 0.5, i will write 0.5. as 05 then if i divide 5 by4 it gives me remainder 1 and quotient 1. i.e.2/39=0.051282051 . 1 3 1 (REMAINDER)
vii) In next step again i am going to combine remainder and quotient i.e. 11 and i will . divide it by 4which gives me 3 as remainder and 2 as quotient and again if i . combine remainder and quotient and divide by 4 gives me 32/4=8,i will stop . up- to last digit 8, if i go on solving the digit i will get recurring values .we can stop . up- to 5 decimal values. i.e. 2/39=0.05128205128 . 1 3 1 3 (REMAINDER) viii)Final answer:2/39=0.05128205128

I love stories and I consider myself a good story teller. As a math teacher I tell stories about mathematics , mathematicians and about doing mathematics. Firstly, I do this because I love it and secondly because my students like it and I think it is a most effective instructional tool in teaching mathematics . Instruction via storytelling creates more realistic, powerful and unforgettable images in the student mind than any other means of delivering the same material. Teaching students by storytelling is an effective way to keep the learners occupied and PREZI offers a great platform to achieve it. In this week’s blog post I am reviewing Prezi, an awesome presentation tool which has a potential to make teaching moments memorable.

underline the ideas presented there, it also supports the use of text, images and videos and also offers a collection of models to choose from and help new users get used to the interface.

Let me acknowledge a fact that as an educator I invest a lot of my time in creating presentations and use it in my classroom but most of the time it will be boring and I end up having a dull presentation that will not keep my students interested and in a math class, as a math teacher I cannot afford this. It takes a lot of creativity and in-depth knowledge of using software’s to create interactive and informative presentation but Prezi offers a great solution for educators with its unique platform of making and delivering presentations. As an alternative of slides and boring animation, Prezi infuse fun and creativity into our presentation.

After reviewing Prezi, I found it very straightforward presentation and easy to use, even for a first timer like me. We have total control over the details such as font size, adding audio files and images, importing PowerPoint slides, Embed PDF’s, sharing in social media etc. It also comes with a number of templates which we can use and still gives us flexibility to relate it to our topic and make it interesting. Prezi creates splendid visual impact in the classroom with its cinematic experience of the zooming function and students will feel ecstatic with the topic which we have designed. The presentation can also be easily shared on the internet and can be accessed by our students at home and they can navigate with Prezi and play around with it observing the connections of ideas and visualize concepts.

The strength’s and advantages of Prezi are enormous yet it has some weaknesses too. The first issue which caught my attention is that it lacks auto correction feature which can be a hassle for users. Prezi is completely web based, even thought it is an advantage but in few circumstances, it can be disadvantage when we have a limited internet access. It is not printer friendly so printing out our presentation may become a challenge to students.

Overall, Prezi is a great tool that most of us would find very useful in the classroom, especially for Math teachers like me. It is a visual storytelling software alternative which will be great advantage to educators and embeds cinematic experience in classrooms which makes teaching moments memorable. I believe that storytelling in mathematics is an excellent approach for creating a safe-learning environment where a student might openly appreciate, comprehend and enjoy mathematics. I feel that storytelling in mathematics classroom creates an atmosphere of imagination, excitement, and thinking which make math fun and memorable for students and by the help of Prezi we can create such an atmosphere in our classroom.

Emerging new tools and multimedia can definitely make storytelling easier, especially if the concept we want to share is complicated like mathematics. What do you guys think? will it be another distraction to students? I fell storytelling is an exceptional tool in mathematics education. How could you see this used in your own subject area?

There are situations in our day to day life when we need tocalculate percentagein our mind and these scenarios are quite common. When was the last time we talked to our friends, colleagues or students about percentage? probably in restaurants, parting things with friends or while discussing with students about their grades, paying rent, mortgage these real-life situations are common and what if we can do this quicker andin our mind easily. Vedic mathematicsoffers a neat solution to these problems.

In my previous blog ,we learned division method when the divisor is 9 and also Nikhilam formula,today in this post we are going to learn how to find percentage of any number using vedic maths.

Below image is a simple illustration of the difference between usual method and Vedic method

To find the percentage we are going to apply the sutra or formula i.e. Vertical and cross wise multiplication .which we have learned inmultiplication method . By using above formula i am going to solve some examples. 1)Lets us find the percentage of36⸓ of 85 STEPS i) Write the given values in two rows 3 6 8 5 ii)Now we will apply the technic of Vertical and cross wise multiplication , consider the 1st pair where cross multiplication is not possible ,we will solve it by vertical multiplication. 36
↓ (3×8=24) 85 24 iii)Consider 2 pairs, here cross multiplication is possible so we will apply the same rule. 3 6 . Χ 85 (15 + 48=63) 6 (CARRIED FORWARD) (24) 3 iv) Next consider the last pair there is only one pair left , so we will apply vertical multiplication method. 36 . ↓ (6×5=30) 8 5 6 3 (CARRIED FORWARD)
(24) 3 0 v) Add all the columns. 3 6 8 5 6 3 (CARRIED FORWARD) (24) 3 0 30 6 0 (ANSWER) vi) Divide the Answer by 100 we will get the final answer 3060÷100=30.60 36⸓ of 85 = 30.60

2)Lets us find the percentage of 52⸓ of 640 STEPS i) Write the given values in two rows 0 52 6 4 0 ii)Now we will apply the technic of Vertical and cross wise multiplication, consider the 1st pair,cross multiplication is not possible , we will solve it by vertical multiplication. 0 52
↓ (0×6=0) 6 4 0 0 iii)Consider 2 pairs, here cross multiplication is possible so we will apply the same rule. 0 5 2 . Χ (0+30=30) 6 4 0 3 (CARRIED FORWARD) 0 0 iii)Consider 3 pairs, here cross multiplication is possible,Multiply extreme digits by cross multiplication and middle digit by vertical method. 0 5 2 . Χ (0+12+20=32) 64 0 3 3 (CARRIED FORWARD) 0 0 2 iv) Next consider the last 2 pairs, cross multiplication is possible so we will apply same rule. 0 5 2 . Χ (0+8=8) 6 40 3 3 (CARRIED FORWARD) 0 0 2 8 v) Next consider the last pair there is only one pair left , so we will apply vertical multiplication method. 0 5 2 . ↓ (0×2=0) 6 4 0 3 3 (CARRIED FORWARD) 0 0 2 8 0 vi)Add all the columns 0 5 2 640 3 3 (CARRIED FORWARD) 0 0 2 8 0 3 3 2 8 0 (ANSWER)

vi) Divide the Answer by 100 we will get the final answer 33280÷100=332.80 52⸓ of 640= 332.80

Nowadays, it appears as if everyone is on internet, writing and reading blogs, especially in our teaching community blogs are extremely popular platform where we find information posing new ideas, instruction and motivation.for educators blogging offers the greatest opportunity to share their work and ideas with the world. Just blogging doesn’t suffice, but doing it well is totally a different story, one that involves a lot of time and commitment, yet sometime on several occasions we will be provoked with some thoughtful questions and criticisms about the whole blogging thing. In this blog post I am sharing one of my thoughts and concerns on sharing and blog post.

The first thing that pinches a relatively young, novice teacher and blogger like me is that critic of no one reads your blog or just cares on what you have to say? Or our work and ideas, plans will not be of benefit to others. Surely many of us might have heard of this one. I guess it is easily the very first criticism against the idea of blogging and sharing resources online and I think that this is the fear which keeps some teachers away from sharing their work online. In Sharing: The Moral Imperative. Dean Shareski talks that we all seek recognition’s for our contributions but the moment we focus on protecting our work we are in some ways antithesis of a teacher, I totally agree with it. No matter even if nobody is paying attention, the course of blogging and sharing our work, thoughts and ideas helps us find and evolve our inner voice and ideas. As DeanShareski agrees even I feel that it is our work as a teacher to share online and not confine our teaching practices and ideas in a building. We teachers as a prolific member of the society must not only teach our students in a confined space but have to go beyond it and by sharing our work, ideas online we can accomplish this task . After hearing what Dean Shareski says I agree that if there is no sharing then there is no education. The purpose of a teaching and a teacher is to educate irrespective of who the audience is.

The benefits of sharing work online are bountiful. I believe by sharing our work Motivates learners and readers to do their very best. In his video Dean Shareski
states few awesome examples and as a math teacher the one which attracted me was Dan Meyer’s MathStories. A relatively young mathematics teacher Dan Meyers from California who created some outstanding math resources which he shared for free, these resources are phenomenal for every math teacher and has helped numerous educators across the globe. Sharing our work online will not only help other educators to improve their teaching and learning practices but it also helps us to build a strong writing, editing, and revision skills. It also gives other learners the opportunity to acquire experience and helps to build their confidence and self-esteem.

sharing the work of our students gives them good motivation to improve their work. Seeing their work published online can inspire a sense of self-esteem and confidence in students, it help’s them develop their self-esteem because they have the opportunity to be recognized for their hard work and commitment. Students will be encouraged by the fact that their work is online for everyone to see, which can give them the confidence boost they need to give the best of themselves.

Sharing the student’s work online also works for the benefit of other students by offering a valuable experience. If students can see a good example, they are more likely to improve their skills and knowledge base. In addition, students can better understand how to perform a well-qualified task and what mistakes to avoid seeing the work of others and use this knowledge for their future projects.

Finally, it leaves me with a question that how many of us truly believe in the culture of sharing? Most of us are harvesting the fruit of sharing culture and are we giving it back? How do we get teachers to share their work online? Do you agree with the idea of sharing our students work online? And will it help them?

Open Educational resources (OER) are an exceptional way for educators to utilize digital content in classroom. Open Educational resources are learning and teaching materials which are freely accessible for everyone online. As I discussed in my previous blog post these material can be remixed, reorganized and revised and can be effectively used to deliver the knowledge to students. These collections of scrutinized, grade-level, content specific resources will be an asset for teachers and elevate our teaching standards and effectively use the digital content in our classrooms. New forms of communications are illuminating new opportunities for educators and students, this opportunity is why open education resources matter.

As a mathematics lecturer one of the big challenge is to keep our students engaged and excited about math. I must say that advances in technology has made our life easier and I must acknowledge that Open Educational resources has made math lessons fun in classroom and made my job as a teacher easier. It provides pre-made activities, options for distinguishing and even aiding with pre- and post- valuations, as well as grading.

In mathematical education , a manipulative is an object that is designed so that a student can perceive a mathematical concept by manipulating it, so its name. The use of manipulative provides a way for students to learn concepts through a practical experience appropriate to development. Manipulative ‘s aid learning math concepts

exciting and engaging for students but for teachers buying these supplies for classroom can be expensive. OER provide us a free solution, by using it we can create manipulative’s ourselves.
Instead of investing a lot of time trying to format everything the resources below provide an excellent platform for teachers to get pre-made math goodies.

Black Line Masters , Provides a range of printable tools, materials and resources which is mainly focused for children’s in their Pre-K to ninth grade. The resources which are available here are extremely helpful for teachers, it provides an ocean of materials which can be easily downloaded and printed as per our needs and requirement.
SEN Teacher has a load full of printable materials which are focused towards elementary level. It has polygons, clocks, coins, Nets (3D Models), arithmetic makers, number squares, Mazes etc.

I AmHomeschooling.com also offers ample of ready to use manipulative’s and tools which will help teachers for Pre-K to High school. It provides abundant resources for every student needs for different levels. We can print materials right from algebra tiles to cuisenaire style rods and fraction circles.
These Open educational resource repositories are very easy to access and extremely user friendly, the mathematical contents and material are very well organized and has high printing quality. It aids teachers prepare resourceful course contents for students and make mathematics fun.

ONLINE LESSONS AND VIDEOS Our classrooms today’s are embedded with remarkable technology and it has helped us deliver course content efficiently and has enhanced classroom learning. There are many resources especially for mathematics teachers by which we can eradicate the long standing math fear phobia from students. The web resources below have some great resources for math teachers and most of them are free, these sites offer best lesson plans videos and cover entire course.

Khan Academy Is an excellent pit stop for teacher’s and it is completely free. It contains exceptional collections of videos presenting difficult mathematics topics in a way which can be easily understood. There are numerous drill questions which can be used in conjunction with the videos. The resources are very well organized and it also

easy to play around with the website. The grade level mathematics resources in Khan academy address all most all core mathematics standards and helps focused learning and the videos aid vibrant instruction.

DiscoveryEducation Provides a series of new and remarkable pre-made lesson plans for math teachers. It is regularly updated with new materials which covers levels from pre-K to K-12.

Where as the National Council of Teachers ofMathematics Showcases that the math lessons can be fun and more than just a “drill and kill!. It has the outstanding accumulation of lesson plans and activities. They are free and accelerated towards different levels, it works to aid us by collective access to quality standards-based resources for learning and teaching mathematics. It also provides collaborative tools for students and instructional support for educators.

Learn Zillion is an excellent stop for math teachers if they are looking to improve classroom discussions with the help of videos and lesson plans complex mathematical subjects can be easy to understand in engaging way. The website has to offer math educational contents in fun and exciting way which is suitable for every level of students. Teachers can use these videos to start a lesson and introduce new ideas which is very important to engage students in math class.

With Mathalicious, Math can be fun and exciting, Using this resource Math teacher can spice up the middle and high school students by making math lesson a whole lot easier. These remarkable assortments of lessons will not only get the students thinking and problem solving, but they will have a lot fun while solving it ! The lessons are classically geared towards using math in practical applications, which really helps students see the significance of learning and understating math

Even though we were supposed to choose one major Open education resource repositories and evaluate its resource but mathematics has such a wide scope and enormous availability resources made me explore quite a few repositories. These resources are very well organized and is extremely user friendly and a great asset for educators to elevate their teaching practices.

The efficient use of these open educational resources embeds a new style to learning which can be implemented to every level of the education sector., letting learning and teaching materials to be produced and then used and reused, remixed by everyone. It will be a boon to novice educators to guide them into perfection. The mathematics teaching fraternity has been extremely enriched by these resources, it has simplified the instructional process and made math class fun and engaging for students. OER’S has increased flexibility in how we use the course content and emphasizes learning communities, learner engagement. The advantages of OER’s are bountiful yet it has some challenges to offer. If not properly trained and educated on using OER’s and finding the right course content teachers tend to invest a lot of time planning their lessons and selecting alternate readings and other materials for class.

With that being said finally it leaves me with questions on how many of us are efficient utilizing OER’s? and contributing to updated the current content? how do we deliver better learning experiences to more students? And How do we get the most up-to-date content when we want it? How to overcome the accessibility issues when file formats that are not always compatible with our computer’s software or version? What do you guy’s think?

Hello Everyone, Vedic mathis an awesome technique for solving arithmetic problems in quicker and simpler way. The cool thing about this technic is that every math checks out. Whether you want to painlessly split a check, to amaze your friends, or to learn a different way to quickly divide numbers, this easy method can be learned within no time. In this week blog post let us learn how to finddivision of 3 or more digits using Vedic mathematics.

Below image is a simple illustration of the difference between usual method and Vedic method

Finding the division of 3 or more digit number when thedivisor is 9.
1)Let us solve example on 3 digit number
Let us consider the number 102÷9 102÷9 STEP 1: Splits the digit102 into 2 parts as a divisor has only one digit as 2goes in remainder part and rest all digits will goes in Quotient part i.e. Q R 1 0 ǀ 2

STEP 2: To find the 1st digit of a quotient put 1 as it is i.e. Q R 1 0 ǀ 2
↓ 1 STEP 3: i)To find the 2nd digit of a quotient ,add quotient and divident digit i.e. (1+0=1) ii)To find the remainder ,add remainder and quotient i.e.(2+1=3) i.e. Q R 1 0 ǀ 2
↓ 1 1 ǀ 3 Answer Quotient=11 Remainder=3

2) Let us solve examples of 4 digit number Let us consider the 1232÷9 1232÷9 STEP 1: Split the digit1232into 2 parts as a divisor has only one digit as 2goes in remainder part and rest all digits will goes in Quotient part i.e. Q R 123ǀ 2 STEP 2: Q R 123 ǀ 2
↓ 1 STEP 3: i)To find the 2nd digit of a quotient add quotient and the divident i.e.(1+2=3) , ii) To find the 3rd digit of a quotient add 2nd digit of quotient with the divident digit i.e. (3+3=6) iii)To find the remainder add 3rd digit of quotient with remainder i.e. (6+2=8) i.e. Q R 1 2 3 ǀ 2
↓ 1 36 ǀ 8 Answer : Quotient =136 Remainder=8

Division method when divisor is closer and smaller than power of 10. By understanding the Concept ofNikhilam formula i.e.all from 9 & last from 10,(i.e. Subtract all digits from 9 & last digit from 10)

Below image is a simple illustration of the difference between usual method and Vedic method

Lets us consider the value 234÷87 STEP 1: Splits the digit234into 2 parts as a divisor has two digit as34goes in remainder part and rest all digits will goes in Quotient part i.e. QR 2ǀ34 STEP 2: Apply Nikhilam formula on 87 and we will get the compliment 13(i.e. 9-8=1, 10-7=3) Q R 2ǀ 34 13↓ 2 STEP 3: i)Multiply 2 with individual digit of 13 i.e. Q R 2 ǀ 34 13 ↓ ǀ 2 6 2 ǀ 6 0 Answer : Quotient =2 Remainder=60

Lets us consider the value 113401÷997 STEP 1: Splits the digit113401into 2 parts as a divisor has three digit so remainder will also have 3 digit ,as401goes in remainder part and rest all digits will goes in Quotient part i.e. QR 113ǀ401 STEP 2: Apply Nikhilam formula on 997 and we will get the compliment 003 (i.e. 9-9=0, 9-9=0,10-7=3) Q R 113 ǀ 401 003 ↓ 1 STEP 3: i)Multiply 1 with individual digit of 003 i.e. Q R 1 1 3ǀ 401 003 ↓ 0 0ǀ 3 1 1 ǀ ii)Multiply 2nd quotient with individual digit of 003 ,and add the third column,to get 3rd quotient Q R 1 1 3ǀ 401 003 ↓ 0 0ǀ 3 0ǀ 0 3 11 3ǀ
iii)Multiply 3rd quotient with individual digit of 003 ,and add the 4th,5th and 6th column,to get the remainder. (Note: Have to repeat the function till we get the number in last column) Q R 1 1 3ǀ 4 0 1 003 ↓ 0 0ǀ 3 0 ǀ 0 3 ǀ 0 0 9 11 3ǀ 7 3 (10) (1 is carried forward) 11 3 ǀ 7 4 0 Answer : Quotient =113 Remainder= 740

A quick update on answering questions from my previous blog post.
One of my fellow classmate Kara Tylor had a question on one of my blog post on how to find cube of two numbers using vedic technic. She asked me how to find cube of more than 2 digit using Vedic math, below I have solved few examples based on the method and my video will further aid the understanding the concept.

Finding cube of 3 digit number using vedic maths. Lets us consider the number (153)^{3 }

STEPS: i)Split the given digits into two part 15 3 ii)consider the 1^{st} part i.e. 15 from left to right go on reducing power i.e 15^{3 } 15^{2 } 15 iii) Next consider the 2^{nd} part i.e. 3 from right to left go on reducing the power i.e. 15^{3 } 15^{2}.315.3^{2} 3^{3
}iv) Find the values and write in 2^{nd} Row 15^{3 }15^{2}.3 15.3^{2}3^{3
}3375 675 135 27 v) Multiply middle terms by 3 i.e. 15^{3 }15^{2}.3 15.3^{2} 3^{3
}3375 675 135 27 ×3 ×3 3375 2025 405 27 vi)To get the final answer we are going to keep unit digit as it is rest other digit are carried forward 15^{3 }15^{2}.315.3^{2}3^{3
}3375 675 135 27 ×3 ×3 3375 2025 405 27 (206) (40) (2) (CARRIED FORWARD) 3581 5 7 7 ANSWER
(153)^{3} = 35,815,77

I hope you like this post and found this technic interesting .

Hi Eci831,
In My Previous blog , we learned how to find cube of any number using Vedic maths. In this Post we are going to study how to find cube root. of any perfect cube easily by using Vedic maths. This is an remarkable trick and is always quite easy and appreciated by many This awe inspiring technique helps us to easily find cube root of 5 or 6 digits number mentally

Before going ahead on this method to find cube root, please make a note and memorize the below table.

To calculate cube root of any Perfect Cube easily, we need to Memorize the cubes, cube of unit digit and its relations from 1 to 10.

Number

Cube

Cube of unit digit

Relations

1^{3}

1

1

1⇒1

2^{3}

8

8

8⇒2

3^{3}

27

7

7⇒3

4^{3}

64

4

4⇒4

5^{3}

125

5

5⇒5

6^{3}

216

6

6⇒6

7^{3 }

343

3

3⇒7

8^{3}

512

2

2⇒8

9^{3}

729

9

9⇒9

10^{3}

1000

0

0⇒0

Mathematics is very interesting and easy to learn when we use tricks to make it simple,now let’s see how easily we can solve cube root of perfects cube by using Vedic method and also by viewing the table above.

Example 1: Find cube root of 19683 Check the last digit number & make a groups of three, three digits from right side i.e. 19683 Can be written as 19, 683 STEP 1

Take the last group which is 683, in this case the last digit of 683 is3

Now compare the last digit number i.e. 3 in the above table in the relation we can see that if last digit of perfect cube is 3 then last digit of cube root is 7, 3⇒7

So we replace 3 by 7 at the right most digit of the cube root 19, 683 7 STEP 2

Now check the remaining digit which is left i.e. 19

We have to check the cube which is less then or equal to the digit 19

In this case the number2^{3}=8which is less than 19

We will replace 19by 2

i.e. 19 , 683 2 7 Hence the answer for cube root of 19683 is27

Example 2: Find cube root of 551368 Check the last digit number & make a groups of three, three digits from right side i.e. 19683 Can be written as 551 , 368

STEP 1

Take the last group which is 368, in this case the last digit of 368 is 8

Now compare the last digit number i.e. 8 in the above table in the relation we can see that if last digit of perfect cube is 8 then last digit of cube root is 2 i.e 8⇒2

So we replace 8 by 2 at the right most digit of the cube root 551 , 368 2 STEP 2

Now check the remaining digit which is left i.e. 551

We have to check the cube which is less then or equal to the digit 551

In this case the number 8^{3}=512 which is less than 551

We will replace 551 by 8
i.e. 551 , 368 8 2
Hence the answer for Cube root of 551,368 =82

Example 3: Find cube root of 148,877 Check the last digit number & make a groups of three, three digits from right side i.e. 148,877 Can be written as 148 , 877 STEP 1

Take the last group which is 877, in this case the last digit of 877 is 7

Now compare the last digit number i.e. 7 in the above table in the relation we can see that if last digit of perfect cube is 7 then last digit of cube root is 3 i.e 7⇒3

So we replace 7 by 3 at the right most digit of the cube root 148 , 877 3

STEP 2

Now check the remaining digit which is left i.e. 148

We have to check the cube which is less then or equal to the digit 148

In this case the number 5^{3}=125 which is less than 148

We will replace 148 by 5 i.e 148 , 877 53

Hence the answer for Cube root of 19683 =53 In the below video i have solved few more problems based on the vedic technics

I hope you liked the video and found the Technic interesting and it simplified the process of finding cube root in a faster way, I appreciate your time for reading my blog post .

As the famous Quote states “Education is not the answer to the question. Education is the means to the answer to all questions.” Therefore, I feel that a well-educated individual has one of the life’s most valuable possessions that too as a result of one’s own hard work. Education is the only companion who will be by your side at all times and will never let you down. That is why; a concrete foundation is indispensable in shaping one’s future.

An educated and qualified person has always something to fall back upon. I strongly believe that one cannot perform better in any other field beside the field that deeply fascinates and captures one’s interest. Determination and persistent hard work follow one’s decision in undertaking a challenging task, there was a time where one was supposed to get this education/knowledge by enrolling to the educational institutions by meeting their academic requirement, it was reachable for only privileged class, but today the whole arena has changed ,the concept of openeducationhas revolutionized the field of education it has replaced the traditional class room and made education free and accessible to all.

Web-based educative platforms have allowed us to remotely replace or facilitate many interactive processes such as purchases, booking appointments, and banking activities, buying flight tickets etc. Getting an education online is the next frontier for web-based convenience, and there are a number of reasons why this is an effective and attractive platform. Some of them are,

Accessibility: Open Education allows students to take courses at their own convenience, on their own time and place. Open learning systems are very effective and importantly Accessible when compared to its traditional counterparts. The rich resources in online communities help students to overcome hurdles. Open education has attracted many students who were very unlikely to gain higher education.

Flexibility:Open education offers students flexibility to choose what to learn and when to learn and how much to learn, in return putting them in control. It provides ocean of resources for students in-comparison with the traditional classrooms.

This week’s concept and videos were very interesting and covered the concept of open education and knowledge sharing. As a mathematics major I had a strong mathematical knowledge but teaching was something very new to me in my initial days of my teaching career. AsAshleymentions in her blog post We are not responsible for creating the material that we provide to students, I think I have exploited most of the resources online which helped me to effectively deliver knowledge to students and thanks to all those who believed in the concept of sharing culture, it made my initial days a lot easier.The resources likeKarnataka education ,Open Educational Resources for Mathematics has helped me a lot in developing material to facilitate my course. It has improved my teaching material and delivery.

And Finally, as an educator I believe that the ultimate goal of teaching is to promote learning and education takes place in different circumstances and contexts. I totally agree with message ofKirby Ferguson in his video series that we are all leaders of remix

Kirby Ferguson’spopular videos illustrates that everything is remix and everything we have created takes motivation from something which is already in existence. The video showcases number of evidences that supports his claims. I was convinced and it kept me thinking that most of things I do in my teaching practices might be exactly the same.

I believe that in order to lay a strong foundation of knowledge we need to copy and when we relate this to our profession we take ideas, material, technic’s from resources and effectively deliver knowledge to students and I think it has made our life as a teacher much easier and effective. Just Imagine the amount of time required if we had to create everything original ..??

After reading the blog post of Joe Marine his question kept me thinking and I wanted to share the same with everyone , Is there a need to reform intellectual property laws to better represent how ideas actually spread? Is it possible to have original work, or is everything truly a remix?

Hello Everyone, It was indeed fun and interesting journey in EC&I 831, I have learned a lot by reading my fellow classmates blog post, every week was so interesting and now we are almost halfway through the course. I have always been fascinated by mathematical studies and having a flair for the subject, there was never any doubt that I would choose concepts in mathematics as my major learning project. I would call this as journey of learning, knowledge sharing. My project was focused on learning how to make mathematics simple, easy and fun for everyone and the tricks of Vedic mathematics

would be perfect to answer to it. In myfirst blog postwe learned what isVedic mathematicsand itssutra’s(Formula’s), the idea behind this was to introduce the concept of Vedic math to my classmates. Inmysecond blog postI focused on learning how to implement this technic in various mathematical concepts, applying Vedic math in solving multiplication problems of complex numbers made multiplication so easy, we could solve 4-5 digit problems in less than a minute and that too without usingcalculator. My next postfocused on learning how to find square of any number usingVedic trick, followed byanother week blog postwhere we learned to find square root of any number using this trick.

I feel happy regarding my progress after reading the comments about my journey by my fellow classmates. Initially it was not so easy as I thought it would be, I had to understand the whole technic first and solve problems related to it. After searching a lot over internet I understood the best way to showcase this concept and learn with my classmates is by making a video of my own where I solve problems, which in-return help everyone understands the Vedic tricks and it applications. Going forward in the second half of our course we learn few more applications of Vedic technic in mathematics and by the end of our journey in EC&I 831 we will make Mathematics simpler and joyful using the tricks of Vedic Mathematics.

Now that we have learned how to findsquare root of any numbers using Vedic math , we will go ahead and use this technic to findcube of numbers. Finding cube of the number plays a vital role in Mathematics, so it is useful to find quick ways to find the cube of the number. We will learn the Vedic trick which will help us finding the cube of numbers in quick way. To find the cube of numbers we are going to learn four types Vedic Tricks, In the first we are going to find the cube of a number which starts with 1 i.e. 11,12,13……. In second type we are going to find the cube of a number which ends with 1 i.e.21,31,41……In the third type we are going to find the cube of a number having same digit i.e.22,33,44……and Finally in the fourth type we are going to find the cube of a number having different digit.

Type 1:Numbers starting with 1 (16)^{3
}1) We consider1as 1^{st} term and6as 2^{nd} term, we will write the given term as it is i.e.1 6 2) Square the 2^{nd }term i.e. 6^{2}=36and also cube the 2^{nd} term i.e.=6^{3}=216and write the values in 1^{st} row i.e. 16 36 216 3) In 2^{nd} row double the 2 middle terms (. i.e. is 2^{nd} term and 3^{rd} term) & write just below 2^{nd} & 3^{rd} term. 1636 216 12 72 (Doubled the value) 4) Add them vertically in column. carry forward the 10^{th} places digit to next column 16 36 216 + 12 72 3 12 21 (Carried forward) 4 0 9 6 (Answer) (16)^{3}=4,096

Type 2: Numbers Ending with 1 (61)^{3} 1) We consider6as 1^{st} term and 1as 2^{nd} term, we will write the given term as it is but in reverse order i.e.61 2) Square the 1^{st }term i.e. 6^{2}=36and also cube the 1^{st }term i.e.=6^{3}=216and write the values in 1^{st} row i.e. 21636 61 3) In 2^{nd} row double the 2 middle terms (. i.e. is 2^{nd} term and 3^{rd} term) & write just below 2^{nd} & 3^{rd} term. 216 36 6 1 72 12 (Doubled the value) 4) Add them vertically in column. carry forward the 10^{th} places digit to next column 216 366 1 + 72 12 10 1 (Carried forward) 226 9 8 1 (Answer) (61)^{3}=226,981

Type 3: Numbers having same digit (55)^{3}^{
}1) We consider5 as 1^{st} term and5as 2^{nd} term, here both the digit is same so we take any one digit .Cube of5 =125and write 4 times i.e.125125 125125 2) In 2^{nd} row double the 2 middle terms (. i.e. is 2^{nd} term and 3^{rd} term) & write just below 2^{nd} & 3^{rd} term. 125 125 125 125 250 250 (Doubled the value) 4) Add them vertically in column. carry forward the 10^{th} places digit to next column 125 125 125 125 + 250 250 41 38 12 (Carried forward) 166 3 75 (Answer) (55)^{3}=166,375

Type 4:Numbers having different digit (32)^{3}^{
}1) We consider3as 1^{st} term and2as 2^{nd} term, Cube the 1^{st} term and 2^{nd} term i.e. (3)^{3 }=27 and (2)^{3}=8 27 8 2) Square the 1^{st} term i.e.3^{2}=9then multiply by 2^{nd} term i.e.9×2=18 27 188 3) Square the 2^{nd} term i.e.2^{2 }=4then multiply by 1^{st} term i.e.4×3=12 27 18 12 8 4) In 2^{nd} row double the 2 middle terms (. i.e. is 2^{nd} term and 3^{rd} term) & write just below 2^{nd} & 3^{rd} term. 27 18 12 8 36 24 (Doubled the value) 5) Add them vertically in column. carry forward the 10^{th} places digit to next column 27 18128 + 36 24 5 3 (Carried forward) 32 7 6 8 (Answer) (32)^{3}=32,768

I hope you found the technic interesting and it simplified the process of finding cube of a number in a faster way, I appreciate your comments and suggestions.