**Vedic Mathematics Newsletter No. 98**

A warm welcome to our new subscribers.

Vedic Mathematics is becoming increasingly popular as more and more people are introduced to the beautifully unified and easy Vedic methods. The purpose of this Newsletter is to provide information about developments in education and research and books, articles, courses, talks etc., and also to bring together those working with Vedic Mathematics. If you are working with Vedic Mathematics - teaching it or doing research - please contact us and let us include you and some description of your work in the Newsletter. Perhaps you would like to submit an article for inclusion in a later issue or tell us about a course or talk you will be giving or have given. If you are learning Vedic Maths, let us know how you are getting on and what you think of this system.

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This issue’s article is “Fifth Power of Two-Digit numbers” by Sanjay Dixit. Sanjay is a VM enthusiast from India who loves practising VM and exploring new areas.

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**2015 COURSES – Starting January 3rd**

Enrolments are also being taken now for the three online courses starting on 3rd January 2015:

1) Teacher Training Course – 9 weeks.

2) Certificate Course – now a 3-week course – the trainer for this course is Dr Arvind Prasad of Queensland University.

3) The Crowning Gem (this is a new course) – 4 weeks.

Full details and application forms can be found here (see courses 1, 2, 3):

http://www.vedicmaths.org/community/calendar-of-events

Also, the next Diploma course will start on 28th January (follows on from the Certificate course).

The next Advanced Diploma course starts on 4th April (follows on from the Teacher Training Course or Diploma Course)

And in May/June we hope to offer another new course: on Applied Mathematics.

**REGISTRATION FOR ONLINE CONFERENCE**

Registrations are now being taken for the Conference on 14th / 15th March 2015. You can see details and the link to the Registration Page here:

http://conference.vedicmaths.org/2015-03%20Online%20Conference.asp

**PEBBLE MATHS PRESENTATIONS**

Yesterday [29th November] at Torakina Beach, Australia, in glorious sunshine we held our 2014 Pebble Maths end of year certificate presentation ceremony. Each child received a well-earned award and a Pebble Maths pen. Everyone has achieved amazingly this year and deserved the accolade that they were given.

Congratulations everybody! And a big thank you to the parents and grandparents for their incredible support.

Have a wonderful summer holiday.

Details and photos of the event can be seen at:

https://www.facebook.com/pebblemaths

**Summer Holiday Workshops! by Vera Stevens**

Vedic Maths Workshop.

Tuesday January 6th - Friday 9th (inclusive)

9:30AM - 11:30AM

Cost: $100

Preparation for High School Workshop.

This workshop is primarily for children who will be entering high school in 2015 to ensure that their foundation in numeracy and basic geometry is solid.

Tuesday January 20th - Friday 23rd (inclusive)

9:30AM - 11:30AM

Cost: $100

Please ring to book a place for your child: 02 66805524

**ARTICLE IN THE SPEAKING TREE**

See this excellent article by James Glover:

http://www.speakingtree.in/spiritual-articles/new-age/fun-with-numbers

**RESEARCH IN SHIV PURAN**

The Shiv Puran is the fourth of the eighteen Vedic Puranas. Dr Kapoor’s research includes interpreting the verses of the Shiv Puran in terms of multidimensional spaces. Please see the e-newsletters at www.vedicganita.org – newsletters from number 34 address the Shiv Puran. For further details contact Dr Kapoor at

**MULTIPLYING BY 9, 99 AND 999**

Here's a quick shortcut for multiplying numbers by 9.

For AB x 9, subtract A + 1, then append 10-B.

For example, with 68 x 9, simply do 68 - 7 = 61, then append 10 - 8 = 2 to get 612.

For 94 x 9, you merge 94 - 10 = 84 with 10 - 4 = 6 to obtain 846.

The underlying algebra is that for a given number 10A + B, with 0 < B <9, we are computing 10[(10A + B) - (A + 1)] + (10 - B) = 90A + 9B = 9(10A + B).

Consequently, this can be applied to larger numbers. For instance, to do

123 x 9, 123 - 13 = 110 is merged with 10 - 3 = 7 to get 1107.

A similar approach can be taken when multiplying by 99. For AB x 99, subtract A+1 from 10 times AB, and then append 10 - B.

For example, with 87 x 99, simply do 870 - 9 = 861, and then append 10 - 7 = 3 to get 8613.

For 37 x 99, you merge 370 - 4 = 366 with 10 - 7 = 3 to obtain 3663.

The underlying algebra is that for a given number 10A + B, with 0 < B <9, we are computing 10[10(10A + B) - (A + 1)] + (10 - B) = 990A + 99B = 99(10A + B).

And finally, for multiplying with 999, for AB x 999, subtract A+1 from 100 times AB, and then append 10 - B.

For example, with 46 x 999, simply do 4600 - 5 = 4595, and then append 10 - 6 = 4 to get 45954.

For 77 x 999, you merge 7700 - 8 = 7692 with 10 - 7 = 3 to obtain 76923.

The underlying algebra is that for a given number 10A + B, with 0 < B <9, we are computing 10[100(10A + B) - (A + 1)] + (10 - B) = 9990A + 999B = 999(10A +B).

Rohan Chandra,

Delhi Technological University, New Delhi, India

**PLANNING FOR THE 2015 CELEBRATIONS?**

Please keep us informed of events you are planning to mark this anniversary so that we can let everyone know.

*******************************************ARTICLE for VM Newsletter 98**

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**Fifth Power of Two digit numbers**

Swami Thirthaji in his Vedic Mathematics book (Chapter 32 Elementary Squaring page 289) explained briefly how to find fourth power of a number. It is possible to extend the application of binomial theorem to find fifth power of a number.

(a+b)^{5} = a^{5} + 5a^{4}b + 10a^{3}b^{2} + 10 a^{2}b^{3} + 5ab^{4} + b^{5}

Coefficients of parts are 1, 5, 10, 10, 5, 1

Let us see how we can find fifth power of a two digit number.

E.g. 21. We will compute from right to left.

Steps:

1. Find ratio of two digits. Left digit divided by right digit. Here it is 2 divided by 1.

2. Find number of parts which is always one more than the power. Since power is five it will have 6 parts

3. We need to know the fifth power of right digit. Here it is one. This will occupy right most part.

4. Multiply each preceding part by ratio starting from fifth part (Anurupyena sutra).

5. Multiply each part by its corresponding coefficient as per above binomial expansion.

6. Keep only one digit in each part and carry over extra digits to left.

Step Part: 1 2 3 4 5 6

4 Ascending Ratio (X) 32 16 8 4 2 1

Coefficient (Y) 1 5 10 10 5 1

5 X * Y 32 80 80 40 10 1

6 40 8 4 1 0 1

Hence 21^{5} = 4084101

If the right digit is greater than 5 then it can be converted to its vinculum form.

Let’s find fifth power of 28.

Its vinculum form is 32 in which you have to imagine there is a bar over the 2 to show it is negative.

Here ratio will be 3 divided by (-2).

Follow steps as above.

After getting answer in vinculum form, convert it to normal form.

Step Part: 1 2 3 4 5 6

4 Ascending Ratio (X) 243 -162 108 -72 48 -32

Coefficient (Y) 1 5 10 10 5 1

5 X * Y 243 -810 1080 -720 240 -32

6 173 -9 1 -6 -3 -2

Change to Normal Form 172 1 0 3 6 8

Hence 28^{5} = 17210368

End of article.

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Editor: Kenneth Williams

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29th December 2014