Vedic Maths1@July2007

Let Noble Thoughts come from all Let Noble Thoughts come from all directions. Rig Vedadirections. Rig Veda

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Vedas one of the oldest BooksVedas one of the oldest Books

Vedas are among the oldest literatures of Vedas are among the oldest literatures of mankind.mankind.

Several millennium years BCSeveral millennium years BC

Vedas contain Science, Maths, Vedas contain Science, Maths, Astronomy, Psychology, Religion, Poetry, Astronomy, Psychology, Religion, Poetry, Philosophy and many more. Philosophy and many more.

Vedas do not denigrade human beings as Vedas do not denigrade human beings as Heathens & Kafirs. It does not talk of Heathens & Kafirs. It does not talk of Crusades, Jihads, Gazi etc. They talk of Crusades, Jihads, Gazi etc. They talk of Universal Happiness, Love and Peace.Universal Happiness, Love and Peace.


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Vedic PhysicsVedic PhysicsMeasurement of TimeMeasurement of Time

Vedic Units of TimeVedic Units of Time(a)(a) Smaller Units of TimeSmaller Units of Time TRUTI TRUTI = 33,750th fraction of a second is the smallest unit of = 33,750th fraction of a second is the smallest unit of timetime

100 Truti 100 Truti = 1 = 1 TatparaTatpara45 Tatpara 45 Tatpara = 1 = 1 NimeshaNimesha

30 Nimesha 30 Nimesha = 1 = 1 Prana Prana = 4 secs = 4 secs 3 Nimesh 3 Nimesh = 1 = 1 VipalaVipala = 0.4 seconds= 0.4 seconds60 Vipalas 60 Vipalas = 1 = 1 PalaPala = 24 seconds= 24 seconds60 Palas 60 Palas = 1 = 1 GhatikaGhatika = 24 Minutes= 24 Minutes60 Ghatikas60 Ghatikas = 1 = 1 Divas Divas = 1 day or 24 Hours= 1 day or 24 Hours7 Divas 7 Divas = 1 Saptah= 1 Saptah = 1 week= 1 week15 Divas 15 Divas = 1 Paksha = 1 Paksha = 1 Fortnight= 1 Fortnight2 Paksha 2 Paksha = 1 Maas= 1 Maas = 1 Month= 1 Month 2 Maas 2 Maas = 1 Ritu= 1 Ritu = 1 Season= 1 Season

6 Maas6 Maas = 1 Ayana (Uttarayan & Dakshinayan) = 1 Ayana (Uttarayan & Dakshinayan) 6 Ritu 6 Ritu = 2 Ayanas = 12 Maas = 2 Ayanas = 12 Maas = 1 Varsha= 1 Varsha = 1 Year= 1 Year

12 Years12 Years = 1 Kumbha= 1 Kumbha 60 Years 60 Years = Sashtiyapta Purti= Sashtiyapta Purti

100 Years100 Years = 1 Shatabda= 1 Shatabda 10 Shatabda10 Shatabda = 1 Sahasrabda = 1 Sahasrabda = 1 Millenium = 1 Millenium = 1,000 years = 1,000 years


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Vedic PhysicsVedic PhysicsMeasurement of TimeMeasurement of Time

Vedic Units of TimeVedic Units of Time(b)(b) Higher Units of TimeHigher Units of Time

432 Sahasrabda = 1 Kali Yug or Yug432 Sahasrabda = 1 Kali Yug or Yug= 432,000 years= 432,000 years2 Yug2 Yug = 1 Dwapar Yug = 1 Dwapar Yug = 864,000 years = 864,000 years 3 Yug3 Yug = 1 Treta Yug = 1 Treta Yug = 1296,000 years= 1296,000 years4 Yug4 Yug = 1 Satya Yug = 1 Satya Yug = 1728,000 years= 1728,000 years10 Yug10 Yug = 1 Maha Yug = 1 Maha Yug = 4.32 Million Years= 4.32 Million Years

1000 Maha Yug = 1 Kalpa1000 Maha Yug = 1 Kalpa = 4.32 Billion Years = Morning of Brahma= 4.32 Billion Years = Morning of Brahma2 Kalpa = 1 Day of Brahma2 Kalpa = 1 Day of Brahma = 2,000 Maha Yug = 8.64 Billion Years = 2,000 Maha Yug = 8.64 Billion Years 360 Days of Brahma 360 Days of Brahma = 1 Year of Brahma = 1 Year of Brahma = 3110.4 Billion Years = 3110.4 Billion Years = 3.1104 Trillion Years= 3.1104 Trillion Years

1 Maha Kalpa1 Maha Kalpa or or Brahma AyuBrahma Ayu = 100 Years of Brahma= 100 Years of Brahma = = 311.04 Trillion Years311.04 Trillion Years

= = 3.1104 X 1014 Solar Years3.1104 X 1014 Solar Years == 311,040,000,000,000 solar years311,040,000,000,000 solar years..

Thus the Vedic Seers had thought of the smallest and the largest units of time namely, Thus the Vedic Seers had thought of the smallest and the largest units of time namely, TRUTI TRUTI , , the smallest Unit of Timethe smallest Unit of Time = = 33,750th part of a Second33,750th part of a Second

Maha Kalpa or Brahma AyuMaha Kalpa or Brahma Ayu, , = = 311,040,000,000,000 solar years311,040,000,000,000 solar years,, the largest Unit of Timethe largest Unit of Time


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Vedic PhysicsVedic PhysicsSpeed of LightSpeed of Light

1300 AD1300 AD–SayanacharyaSayanacharya

PM of Vijayanagar Emperor Bukka IPM of Vijayanagar Emperor Bukka I

Speed of Light Speed of Light 2,202 Yojanas in half Nimesha2,202 Yojanas in half Nimesha

= 186,536 miles per second= 186,536 miles per secondWest only in 17 century !!West only in 17 century !!


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Quotes on Vedic SciencesQuotes on Vedic Sciences

From the From the VedasVedas (ancient Indian (ancient Indian Scriptures), we learn a practical art of Scriptures), we learn a practical art of surgery, medicine, music, house surgery, medicine, music, house building under which mechanised art is building under which mechanised art is included. They are encyclopedia of included. They are encyclopedia of every aspect of life, culture, religion, every aspect of life, culture, religion, science, ethics, law, cosmology and science, ethics, law, cosmology and meteorology.meteorology.

Sir William Jones (Sir William Jones (1746-1794)1746-1794)


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Quotes on Indian MathematicsQuotes on Indian Mathematics

We owe a lot to We owe a lot to Indians, who taught Indians, who taught us how to count, us how to count, without which no without which no worthwhile scientific worthwhile scientific discovery could discovery could have been made.have been made.

Albert EinsteinAlbert Einstein


1414

Tatitreeya Samhita 7-2-20-1Tatitreeya Samhita 7-2-20-1 in 100 BCEin 100 BCE

11 Ekam Ekam

1010 Dasham Dasham

100 Shatam100 Shatam

101033 Sahasram Sahasram

101055 Lakshaha Lakshaha

101077 Kotihi Kotihi

101099 Ayutam Ayutam

10101111 Niyutam Niyutam

10101313 Kankaram Kankaram

10101515 Vivaram Vivaram

10101717 Parardhaha Parardhaha

10101919 Nivaahaha Nivaahaha

10102121 Utsangaha Utsangaha

10102323 Bahulam Bahulam

10102525 Naagbaalaha Naagbaalaha

10102727 Titilambham Titilambham

10102929 VyavasthaanaVyavasthaana

PragnaptihiPragnaptihi

10103131 Hetuheelam Hetuheelam

10103333 Karahuhu Karahuhu


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Tatitreeya Samhita 7-2-20-1Tatitreeya Samhita 7-2-20-1 in 100 BCEin 100 BCE

10103535 Hetvindreeyam Hetvindreeyam

101037 37 Samaapta Samaapta lambhahalambhaha

10103939 Gananaagatihi Gananaagatihi

10104141 Niravadyam Niravadyam

10104343 Mudraabaalam Mudraabaalam

10104545 Sarvabaalam Sarvabaalam

10104747 Vishamagnagatihi Vishamagnagatihi

10104949 Sarvagnaha Sarvagnaha

10105151 Vibhtangamaa Vibhtangamaa

10105353 Tallaakshanam Tallaakshanam

In Anuyogdwaar In Anuyogdwaar SutraSutra

100 BCE one 100 BCE one numeral is raised as numeral is raised as high as 10high as 10140140

The highest prefix The highest prefix used for raising 10 to used for raising 10 to a power in a power in Today’s Today’s MathsMaths is D for is D for 10103030


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Mathematical PuzzlesMathematical Puzzles1 Rupee 1 Rupee = 100 Paise= 100 Paise

= 10 X 10 Paise= 10 X 10 Paise = 1/10 Re X 1/10 Re= 1/10 Re X 1/10 Re = 1/100 Re= 1/100 Re = 1 Paise= 1 Paise

1 Rupee 1 Rupee = 100 Paise= 100 Paise √1 Rupee √1 Rupee = √100 Paise = √100 Paise = 10 Paise= 10 Paise

1 Rupee1 Rupee = 100 Paise= 100 Paise

¼ Rupee¼ Rupee = 100/4 = 25 Paise= 100/4 = 25 Paise

√ √1/4 Re = ½ Re1/4 Re = ½ Re = 5 Paise= 5 Paise


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Mathematical PuzzlesMathematical Puzzles1.1. Ram, Bharat and Laxman are standing one behind the other. There are 3 Ram, Bharat and Laxman are standing one behind the other. There are 3

black and 2 white caps in the room behind them. Vasishtha brings 3 caps black and 2 white caps in the room behind them. Vasishtha brings 3 caps and places on their heads. Bharat and Laxman cannot guess the color of and places on their heads. Bharat and Laxman cannot guess the color of the cap on their heads. But Ram guesses it. How?the cap on their heads. But Ram guesses it. How?

2.2. Some birds on a tree saw a few flowers in the pond beneath. They flew and Some birds on a tree saw a few flowers in the pond beneath. They flew and sat one on each flower. One bird did not find a flower. They then sat 2 birds sat one on each flower. One bird did not find a flower. They then sat 2 birds to a flower, one flower remained extra. How many birds and flowers were to a flower, one flower remained extra. How many birds and flowers were there?there?

3.3. There are 9 balls of identical size and weight. But one of them being There are 9 balls of identical size and weight. But one of them being defective weighs a few grams less. Using a balance only twice, how will you defective weighs a few grams less. Using a balance only twice, how will you identify the defective ball?identify the defective ball?

4.4. There are 10 bags with coins in them. All the coins weigh 10 grams each. There are 10 bags with coins in them. All the coins weigh 10 grams each. One of the bags contains coins with only 9 grams each. Using a weighing One of the bags contains coins with only 9 grams each. Using a weighing machine only once, how will you identify which bag contains the defective machine only once, how will you identify which bag contains the defective coins?coins?

5.5. There are 10 liters of milk in a can. There are also 3 liter and 7 liter empty There are 10 liters of milk in a can. There are also 3 liter and 7 liter empty measuring jars. Without using any other vessel how will you divide the milk measuring jars. Without using any other vessel how will you divide the milk into 5 liters.into 5 liters.


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Mathematical PuzzlesMathematical Puzzles6.6. A temple in a village has 3 shrines, for Lord Ganesh ji, Lord A temple in a village has 3 shrines, for Lord Ganesh ji, Lord

Vishnu ji and Lord Shiva ji. The temple priest generally plucks 24 Vishnu ji and Lord Shiva ji. The temple priest generally plucks 24 flowers from a the temple tree and offers 8 flowers to each of the flowers from a the temple tree and offers 8 flowers to each of the deity while doing puja. On a particular rainy day he could not find deity while doing puja. On a particular rainy day he could not find enough flowers to do the puja. He therefore dipped the flowers in enough flowers to do the puja. He therefore dipped the flowers in a well once. They became double. He then did the puja in Ganesh a well once. They became double. He then did the puja in Ganesh ji’s shrine with 8 flowers, dipped the remaining flowers in holy ji’s shrine with 8 flowers, dipped the remaining flowers in holy water, the flowers became double, performed the puja in Lord water, the flowers became double, performed the puja in Lord Vishnu ji’s shrine, dipped the remaining flowers in the holy water Vishnu ji’s shrine, dipped the remaining flowers in the holy water for the third time. The flowers once again became twice the for the third time. The flowers once again became twice the amount and now he completed the puja at the shrine of Lord amount and now he completed the puja at the shrine of Lord Shiva ji. Lo there were no flowers left. How many flowers did the Shiva ji. Lo there were no flowers left. How many flowers did the priest pluck from the tree?priest pluck from the tree?

7.7. At one cross-road junction, one road goes to Bangalore and the At one cross-road junction, one road goes to Bangalore and the other road leads to Chennai. At the junction there are two other road leads to Chennai. At the junction there are two persons selling tea and snacks, one who always tells truth and persons selling tea and snacks, one who always tells truth and the other always tells lies. To any question asked they will say the other always tells lies. To any question asked they will say "YES" or "NO". By asking just one question to any one person, "YES" or "NO". By asking just one question to any one person, how to find the right choice to reach Bangalore. What could be how to find the right choice to reach Bangalore. What could be that question?that question?


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Mathematical PuzzlesMathematical Puzzles1.1. Ram, Bharat and Laxman are standing one behind the other. There are 3 black and 2 white caps Ram, Bharat and Laxman are standing one behind the other. There are 3 black and 2 white caps

in the room behind them. Vasishtha brings 3 caps and places on their heads. Bharat and Laxman in the room behind them. Vasishtha brings 3 caps and places on their heads. Bharat and Laxman cannot guess the color of the cap on their heads. But Ram guesses it. How?cannot guess the color of the cap on their heads. But Ram guesses it. How?

2.2. Some birds on a tree saw a few flowers in the pond beneath. They flew and sat one on each Some birds on a tree saw a few flowers in the pond beneath. They flew and sat one on each flower. One bird did not find a flower. They then sat 2 birds to a flower, one flower remained flower. One bird did not find a flower. They then sat 2 birds to a flower, one flower remained extra. How many birds and flowers were there?extra. How many birds and flowers were there?

3.3. There are 9 balls of identical size and weight. But one of them being defective weighs a few There are 9 balls of identical size and weight. But one of them being defective weighs a few grams less. Using a balance only twice, how will you identify the defective ball?grams less. Using a balance only twice, how will you identify the defective ball?

4.4. There are 10 bags with coins in them. All the coins weigh 10 grams each. One of the bags There are 10 bags with coins in them. All the coins weigh 10 grams each. One of the bags contains coins with only 9 grams each. Using a weighing machine only once, how will you identify contains coins with only 9 grams each. Using a weighing machine only once, how will you identify which bag contains the defective coins?which bag contains the defective coins?

5.5. There are 10 liters of milk in a can. There are also 3 liter and 7 liter empty measuring jars. There are 10 liters of milk in a can. There are also 3 liter and 7 liter empty measuring jars. Without using any other vessel how will you divide the milk into 5 liters.Without using any other vessel how will you divide the milk into 5 liters.

6.6.

7.7. A temple in a village has 3 shrines, for Lord Ganesh ji, Lord Vishnu ji and Lord Shiva ji. The A temple in a village has 3 shrines, for Lord Ganesh ji, Lord Vishnu ji and Lord Shiva ji. The temple priest generally plucks 24 flowers from a the temple tree and offers 8 flowers to each of temple priest generally plucks 24 flowers from a the temple tree and offers 8 flowers to each of the deity while doing puja. On a particular rainy day he could not find enough flowers to do the the deity while doing puja. On a particular rainy day he could not find enough flowers to do the puja. He therefore dipped the flowers in a well once. They became double. He then did the puja in puja. He therefore dipped the flowers in a well once. They became double. He then did the puja in Ganesh ji’s shrine with 8 flowers, dipped the remaining flowers in holy water, the flowers became Ganesh ji’s shrine with 8 flowers, dipped the remaining flowers in holy water, the flowers became double, performed the puja in Lord Vishnu ji’s shrine, dipped the remaining flowers in the holy double, performed the puja in Lord Vishnu ji’s shrine, dipped the remaining flowers in the holy water for the third time. The flowers once again became twice the amount and now he completed water for the third time. The flowers once again became twice the amount and now he completed the puja at the shrine of Lord Shiva ji. Lo there were no flowers left. How many flowers did the the puja at the shrine of Lord Shiva ji. Lo there were no flowers left. How many flowers did the priest pluck from the tree?priest pluck from the tree?

8.8. At one cross-road junction, one road goes to Bangalore and the other road leads to Chennai. At At one cross-road junction, one road goes to Bangalore and the other road leads to Chennai. At the junction there are two persons selling tea and snacks, one who always tells truth and the the junction there are two persons selling tea and snacks, one who always tells truth and the other always tells lies. To any question asked they will say "YES" or "NO". By asking just one other always tells lies. To any question asked they will say "YES" or "NO". By asking just one question to any one person, how to find the right choice to reach Bangalore. What could be that question to any one person, how to find the right choice to reach Bangalore. What could be that question?question?

Let Noble Thoughts come from all directions. Rig Veda

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Multiplying by 5

• Case 1: All Even Digits

• Case 2: Odd and Even Digits {last Digit is even}

• Case 3: Odd and Even Digits {last Digit is odd,

including all odd digits}


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Multiplying by 5– All Even Digits

• Step 1: Divide by 2• Step 2: Add a Zero

• Example: 5 x 4 • Step 1: Divide 4 by 2 = 2• Step 2: Add Zero = 20

• Example: 5 x 12 = 6 / 0 = 60


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Multiplying by 5 – All Even Digits

• Solve: • 5 X 4 8 2 0• 5 X 2 4 8 2 4 • 5 X 4 8 0 0 0 6 6 • 5 X 6 2 2 2 4 8 0

• Example: 5 X 2 4 6

• Step 1: (Divide 246 by 2) =1 2 3

• Step 2: (Add Zero) 0

• Answer is 1 2 3 0

• Solve:

• 5 X 2 8 6

• 5 X 4 6 4 8 4 4

• 5 X 2 0 0 4 6 8 2 4

• 5 X 8 0 6 4 0 6 8 2


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Multiplying by 5 –Even and Odd Digits

(Last Digit – Even)• Example: 5 X 2 3 6 • Step 1 : Put a dot below all odd numbers

• Step 2 : Subtract one from all odd Digits to make them EVEN and write this Even Digit above the Odd Digit

• Step 3 : Carry Forward this one to the Digit on the Right of the Odd Digit

• Step 4 : Divide each Digit by 2 and Add Zero


24

Multiplying by 5 –Even and Odd Digits(Last Digit – Even)

• Example: 5 X 2 3 6 9 4• Step 1 : Put a dot below all odd numbers

• 2 3.. 6 9.. 4 • Step 2 : Subtract one from all odd Digits to

make them EVEN and write this Even Digit above the Odd Digit

• 2 23.. 6 89.. 4


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• Step 3 : Carry Forward this one to the Digit on the Right of the Odd Digit

• 2 23. 16 89. 14

• Step 4 : Divide each Digit by 2 and Add Zero

• 1 1 8 4 7 0 Answer


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• Solve

• 5 X 2 3 4

• 5 X 2 0 3 0 4

• 5 X 2 0 0 3 0 0 4

• 5 X 4 3 5 6 7 7 8

• 5 X 1 2 3 4 5 6 7 8


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Multiplying by 5 –All Odd Digits

• Example :3 5 7 • Step 1 : Subtract one from all Odd

Digits and make them Even

• Step 2 : Carry forward this one to the Digit on Right side

• Step 3 : Divide by 2 and add “5”


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• Example :3 5 7

• Step 1 : Subtract one from all Odd Digits and make them Even

• 2 4 6

• Step 2 : Carry forward this one to the Digit on Right side

• 2 14 16

• Step 3 : Divide by 2 and add “5”

• 1 7 8 5 Answer


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• Solve• 5 X 3 5 7 9 • 5 X 9 7 5 3• 5 X 1 3 0 7 9 • 5 X 3 0 5 5 7 7 • 5 X 999 777 555 • 5 X 303 507 7009


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Multiplying by 5 – Any Number

• Solve

• 5 X 2 3 5 8

• 5 X 1 2 3 0 5 6

• 5 X 2 3 4 5 6 7 8 9

• 5 X 1 2 3 4 5 6 7 8 9

• 5 X 9 8 7 6 5 4 3 2 1

Let Noble Thoughts come from all Let Noble Thoughts come from all directions. Rig Vedadirections. Rig Veda 3131

MATHEMATICIAN TIME PERIODMATHEMATICIAN TIME PERIOD Baudhayana (700 B.C.E. Baudhayana (700 B.C.E.

11stst to explain Pythagoras Theorem to explain Pythagoras Theorem) ) Apastamba (600) Apastamba (600)

Katyayana (500) Katyayana (500) Umaswati (150 B.C.E.)Umaswati (150 B.C.E.)

Aryabhata (476-c. 550 C.E.) Aryabhata (476-c. 550 C.E.) Varahamihira (c. 505-c. 558) Varahamihira (c. 505-c. 558) Brahmagupta (c. 598-c. 670) Brahmagupta (c. 598-c. 670) Govindaswami (c. 800-850) Govindaswami (c. 800-850)

Mahavira (Mahaviracharya) (850) Mahavira (Mahaviracharya) (850)


MATHEMATICIAN TIME PERIODMATHEMATICIAN TIME PERIOD Pruthudakaswami (850) Pruthudakaswami (850) Sridhara (900) Sridhara (900) Manjula (930) Manjula (930) Aryabhata II (950) Aryabhata II (950) Prashastidhara (958) Prashastidhara (958) Halayudha (975) Halayudha (975) Jayadeva (1000) Jayadeva (1000)


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What is Vedic Mathematics?

• It is an ancient technique, which simplifies multiplication, divisibility, complex numbers, squaring, cubing, square and cube roots. Even recurring decimals and auxiliary fractions can be handled by Vedic mathematics.

• It is useful in Arithmetic, Algebra, Geometry, Trigonometry, Solving of Equations, Calculus and in most branches of modern Maths.


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Who Brought Vedic Maths to limelight?

• The subject was revived largely due to the efforts of Jagadguru Swami Bharathikrishna Tirthaji of Govardhan Peeth, Puri Jaganath (1884-1960). Having researched the subject for years, even his efforts would have gone in vain but for the enterprise of some disciples who took down notes during his last days. That resulted in the book, Vedic Mathematics, in the 1960s.

• These are now available in a book called

• “VEDIC MATHEMATICS”• by H.H. Jagadguru Swami Sri Bharati Krishna Tirthaji

Maharaj.• Publishers Motilal Banarasidass, Bunglow Road, Jawahar

Nagar, Delhi –110 007; or • Chowk, Varanasi (UP); or Ashok Raj Path, Patna,

(Bihar)


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To Purchase Book Vedic Mathematics

• “VEDIC MATHEMATICS”• by H.H. Jagadguru Swami Sri Bharati

Krishna Tirthaji Maharaj.

• Publishers: Motilal Banarasidass, Bunglow Road, Jawahar Nagar, Delhi –110 007; or

• Chowk, Varanasi (UP); or Ashok Raj Path, Patna, (Bihar)

• Available in leading Book-shops in major cities in Bharat. Rs. 125/- Paper Back

• Rs. 250/- Bound Volume


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To Purchase Book Vedic Mathematics

• “Introduction to VEDIC MATHEMATICS”• Volume I and II• by V.G. Unkalkar and S.Seshachala Rao.

• Publishers: Dandeli Education Society, Bangurnagar, Dandeli (UK) 581 325, Karnataka.

• Available in leading Book-shops in Bangalore. Rs. 50/- each volume

•


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Square of Numbers

Yavd Unena, As much the DifferenceTavad Unikrut, So much further DifferenceVargam Cha Yojayet Square the Difference

• Case 1:• Numbers near the Base 10, 100, 1000 • Case 2 : • Numbers near the base 20, 30, 40, 200, 300,

400, 700, 900, ….


40

Square of NumbersNear the Base 100, 1000 etc

• Example: 98 x 98

• Step 1: Difference from 100 is - 02

• Step 2: So much Further Difference means subtract 02 from 98 which is 98 - 02 = 96

• Step 3: Square the Difference ie 02 x 02 = 04

• Step 4: Answer is 96 04


41


• Example: 96 x 96


• Step 2: So much Further Difference means subtract 04 from 96 which is 96 - 04 = 92




42


• Example: 103 X 103

• Step 1: Difference from 100 is + 03

• Step 2: So much Further Difference means ADD 03 to 103 which is 103 + 03 = 106




43


• Solve• 97 X 97 95 X 95 92 X 92• 89 X 89 88 X 88 91 X 91• 101 X 101 105 X 105 106 X 106• 109 X 109 111 X 111 112 X 112 • 999 X 999 995 X 995 991 X 991• 1005 X 1005 1012 X 1012 9999 X

9999


44

Square of NumbersNear the Base 20, 30, 70 etc


• Step 1: Difference from 20 is + 2

• Step 2: So much Further Difference means add 2 to 22 which is 22 + 2 = 24

• Step 3: Multiply by base 2 ie 24 x 2 = 48


• Step 5: Answer is 48 / 4 = 484


45


• Example: 79 X 79 = 624 1


• Step 2: So much Further Difference means Subtract 1 from 79 which is 79 – 1 = 78



• Step 5: Answer is 624 / 1 6241


46


• Example: 799 X 799 = 6384 01


• Step 2: So much Further Difference means Subtract 01 from 799 which is 799 – 01 = 798



• Step 5: Answer is 6384 / 01 =638,401


4747

Quotes on MathematicsQuotes on Mathematics

Bhaskara’s (1120 A.D.)Bhaskara’s (1120 A.D.) Cakravaal method is beyond all Cakravaal method is beyond all praise; it is certainly the finest praise; it is certainly the finest thing achieved in the theory of thing achieved in the theory of numbers before Lagrange.numbers before Lagrange.

Hankel,Hankel, Germany GermanyAre the countrymen of Bhaskara Are the countrymen of Bhaskara and Sankaracharya inferior to and Sankaracharya inferior to those of Newton and Darwin? those of Newton and Darwin? Sister NiveditaSister Nivedita


4848

Quotes on MathematicsQuotes on Mathematics"... the world owes most to India in the realm "... the world owes most to India in the realm of mathematics, which was developed to a of mathematics, which was developed to a stage more advanced than that reached by stage more advanced than that reached by any other nation of antiquity. The success of any other nation of antiquity. The success of Indian mathematics was mainly due to the fact Indian mathematics was mainly due to the fact that Indians had a clear conception of the that Indians had a clear conception of the abstract number (Algebra) as distinct from the abstract number (Algebra) as distinct from the numerical quantity of objects or spatial numerical quantity of objects or spatial extension." extension."

The Wonder That was India The Wonder That was India - - Australian Indologist A.L. BashamAustralian Indologist A.L. Basham


4949

Quotes on MathematicsQuotes on Mathematics

The Theorem of Pythogorus was The Theorem of Pythogorus was known to the ancient Indian known to the ancient Indian mathematicians in all its generalities. mathematicians in all its generalities. The much travelled Pythogorus The much travelled Pythogorus probably got this theorem from India probably got this theorem from India and publicised it all over the world. and publicised it all over the world. BurkeBurke

(A German Historian of Maths)(A German Historian of Maths)


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To Multiply by 11• Clue: Add adjacent numbers

• Step 1: write the the first and last digit as they are

• Step 2: Add two adjacent digits from the right and fill the gaps

• Step 3: Carry forward 1 if need be to the left


51

To Multiply by 11 – No Carry Forward


• Step 1: write the the first and last digits as they are

• 2 N 4


• 2 6 4 (2 + 4 = 6)


52

To Multiply by 11• Example: 11 X 17


• 1 N 7


• 1 8 7 (1 + 7 = 8)


53

To Multiply by 11

•Solve any five:

•11 X 22 11 X 33 11 X 27 11 X 36

•11 X 44 11 X 54•11 X 62 11 X 72•11 X 81 11 X 71


54

To Multiply by 11 – Carry Forward• Example: 11 X 38

• Here 3 + 8 = 11. Hence Carry forward


• 3 N 8


• 4 1 8 (3 + 8 = 11)


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To Multiply by 11 – Carry Forward

•Solve any five:

11 X 29 11 X 39 11 X 48 11 X 68

•11 X 99 11 X 88•11 X 76 11 X 67•11 X 85 11 X 58


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To Multiply by 11 – 3 or more Digits

• Example: 11 X 2 3 4


• 2 N M 4


• 2 5 7 4 (2 + 3 & 3 + 4)


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To Multiply by 11 – 3 or more Digits

• Example: 11 X 2 6 2 3 4


• 2 N M P Q 4


• 2 8 8 5 7 4


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To Multiply by 11 – 3 or more Digits with Carry Forward

• Example: 11 X 2 8 1 9 4


• 2 N M P Q 4


• 3 1 0 1 3 4


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To Multiply by 11 – Carry Forward

•Solve any five:

•11 X 2709 11 X 3509 •11 X 45318 11 X 62815•11 X 8272 11 X 9292•11 X 72635 11 X 1234567•11 X 987654 11 X 9080706


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Vedic MathematicsVedic Mathematics Vedic Mathematics is an ancient system of Vedic Mathematics is an ancient system of mathematics originated in India. Sankracharya Sri mathematics originated in India. Sankracharya Sri Bharati Krishna Tirthaji Maharaja rediscovered Bharati Krishna Tirthaji Maharaja rediscovered Vedic Mathematics between 1911 and 1918. This Vedic Mathematics between 1911 and 1918. This knowledge was passed down by oral tradition until knowledge was passed down by oral tradition until writing was invented. The Vedic system methods are writing was invented. The Vedic system methods are not only simple and comprehensible but inspiring to not only simple and comprehensible but inspiring to interested students. With the Vedic system one can interested students. With the Vedic system one can solve difficult problems and huge sums immediately. solve difficult problems and huge sums immediately. In fact, Vedic Mathematics has formulae and In fact, Vedic Mathematics has formulae and techniques for almost every field of Mathematics. techniques for almost every field of Mathematics. The entire Vedic Mathematics system consists of The entire Vedic Mathematics system consists of hundreds of tricks that students can use throughout hundreds of tricks that students can use throughout their educational career. They are further divided their educational career. They are further divided into 16 sutras and 14 sub-sutras.into 16 sutras and 14 sub-sutras.


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16 16 SutrasSutras1.1. Ekadhikena PurvenaEkadhikena Purvena

2.2. Nikhilam Navata scaramam DarsatahNikhilam Navata scaramam Darsatah

3.3. Urdhva TiryagbhyamUrdhva Tiryagbhyam

4.4. Paravartya YojayetParavartya Yojayet

5.5. Sunyam SamyasamucchayeSunyam Samyasamucchaye

6.6. Anurupye SunyamanyatAnurupye Sunyamanyat

7.7. Sankalana-vyavkalanabhyamSankalana-vyavkalanabhyam

8.8. PuranpuranabhyamPuranpuranabhyam

9.9. Chalana-KalanabhyamChalana-Kalanabhyam

10.10. YavdunamYavdunam

11.11. VyastisamastihVyastisamastih

12.12. Sesanyankena CharmenaSesanyankena Charmena

13.13. Sopantyadyaym UantyamSopantyadyaym Uantyam

14.14. Ekanyunena PurvenaEkanyunena Purvena

15.15. GunitasamuchachayahGunitasamuchachayah

16.16. GunaksamuchchayahGunaksamuchchayah


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Sub-SutrasSub-Sutras

1.1. AnurupyenaAnurupyena2.2. Sisyate SesasamjnahSisyate Sesasamjnah3.3. AdyamadyenantyamantyenaAdyamadyenantyamantyena4.4. Kevalaih Saptakam GunyatKevalaih Saptakam Gunyat5.5. VestanamVestanam6.6. Yavdunam TavdunamYavdunam Tavdunam7.7. Yavdunam Tavdunam Varganchya YojayetYavdunam Tavdunam Varganchya Yojayet8.8. AntyayordasakepiAntyayordasakepi9.9. AntyayorevaAntyayoreva10.10. SamuchchyagunitahSamuchchyagunitah11.11. LopanasthapanabhyamLopanasthapanabhyam12.12. VilokanamVilokanam13.13. Gunitsamuchchayah SamuchchayagunitahGunitsamuchchayah Samuchchayagunitah14.14. DhvajankaDhvajanka


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General Multiplication

•Sutra is “URDHVA TIRYAK”

Vertical and Cross-wise

•General Multiplication of 2 digits, 3 digits, more digits


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General Multiplication of 2 digits

• 1 2• X 1 3• ↑ X ↑

• Answer: Unit Place : 6 – (vertically multiplying 2 and 3 – both are units)

• Tens place: 5– Cross- multiply 1X3 + 1X2 ie 3 + 2 = 5

• Hundred Place: 1– (Vertically multiply 1 and 1 which both are in tens

place)• Hence answer is 1 5 6


65


•Solve any Five:

• 12 X 13 22 X 13 22 x 23• 14 X 23 45 X 45 17 X 11

• 26 X 51 34 X 43 16 X 41

• 27 X 52 35 X 45 24 X 42


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• 1 2 1• X 1 3 2• ↑ X *X ↑

• Answer: Unit Place : 2– (vertically multiplying 1 and 2 – both are units)

• Tens place: 7– Cross- multiply 1X3 + 2X2 ie 3 + 4 = 7

• Hundred Place: 9– (Vertically & Cross multiply 1X2 + 1X1 + 2X3

• Thousand Place: 5– (Cross multiply 1X3 + 1X2

• Ten Thousand Place: 1– (Vertically multiply 1X1

• The Answer is: 15,972


67


• 1 9 9 _• X 1 3 2 this is same as 2 0 1• ↑ X *X ↑ x 1 3 2• ___• Answer: Unit Place : - 2

– (vertically multiplying 1 and 2 – both are units)• _• Tens place: - 3

• Hundred Place: 4 – 1 = 3

• Thousand Place: 6

• Ten Thousand Place: 2• 2 6 3 0 0 - 32 = 2 6, 2 6 8

• Pl note that all big digits like 6, 7, 8, 9 can be transformed _ _ _ _into small digits when subtracted from 10 to 4, 3, 2, 1


Quotes on AstronomyQuotes on Astronomy

There were very advanced Hindu There were very advanced Hindu astronomers in 600 B.C. The Vedas astronomers in 600 B.C. The Vedas contain an account of the dimensions of contain an account of the dimensions of Earth, Sun, Moon, Planets and Galaxies. Earth, Sun, Moon, Planets and Galaxies.

Emmelin Plunret, Emmelin Plunret, in “Calendars and Constellation”in “Calendars and Constellation”


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To Create Tables for 9, 19, 29, ….99 etc

• _• 9 is 10 – 1 = 1 1• _• 19 is 20 – 1 = 2 1• _• 99 is 100 – 1 = 10 1

• And so on…..


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To Create Tables for 9

X 9 Here 9 is 10 - 1 _ _

1 1 1 1

1 X 9 0 9 0 9 0 9

2 X 9 8 1 1 8

3 X 9 7 2 2 7

4 X 9 6 3 3 6

5 X 9 5 4 4 5

6 X 9 4 5 5 4

7 X 9 3 6 6 3

8 X 9 2 7 7 2

9 X 9 1 8 8 1

10 X 9 0 9 0 9 0


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X 1 9 Here 19 is 2 0 - 1 _ _

2 1 2 1

1 X 1 9 1 9 1 9 1 9

2 X 1 9 8 3 3 8

3 X 1 9 7 5 5 7

4 X 1 9 6 7 7 6

5 X 1 9 5 9 9 5

6 X 1 9 4 11 11 4

7 X 1 9 3 13 13 3

8 X 1 9 2 15 15 2

9 X 1 9 1 17 17 1

10 X 19 0 19 0 19 0


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X 2 9 Here 29 is 3 0 - 1 _ _ Pl. Complete this …

3 1 3 1

1 X 2 9 2 9 2 9 2 9

2 X 2 9 8 5 ? 5 8

3 X 2 9 7 8 ? 8 7

4 X 2 9 6 11? 11 6

5 X 2 9 5 ? 14? 14 5

6 X 2 9 4 ? 17? 17 4

7 X 2 9 3 ? 20? 20 3

8 X 2 9 2 ? 23? 23 2

9 X 2 9 1 ? 26? 26 1

10 X 29 0 290 2 9 0


73


Create Tables for

39 49 59 69 79 89 99 109

199 999 etc


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Ekadhikena PurvenaEkadhikena Purvena means “By One More Than the Previous One”. This formula is commonly used in multiplication and division. The Ekadhikena Purvena can also be used to square certain numbers ending in 5 or where the sum of the ones digits is 10. For example, use the Ekadhikena Purvena to square the number 135.

Example 1: 95 x 95 = ?

Step 1: Multiply the last digits of the numbers. 5 x 5 = 25Step 2: Multiply the previous by one more than itself. 9 x (9 + 1) = 9 x 10 = 90Step 3: Put these numbers together. 9,025

Example 2: 135 x 135 = ?

Step 1: Multiply the last digits of the numbers. 5 x 5 = 25Step 2: Multiply the previous by one more than itself. 13 x (13 + 1) = 13 x 14= 182Step 3: Put these numbers together. 18,225 Example 3: 133 x 137 = ?Step 1: Multiply the last digits of the numbers. 3 x 7 = 21Step 2: Multiply the previous by one more than itself. 13 x (13 + 1) = 13 x 14= 182Step 3: Put these numbers together. 18,221


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Sunyam SamyasamucchayeSunyam Samyasamuchchaye means “If the Samuccaye is the same, it is zero”. Under this sutra

there are many formulae you can use in the branch of Algebra. Below, I will show two examples of the Sunyam Samyasamucchaye.

Example 1: 7x + 5 = 9x + 7 9x – 5 7x +17 In this case, the numerator of the first part of the equation plus the numerator of the

second part is equal to the denominator of the first part plus the denominator of the second part,(7x + 5) + (9x + 7) = (9x – 5) + (7x +17)16x + 12 = 16x + 12According to the sutra, if both sides of the equation are equal, you can conclude that either

side is equal to zero. Therefore 16x + 12 = 016x = -12x = -3/4 Example 2: 16x – 3 = 2x - 15 7x + 7 11x - 25 (16x – 3) + (2x – 15) = (7x + 7) + (11x - 25)18x – 18 = 18x – 1818x – 18 = 018x = 18x = 1


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Urdhva TiryagbhyamUrdhva Tiryagbhyam means “Vertically and Crosswise”. This sutra is commonly used for the multiplication of two positive integers with any number of digits. By this method, you can multiply any two numbers that have the same number of digits.

Example 1: 37 x 33 = ? Step 1: Write 33 under 37 37 33Step 2: Multiply the left most column and write it below. 37 33 9Step 3: Multiply diagonally and add. (3 x 3) + (3 x 7) = 9 + 21 = 30 Since 30 is a 2 digit number, I keep the last digit, 0, next to 9 and put the tens digit below 9. 37 33 90 3Step 4: Multiply the right most column and put the ones digit next to 0 and the tens digit below 0. 7 x 3 = 21 37 33 901 32_Step 5: Add the two numbers you have left. 901 + 320 1221


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Paravartya Yojayet Paravartya Yojayet means “Transpose and Apply”. This sutra is mainly used in solving simultaneous simple equations. With the Paravartya Yojayet you can solve for x and y instantly in your mind. Example 1: 2x + 3y = 8 4x + 5y = 14 Step 1: For the value of x, start by cross multiplying the y-coefficients with the constants. 3 x 14 = 42 8 x 5 = 40 Step 2: Subtract these values. 42 – 40 = 2. This is the numerator for the value x.Step 3: Now cross multiply the coefficients of x and y. But start with 3 x 4, not 2 x 5. 3 x 4 = 12 2 x 5 = 10 12 – 10 = 2. This is the denominator of x. x = 2/2 = 1Step 4: To get y, repeat the above steps, but replace the coefficients of y with the coefficients of x. y = (32 – 28)/(12 – 10) = 2


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Multiplying Numbers Near a BaseOnce while working on Vedic Mathematics, I could not find any simple method to multiple two numbers equidistant from a common base. So while working along with my son, I discovered a very simple and easy method to do that. Here is how it works. Example 1: 56 x 64 = ?Step 1: Both 56 and 64 are near a common base 60. 56 + 4 = 60 60 – 4 = 60Step 2: Multiply 60 x 60 instead of 56 x 64. 60 x 60 = 3600 We are not at our answer yet. Step 3. Subtract 42 from 3600. 42 or 4 x 4 = 16 3600 – 16 = 3584 (Answer) Example 2: 77 x 83 = ?Step 1: Both 77 and 83 are near 80Step 2: Multiply 80 x 80 instead of 77 x 83. 80 x 80 = 6400Step 3: Subtract 32 from 6400. 6400 – 9 = 6391 (Answer)


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Nikhilam Navatashcaramam DashatahThis formula is used in special cases on multiplication. The meaning of this sloka is all from nine and last from ten. Let’s compute the square of a number.Step 1. Find the nearest base from the number, i.e, 10, 100.Step 2. If the number is less than the base, take the deficiency of the number from

the base. If the number is greater than the base take the surplus.Step 3. The square of the number = (number -/+ deficiency /

surplus) (square of deficiency/surplus)

Examples : Number = 8base = 10, deficiency = 2. Put these in the formula abovesquare of 8 = (8 - 2) (square of 2) = 64Number = 105base = 100, surplus = 5square of 105 = (105 + 5) (square of 5) = 11025


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Multiplications of Numbers Near BaseLet's do 899x995, 45x47, 12x14. You are right it's hard without calculator. But using Vedic math you can do it on your finger tips. Here is how it works.

Select a base close to the numbers. Do cross subtraction/addition depending upon deficiency or surplus respectively. Then multiply the deficiencies or surpluses. In case of half base devide the the first part of the answer by 2. 899 -101 45 - 05(base/2 50) 12 + 2995 - 005 47 - 03 14 + 4----------------- -------------------- ----------884, 505 42, 15 =>2115 16, 8

Multiplication by 11, 111, 111…Put the sum of the two digits in between those two digits to get the result.

24 x 11 = 2 ( 2 + 4 ) 4 = 2 ( 6 ) 4 = 26485 x 11 = 8 ( 8 + 5 ) 5 = 8 ( 13 ) 5 = 935

Let’s take three digits number and multiply by 11. Put the sum of the first two digits after first digit and sum of the last two digits before last digit.

243 x 11 = 2 ( 2 + 4 ) ( 4 + 3 ) 3 = 2673

Dividing by 9The first digit is the quotient and the sum of the two digits is the remainder.

25 / 9 = 2 remainder ( 2 + 5 ) = 2 remainder 751 / 9 = 5 remainder 6

Let’s take three digit number. The sum first two digit is the quotient and the sum of the three digits is the remainder.

259 / 9 = 27 remainder 16 = 28 remainder 7


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Dividing by 19, 29, --99 etc• 1st Example - 1 Divided by 19, 29, 39, …. 129 etc

• To Divide 1 by numbers ending in 9 like 1 divided by 19, 29, 39, ….. 119 etc.• Some of these numbers like 19, 29, 59 are prime numbers and so cannot be

factorised and division becomes all the more difficult and runs into many pages in the present conventional method and the chances of making mistakes are many.

• The Vedic Solution is obtained by applying the Sutra (theorem) Ekadhikena Purvena which when translated means “One more than the Previous”

• Take for example 1 divided by 19. In the divisor 19, the previous is 1 and the factor is obtained by adding 1 to it which is 2. Similarly when we have to divide by 29, 39, … 119 the factors shall be 3,4,… 12 respectively. (Add 1 to the previous term in the divisor).

• After this divide 1 by the factor in a typical Vedic way and the answer is obtained in 1 step. Thus

• 1 divided by 19 = 0.0 5 2 6 3 1 5 7 8 9 4 7 3 6 8 4 2 1 (18 Digits)

• 1 divided by 29 = 0.0 3 4 4 8 2 7 5 8 6 2 0 6 8 9 6 5 5 1 7 2 4 1 3 7 9 3 1 • (28 Digits)


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Multiplier-digits consist entirely of nines

• The Sutra : Ekanyunena Purvena, “one less”

• 777 multiplied by 999 = 776,223• * 776 is one less than multiplicand 777

223 is the compliment (PURAK) of 776 from 999)

• 120 35 79 multiplied by 999 99 99 = 120 35 78, 879 64 21

• 1234 5678 09 multiplied 9999 9999 99 = 1234 5678 08 8765 4321 91

• Such multiplications come up in advanced astronomy.


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• Solve any Five

• 45 X 99 54 X 99 99 X 99

• 123 X 999 234 X 999 999 X 999

• 9876 X 9999 9753 X 9999 8502 X 9999

• 97531 X 99999 24608 X 99999 10203 X 99999

• 135 792 468 X 999 999 999


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• Special Case when number of 9 in the multiplier is MORE than those of multiplicand

• 35 X 999 = 34/9 65 = 34 965• 135 X 99 999 = 134/99 865 = 13499,865• 97531 X999 99999 = 97530/999 02469

• Special Case when number of 9 in the multiplier is LESS than those of multiplicand

• 1 35 X 99 = 134 {-1}/ 65 = 133 65• 13 579 X 999 = 13 578 {-13} / 421 = 13 565 421• 2 222 X999 = 2, 221 {-2}/ 778 = 2 219 778

Let Noble Thoughts come from all directions. Rig Veda 85

Algebraic DivisionsAlgebraic Divisions• Sutra: Sutra: Paravartya YojayetParavartya Yojayet “Transpose and Apply”“Transpose and Apply”

• Divide Divide (12 x(12 x22 – 8x – 32) by – 8x – 32) by (x - 2),(x - 2), factor is factor is +2+2

• (X – 2)/2(X – 2)/2 12x12x22 - 8x -32 - 8x -32• +24x + 32+24x + 32• ----------------------------------------------------• 12x + 1612x + 16 R = 0R = 0

• Divide Divide 7 x7 x22 + 5x + 3 + 5x + 3 by by x – 1x – 1 Q = 7x + 12 Q = 7x + 12 R = 15R = 15

• DivideDivide 7 x7 x22 + 5x + 3 + 5x + 3 byby x + 1x + 1 Q = 7x –2Q = 7x –2 R = 5R = 5

• DivideDivide xx33 + 7 x + 7 x22 + 6x + 5 + 6x + 5 byby x – 2 x – 2 Q = xQ = x22 + 9x + 24 + 9x + 24 R = 53R = 53

• DivideDivide xx44 – x – x33+ x+ x22 + 3x +5 + 3x +5 byby x x22 – x – 1 – x – 1 Factors are Factors are x + 1x + 1• Q = xQ = x22 + 0x + 2 + 0x + 2 R = 5x + 7R = 5x + 7

• DivideDivide 6x6x44 + 13 x + 13 x33 + 39 x + 39 x22 + 37x + 45 + 37x + 45 byby xx22 – 2x – 9 – 2x – 9• Factors 2x + 9Factors 2x + 9 Q = 6 xQ = 6 x22 + 25x + 143 + 25x + 143 R R

= 548x + 1332= 548x + 1332•


Algebraic DivisionsAlgebraic Divisions• DivisionDivision•• DivideDivide x4 + xx4 + x22 + 1 + 1 byby x2 – x +1x2 – x +1 (add 0 x(add 0 x33 & &

0x)0x)• x4x4 + 0 + 0 xx33 + x + x22 + 0x + 1 + 0x + 1 Factors are:Factors are: x - 1x - 1• Q = xQ = x22 + x + 1 + x + 1 R = 0R = 0

• DivideDivide 1 2 3 4 1 2 3 4 by by 1 1 21 1 2 Factors -1 –2Factors -1 –2 Q = 11 R = 2 Q = 11 R = 2

• DivideDivide 1 3 4 5 61 3 4 5 6 byby 1 1 2 31 1 2 3 Q = 12 R = - 20 Q = 12 R = - 20• The Reminder cannot be negative. Hence Q = 11 The Reminder cannot be negative. Hence Q = 11

R = 1 1 0 3 R = 1 1 0 3

• DivideDivide 1 3 9 0 5 1 3 9 0 5 byby 1 1 31 1 3 Q = 1 24Q = 1 24 R = -R = -107107

• Q = 123Q = 123 R = 6R = 6


Shortest Proof of Pythagorus Shortest Proof of Pythagorus TheoremTheorem

• Take a square of length A+B.Take a square of length A+B.• Draw another square inside it of Z Draw another square inside it of Z

dimensiondimension

• (A+B)(A+B)22 = Z= Z22 + 4x1/2 AB + 4x1/2 AB

• AA2 2 + B+ B22 + 2AB + 2AB = Z= Z22 + 2AB + 2AB

• AA2 2 + B+ B22 = Z= Z22


Quotes on Pythagoras Quotes on Pythagoras TheoremTheorem

• There are over 400 different proofs for Pythogorus Theorem. But the shortest was given by ancient unknown Indian scholar in just two lines.

• Iswar Chakravartty• (Prof Emeritus, canada)• [email protected]


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Quotes on Vedic SciencesQuotes on Vedic SciencesTwo Line Proof of PythogorusTwo Line Proof of Pythogorus

(x+y)(x+y)22 = z = z22 + 4 x 1/2 xy + 4 x 1/2 xy

xx22 + y + y22 + 2xy = z + 2xy = z22 + 2 xy + 2 xy

xx22 + y + y22 = z = z22

Iswar ChakravarttyIswar Chakravartty(Prof Emeritus, canada) [email protected](Prof Emeritus, canada) [email protected]


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Factorisations of Quadratics

• Sutras:

• Anurupyena “Proportionately”

• Adhyam-Adhyena, Antyam-antyena

• “first by the first and the last by the last”


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Verifying Correctness of answers• A Sub-sutra of immense utility for the purpose of verifying the

correctness of our answers in multiplications, divisions and factorisations:

• Gunita-Samuchhayah Samuchhaya-gunitah means

• “The product of the sum of the coefficients in the factors is equal to the sum of the coefficients in the product”

• Product of the Sc in the factors = Sc of the product

• Example (x+7) (x+9) = (x2 + 16x + 63)• (1+7) (1+9) = (1 + 16 + 63) = 80• Or (x+1) (x+2) (x+3) = (x3 + 6X2 + 11x +

6)• (1+1) (1+2) (1+3) = (1 + 6 + 11 + 6) = 24


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Factorisations of Harder Quadratics

• Lopana-stapana-bhyam “by (alternate) Elimination and Retention”

• It is very difficult to factorise the long quadratic (2x2 + 6y2 + 3z2 + 7xy + 11yz + 7zx)

• But “Lopana-Sthapana” removes the difficulty. Eliminate z by putting z = 0.

• Hence the given expression Ez = 2x2 + 6y2 + 7xy = (x+2y) (2x+3y)

• Similarly, if y=0, then Ey = 2x2 + 3z2 + 7zx = (x+3z) (2x+z)

• Hence E = (x+2y+3z) (2x+3y+z)


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Factorisations of Harder Quadratics

• Factorise 2x2 + 2y2 + 5xy + 2x- 5y –12 =

Ey = (x+3) (2x-4) and

• Ex = (2y+3) (y-4)• Hence, E = (x+2y+3) (2x+y-4)

• * This “Lopana-sthapana” method (of alternate elimination and retention) will be found highly useful in HCF, in Solid Geometry and in Co-ordinate Geometry of the straight line, the Hyperbola, the conjugate Hyperbola, the Asymptotes etc.


Factorisations of Harder Factorisations of Harder Quadratics – Special Quadratics – Special CasesCases

• Sunyam Samya samuccaye Sunyam Samya samuccaye

• (when Samuccaya is the same, that (when Samuccaya is the same, that Samuccaya is zero)Samuccaya is zero)

• Samuccaya is a technical term which Samuccaya is a technical term which has several meanings.has several meanings.



• (when Samuccaya is the same, that Samuccaya is zero)(when Samuccaya is the same, that Samuccaya is zero)

• Special Types of seeming CubicsSpecial Types of seeming Cubics

(x- 3)(x- 3)33 + (x –9) + (x –9)33 = 2(x –6)= 2(x –6)33

• current method is very lengthy, but Vedic method says, current method is very lengthy, but Vedic method says, (x-3) + (x-9) = 2x – 12 Hence x = 6(x-3) + (x-9) = 2x – 12 Hence x = 6

•

• (x-149)(x-149)33 + (x-51) + (x-51)33 = 2(x-100) = 2(x-100)33

HenceHence 2x-200 =0 &2x-200 =0 & x = 100x = 100

• (x+a+b-c)(x+a+b-c)33 + (x+b+c-a) + (x+b+c-a)33 = 2(x+b)= 2(x+b)33

Hence Hence x = -bx = -b



• Solve the following equation by both Vedic Maths and conventional Solve the following equation by both Vedic Maths and conventional methodsmethods

• Vedic Maths Method Vedic Maths Method Vedic Sutra is: Vedic Sutra is: Shunya Samya SamuccayaShunya Samya Samuccaya

• When we open the LHS and RHS without raising to the power of 3 we getWhen we open the LHS and RHS without raising to the power of 3 we get

• LHS : LHS : X -1 + X-3 X -1 + X-3 = 2X - 4= 2X - 4

• RHS:RHS: 2 (X-2) 2 (X-2) = 2X - 4= 2X - 4

• Hence LHS = RHS = Samuccaya = 2x-4 = 0Hence LHS = RHS = Samuccaya = 2x-4 = 0

• Hence Hence x = 2x = 2

333 2231 XXX



• SolveSolve

• Conventional Method by Opening the bracketsConventional Method by Opening the brackets

• LHS:LHS:

• = =

• RHS:RHS: 2 ( -6 )2 ( -6 )

• = 2 -12 = 2 -12

• Hence the equation is Hence the equation is

• 2 -12 2 -12 = 2 -12 = 2 -12

• Hence Hence - = - =

• HenceHence = =

• Hence Hence = =

• Which is the same as in Vedic MethodWhich is the same as in Vedic Method

•

• VerificationVerification

• Put = 2 in both LHS and RHSPut = 2 in both LHS and RHS

• LHS: LHS: (2 – 1) + (2 – 3) (2 – 1) + (2 – 3)

• = 1 – 1 = 1 – 1 = 0= 0

• RHS: 2 (2 – 2) RHS: 2 (2 – 2) = 0= 0

• Hence Hence = =

•

•

333 2231 XXX

33 31 XX



• Solve the following equation by both Vedic Maths and conventional methodsSolve the following equation by both Vedic Maths and conventional methods• + = 2 + = 2 • Vedic Maths MethodVedic Maths Method• Vedic Sutra is: Vedic Sutra is: Shunya Samya SamuchayaShunya Samya Samuchaya• Meaning:Meaning: If Samuchaya is same, then Samuchaya is Zero.If Samuchaya is same, then Samuchaya is Zero.• (Samuchaya is a special Technical Term and has different meanings at different places)(Samuchaya is a special Technical Term and has different meanings at different places)• When we open the LHS and RHS without raising to the power of Three we getWhen we open the LHS and RHS without raising to the power of Three we get• LHS : LHS : X -1 + X-3 X -1 + X-3 = 2X - 4= 2X - 4• RHS:RHS: 2 (X-2) 2 (X-2) = 2X - 4= 2X - 4• Hence LHS = RHS = Samuchaya = 2x-4 = 0Hence LHS = RHS = Samuchaya = 2x-4 = 0• Hence Hence x = 2x = 2• Conventional Method by Opening the bracketsConventional Method by Opening the brackets• LHS:LHS: - 3 - 3 ++ - 9 - 9 • = 2 -12 = 2 -12 • RHS:RHS: 2 ( -6 )2 ( -6 )• = 2 -12 = 2 -12 • Hence the equation is Hence the equation is • 2 -12 2 -12 = 2 -12 = 2 -12 • Hence Hence - = - = • HenceHence = = • Hence Hence = = • Which is the same as in Vedic MethodWhich is the same as in Vedic Method• • VerificationVerification• Put = 2 in both LHS and RHSPut = 2 in both LHS and RHS• LHS: LHS: (2 – 1) + (2 – 3) (2 – 1) + (2 – 3) • = 1 – 1 = 1 – 1 = 0= 0• RHS: 2 (2 – 2) RHS: 2 (2 – 2) = 0= 0• Hence Hence = = • •

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