The Power of Vedic Maths with Trigonometry

THEPOWEROFVEDICMATHS2ndEDITION

ATULGUPTA

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THEPOWEROFVEDICMATHSISBN81-7992-357-6

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PREFACEMathematicsisconsideredtobeadryandboringsubjectbyalargenumber

ofpeople.Childrendislikeand fearmathematics foravarietyof reasons.Thisbook is written with the sole purpose of helping school and college students,teachers, parents, common people and people from non-mathematical areas ofstudy,todiscoverthejoysofsolvingmathematicalproblemsbyawonderfulsetoftechniquescalled‘VedicMaths’.

Thesetechniquesarederivedfrom16sutras(verses)intheVedas,whicharethousands of years old and among the earliest literature of ancient Hindus inIndia.Theyareanendlesssourceofknowledgeandwisdom,providingpracticalknowledge in all spheresof life. JagadguruSwamiSriBharatiKrshnaTirthaji(1884-1960)wasabrilliant scholarwhodiscovered the16 sutras in theVedasandspent8yearsintheirintensestudy.Hehasleftaninvaluabletreasureforallgenerationstocome,consistingofasetofuniqueandmagnificentmethodsforsolving mathematical problems in areas like arithmetic, algebra, calculus,trigonometryandco-ordinategeometry.Thesetechniquesareveryeasytolearnand encapsulate the immense and brilliantmathematical knowledge of ancientIndians,whohadmadefundamentalcontributionstomathematicsintheformofthedecimalnumerals,zeroandinfinity.

Ihavetrainedthousandsofchildrenofallagegroupswiththesetechniquesand I find that even young children enjoy learning and using them. Thetechniques reduce drastically, the number of steps required to solve problemsand inmany cases, after a little practice,manyof theproblems canbe solvedorally. Itgives tremendousself-confidence to thestudentswhich leads themtoenjoymathematicsinsteadoffearinganddislikingit.

Ihavewrittenthisbookintheformofacookbook,whereareadercangraspa technique quickly, instead of reading through a largemass of theory beforeunderstandingit.Ihaveconsideredtechniquesformajorarithmeticaloperationslike multiplication, division, computation of squares and square roots andcomplexfractions,besidesawhole lotofother techniques.Eachtechniquehasbeenexplainedindetailwiththehelpofsolvedexamples,usingastep-by-step

approachandIamsurethatthereaderwillbeabletounderstandandmasterthecontentseasily.Everychapterhasa largenumberofproblemsforpracticeandthebookcontainsover1000suchproblems.Theanswersaregivenalongsidesothat the reader can either solve the problems orally or use paper and pen andcompare with the given answer. The chapters should be read sequentially toabsorbthematerialandthencanbeusedforreferenceinanydesiredorder.

IhavealsoincludedaspecialchapterinwhichIhaveshowntheapplicationofthetechniquestosolveproblems,collectedfromseveralcompetitiveexams.This is a unique feature of the book and should add to the popularity of thetechniques.

I have tried to make all the examples and answers error-free but if anymistakeisdiscovered,IwillbeobligedifIaminformedaboutthesame.

Constructive criticism and comments can be sent to me [email protected]

ATULGUPTAMobile:9820845036

mailto:[email protected]

Contents

Preface

Chapter1TwoSimpleTechniques

Subtractionfrom100/1000/10000

NormalMethod

VedicMethod

Multiplicationwithaseriesof9s

Multiplicationofanumberbysamenumberof9s

Multiplicationofanumberbygreaternumberof9s

Multiplicationofanumberbylessernumberof9s

Chapter2Operationswith9

Computationofremainder(Navasesh)ondivisionby9

Basicmethod

Firstenhancement

Secondenhancement

Finalcompactmethod

Verificationoftheproductoftwonumbers

Verificationofthesumoftwonumbers

Verificationofthedifferenceoftwonumbers

Verificationofthesquareorcubeoftwonumbers

Limitationintheverificationprocess

Computationofthequotientondivisionby9

Method1

Method2

Chapter3Operationswith11

Multiplication

Divisibilitytestofnumbersby11

Multiplicationwith111

Chapter4Multiplication(Nikhilam)

Secondarybasesof50

Secondarybaseof250

Secondarybaseof500

Secondarybaseslike40,60

Secondarybaseof300

Chapter5Multiplication(UrdhvaTiryak)

2-Digitmultiplication


Multiplying3-digitand2-digitnumbers


Multiplying4-digitand3-digitnumbers

Chapter6Division

Divisionbyaflagofonedigit(noremainder)

Divisionbyaflagofonedigit(withremainder)

Divisionwithadjustments

Divisionwithaflagof2digits

Divisionwithaflagof3digits

Chapter7SimpleSquares

Chapter8SquareofAnyNumber

Definition-DwandwaorDuplexSquareofanynumber

Squareofanynumber

Chapter9SquareRootofaNumber

Chapter10CubesandCubeRoots

Computingcubesof2-digitnumbers

Cuberootsof2-digitnumbers

Computingfourthpowerof2-digitnumbers

Chapter11Trigonometry

Triplet

Computingtrigonometricratios

Computingtrigonometricratiosoftwicetheangle

Computingtrigonomegtricratiosofhalftheangle

Chapter12AuxiliaryFractions

Divisorsendingwith‘9’




Numbersendingwith‘6’

Otherdivisors

Chapter13MishrankorVinculum

ConversiontoMishrank

ConversionfromMishrank

Applicationinaddition

Applicationinsubtraction

Applicationinmultiplication

Applicationindivision

Applicationinsquares

Applicationincubes

Chapter14SimultaneousEquations

Chapter15Osculator

Positiveosculators

Negativeosculators

Chapter16ApplicationsofVedicMaths

Sampleproblems

SolutionsusingVedicmaths

Problemsforpractice

Answers

TWOSIMPLETECHNIQUES

Wewill begin our journey into the fascinatingworld ofVedicmathswithtwosimpletechniqueswhichwilllaythefoundationforsomeofthetechniquesinthefollowingchapters.

I.Subtractionfrom100/1000/10000

Wewillstartwithaverysimpletechniquewhereinwewillseetheuseofthesutra‘Allfrom9andlastfrom10’.Thisisusedtosubtractagivennumberfrom100, 1000, 10000 etc. It removes the mental strain which is existent in themethodtaughtinschools.

ThismethodisalsousedlateronintheNikhilammethodofmultiplication.

Considerthesubtractionof7672from10000.

a)NormalMethod

Thenormalmethodis

We carry ‘1’ from the left side and continue doing so till we reach therightmostdigit,leavingbehind9ineachcolumnand10inthelastcolumn.

Then,wesubtracttherightmostdigit‘2’from10andwritedown‘8’.Next,wesubtractthedigit‘7’from‘9’andwritedown‘2’.

Werepeatthisprocessforalltheremainingdigitstotheleft.

Throughthisoperation,thefinalresultisalwaysobtainedfromrighttoleft.

Mentally,thereisacarryoperationforeverydigit,whichistimeconsumingandslowsdowntheoverallprocess.

b)VedicMethod

TheVedicmethodusesthesutra‘Allfrom9andlastfrom10’andgivesaverysimpleandpowerfultechniquetoachievethesameresult.

Theresultcanbeobtainedfrombothlefttorightaswellasrighttoleftwithequalease. Itstates that theresultcanbeobtainedbysubtractionofeachdigitfrom‘9’andthelastdigitfrom‘10’.

Hence,inthegivenexample,

Wecanget the result from left to right or viceversa from right to left as

i.e.alldigitsexceptthelastonearesubtractedfrom9,thelastdigitissubtractedfrom10andtheresult(2328)iswrittendowndirectly.Thementalburdenofacarryforeachcolumnvanishesandtheanswercanbeobtainedeasily,inajiffy.

Thesametechniquecanbeappliedfordecimalsubtractionalso,e.g.2.000-0.3436.

Thecoreoperationhereissubtractionof3436from10000where1isacarryfromleft.

1.Examplesforsubtractionsfrombasesof100,1000etc.

Thissimplesubtractionmethodwillbeusedfurtherinmultiplication(Ch.4)andmishrank(Ch.13).

II.Multiplicationwithaseriesof9s

a)Multiplicationofanumberbysamenumberof9s

Letususethistechniquetoseehowtocarryoutmultiplicationofann-digitnumberbyanumberwith‘n’numberof9seg.any3-digitnumberby999ora4-digitnumberby9999.

Letusseeanexampleviz.533×999

Thesolutioncanbeobtainedas533×(1000-1)

=533000-533

=532467.

Theanswerconsistsoftwo3-partnumbersviz.‘532’and‘467’.

Thefirstpart‘532’is1lessthanthegivennumber,i.e.533-1.

Thesecondpart‘467’isequalto(1000-533)or(999-532).Itissimplythe‘9’s complement of the first 3 digits of the result; i.e. 9s complement of 532wheretherequireddigitscanbeobtainedas:4=9-5

6=9-3

7=9-2

Similarly,3456×9999willconsistoftwoparts

3455/6544

i.e.34556544(where,6544is‘9’scomplementof3455).

This simple technique can be used to get the result orally whenever anynumberismultipliedbyanumberconsistingofanequalnumberof9s.

Thesametechniquecanbeusedformultiplicationofdecimalnumbers.

Considertheexample4.36×99.9

Firstly,multiply436by999andwritedowntheresultas435564.

Now,since thenumbershave twoandonedecimalplaces respectively, thefinal resultwill have three decimal places andwe put the decimal point threepointsfromtheright.So,thefinalanswerwillbe435.564.

b)Multiplicationofanumberbygreaternumberof‘9’s

Let us now see how to carry outmultiplication of an n-digit number by anumberwithgreaternumberof‘9’sE.g.235by9999or235by99999

i)Consider235by9999

Sincethenumberofdigitsin235isthreewhiletherearefourdigitsin9999,wepad235withonezerotoget0235andcarryoutthemultiplicationasbefore.

0235×9999=0234/9765andthefinalresultis2349765.

ii)Consider235by99999

Padding235withtwozeroes,weget

00235×99999=00234/99765andthefinalresultis23499765.

c)Multiplicationofanumberbylessernumberof9s

Let us now see how to carry outmultiplication of an n-digit number by anumberwithlessernumberof9s.

i)Consider235by99

Sincetherearetwodigitsin99,weplaceacolontwodigitsfromtherightinthegivennumbertoget2:35

Weenclosethisnumberwithinacolonontherighttoget

2:35:

Wenowincreasethenumbertotheleftofthecolonby1andderiveanewnumberas3:35

We align the colon of the new number with the enclosing colon of theoriginalnumbertoget2:35:

3:35

Wenow subtract 3 from235 and35 from100, to get the final answer as

ii)Consider12456by99

Sincetherearetwodigitsin99,weplaceacolontwodigitsfromtherightinthegivennumbertoget124:56Weenclosethisnumberwithinacolonontherighttoget124:56:

Wenowincreasethenumbertotheleftofthecolonby1andderiveanewnumberas125:56

We align the colon of the new number with the enclosing colon of theoriginalnumbertoget124:56:

125:56Wenowsubtracttogetthefinalansweras

2.Examplesformultiplicationbyseriesof9s

In examples where decimal numbers are involved, we multiply the twonumbersassumingthattherearenodecimalsandthenplacethedecimalintherightplace,asexplainedabove.

OPERATIONSWITH9

The number 9 has a special significance inVedicmaths.Wewill start bylookingatasimplebutpowerfultechniquewhichisusedtogettheremainderondividing any number by 9. This remainder is given several names like‘Navasesh’,‘Beejank’ordigitalroot.Ithaspowerfulpropertieswhichareusedto check the correctness of arithmetic operations like addition, subtraction,multiplication,squaring,cubingetc.Itcanalsobeusedtocheckthedivisibilityofanygivennumberby9.

Wewillbeginbyseeinghowtocomputethenavaseshofanygivennumberandthenproceedtoseeitsapplications.

I.Computationofremainder(Navasesh)ondivisionby9

Thenavaseshofanumberisdefinedastheremainderobtainedondividingitby9.

Example:Thenavaseshof20is2Thenavaseshof181is1Thenavaseshof8132is5.Thenavaseshof2357is8

Letusseehow touse techniquesprovidedbyVedicmaths tocompute thenavaseshforanynumber.Wewilldenoteitby‘N’.

a)BasicMethod

Tocomputethenavasesh(N)addallthedigitsinthenumber.

Ifwegetasingledigitastheanswer,thenthatisthenavasesh(ifthesumis0or9,thenthenavaseshis0).If there ismore thanonedigit in the sum, repeat the process on the sumobtained.Keeprepeatingtillasingledigitisobtained.

Examples:

Navaseshof20canbecomputedbyaddingthedigits2and0togettheresultas2.

Navaseshof8132issimilarlyobtainedasfollows:Pass1:8+1+3+2gives14,whichhasmorethanonedigit.

Pass2:1+4gives5whichistherequirednavasesh.

b)FirstEnhancement

If there isoneormore‘9’ in theoriginalnumber, theseneednotbeaddedand can be ignored while computing the sum of the digitsExample : If thenumberis23592,then‘N’canbecomputedasfollows.

Pass1:2+3+5+2=12(ignore9)Pass2:1+2=3.Sothenavaseshis3.

c)SecondEnhancement

Whilesummingupthedigits,ignoregroupswhichaddupto9.

Example:Ifthenumberis23579,then‘N’canbecomputedasfollows.

Ignore‘9’asexplainedabove

Ignore2+7(ie1stand4thdigitswhichaddupto‘9’)andaddonly‘3’and‘5’togetthe‘N’as8.

d)FinalCompactMethod

Asbefore,ignoreall‘9’sandgroupswhichaddupto9.

Whilesumminguptheremainingdigits,ifthesumaddsuptomorethantwodigitsatanyintermediatepoint, reduce it toonedigit there itselfbyaddingup

thetwodigits.Thiswilleliminatetheneedforthesecondpasswhichisshownintheexamplesabove.

Example:Ifthenumberis44629,then‘N’canbecomputedasfollows.

Ignore‘9’inthelastposition.

Addtheremainingdigitsfromtheleftwherethefirstthreedigits,viz4+4+6addupto14.

Since,thishastwodigits,weaddthesedigitshereitselftogetasingledigiti.e.1+4=5

Nowadd‘5’totheremainingdigit‘2’togetthe‘N’as7.

Allmethodsgivethesameanswer,buttheresultcanbeobtainedfasterandwithlessercomputationwhenweusetheimprovedmethods.

II)Verificationoftheproductoftwonumbers

Thecorrectnessofanyarithmeticoperationcanbeverifiedbycarryingoutthe sameoperationon thenavaseshof thenumbers in theoperation.Oncewehave learnthow tocompute thenavaseshofanumber,wecanuse it tocheckwhethertheresultofoperationslikemultiplication,additionandsubtractionontwo(ormore)numbersiscorrectornot.

Letusseeanexample.

Let’staketheproductof38×53whichis2014.

Howdoweverifythecorrectnessoftheanswer‘2014’?

Let’stakethe‘N’ofeachofthemultiplicandsandoftheproductandseetherelationbetweenthesameNow,N(38)=2..Navaseshofthefirstnumber

N(53)=8..Navaseshofthesecondnumber

N(2014)=7..Navaseshoftheproduct

Considertheproductofthe‘N’ofthe2multiplicandsi.e.N(38)×N(53)=2×8=16whosenavaseshinturnis7.

The‘N’oftheproductisalso‘7’!!

Thus theproduct of the navaseshof the twomultiplicands is equal to thenavaseshoftheproductitself.Thissimpletestcanbeusedtoverifywhethertheansweriscorrect.

AnotherExample

22.4×1.81=40.544

Onignoringthedecimalpoints,thenavaseshofthetwonumbersareN(224)=8and

N(181)=1andtheproductofthenavaseshis8.

Thenavaseshof theproduct isalsoN(40544)=8andhence theanswer isverified.Thepositionofthedecimalpointcanbeverifiedeasily.

III)Verificationofthesumoftwonumbers

Similarly,let’sseethecaseofthesumofthetwonumbers38+53whichis91.

WeseethatN(38)+N(53)=2+8=10whichinturnhasanavaseshof1.

Thereforethenavaseshofthesumofthenavaseshis1.

The‘N’ofthesum(91)isalso1!!

Thusthesumofthenavaseshofthetwonumbersisequaltothenavaseshofthesumitself.

IV)Verificationofthedifferenceoftwonumbers

Let’sseethecaseofthedifferenceofthetwonumbers.

53-38=15N(53)–N(38)=8-2=6N(15)=6

Thus the difference of the navasesh of the two numbers is equal to thenavaseshofthedifferenceitself.

Sometimes,thedifferenceofthenavaseshmaybenegative.Then,itcanbeconvertedtoapositivevaluebyadding9toit.

ThusN(–5)=4,N(–6)=3.

Letustakeanotherexample,viz

65-23=42N(65)=2N(23)=5N(65)–N(23)=–3,

Asexplainedabove,N(–3)=6

Thenavaseshoftheresultshouldbe6.

Weseethatthenavaseshof42isindeed6.

Hence,wecanusethismethodtoverifythedifferenceoftwonumbers.

V)Verificationofthesquareorcubeoftwonumbers

Let’stakethecaseofsquaringanumber,say,34.

342=1156

Nowtakethenavaseshonbothsides,i.e.navaseshof34is7andnavaseshof1156is4.

Letussquarethenavaseshoftheleftsidei.e.72=49whosenavaseshisalso4.

Thusthesquareofthenavaseshofthegivennumberisequaltothenavaseshof the square. The reader can try verifying for the cube of any number byhimself.

VI)Limitationintheverificationprocess

Thisverificationmethodhasalimitationwhichhastobekeptinmind.

Consideragaintheexampleof38×53=2014

WeknowthatN(38)×N(53)shouldequalN(2014)

Ontheleftside,wehave2×8=16which,inturn,givesanavaseshof7.

Now, if theproducton theRHSwasassignedas2015, its ‘N’wouldbe8fromwhichwewouldcorrectlyconcludethattheresultiswrong.

Butiftheanswerwaswrittenas2041insteadof2014,the‘N’wouldstillbe7,leadinguswronglytotheconclusionthattheproductiscorrect.

So, we have to remember that this verification process using navasesh,wouldpointoutwronglycomputedresultswith100%accuracybuttherecanbemore than one result which would pass the ‘correctness’ test.We have to becarefulaboutthis.

Inspiteofthislimitation,itremainsaverypowerfulmethodforverificationofresultsobtainedbyanyofthethreemainarithmeticoperations.

VII)Computationofthequotientondivisionby9

Letusnowseehowtogetthequotientondividinganumberby9.InVedicmaths,thedivisionisconvertedtoasimpleadditionoperation.

a)Method1

Thefirstmethodconsistsofaforwardpassfollowedbyabackwardpass.

Example:Divide8132by9

Steps:

Placeacolonbeforethelastdigit,whichshowsthepositionseparatingthequotientandtheremainder.

Startfromtheleftmostdigitandbringitdownasitis.

Carryit(8)tothenextcolumnandaddittothedigitinthatcolumni.e.1+8andbringdown9.

Carry9tothenextcolumn,additto3andbringdown12.

Repeatbycarrying12tothenext(last)column,addto2andbringdown14asshown.

After reaching the end,wework backwards from the right to the left bycomputing the remainder in the last column and by carrying the surplusdigitstotheleft.Wehavetoretainonlyonedigitateachlocation.

Examinethelastcolumn,whichshouldholdtheremainder.Ifthenumberismorethan9,wesubtractthemaximummultiplesof9fromit,whichgivesthequotientdigittobecarriedtotheleft,leavingtheremainderbehind.

Since the number in the last column (14) contains onemultiple of 9,wecarry1tothelefthanddigit(12)andretain5(14-9)asfinalremainder.

Attheendoftheforwardpass,thedigitslookedasfollows:8.9.12:14.

whichnowbecome

8.9.13:5.

Nowretain3andcarry1tothelefttoget10.

8.10.3:5.

Next,retain0andcarry1tothelefttoget9.

9.0.3:5.

Sothefinalcomputationwouldappearasfollow

The quotient is 903 and the remainder is 5.Notice that the steps are verysimpleandverylittlementaleffortisrequiredtoobtaintheanswer.

b)Method2

Here,wewillcomputetheresultintheforwardpassitself.Duringcolumn-wisecomputation,ifthedigitinanycolumnbecomes9ormore,wecantransferthemultiplesof9totheleftandretaintheremainderinthecolumn.

Intheaboveexample,atthesecondcolumn,whenweget9,weretainzero(9 - 9) in the column and carry one to the left. We then take 0 to the nextcolumn.

Here,wewillsubtract9from9toretain0andcarry1totheleft.Onadding1to8,wewillget9andthecomputationwilllookasshownbelow.

This eliminates the need for the backward pass and increases the overallspeedofcomputation.

AnotherExample

Considerthedivisionof8653by9

So,thequotientis961andtheremainderis4.

3.Examplesfornavasesh

OPERATIONSWITH11

Inthischapterwewillconsidertwodistinctoperationswiththenumber11.Firstly,wewillseehowtomultiplyanynumberofanylengthby11andwritethe result, orally in a single line. Secondly, we will see how to test thedivisibility of a given number by 11, i.e. whether a given number is fullydivisible by 11 or not. These are useful techniques and if while solving aproblem, one of the intermediate steps leads to multiplication by 11 or itsmultiple,thistechniquecanbeusedtosolvetheproblemquickly.

Oncethetechniqueformultiplicationby11ismastered,multiplicationofanumberby22,33,44etc,amultipleof11,canbecarriedoutquicklybysplittingthemultiplicandas11×2,11×3or11×4.Thegivennumbercanbemultipliedby11andthentheproductcanbemultipliedfurtherbytheremainingdigit,i.e.2or3,asthecasemaybe.

Multiplication

I.Letusconsiderthemultiplicationof153by11.

Thetraditionalmethodiscarriedoutasfollows:

Thefaster,onelinemethodisasfollows:

Weobtaineachdigitof theresultbyaddingpairsofnumbersasexplainedbelow:Steps

Startfromtherightandwritedownthedigit3directly

Formpairsofconsecutivedigits,startingfromtherightandwritedownthesum,oneatatime.

Startfromtheright,formthefirstpairconsistingof3and5,addthemandwritedowntheresult,i.e.writedown8

Movetotheleft,formthenextpairbydroppingtheright-mostdigit3andincludingthenextdigiti.e.1.

Add5and1andwritedowntheresult6.

Now,ifwemoveleftanddropthedigit5,weareleftonlywithonedigit,i.e.theleft-mostdigit.

Writedowntheleftdigit1directly.

Thusthefinalansweris1683.Isn’tthatrealfast?

II.Letusnowmultiply25345by11

Theresultisobtainedstepbystepasshownbelow:Steps

Thisgivestheresultof25345×11=278795

III.Let’stryanotherexample:157×11

Wewillusethesametechniqueasfollows:

Here,theadditionof7and5gives12,whereinweretain2andcarry1totheleft.Thenextpairconsistsof5and1whichonadditiongives6.Addthecarrydigit1toittoget7.

Sothefinalansweris1727,whichcanbeverifiedbythetraditionalmethod.

If the multiplication is between decimal numbers, we carry out themultiplication as if there are no decimals and then simply place the decimalpointattherightplace.

4.Examplesformultiplicationby11

Divisibilitytestofnumbersby11

Howdowetestwhetheranumberisdivisibleby11?

Example:Is23485divisibleby11?

Steps

Startfromtheright(unit’splace)andaddallthealternatenumbersi.e.allthedigitsintheoddpositionsviz.5,4and2whichgives11.

Now,startfromthedigitinthetensplaceandaddallthealternatenumbersi.e.allthedigitsintheevenpositionsviz.8and3whichgives11.

Ifthedifferenceofthetwosumsthusobtainediseitherzeroordivisibleby11,thentheentirenumberisdivisibleby11.

In the example above, the difference is zero and hence the number isdivisible.

Let’stakeanotherexampleandcheckthedivisibilityof17953.

Thesumofthedigitsintheoddpositionsis3+9+1whichisequalto13.

Thesumofthedigitsintheevenpositionsis5+7whichisequalto12.

Sincethedifferenceof thesetwosumsis1whichisneither0nordivisibleby11,thegivennumberisnotdivisibleby11.

Ifthegivennumberis17952insteadof17953,thetwosumsare12(2+9+1)and12(5+7).Thedifferenceof the twosums iszeroandhence thegivennumberisdivisibleby11.

Exercises

1.Is8084divisibleby11?Ans(No)2.IfX381isdivisibleby11,whatisthesmallestvalueofXAns(7)

Multiplicationwith111

Wecanusethesametechniquetomultiplyanumberby111.Aswith11,weproceed with the computation from the right starting with 1 digit andprogressivelyincreasingthedigitstillwereach3digits.Anexamplewillmakeitclear.

E.g.4132×111

Theresultisobtainedstepbystepasshownbelow:

Steps

Thisgivestheresultof4132×111=458652

Thetechniquecanbeextendedfornumberslike1111,11111etc.

MULTIPLICATION(NIKHILAM)

Multiplicationofnumbersplaysaveryimportantroleinmostcomputations.Vedicmaths offers twomain approaches tomultiplying2 numbers.These arethe

NikhilammethodandUrdhvaTiryakmethod

IntheNikhilammethodwesubtractacommonnumber,calledthebase,fromboth the numbers. The multiplication is then between the differences soobtained.

Inthenextchapterwewilltakeupthe‘UrdhyaTiryak’method,whichusescross-multiplication.

Let us start the study of the Nikhilam method now by consideringmultiplication of numberswhich are close to a base of 100, 1000, 10000 etc.Lateron,wewillseethemultiplicationofnumbersclosetobaseslike50,500,5000etc.andfinallyanyconvenientbase.

Forabase100,thenumberswhicharebeingmultipliedmaybe

lessthanthebase(e.g.96,98)morethanthebase(e.g.103,105)mixed(e.g.96&103)

Case1:Integerslessthanthebase(100)

Consider the multiplication of 93 × 98. Let’s look first at the traditionalmethodandthenatVedicmethod.

VedicMethod

Write the twonumbersonebelowtheotherandwrite thedifferenceof thebase (100) from each of the number, on the right side as shown:

Takethealgebraicsumofdigitsacrossanycrosswise-pair.e.g.93+(-2)or98+(-7).

Bothgivethesameresultviz.91,whichisthefirstpartoftheanswer.

Thisisinterpretedas91hundredsor9100,asthebaseis100.

Now, justmultiply the twodifferences (–2)× (–7) to get 14,which is thesecondpartoftheanswer.

Thefinalansweristhen9114,whichcanalsobeinterpretedas9100+14.

ImportantPoints

Sincethebaseis100,weshouldhavetwodigits(equaltonumberofzeroesinthebase)inthesecondpartoftheanswer.

If thenumberofdigits isone, thenumber ispaddedwithonezeroontheleft.

If the number of digits ismore than two, then the leftmost digit (in thehundred’sposition)iscarriedtotheleft.

Themethodisequallyeffectiveevenifonenumberisclosetothebaseandtheotherisnot.

Case2:Integersabovethebase(100)

Considerthemultiplicationof103×105.Thetechniqueremainsexactlythe

sameandcanbewrittenasfollows:

Noticethatherethedifferencesarepositive,i.e.+3(103-100)and+5(105-100).

Theadditioninthecrosswisedirectiongives108(103+5or105+3).

Theproductofthedifferencesis15andsothefinalresultis10815.

Onceagain,wehavetoensurethatwehaveexactlytwodigitsinthesecondpartoftheanswer,whichisachievedbyeitherpaddingwithzeroesorbycarrytotheleft.

E.g.Considerthefollowingtwocases

Here, 13 and15havebeenmultipliedbyusing the same technique,wherethebasehasbeentakenas10.

Thissametechniquecanalsobeusedtofindthesquaresofnumbersclosetothebase,e.g.962or1042.

Case3:Integersbelowandabovethebase(100)

Considerthemultiplicationof97×104.

Here97isbelow100while104isabove100.Usingthesametechnique,weget

Herethe2ndpartoftheansweris-12.Wenowcarry1fromtheleftwhichisequivalent to 100 (since the base is 100) and the final answer is obtained asfollows:101(–12)=100(100-12)

=100/88=10088.

Thiscanalsobeinterpretedas10100-12=10088.

Use the technique explained in chapter 1 (all from 9 and last from 10) tosubtractfrom100.

Onceagain,the2ndpartofthenumbershouldhaveexactlytwodigits,sincethebasehas2zeroes.

Considerthefollowing2examples:

132 can be obtained by multiplying 12 by 11 as explained in Chapter 2.Since 132 ismore than 100, we have to carry 2 from the left, whichwill beequivalentto200(2×100).

Hencetheresultwouldbe

=99/(200-132)=9968

Orasbefore,wecangettheanswerbyinterpretingitas

10100-132=9968.

Case4:Examplesofdecimalnumbers

Consider9.8×89.Wewillassumethenumbers tobe98and89andcarry

outthemultiplicationasshownbelow:

Now,sincetheoriginalnumbershaveonedigitinthedecimalplace,weputthedecimalpointonedigitfromtherightandgetthefinalansweras872.2.

Case5:Exampleswithbasesof1000and10000

Considertheexample997×993,wherethebaseis1000.

Theprocedureremainsthesame.

Thefirstpartoftheansweri.e.990isobtainedbythecrosswisesum,993-3or997–7.

Thesecondpartisobtainedbymultiplying3and7andpadding21with‘0’sinceweshouldhave3digits.

Hencethefinalansweris990021.

Now,considertheexample9998×10003,wherethebaseis10000.

Thefirstpartoftheansweri.e.10001isobtainedbythecrosswisesum,9998+3or10003-2.

Thesecondpart isobtainedbymultiplying–2and+3andpadding–6with‘000’sinceweshouldhave4digits(numberofzeroesinbase10000).Wenowcarry1fromtheleft,(equivalentto10000)andsubtract6fromittoget9994.

Asbefore,thisis100010000-6.

Hencethefinalansweris10000/9994i.e.100009994.

The method for multiplying decimal numbers has been explained abovewherein we first consider the numbers without decimal points, multiply themusingthistechnique,andthenplacethedecimalpointattheappropriatelocation.

Case6:Otherbases—50,250,500etc

Let us now consider themultiplication of numberswhichmay be close tootherbaseslike50,250,500etc.whicharemultiplesof50.Later,wewillseehowthemethodcanbeextendedtobaseslike40,whicharenotmultiplesof50.Thereaderwillthenbeabletoextendthemethodtoanybaseofhischoice.

Thesolvedexamplesaregroupedunder:

i.Secondarybaseof50ii.Secondarybaseof250iii.Secondarybaseof500iv.Secondarybaseof40and60v.Secondarybaseof300

I.Secondarybaseof50

Example1-Considerthemultiplicationof42×46

Here,wecantakethesecondarybaseas50.Thecorrespondingprimarybasecan be considered as 10 or 100. The scaling factor between the primary andsecondarybasewouldbeasfollows:Factor=Secondarybase/Primarybase

5=50/10(ifprimarybase=10)

1/2=50/100(ifprimarybase=100)

We can use either of these scaling operations and compute the requiredanswer.Let’sseeeachcaseindetail.

A)Case1-Primarybase10andsecondarybase50for42×46

Secondarybaseis50Scalingoperation:Factor=5=50/10

(secondarybase÷primarybase)One digit to be retained in the right hand part of the answer becauseprimarybasehasonezero.

Theinitialpartofthemethodremainsthesameandisasfollows:Consider:42×46(subtract50fromeachnumber)

Now, we have to scale up the left part of this intermediate result bymultiplyingby5.

Since38×5=190,theanswerbecomes190/32.

Sincetheprimarybase(10)containsonezero,wewillretainonedigitintherightpartoftheanswerandcarry3totheleft.

So,thefinalanswerbecomes1932.

B)Case2-Primarybase100andsecondarybase50for42×46

Secondarybaseis50Scalingoperation:Factor=1/2=50/100

(secondarybase÷primarybase)Twodigitstoberetainedintherighthandpartoftheanswer.

Thetechniqueremainsthesameandisasfollows:

Consider:42×46(subtract50fromeachnumber)

Now, we have to scale down the left part of this intermediate result bymultiplyingitby1/2.

Since38/2=19,theanswerbecomes19/32=1932.

Sincetheprimarybaseis100,whichcontainstwozeroes,wewillretaintwodigitsintherightpartoftheanswer.

So,thefinalanswerbecomes1932.

Wow,isn’tthatwonderful?Boththeoperationsgiveusthesameresult.

SpecificSituations

Letusnowconsiderafewspecificsituations:

Afractionisobtainedduringthescalingoperationand/orTherightpartisnegative

Wewilltakeanexamplehavingboththesesituations.


C)Case1-Primarybase10,secondarybase50for52×47


(secondarybase÷primarybase)Onedigittoberetainedintherighthandpartoftheanswer


Wewillscaleupthisintermediateresultbymultiplying49by5.

Since49×5=245,theanswerbecomes245/(–6)

Sincetheprimarybaseis10,wewillcarryonefromtheleftwhichisequalto10andsubtract6fromit.

Sothefinalanswerbecomes244/4=2444.

Thiscanalsobeconsideredas2450-6=2444.

D)Case2-Primarybase100andsecondarybase50for52×047

Secondarybaseis50ScalingOperation:Factor=1/2=50/100

(secondarybase÷primarybase)Twodigitstoberetainedintherighthandpartoftheanswer

Consider:52×47(Subtract50fromeachnumber)

Wewillscaledownthisintermediateresultbydividing49by2.

Since49/2=24.5,theanswerbecomes24½/(–6)

Sincetheprimarybaseis100,wewillcarry1/2fromtheleft.Thisisequaltoacarryof50(primarybase100/2).Wewillsubtract6from50giving44.


Since 24.5 hundreds is 2450, this can be interpreted also as 2450 / (–6) =2444

Wegetthesameanswerirrespectiveoftheprimarybase.Thefractionalpartcanbeavoidedifwechooseascalingoperationinvolvingmultiplicationinsteadofdivision.

II.Secondarybaseof250


E)Primarybase1000andsecondarybase250for235×266


(secondarybase÷primarybase)Threedigitstoberetainedintherighthandpartoftheanswer


Considerthesteps


Since251/4=62¾,theanswerbecomes62¾/(–240).

Sincetheprimarybaseis1000,wewillcarry3/4fromtheleftwhichisequaltoacarryof750(primarybase1000*¾)andsubtract240from750giving510.


Or,itcanbeconsideredas62750-240=62510.

III.Secondarybaseof500


F)Primarybase100andsecondarybase500for532×485



Consider:532×485(Subtract500fromeachnumber)

Considerthesteps


Since517×5=2585

Sotheanswerbecomes2585/(–480).

Sincetheprimarybaseis100,wewillcarry5fromtheleftwhichisequaltoacarryof500(primarybase100*5)andsubtract480from500giving20.


Or,itcanbeconsideredas258500-480giving258020.

G)Primarybase1000andsecondarybase500for532×485


(secondarybase÷primarybase)Threedigitstoberetainedintherighthandpartoftheanswer

Considerthesteps


Since517/2=258½,theanswerbecomes258½/(–480).

Sincetheprimarybaseis1000,wewillcarry½fromtheleftwhichisequaltoacarryof500(primarybase1000*½)andsubtract480from500giving20.


Theleftpartoftheansweris258.5hundredswhichisequalto258500.

So,itcanbeconsideredas258500-480giving258020.

IV.Secondarybaseslike40,60


H)Primarybase10andsecondarybase40for32×48





Since40×4=160,theanswerbecomes160/(–64).


Thisisalsoequivalentto1600-64.

So,thefinalansweris1536.


I)Primarybase10andsecondarybase60for64×57

Secondarybase=60Scalingoperation:Factor=6=60/10




Since61×6=366,theanswerbecomes366/(–12)=364/8.


Thisisalsoequivalentto3660-12.


V.Secondarybaseof300

Example-Considerthemultiplicationof285×315

J)Primarybase100andsecondarybase300for285×315

Secondarybase=300Scalingoperation:Factor=3=300/100




Since300×3=900,theanswerbecomes900/(–225).



Conclusion

Thissectiongivesanideaofhowavarietyofcombinationsofprimaryandsecondary bases can be used to arrive at the final answer quickly. The readershould choose the secondary base carefully which can reduce the time forcomputationdrastically.

5.Examplesformultiplicationofnumbersbelowthebase(100)

6.Examplesformultiplicationofnumbersabovethebase(100)

7. Examples for multiplication of numbers above and below the base(100)

8.Examplesforsquaresofnumbersbelowthebase(100)

9.Examplesforsquaresofnumbersabovethebase(100)

10. Examples for multiplication of numbers (3-digits) below the base(1000)

11. Examples for multiplication of numbers (3-digits) abov the base(1000)

12.Examplesformultiplicationofnumbers(3-digits)above&belowthebase(1000)

13.Examplesforsquaresofnumbers(3-digits)belowthebase(1000)

14.Examplesforsquaresofnumbers(3-digits)abovethebase(1000)

15. Examples for multiplication of numbers (4-digits) below the base(10000)

16. Examples for multiplication of numbers (4-digits) above the base(10000)

17.Examplesforsquaresofnumbers(4-digits)belowthebase(10000)

18.Examplesforsquaresofnumbers(4-digits)abovethebase(10000)

19.Examplesformultiplicationofmixednumbers

MULTIPLICATION(URDHVATIRYAK)

VerticallyandCrosswise

Wewill now consider the othermethod ofmultiplication viz. the ‘UrdhvaTiryak’ or cross-multiplication technique. It consists of vertical and crossmultiplication between the various digits and gives the final result in a singleline.

Wewillconsiderthemultiplicationofthefollowing:

Two2-digitnumbersTwo3-digitnumbersTwo4-digitnumbers2-digitwith3-digitnumber2-digitwith4-digitnumber3-digitwith4-digitnumber

Once the technique isclear, itcanbeextended to themultiplicationofanynumberofdigits.

Case1:2-digitmultiplication

Considerthefollowingmultiplicationasdonebythetraditionalmethod.

TheVedicmathsmethod consists of computing three digits a, b and c as

showninthediagram

Steps:

Digit(a):Multiplytherightmostdigits8and2toget16.

Digit(b):Carryoutacrossmultiplicationbetween3and2,and8and5toget6and40respectivelyandonaddingthem,weget46.

Digit(c):Multiplytheleftmostdigits3and5toget15.

We have to retain one digit at each of the three locations and carry thesurplustotheleft.

Hencein

154616

westartfromtheright,retain6andcarry1totheleft.Thisgives47inthemiddlelocation.

Now,weretain7andcarry4totheleft.Weadd4to15toget19.

Thisgives1976i.e.1976whichistherequiredanswer.

To speed up this operation, the carry digit can be written just below thenumber on the left. It can then be added to the number to it’s left at eachoperationitself.Let’sseethemultiplicationagain.

Here,therightmostverticalmultiplicationiscarriedoutfirsttoget8×2=16.Wewrite6andwritethecarry1asshown.

Nextwecarryoutthecrossmultiplication3×2+5×8=46,addthecarry(1)toitandget47.Wewritedown7andthecarrydigit4asshown.

The3rdproductis3×5whichgives15towhichweaddthecarrydigitof4andget19.

Asyoucansee,themethodgivesyouthefinalanswerfasterinjustoneline.

Let’stryanotherexample.

Steps:

Digit(a):Multiplytherightmostdigits6and3toget18.Wewrite8andshow1asthecarryonit’sleft.

Digit(b):Carryoutacrossmultiplicationbetween7and3,and6and4toget21and24respectivelyandaddthemtoget45.Addthecarrydigit1andget46.Writedown6andshowthecarryas4.

Digit (c) :Multiply the left-most digits 7 and 4 to get 28.Add the carrydigit4toittoget32.

20.Examplesfor2digitcrossmultiplication


Letusnowconsidermultiplicationoftwo3digitsnumbers.

Considerthefollowingexample

TheVedicmathsmethodconsistsof computing fivedigits a, b, c, d, e as

showninthediagram

Wewill use themethod explained in the previous case inwhich the carrydigitateachstepwasaddedtothenumbertoitsleft.

Steps:

Digit(a):Multiplytherightmostdigits6and8toget48.Wewrite8andshow4asthecarry,onitsleftside.

Digit (b) :Carryout a2-digit crossmultiplicationbetween2 and8 (16),and6and5(30).Addtheproducts16and30toget46.Addthecarrydigit4andget50.Writedown0andshowthecarryas5.

Digit (c) :Carry out a 3-digit crossmultiplication between3 and 8 (24),and6and2(12)and2and5(10).Addthethreeproducts24,12and10toget46.Addthecarrydigit5andget51.Writedown1andshowthecarryas5.

Digit(d):Carryouta2-digitcrossmultiplicationbetween3and5(15)and2and2(4).Add15and4toget19.Addthecarrydigit5toget24,Writedown4andshow2asthecarry.

Digit(e):Multiplytheleftmostdigits3and2,toget6.Addthecarrydigit2toittoget8.

Finally,themultiplicationlooksasfollows:

Let’stryanotherexample;thecomputationisasshown.

Case3:Multiplying3-digitand2-digitnumbers

Ifwehave tomultiplya3-digitwitha2-digitnumber,wecan treat the2-digitnumberasa3-digitnumberbypaddingitwithazeroontheleft.

E.g. 354×45canbewrittenas354×045and the same techniquecanbeused.

21.Examplesfor3digitcrossmultiplication

22.Moreexamples3digits×2digits


The techniques discussed till now can be extended further for four digits.

Thediagrambelowcanbeused

Let’sconsiderthefollowingexample

Thefinalansweris9073792.

Steps:

Digit(a):Multiplytherightmostdigits2and6toget12.Writedown2

andshowthecarryas1.

Digit (b) :Carryout a2-digit crossmultiplicationbetween3 and6 (18),and2and5(10)andaddthemtoget28.Addthecarrydigit1andget29.Writedown9andshowthecarryas2.

Digit (c) :Carryouta3-digitcrossmultiplicationbetween1and6 (6),2and2(4)and3and5(15)andaddthemtoget25.Addthecarrydigit2andget27.Writedown7andshowthecarryas2.

Digit(d):Carryouta4-digitcrossmultiplicationbetween2and6(12),2and4(8),1and5(5)and3and2(6)andaddthemtoget31.Addthecarrydigit2toget33.Writedown3andshow3asthecarry.

Digit(e):Carryouta3-digitcrossmultiplicationbetween2and5(10),3and4(12)and1and2(2)andaddthemtoget24.Addthecarrydigit3andget27.Writedown7andshowthecarryas2.

Digit(f):Carryouta2-digitcrossmultiplicationbetween2and2(4),and1and4(4)andaddthemtoget8.Addthecarrydigit2andget10.Writedown0andshowthecarryas1.

Digit(g):Multiplytheleftmostdigits2and4toget8.Addthecarrydigit1toittoget9.

Let’s try another example; the computation is as shown:

Thefinalresultis19745313.

Case5:Multiplying4-digitand3-digitnumbers

Ifwehave tomultiplya4-digitwitha3-digitnumber,wecan treat the3-digitnumberasa4-digitnumberbypaddingitwithazeroontheleft.

E.g.3542×453canbewrittenas3542×0453andthesametechniquecan

beused.

23.Examplesfor4digitand5-digitcrossmultiplication

24.Examplesformultiplication4-digitsand5-digitsw3-digits

DIVISIONThe operation of division is basically the reverse of multiplication. In the

traditionalapproach,itisaslowandtediousmethodwithaconsiderableamountofguesswork.Ateachstep,anapproximatequotientdigit isguessed,which isthenmultipliedbythedivisorandtheremainderiscomputed.Iftheremainderisnegative or contains additional multiples of the divisor, the quotient digit isrecomputed. As the number of digits in the divisor increase, the divisionbecomesprogressivelyslower.

Vedicmathsprovidesaveryelegantmethodwhereinthedivisionisreducedtoadivisionbyasingledigitinmostcasesoratmosttotwosmalldigitslike12,13etc.inafewcases.Ateachstep,thedivisionbya1-digitor2-digitdivisorisfollowedbyamultiplication,basedonthe‘UrdhavTiryak’approachandfinallyasubtraction.

The reduction in both time and mental strain using the Vedic mathstechniqueisastounding.

Considertheexampleofdividing87108by84whichisa2-digitdivisor.The

traditionalmethodisshownbelow:

Ateachstage,wehavetomentallycarryoutatrialdivisionandthenwritedownthequotientwhichgiveseitherapositiveorzeroremainder.Thisisaslowand tedious procedure and hence the operation of division is detested.As thenumber of digits increase in the divisor, the process becomesmore andmoreslowandtimeconsuming.

TheVedicmethodhasbeendevisedbeautifullyinwhichthedivisionshavebeen reduced to a one-digit division. This speeds up the process considerablyandmakesitanenjoyableexperience.

Case1:Divisionbyanumberwithaflagofonedigit(noremainder)

Let’sconsider thesameexampleagainandcarryoutdivisionbytheVedicmethod.

Divide87108by84

Wewillwritethedivisoras84where4iscalledtheflag.Thedivisionwillbecarriedoutbyasingledigitviz.8.

Steps:

Placeacolononedigitfromtheright(=numberofdigitsofflag)

Divide8by8togetquotientas1andremainderas0

Carrytheremainder0tothenextdigit7andget7(calledgrossdividend,GD)

Multiplythequotient1bytheflag4toget4andsubtractitfrom7toget3(callednetdividend,ND)

Divide3by8togetthequotientas0andremainderas3

Carrytheremainder3tothenextdigit1andget31(GD)

Multiplythequotient0bytheflag4toget0andsubtractitfrom31toget31(ND)


Carrytheremainder7tothenextdigit0andget70.

Multiplythequotient3bytheflag4toget12andsubtractitfrom70toget58


Atthispoint,theintegerpartofthefinalquotienthasbeenobtainedsincewehavereachedthecolon.Allthebalancedigitswouldbethedecimalpart

Carrytheremainder2tothenextdigit8andget28


Since the difference is now 0, the division is completed and we get theansweras1037

Importantpoints:

Ateach stage, thenetdividend isobtainedbymultiplying thequotient inthepreviouscolumnbytheflagdigit

Thenetdividendwasalwayspositive

Later,wewill seewhat adjustments have to bemade if the net dividendbecomesnegativeinanycolumn.

SecondExample

Divide43409by83

Steps:









Carry the remainder 0 to the next digit 9 and get 09. At this point, theintegerpartof thefinalquotienthasbeenobtainedsincewehavereachedthecolon.Allthebalancedigitswouldbethedecimalpart.


Since the difference is now 0, the division is completed and we get theansweras523

Trya)4341283andb)4340373

Case2:Divisionbyanumberwithaflagofonedigit(withremainder)


Thequotientis342andtheremainderis48,whichappearsunderthecolumnjustafterthecolon.Thefinalanswerindecimalformis342.623.

Steps:

Thestepsasexplainedabovearecarriedouttillwereachthecolumnafterthecolonwhichshowstheremainderas48.Ifwewanttogettheanswerindecimal form, we have to carry out division further by using the sameprocedure.


Now,weattachazero to thedividendandwrite the remainder to its left,givingthenextnumberas60.



Carrytheremainder4tothenextdigit0(attachedasbefore)andget40



Thisprocesscanbecarriedoutuptoanydesiredlevelofaccuracy.

Case3:Divisionwithadjustments


Steps:




Multiplythequotient3bytheflag7toget21andsubtractitfrom43toget22(ND)



Multiplythequotient3bytheflag7toget21andsubtractitfrom18toget-3(ND)

Now,wehaveapeculiarsituationwhereinthe‘NetDividend’hasbecomenegative.Theprocesscannotgoanyfurther.Wewillnowhavetomakeanadjustment.

Wewillgoonestepbackwardandinsteadofwritingthequotientas3whiledividing 22 by 7, we will write down only 2 as the quotient and carryforwardabiggerremainderof8(ie22-7*2)


Multiplythequotient2bytheflag7toget14andsubtractitfrom88toget74(ND)whichisnowpositive

Carry out the process in the remaining columns, making an adjustmentwheneverthenetdividendbecomesnegative.

Thefinalresultis329withremainder49orindecimalform,itis329.6363.

Case4:Divisionbyanumberwithaflagof2digits

Letusnowconsiderdivisionbya3digitdivisorwheretheflagwouldhavetwodigits.


Steps:

Placeacolontwodigitsfromtherightsincewenowhavetwodigitsintheflag



Now,sincetherearetwodigitsintheflag,wewillusetwoquotientdigitsfromtheprevioustwocolumnstocarryoutacrossmultiplicationwiththetwo digits of the flag. We will then subtract the result from the gross

dividend(GD)tocomputethenetdividend(ND).

Sincethequotient4hasonlyonedigit,wepaditwithazeroontheleftand

multiplythequotient04bytheflag14asfollows:

Wewillsubtract4fromtheGD32toget28(ND)



Now,theprevioustwoquotientdigitsare4and5whicharemultipliedby

theflag14asfollows:

Wewillsubtract21fromtheGD39toget18(ND).

Repeattheprocessfortheremainingcolumns.

Case5:Divisionbyanumberwithaflagof3digits

Letusnowconsiderdivisionbya4-digitdivisorwheretheflagwouldhavethreedigits.


Steps:

Placeacolonthreedigitsfromtherightsincenowtherearethreedigitsintheflag



Now, since there are three digits in the flag, we will use three quotientdigits from the three immediatelypreviouscolumnsandcarryout a crossmultiplicationwith the three digits of the flag.Wewill then subtract theresultfromthegrossdividend(GD)tocomputethenetdividend(ND).

Sincethequotient1hasonlyonedigit,wepaditwithtwozeroesontheleftand multiply the quotient 001 by the flag 524 as follows :

Wewillsubtract5fromtheGD8toget3(ND)



Now,theprevioustwoquotientdigitsare0and1.Wepadthemwithone

zeroandmultiplybytheflag524asfollows:




Now,thepreviousthreequotientdigitsare1,0and5.Wemultiplybythe

flag524asfollows:


Repeattheprocessfortheremainingcolumns.

25.Examplesfordivision-flagof1digit

26.Examplesfordivision-flagof2digits

27.Examplesfordivision-flagof3digits

SIMPLESQUARESInthischapterwewilllookattwosimpletechniques.Inthefirstone,wewill

learnhowtofindthesquareofanumberwhichendsin5,e.g.35.Inthesecondone,wewillseehowtofindtheproductoftwonumberswhichhavethesamestartingdigitandwhichhavethesumofthelastdigitas10e.g.33and37.

Case1:Numbersendingin5

Considerthesquareof35.Thetechniqueisverysimpleandconsistsoftwosteps.

Steps:

Multiplythefirstnumberviz.3bythenumbernexttoiti.e.multiply3by4andwritedowntheresultas12.Next,multiplytheseconddigiti.e.5byitselftoobtain25andattachitattheendoftheresultobtainedinthepreviousstep.

Theresultis1225.

Example-1

Compute652

Step1:6×7=42

Step2:5×5=25

Andthefinalansweris4225.

Example-2

Let’scompute1252

Step1:12×13=156

Step2:5×5=25

Andthefinalansweris15625.

Theproductof12×13canbeobtainedbytheNikhilammethodasexplainedbefore.

28.Examplesforsquaresofnumbersendingin5

Case2:Twonumbersstartingwithsamedigitandendingdigitsaddingupto10

Considertheexample23×27.

Now,23×27=621

The product can be obtained by using the same technique as for squaringnumbersendingwith5.

Steps:

Sinceboththenumbersstartwiththesamedigit,multiplyit(i.e.2)byitsnextdigit(i.e.3)toget6.

Nowmultiplythelastdigitsofboththenumbersviz.3×7toget21.

Thefinalresultisobtainedbytheconcatenationofthetwonumbersi.e.‘6’+‘21’=621.

Example-1

Compute84×86

Theresultwillbeobtainedas

Step1:8×9=72(firstdigit‘8’multipliedbyitsnextdigit‘9’)Step2:4×6=24(productoflastdigitofeachnumber)givingthefinalresultas7224.

Example-2

Compute92×0.98

Theresultcanbeobtainedas

Step1:9×10=90

Step2:2×8=16

whichgivestheintermediateansweras9016.

Now,placethedecimalpointaftertwodigitsfromtherighttogetthefinal

answeras90.16.

Important

Thesametechniquecanalsobeusedforbiggernumberslike318×312or1236×1234,wherethelastdigitsaddupto10andtheremainingdigitsarethesame.

29.Examplesforproductofnumbershavingendingdigitsaddingto10

SQUAREOFANYNUMBERThere are many problems in which we have to compute the square of a

number,i.e.multiplyanumberbyitself.E.g.squareof382,2.3672etc.

Vedic maths provides a powerful method to compute the square of anynumber of any length with ease and get the answer in one line. It uses theconceptof‘dwandwa’orduplexwhichisexplainedinthischapter.

Wewillstartwithsomebasicdefinitions,whicharethebuildingblocksforthesetechniques.

I.Definition-DwandwaorDuplex

Wewilldefineatermcalled‘dwandwa’orduplexdenotedby‘D’.

a)DuplexofasingledigitisdefinedasD(a)=a2

D(3)=32=9,D(6)=62=36etc.

b)DuplexoftwodigitsisdefinedasD(ab)=2abD(37)=237=42D(41)=241=8D(20)=0

c)Duplexof3digitsisdefinedasD(abc)=2ac+b2

ThiscanbederivedbyusingD(a)andD(ab)definedabove.

Wepickupthedigitsatthetwoextremeendsi.e‘a’and‘c’andcomputeitsduplexas2ac.

Wethenmoveinwardsandpickuptheremainingdigit‘b’.

Weaddtheduplexofbi.e.b2totheresulttogettheduplexofthe3digits‘a’,‘b’and‘c’.

ThusD(346)=236+42=36+16=52

D(130)=210+32=9D(107)=217+02=14.

d)Duplexof4digitsisdefinedas

D(abcd)=2ad+2bc=2(ad+bc)Onceagain,westartfromthetwo digits ‘a’ and ‘d’, compute their duplex as 2 a d, then moveinwards,wegetanotherpair‘b’and‘c’andcomputetheirduplexas2bc.

Thusthefinalduplexis2ad+2bcD(2315)=(225)+(231)=20+6=26

D(3102)=(232)+(210)=12+0=12

D(5100)=(250)+(210)=0

e)DuplexoffivedigitsisdefinedasD(abcde)=2ae+2bd+c2

D(21354)=(2×2×4)+(2×1×5)+32=16+10+9=35

D(31015)=(2×3×5)+(2×1×1)+02=30+2+0=32

Summaryofcomputationofduplex

30.Computetheduplexofthefollowingnumbers

II.Squareofanynumber

Once the concept of ‘duplex’ is clear, we can compute the square of anynumber very easily.Let us startwith the square of a 2-digit number and thenextendittobiggernumbers.

The square of a 2-digit number ab is defined as (ab)2 = D(a)D(ab)D(b)Considerthesquareof34.

Then,342=D(3)D(34)D(4)WhereD(3)=9,D(34)=24andD(4)=16

So,342=9/24/16.

Now, we start a backward pass from the right, retain one digit at eachlocationandcarry thesurplus to the left.Theworkingwould lookasfollows:342=92416

=9256..Weretain6andcarry1totheleftgetting25.=1156..Weretain5andcarry2totheleftgetting11.

Sothefinalansweris1156.

Similarly,822=64/32/4=6724

Thetechniquecannowbeextendedtoa3-digitnumberasfollows:(abc)2=D(a)D(ab)D(abc)D(bc)D(c)Thus,2132=441369

=45369

Thesquareofanynumbercannowbeobtainedveryeasilybyjustextendingthesetofduplexes.

31.Examplesforsquaresof2-digitnumbers

32.Squaresof3-digitnumbers

SQUAREROOTOFANUMBERIn the previous chapter we have seen the concept of ‘duplex’ and how it

helpsustocomputethesquareofanynumberveryrapidly.

Thecomputationofthesquarerootalsoinvolvestheuseoftheduplex,onlythistime,wewouldbesubtractingitateachsteptocomputethesquareroot.

We will now study the technique to find the square root of any numberwhich may or may not be a perfect square. The first three examples are ofnumberswhichareperfectsquares,andthenexttwoexamplesareofimperfectsquares including a decimal number. The general steps are explained withspecificexamplesbelow:Steps:

Countthenumberofdigits,n,inthegivennumber.

Ifthenumberofdigitsiseven,startwiththeleftmosttwodigits,elsestartwith the leftmost singledigit.Letuscall thenumber soobtainedas ‘SN’(startingnumber).

Considerthehighestpossibleperfectsquare,say‘HS’,lessthanorequalto‘SN’andcomputeitssquareroot‘S’.

Writedownthesquareroot‘S’belowtheleftmostdigits.Thisbecomesthefirstdigitoftherequiredsquareroot.

Multiplythesquareroot‘S’by2andwriteitontheleftside.Thiswouldbethedivisor,say‘d’.

Forallthebalancedigitsintheoriginalnumber,themethodwouldconsistofdivisionbythedivisor‘d’aftersubtractionoftheduplexofthepreviousdigits(exceptthefirstone).

Thedifferencebetween‘SN’and‘HS’ i.e. theremainder ‘R’ iscarried tothenextdigitandwrittentoitslowerleft,forminganewnumber.Thisnextnumberiscalledthegrossdividend(GD).

The first GD is divided by the divisor ‘d’ to get a quotient Q and aremainder ‘R’. The quotient is written down as the second digit of thesquarerootandtheremainderisagaincarriedtothenextdigittoformthenextGD.

Fromthis stageonwards, theduplexofalldigits starting from theseconddigit of the square root, to the column previous to the column ofcomputation,issubtractedfromeachGDtogettheNetDividend,ND.

Ateachcolumn,theNDisthendividedbythedivisor‘d’andthequotientiswritten downas the next digit of the square root, and the remainder iscarriedtothenextdigittoformthenextGD.

TheprocessiscarriedouttilltheNDbecomeszeroandnomoredigitsareleftinthenumberforcomputationorwhenwehavereachedadesiredlevelofaccuracy.

If theND does not become zero,we can attach zeroes to the end of thegivennumberandcarryouttheprocesstogetthesquareroottoanylevelofaccuracy,i.e.anynumberofdecimalplaces.

Once thesquareroot isobtainedorwhenwedecide tostop,weplace thedecimalpointaftertherequireddigitsfromtheleft.Ifniseven,thedecimalpointisput(n2)digitsfromtheleft.Ifnisodd,thedecimalpointisplaced(n+1)2digitsfromtheleft.

Theprocesswillbecomeveryclearonstudyingtheexamplesgivenbelow.

I.Perfectsquareroot

Considertheexampleofcomputingthesquarerootof2116Theworkingisasfollowswhichisexplainedstepwise.

Steps:

Countthenumberofdigits,n,whichis4.

Sinceniseven,westartwiththetwoleftmostdigitswhere‘SN’is21

Here the highest perfect square (‘HS’) less than 21 is 16, its square root(‘S’)is4andthedifference‘R’is5.

Wewritethesquareroot4belowtheleftmostdigitsandwrite8(=2×4)ontheleftsideasshown.

4isthefirstdigitoftherequiredsquareroot.

Forthebalancedigits,themethodwouldconsistofdivisionbythedivisor8.

Thedifference i.e.5 iscarriedtothenextdigit1andwrittentoit’s lowerleftasshown.ThevalueoftheGDisnow51.

TheveryfirstGDi.e.51isdividedbythedivisor8togetaquotient6andaremainder3.Thequotientiswrittendownaspartofthefinalresultandtheremainderiscarriedtothenextdigit6toformthenextGD36.

Wehavenowreachedthethirdcolumn.WewillsubtracttheduplexofthedigitinthesecondcolumnfromtheGDtogettheNetDividend,ND.

Thedigitinthesecondcolumnoftheresultis6anditsduplexis36.

36 is subtracted from theGDwhich is also 36 to get theND as 0. Thissignalstheendoftheprocess.

Hence,thesquarerootisobtained.Now,wewillplacethedecimalpoint.

Sincenis4andiseven,thedecimalpointisplaced(n/2)digitsfromtheleft.So,here,thedecimalpointwillbeplaced2digitsfromtheleftandthefinalansweris46.

II.Perfectsquareroot

Considertheexampleofcomputingthesquarerootof59049

Steps:

Countthenumberofdigits,n,whichis5.

Sincenisodd,westartwiththeleftmostdigitviz.5

Herethehighestperfectsquarelessthan5is4,itssquarerootis2andtheremainder‘R’is1(5–4).

Wewritethesquareroot2belowtheleftmostdigitandwrite4(=2×2)ontheleftsideasshown.

Theremainder1iscarriedtothenextdigit9givingthevalueoftheGDas19.

19 is dividedby4 toget aquotient 4 and a remainder3.Thequotient iswrittendownandtheremainderiscarriedtothenextdigit0togivethenextGDas30.

Theduplexof4is16whichissubtractedfrom30(GD)toget14(ND).

14 is dividedby4 toget aquotient 3 and a remainder2.Thequotient iswrittendownandtheremainderiscarriedtothenextdigit4togivethenextGD24.

Now, the duplex of the two previous digits 4 and 3, i.e. 2 4 3 = 24 issubtracted from the GD 24 to get the ND as 0. Since we have not yetreachedthelastdigitinthegivennumber,wecannotstophere.Ondividing0by4wegetthequotientas0andtheremainderas0,whichiscarriedtothenextdigitgivingtheGDas09.

Thethreeimmediatelypreviousdigitsare4,3and0.Theirduplexis9(240+32).Whentheduplexissubtracted,theNDbecomesequaltozeroandnomoredigitsareleft.Thissignalstheendoftheprocess.

Sincenis5,thedecimalpointisplaced(n+1)/2digitsi.e.3digitsfromtheleftandthefinalansweris243.

III.Imperfectsquarerootasadecimalnumber

Considertheexampleofcomputingthesquarerootof59064

Since this number is slightly larger than the number taken in the previousexample,itisnotaperfectsquare.Wewillnowseehowtogetitssquareroottoanydesiredlevelofaccuracy.Atanypoint,iftheNDbecomesnegative,wegobackonestepanddecreasethequotientbyonetoincreasetheremainderthatiscarriedforward.Thisprocessofadjustmentisexplainedinthisexample.

Steps:

ThefirstfewstepsarethesametillwereachthefourthcolumnwheretheGDis26andtheNDis2.

2 is divided by 4 to get a quotient 0 and a remainder 2. The quotient iswrittendownandtheremainderiscarriedtothenextdigit4togivethenextGD24.

Theduplexof4,3and0is9whichissubtractedfrom24(GD)toget15

(ND).

15isdividedby4togetaquotient3andaremainder3.Weattachazerotocontinuewiththeprocess.

Thequotientiswrittendownandtheremainderiscarriedtothenextdigit0toformthenextGD30.

Now, the duplex of the previous digits 4, 3, 0 and 3 i.e. 2 4 3 = 24 issubtractedfromtheGD30togettheNDas6.

Ondividing6by4wegetthequotientas1andtheremainderas2,whichiscarriedtothenextdigitgivingtheGDas20.

Theduplexoftheimmediatelypreviousdigits4,3,0,3and1is26(241+2 3 3) which gives a negative ND. Hence, we decrease the previousquotientby1andcarryforwardtheremainder6givingtheGDas60.

Nowtheduplexofthepreviousdigits4,3,0,3and0is18(233).WhenthisduplexissubtractedfromtheGD,wegettheNDas42.

Thisprocesscanbecarriedouttoanydesiredlevelofaccuracy.

Sincen is5, thedecimalpoint isput (5+1) /2=3digits fromthe left,givingthefinalansweras243.03086.

IV.Squarerootwithadjustments

Considerthesquarerootofadecimalnumber59064.78

Thisisadecimalnumberwhichisnotaperfectsquare.

Steps:

Writethenumberwithoutthedecimalpointandcarryoutthesamestepsasexplainedbefore.

The starting number (SN) is obtained from the integer port only of thedecimalnumber.

ThefirstfewstepsarethesametillwereachthefifthcolumnwheretheGDis24andtheNDis15.

As before, 15 is divided by 4 to get a quotient 3 and a remainder 3.Wecarrytheremaindertothenextdigit7toget37(GD).

Now,theduplexofthepreviousdigits4,3,0and3is24(243)whichissubtractedfromtheGD37togettheNDas13.

Ondividing13by4,wegetthequotientas2andtheremainderas5,whichis carried to thenext digit giving theGDas58.Note that thequotient istakenas2andnot3togetapositiveNDinthenextcolumn.

Theduplexoftheimmediatelypreviousdigits4,3,0,3and2is34(242+233)whichissubtractedfrom58givingtheNDas24.

24 is divided by 4 to get a quotient of 4 and a remainder of 8which iscarried forward. A zero is attached at this stage to enable the process tocontinuetohigherlevelsofaccuracy.

Thisprocessiscarriedfurtherasshown.

Sincen is5, thedecimalpoint isput (5+1) /2=3digits fromthe left,givingthefinalansweras243.032467.

V.Perfectsquarerootofadecimalnumber

Now, consider the square root of a decimal number 547.56which is aperfectsquare.

Steps:

Thestepsaresimilartothoseexplainedbeforeandthereadershouldhavenodifficultyincomputingthesame.

Since theoriginalnumberhas3digits, thedecimalpoint isplacedafter2digitsfromtheleft.

Important

IfthestartingnumberSNissmall,saylessthan4,thenthenextgroupof2digitscanbeincludedintheSN.e.g.incomputingthesquarerootof13251,wecantakeSNas132insteadof1.

35.ExamplesforsquarerootsofMiscellaneousnumbers

CUBESANDCUBEROOTS

a)Computingcubesof2-digitnumbers

Let us consider a two digit number say ‘ab’. Further, let us consider theexpansionoftheexpression(a+b)3,whichcanbeusedtofindthecubeofthegivennumber‘ab’.

Weknowthat(a+b)3=a3+3a2b+3ab2+b3

Wenoticethatthe1sttermisa3,the2ndterma2b=a3×(b/a),the3rdtermab2=a2b×(b/a)andthe4thtermb3=ab2×(b/a)Thus,eachofthe2nd,3rdand4thtermscanbeobtainedfromitsprevioustermbymultiplyingitbythecommonratio(b/a).

Wecan also consider the2nd term3a2b= a2b+2a2band the3rd termas3ab2=ab2+2ab2

i.e.wecansplititasthesumoftwoterms.

Hence,ifwecomputea3andtheratio(b/a),wecanderivealltheremainingtermsveryeasily.Thetwomiddletermshavetobethendoubledandaddedasshowntoget(ab)3.

Hence, in order to get the cube of any two digit number, we start bycomputingthecubeofthefirstdigitandtheratioofthe2nddigitbythe1stdigit.Theratiocanbemorethan,equal toor less thanone.Themethodremains thesameinallcases.

Case1:Ratioismorethan1

Letusconsiderasimpleexamplei.e.123

Herethefirstdigitis1anditscubeisalso1.Theratioofthe2ndtothe1stdigitis2(=2/1).

Theremaining3termscanbeobtainedbymultiplyingeachoftheprevioustermsby2asshown:123=1248(cubeoflastdigit)

Wealsonoticethatthe4thtermisthecubeofthe2nddigiti.e.8isthecubeof2whichistheseconddigitinthegivennumber12.

Wehave tonow take twice thevalueofeachof the twomiddle termsandwritethemdown.Onaddition,wegetthecubeofthegivennumber12.

Similarly,letuslookat253

Case2:Ratioisequalto1

Case3:Ratioislessthan1

36.Examplesforcubesofnumbers

b)Cuberootsof2-digitnumbers

Let’sconsiderthecubesofnumbersfrom1to9.13=123=8

33=2743=6453=12563=21673=34383=51293=729

Ifweobserveclosely,wefindthatthelastdigitineverycubeisuniqueanddoesnot repeat foranynumber (1 to9).Thispropertycanbeused to find thecuberootofanytwodigitnumberveryeasily.

Let’sfindthecuberootof438976.

Wegroup it into setsof3digitseach, starting from the right.So,wehavetwogroupsviz.

438976Steps:

Welookatthefirstgroupfromtheright,i.e.976andcompareitslastdigiti.e.6withthecubesofdigitsfrom1to9wherethecubealsoendswith6.Thenumber216endswith6.

Wetakeitscuberootwhichis6.Thelastdigitoftherequiredcuberootis6.

Now, we consider the remaining 3-digit group i.e. 438. We locate thehighestcubelessthan438.Therequiredcubeis343sincethecubeaboveitis512whichisgreaterthan438.

Wetakeitscuberootwhichis7.Thefirstdigitoftherequiredcuberootis7.

Therequiredcuberootis76.

This technique provides an easy and powerful method to compute 2-digitcuberoots.Itcanbeusedonlywhenthegivennumberisaperfectcube.

37.Examplesforcuberootsof2digitnumbers

c)Computingfourthpowerof2-digitnumbers

A similar approach can be used to obtain the fourth power of a two-digitnumber,i.e.(ab)4

Subsequent to the first term, each of the remaining terms is obtained bymultiplyingtheprevioustermbyb/a.Letusconsiderexamplesfromeachofthecasesviz.whentheratioismorethan,equaltoandlessthan1.

Case1:Ratioismorethan1


Herethefirstdigitis1anditsfourthpowerisalso1.Theratioofthe2ndtothe1stdigitis2.

Theremaining4termscanbeobtainedbymultiplyingeachoftheprevioustermsby2asshown:Case2:Ratioisequalto1


Herethefirstdigitis1anditsfourthpowerisalso1.Theratioofthe2ndtothe1stdigitis1.

Theremaining4termscanbeobtainedbymultiplyingeachoftheprevious

termsby1asshown:

Case3:Ratioislessthan1


Herethefirstdigitis2anditsfourthpoweris16.Theratioofthe2ndtothe1stdigitis1/2.

Theremaining4termscanbeobtainedbymultiplyingeachoftheprevious

termsby1/2asshown:

38.Examplesforfourthpowerofnumbers

TRIGONOMETRY

It is that branchofmaths inwhichwe study the relationshipsbetween thesides and angles of triangles.Most of the readers would be familiar with thebasic concepts of trigonometry which are taught in schools and colleges.Wehavelearntaboutthedefinitionsofsine,cosineandtangentofagivenangleinatriangle and how to compute any of these ratios, if any of the other ratios isgiven.

Letus refresh these conceptsbyusing the following right-angled triangle:

Thevarioustrigonometricratiosaredefinedasfollows:1.sinA=b/c2.cosA=a/c3.tanA=b/a4.cosecA=1/sinA=c/b5.secA=1/cosA=c/a6.cotA=1/tanA=a/b

Someoftheproblemsintrigonometryareasfollows:

givenanyoneofthesixratiosforanyangleinatriangle,computeanyoftheremainingratiosgivenanyoneof theratios ina triangle,compute theremainingratiosfortwicethegivenanglee.g.givensinA,computetan2Agivenanyoneoftheratios,computetheremainingratiosforhalfthegivenanglee.g.givensinA,computetanA/2

Standard trigonometric formulaeexist tocarryoutvariouscomputationsofthisnature.

A.Triplet

In Vedic maths we have a concept of a triplet which is very effective insolvingallthesetypesofproblems.Intheabovetriangle,atripletisdefinedasa,b,c

which are the measures of the two sides ‘a’ and ‘b’ followed by thehypotenuse(‘c’)attheend.

Therelationc2=a2+b2holdstrueinthiscase.

Ifanytwovaluesofthetripletaregiven,wecancomputethe3rdvalueandbuildthecompletetriplet.

Theratiosarenowdefinedasfollows

sinA=2ndvalue/lastvaluecosA=1stvalue/lastvaluetanA=2nd

value/1stvalueLetusseesomeexamplesofhowtobuildandusethetriplet.

1.Supposethata=3andb=4.Letusseehowtobuildthetriplet.

Givenpartialtripletis3,4,

Thereforethelastvaluewillbe

Hencethecompletedtripletis3,4,5.

2.Ifwearegivenanincompletetripletas12,__,13,Thecompletedtriplet

wouldbe12,5,13wherethevalue5isobtainedas

Oncethetripletisbuilt,allthesixtrigonometricratioscanbereadoffeasilywithoutanyfurthercomputationoruseofanyformulae.

B.Computingtrigonometricratios

Letusassumethatinatriangle,thevalueoftanAisgivenas4/3.WehavetofindoutthevalueofcosecA.Thetraditionaltrigonometricmethodwouldusethefollowingformula:cosec2A=1+cot2AastanA=43,cotA=34

Onsubstitutingthisvalueinthegivenformula,wegetcosec2A=1+9/16

Therefore,cosec2A=25/16

andcosecA=54,sinA=45

IfwealsowantthevalueofcosA,wecanusetheformulatanA=sinA/cosA

OnsubstitutingthevalueoftanAandsinA,wegetcosA=sinA/tanA=(45)(43)=35

Let us now see how to use the Vedic maths technique of the triplet tocomputethevaluesofcosecA,cosAetc.

Theincompletetripletinthisexampleis3,4,__

tanA=4/3=2ndvalue/1stvalueThecompletetripletwouldnowbe

3,4,5asseenbefore.

Assoonasthistripletisbuilt,wecanreadoffalltheratiosE.g.cosecA=lastvalue/2ndvalue=5/4

cosA=1stvalue/lastvalue=3/5

The reader can see that there is no need to carry out any complexcalculations at all and the triplet can be used to obtain all the six ratios veryeasily.

C.Computingtrigonometricratiosoftwicetheangle

Aswehaveseenabove,thetripletfortheangleAiswrittenasa,b,c

The triplet for the angle 2A can be obtained very easily by using thefollowingcomputation(a2–b2),2ab,c2

Letusseeanexample.

AssumethatwearegiventhevalueofsinA=3/5andwehavetocomputetan2A.

Thenormalwaywouldbeasfollows:

Sin2A+cos2A=1

Thereforecos2A=1–sin2A=1–(9/25)

=16/25

ThereforecosA=4/5

Now,sin2A=2*sinA*cosA=2*(3/5)*(4/5)=24/25

Also,cos22A=1–sin22A=1–(242/252)=49/625

Thereforecos2A=7/25

andtan2A=sin2A/cos2A

=24/7

Considerthetripletmethodnow.

SincesinA=3/5,thepartialtripletforangleAis__,3,5

AndthereforethecompletetripletforangleAwillbe

4,3,5

4,3,5Thetripletforangle2Acanbecomputedbyusingtheformula(a2–b2),2ab,

c2

7,24,25Thevalueoftan2AcannowbereadoffdirectlyastanA=24/7and

cos2A=7/24

Theeaseandeleganceofthemethodisobvious.Thereisnoneedtouseanyformulaeandalltheratioscanbereadoffeasily.

D.Computingtrigonometricratiosofhalftheangle

IfthetripletfortheangleAis

a,b,c

then the triplet for the angle A/2 can be obtained by using the followingcomputation

Let us see the example where the value of sin A = 3 5 and we want tocomputetanA2.

Nowaspertrigonometricformula,

sinA=(2*tanA/2)/(1+tan2A/2)Ifwesolvethisequation,weget

tanA/2=1/3

LetusseethetripletmethodinactionThecompletetripletforangleAis4,3,5

The triplet for angle A / 2 can be computed by using the formula

andhencethetripletforA/2is

Now,wecanreadoffthevalueas

tanA/2=1/3,

Exercise-39

*

AUXILLIARYFRACTIONSWehave seen elegant techniques provided byVedicmaths for division of

any number by any divisor. Themethodsmake the process easy and fast andcoupledwith the techniquesofmishrank(Chapter13), the increase inspeed isamazing.

Wewillnowseeadditional techniqueswhichcan speedup thedivisionoffractionalnumberswhere:

thedividendislessthanthedivisorandthedivisorissmalle.g.having2or3digitsandthelastdigitofthedivisorendswith1,6,7,8and9E.g.ofsuchdivisorsare19,29,27,59,41,67,121etc.

Otherdivisorsendingin2,3,4,5and7canbeconvertedtosuchdivisorsbymultiplyingbyasuitablenumber.

Wewillseeexamplesofdivisorsendingineachofthedigitsfrom1to9togetagoodinsightintotheentireprocess.

I)Divisorsendingwith‘9’

Letusstartwiththecasewherethedivisorendswith9.

E.g.19,29,39,89etc.

GeneralSteps

Add1tothedenominatorandusethatasthedivisor

Remove the zero from the divisor and place a decimal point in thenumeratorattheappropriateposition

Carryoutastep-by-stepdivisionbyusingthenewdividend

Everydivisionshouldreturnaquotientandaremainderandshouldnotbeadecimaldivision

Everyquotientdigitwillbeusedwithoutanychange tocompute thenextnumber to be taken as the dividend. In all other cases, the quotient digitwouldbealteredbyapre-definedmethodwhichwillbeexplainedineachcase.

Example1:

Letusnowseethedetailsfor5/29

Steps

Add1tothedivisortoget30

Themodifieddivisionisnow5/30

Remove thezero from thedenominatorbyplacingadecimalpoint in thenumeratorgivingthemodifieddivisionas0.5/3

Wewillnotwritethefinalansweras0.16666but

Wewillcarryoutastep-by-stepdivisionexplainedbelow

Stepsforstep-by-stepdivision

Divide5by3,togetaquotient(q)=1andremainder(r)=2

Writedownthequotientintheanswerlineandtheremainderjustbelowittoitsleftasshown

Now,thenextnumberfordivisionis21whichondivisionby3givesq=7

andr=0.Theresultnowlooksas

Repeatthedivisionateachdigittogetthefinalanswertoanydesiredlevelofaccuracy!!Thenextfewdigitsareshownbelow.

On dividing 7 by 3, we get q = 2 and r = 1. The result now looks as


Thefinalresultupto8digitsofaccuracyis

andtheresultof5/29=0.17241379.

Example2:

Similarly,88/89willbecomputedas

Example3:Similarly,41/119willbecomputedas

So,41/119=0.34453781

Notes:

For all numbers endingwith 3, we can convert them to numbers endingwith9bymultiplyingby3.

E.g.ifthedivisoris13,wecanconvertitto39bymultiplyingby3.

Hence,3/13=9/395/23=15/69.

Similary,fornumbersendingwith7,wecanmultiplyby7togetanumberendingwith9.

E.g.ifthedivisoris17,wecanconvertitto119bymultiplyingby7.Hence,5/7=35/49,4/17=28/1195/27=35/189.

40.Examplesforauxiliaryfractions-divisorendingwith9

II.Divisorsendingwith‘1’

Now, let us consider the fractions where the divisors are numbers endingwith1.

E.g.21,31,51,71etc.

GeneralSteps

Subtract1fromthenumerator

Subtract1fromthedenominatorandusethatasthedivisor


Carryoutastep-by-stepdivisionbyusingthenewdivisor


Subtractthequotientfrom9ateachsteptoformthenextdividend

Example1:


Steps

Subtract1fromthenumeratortoget3

Subtract1fromthedivisortoget20

Themodifieddivisionnowis3/20

Remove thezero from thedenominatorbyplacingadecimalpoint in thenumerator

Themodifieddivisionis0.3/2


Wewillcarryoutastep-by-stepdivisionasexplainedbelow



Writedownthequotientintheanswerlineandtheremainderjustbelowit

toit’sleftasshown

Subtractthequotient(1)from9andwriteitdown

Now,thenextnumberfordivisionis18andnot11.

Ondividing18by2,weget theq=9andr=0.Theresultnowlooksas

Subtractthequotient(9)from9andwriteitdown

Now,thenextnumberfordivisionis0andnot9.

Repeatthedivisionasexplainedbefore.


The final result upto 8 digits of accuracy is

andtheresultof4/21=0.19047619

Example2


Since the pattern is repeating, we can write down the answer as 7/111 =0.063063063

Notes:

For all numbers endingwith 3, we can convert them to numbers endingwith1bymultiplyingby7.


Hence,3/13=21/915/23=35/161

Similary,fornumbersendingwith7,wecanmultiplyby3togetanumberendingwith1.


Hence,5/7=15/214/17=12/515/27=15/81

Once themethods are clear for divisors endingwith 9 andwith 1,we canmoveovertootherdivisorsendingwith8,7and6.

41.Examplesforauxilliaryfractions

III)Divisorsendingwith‘8’

Letusconsidercaseswherethedivisorendswith8.

E.g.18,28,48,88etc.

GeneralSteps



Carryoutastep-by-stepdivisionasexplainedbefore

Everydivisionshouldreturnaquotientandaremainderandshouldnotbecarriedoutasadecimaldivision

The quotient and the remainder is taken as the next base dividend asexplainedabove

Sincethelastdigitofthedivisoris8whichhasadifferenceof1from9,wemultiply the quotient by 1 and add to the base dividend at each stage tocomputethegrossdividend.

Example1:


Steps





Wewillcarryoutastep-by-stepdivisionasexplainedbelow



Writedownthequotientintheanswerlineandtheremainderjustbelowittoitsleftasshown

Now,thebasedividendis21

Addthequotientdigit(1)totheinitialdividendtoget22.

Thenextnumberfordivisionis22whichondivisionby3givesq=7andr

=1.Theresultnowlooksas


Addthequotientdigit(7)tothebasedividendtoget24



Repeatthedivisiontogetthefinalanswerasshownbelow.

Thefinalresultupto5digitsofaccuracyis

andtheresultof5/28=0.17856.

42.Examplesforauxilliaryfractions-divisorsendingwith8

IV)Divisorsendingwith‘7’


E.g.17,27,37,57etc.

GeneralSteps





The quotient and the remainder is taken as the next base dividend asexplainedbefore

Sincethelastdigitofthedivisoris7whichhasadifferenceof2from9,wemultiply the quotient by 2 and add to the base dividend at each stage tocomputethenextgrossdividend.

Example1:


Steps





Wewillcarryoutastep-by-stepdivisionexplainedbelow




toitsleftasshown

Now,theinitialbasedividendis21

Multiplythequotientdigit(1)by2andaddtothebasedividend21toget23




Multiplythequotientdigit(7)by2toget14andaddittothebasedividend(27)toget41

Thenextnumberfordivisionis41whichondivisionby3givesq=13and

r=2.Theresultnowlooksas

Atthispoint,notethatthequotientdigithasbecomea2-digitnumberviz.13. Like before, it ismultiplied by 2 to get 26which is added as shown

below:

Theremainder‘2’isaddedtothetensdigitof13and26.

Now,wedivide59by3 toget thequotientas19andtheremainderas2.

Theworkingappearsasshownbelow:

19ismultipliedby2toget38,whichisaddedasshownbelow:

Now,wedivide77by3 toget thequotientas25andtheremainderas2.

Theworkingappearsasshownbelow:

25ismultipliedby2toget50,whichisaddedasshownbelow:

Ondividing95by3wegetthequotientas31andtheremainderas2.The

workingappearsasshownbelow:

Wewillnowseehowtobuildthefinalresultbyretainingasingledigitateachlocation.Theanswerconsistsofthefollowingnumbers:0.17(13)(19)(25)(31)

Startfromtheright,retain1,carry3totheleft toget0. 1 7 (13) (19)(28)1

Retain8from28,carry2tothelefttoget0.17(13)(21)81

Repeattogetthefinalresultas0.17(15)1810.185181

Thefinalresultwith6digitsofaccuracyis5/27=0.185181.

43.Examplesforauxiliaryfractions

V)Numbersendingwith‘6’


E.g.26,36,56,86etc.

GeneralSteps





The quotient and the remainder is taken as the next base dividend as

explainedbefore

Sincethelastdigitofthedivisoris6whichhasadifferenceof3from9,ateachstage,wemultiply thequotientby3andadd to thebasedividend tocomputethenextgrossdividend.

Example1:


Steps





Wewillnowcarryoutastep-by-stepdivisionasexplainedbelow




toit’sleftasshown

Now,theinitialbasedividendis60

Multiplythequotientdigit(0)by3andaddtothebasedividend60toget60.This is a redundant stephere since thequotientdigit is zero,but it isshownforuniformity.




Multiplythequotientdigit(7)by3toget21andaddittothebasedividend(47)toget68



Repeatthedivisiontogetthefinalanswerasshownbelow.

The final result upto 8 digits of accuracy is

44.Examplesforauxiliaryfractionsdivisorsendingwith6

VI.OtherDivisorsInthesectionsgivenabove,wehaveseendivisorsendingwith1,6,7,8and

9.Whatdowedowithdivisorsendingwith2,3,4and5?Wecanconvertsuchdivisors,byasuitablemultiplication,todivisorsforwhichwehaveseenVedictechniques.Usethetablegivenbelowforguidance.

We can see from this table that if a proper digit is used to multiply thenumeratorandthedenominator,wecanobtainafractionwhereinwecanapplyoneoftheknowntechniquesandderivetheanswerquickly.Sometimes,wemayhavetocarryoutthemultiplicationtwicetogetitinastandardform.

45.Examplesforauxiliaryfractionsmiscellaneousexercises

MISHRANK(VINCULUM)

Thetechniqueofmishrankisverypowerfulandprovidesuswithamethodtoconvertdigitsinanumberwhicharegreaterthan5,todigitslessthan5.Afterthe conversion, all arithmetic operations are carried out using the convertednumber,whichmaketheoperationverysimpleandfast.

It is not necessary to convert all the digits but a judicious conversion ofcertain digits (above 5) can decrease the computation effort considerably.Thereaderhastobepatientwhilelearningthistotallynewandfascinatingtechniqueandgainexpertiseinordertousethismethodeffectivelytogreatadvantage.

We will begin by seeing the method to convert any given number to itsmishrankequivalentandbacktoitsoriginalvalue.

a)ConversiontoMishrank

Mishrankdigitsarecomputedforthosedigitswhicharegreaterthan5.

Thestepsinvolvedare:

Subtractthegivendigitfrom10andwriteitwithabaraboveit

Addonetothedigitontheleft

Examples

Themishrankdigithasanegativevalueinthenumber.

Thus,inthefirstexampleabove,

15 isequivalentto150-2i.e.148

Similarly, 2 3 is equivalent to 2305 - 20 = 2285while 1 isequivalentto10000-1112=8888

b)ConversionfromMishrank

Onlymishrankdigits,whicharemarkedwithabar,arereconvertedbacktotheiroriginalvalues,asfollows:

Subtract 1 from the non-mishrank digit to the immediate left of themishrankdigitandwriteitdown.Subtractthemishrankdigitfrom10andwriteitdown.Repeatforallmishrankdigits.

Inexample4,wehaveaseriesofmishrankdigitsandwecanusethesutra‘Allfrom9andlastfrom10’toconverttotheoriginalnumber.

Mishrank can be used very effectively in operations like addition,subtraction,multiplication,division,findingsquaresandcubesofnumbers.Wewilllookatsomeexamplestoseeitstremendouspower.

c)Applicationinmultipleadditionandsubtraction

Letusconsiderthefollowingproblem:

232+4151–2889+1371

This could either be solvedby first adding232 and4151, then subtracting2889fromthesumandthenadding1371totheresult.

Or,wecouldadd232,4151and1371andthensubtract2889fromthesum.

Instead,wecouldspeeduptheoperationbyconvertingthesubtractiontoanadditionoperationasshownbelow:Letuswrite–2889as+

Here,eachdigitiswrittenasanegativedigit.

Now,theproblemcanberewrittenas

232+4151+ +1371

orwrittenvertically,itappearsas

Sincethisisasingleoperation,itisfasterandcauseslessstraintothemind.Themishrankdigitsintheresultcanbeconvertedtonormaldigitstoobtainthefinalansweras2865.

d)Applicationinsubtraction

Example1:Considerthesubtractionof6869from8988

Thesubtractionoperationwouldbeslowandmentallytiring,withacarryatalmost every step. The operation can be simplified and speeded up by usingeitherofthefollowingtechniques:

Convert the subtraction toadditionofanumberas shown in thepreviousexample.

Carryoutadigit-by-digitsubtraction,withoutanycarry.

Let’sseethefirsttechnique.

6869iswrittenwithabarontopofeachdigittosignifythatitisanegativedigit.

whichisequalto2119.Here,intheunit’splace,8+ gives .

Let’sseethesecondtechnique.

Thesameresultcanbeobtained ifwesimplysubtracteachdigitwithoutacarry.

Let’sseetheoperation:

Here, in the unit’s place, 8 – 9 gives ,which gives 2119 on conversionfrommishrank.

e)Applicationinmultiplication

Example1:Considerthemultiplicationof69with48.

The operation can be carried out by the cross-multiplication techniqueexplainedinChapter5.Thecomputationcanbesimplifiedfurtherifweconvertthenon-mishrankdigitstomishrankdigits.Wewillseethemethodbelow.

69=7

48=5

Now,wewillcarryoutacrossmultiplicationofthesetwonumbers.

Example2:Considerthemultiplicationof882by297

Thecorrespondingmishrankdigitsare

882=9 2(Weconvertthe8intheten’splaceanddonotconvertthe8inthehundred’splace)297=30

Now,wewillcarryoutacrossmultiplicationofthesetwonumbers.

f)Applicationindivision

ExampleI:Considerthedivisionof25382by77

77canbewrittenas8 ,whereweconvert7inunit’splacetoitsmishrankdigit.

Divisioniscarriedoutbythesametechniqueasexplainedbefore.Itwillbeseen that there is no need to make any adjustment at any column and theprocedurecanprogressveryfast.

Thefinalresultis329withremainder49,orindecimalformitis329.6363.

g)Applicationinsquares

Inordertogetthesquareofanynumber,weneedtocomputeduplexvaluesof a lot of numbers. If someof thedigits are above5, the computationof theduplex becomes tedious. Consider the square of 897. The computation of theduplexisquitetime-consumingsincethedigitsarebig.Wewillseethemagicofthemishrankdigitsnow.

Botharevalidmishrankequivalentnumbers.

Wewillnowcomparethecomputationofthesquareofthegivennumberforallthesenumbers.

Let’slookatthenextconversion,i.e.1 0

asbefore.

Wecanseethatthecomputationsbecomeincreasinglysimpler.

h)Applicationincubes

Considerthecubeof89where893=704969

The computation without using the mishrank would appear as follows :

Thecomputationisverytediousandcumbersome.

Let’sconverttomishrankandseewhatistheresult.

Thisiseasierandfaster.

In each of the operations it can be seen that the operation on thecorrespondingmishranknumberisbotheasyandfast.Thereaderhastopracticetoconvertnumberstotheirmishrankequivalentsandbackagain, tofullyavailofthepowerofthistechnique.

46.Examplesinuseofmishrank

SIMULTANEOUSEQUATIONSEveryoneisfamiliarwithsimultaneousequationswhereinwehavetosolve

twoequationsintwounknownssayxandy.

The normal method consists of multiplying each equation by a suitablenumber so that the coefficients of either x or y become same in both theequations,which can then be subtracted out of the equations, leaving a singleequationinoneunknown.

Vedicmathsprovidesasimplemethodtosolvetheequationswithoutgoingthroughthislengthyprocessandgivesthevaluesofxandyinasingleline.Thiswouldbesimilarto‘Cramer’sRule’butiseasiertorememberanduse.

Wewillconsiderpairsofequationseachwithtwounknownssayxandy.

Example4x+7y=5…(1)3x+9y=10…(2)

Thenormalmethodtosolvesuchsimultaneousequationsistoeliminateanyoneunknownbysuitablemultiplicationofeachequationbyaconstantandthensolvingfortheotherunknownvariable.Intheaboveexample,multiplyequation(1)by3andequation(2)by4andsubtract.

Whichgivesy=5/3.

Onsubstitutingthisvalueofyinequation(1),weget4x+7*5/3=5

Giving4x=-20/3∴x=-5/3

WewillnowseethetechniquesprovidedbyVedicmaths.

Equationsin2unknownscanbedividedinto3categoriesandwewillseethetechniquesforeachofthem.

Whentheratioofthecoefficientsofeitherxoryisthesameasthatoftheconstants

Whentheratioofthecoefficientsofxandyareinterchanged

Allothertypesnotfallinginthesetwocategories

Category1

Considertheexample:

3x+4y=10

5x+8y=20

Inthisexample,theratioofthecoefficientsofyis4/8=1/2.Theratiooftheconstantsis10/20=1/2.

Since both are equal, the value of x in such a case is equal to 0. If thecoefficientsofxwereinthesameratio,valueofywouldbezero.

Thevalueofycanbeobtainedfromthefirstequationbysubstitutingx=0,givingtheequationas4y=10

ory=5/2

Category2

Now,letusconsiderthemethodtosolvesimultaneousequationswherethecoefficients of x and y are interchanged. Let us consider the example givenbelow.

45x–23y=113…(1)

23x–45y=91…(2)

Wewilladdandthensubtractthegivenequationstogetasimplifiedsetasshownbelow68x–68y=204…(3)

22x+22y=22…(4)

Dividingeq(3)by68andeq(4)by22,wegetx-y=3…(5)

x+y=1…(6)

Thisisatrivialset,whichgivesthevaluesofx=2andy=-1.

Category3

Letusconsiderthemethodtosolveanygeneralsimultaneousequation.Letusconsidertheexamplesolvedatthebeginningofthischapter.

4x+7y=5…(1)

3x+9y=10…(2)

Letusrepresentthisasasetofcoefficientsasfollowsabc

pqr

Wherea=4,b=7,c=5,p=3,q=9,r=10

Westartwiththecoefficientatthecentreintheupperrowviz‘b’andgetthe

valueof‘x’byusingtheformula

Notice the cross multiplication which is used extensively here in thisformula.

b*randc*qarecross-multiplications,fromthecentretotheright.

Similarly,b*p anda*q are crossmultiplications, from the centre to theleft.

Togetthevalueofy

Here,again,c*panda*rarecross-multiplicationswhilethedenominatoristhesameasbefore.

47.Exercisesforpractice

OSCULATORSThe concept of ‘Osculator’ is useful to check the divisibility of a given

numberbydivisorsendingwith9or1oramultiplethereof.

E.g.Checkthedivisibilityof52813by59orof32521by31orof35213by13or17.

An osculator is a number defined for any number ending in 9 or 1 and isobtainedfromthenumberbyasimplemechanismdescribedinthischapter.

Theuseofosculatorswouldbeseverelylimitedifonlythesetwocategoriesof numbers viz. ending with 9 or 1 were considered. Interestingly, numbersending with 3 and 7 can also converted to numbers ending with 1 or 9 by asuitablemultiplication.The same techniques canbeused for all suchnumberstoo.

Osculators are categorized into two main types viz. positive and negativedependingonwhetherthenumberendswith9orwith1.

PositiveOsculators

Theosculatorofanumberendingin9isconsideredpositive.

Itisobtainedby

Dropping9

Adding1totheremainingnumber

Theosculatorfor19is2,for29itis3andfor59,itis6.

Ifthegivennumberdoesnotendin9butin1or3or7,wecanmultiplyitby9, 3 or 7 respectively, to convert it to a number which ends in 9 and thencomputetheosculatorbydroppingthe9.

Examples are given below of how to compute the osculators for variousnumbers.

Oncetheosculatorofanumberhasbeencomputed,itisasimplemattertocheckthedivisibilityofanygivennumberbyit.

Example:Letuscheckthedivisibilityof228by19.

Steps

Computetheosculatorofthedivisor(19)whichis2.

Digit1

Startwiththecomputationfromtherighthanddigit(8)ofthedividend

Multiplyitbytheosculatortoget16(8×2)

Digit2

Addittothedigittoitslefttoget18(16+2)


Subtractmaximumpossiblemultiplesofthedivisorfromit.Hence,19canbesubtractedfromittoget17.

Digit3

Addittothedigittoitslefttoget19(17+2)

Since19isdivisibleby19,thegivennumberisalsodivisibleby19.

Computationofpositiveosculatorsfordifferentnumbers

48.Examplesforcheckingthedivisibilityofthefollowingby19


51.Examplesforcheckingdivisibility–miscellaneousexamples

NegativeOsculators

Let us now consider negative osculators, which are defined for numbersendingwith1.

Togetthenegativeosculatorsofanumberendingin1

Drop1

Theremainingnumberistheosculator

Theosculatorfor11is1,for21itis2andfor61,itis6.

Example:Letusworkoutthedivisibilityof2793by21.

Steps

Startwiththeosculatorofthedivisor(21)whichis2

Markallalternatedigitsfromthesecondlastdigitasnegativebyplacingabarontop

Digit1

Startwiththecomputationfromtherighthanddigit(3)ofthedividend


Digit2

Addittothedigittoitslefttoget3(6+9)Multiplyitbytheosculatortoget6(3×2)Subtract maximum possible multiples of the divisor from it. We cannotsubtractanythinghere.

Digit3

Addittothepreviousdigittoget1(7+6)Multiplyitbytheosculatortoget2(1×2)

Digit4

Addittothepreviousdigittoget0(2+2)Multiplyitbytheosculatortoget0(0×2)

Since0isdivisibleby21,thegivennumberisalsodivisibleby21.

52.Examplesforcheckingthedivisibilityby21

53.Examplesforcheckingthedivisibilityby31

54.Miscellaneousexamples

APPLICATIONSOFVEDICMATHS

Wenowcometothemostinterestingchapterofthisbook,inwhichwewilllearn how to apply the techniques learnt so far.After reading and solving theproblems in this chapter, the reader will realize the immense benefit that isderived by using the simple and elegantmethods ofVedicmaths in solving avarietyofproblems.

Ihavedividedthischapterinto3parts.Inthefirstpart,Ihavelistedtwentyquestionsselectedfromvariouscompetitiveexams.Thereaderisencouragedtosolve these problems independently and track the time taken for solving.Thispart is followedby thesolutions to theproblems,using techniquesfromVedicmaths,includingareferencetotherelevantchapter.Thesolutionsarebriefbutsufficient to follow and the reader will easily grasp the application of thetechniqueslearntsofar.Theeaseandtheefficiencywouldbeappreciated.Inthethird part, I have included somemore problems for practicewhich the readershouldsolveandgainexpertiseinthesubject.

SampleProblems

1)Whichisthegreatest6-digitnumberwhichisaperfectsquare?

2) If the length and the breadth of a square is increased and reduced by 5%respectively,whatisthepercentageincreaseordecreaseinthearea?

3)Compute896×896–204×204

4) If all the digits in numbers 1 to 24 are written down, would the number soformedbedivisibleby9?

5)Whatshouldbethevalueof*inthefollowingnumbersothatitisdivisibleby11?

8287*845*381

6)Whichofthefollowingnumbersisdivisibleby99?135792,114345,3572404,913464

7)Compute781×819.

8)Agardenerplanted103041treesinsuchawaythatthenumberofrowswereasmanyastreesinarow.Findthenumberofrows.

9)Findthevalueof5112.

10)Packetsofchalkarekeptinaboxintheformofacube.Ifthetotalnumberofpacketsis2197,findthenumberofrowsofpacketsineachlayer.

11)Givena+b=10,ab=1.Computethevalueofa4+b4.

12) The population in a village increases at a rate of 5% every year. If thepopulationinagivenyearis80000,whatwillbethepopulationafter3years.

13)Whatisthelengthofthegreatestrodwhichcanfitintoaroomoflength=24mt,breadth=8mtandheight=6mt.

14)Thedifferenceonincreasinganumberby8%anddecreasingitby3%is407.Whatistheoriginalnumber?

15)Rs.500growstoRs.583.20in2yearscompoundedannually.Whatistherateofinterest?

16)Anumber19107isdividedinto2partsintheratio2:7.Findthetwonumbers.

17)AsumofRs.21436isdividedamong92boys.Howmuchmoneywouldeachboyget?

18)Evaluate13992.

19)Evaluate397*397+397104+104104+397*104

20)Ifthesurfaceareaofthesideofacubeis225cm2.Whatisthevolumeofthecube?

SolutionsusingVedicmaths

1)Whichisthegreatest6-digitnumberwhichisaperfectsquare?

Solution:A6-digitnumberwillhaveasquarerootofexactly3digits.

Thelargest3-digitnumberis999.Hence,thesquareof999istherequirednumber.

999×999=998001(ReferChapter1)2)Ifthelengthandthebreadthofa square is increased and reduced by 5% respectively, what is thepercentageincreaseordecreaseinthearea?

Solution:Letthelengthandbreadthbe‘l’and‘b’respectively.Thenewlengthwillbe1.05landthenewbreadthwillbe0.95b.Hence,thenewareawillbe1.05l×0.95b.Bynikhilam,weget1.05×0.95=100/-25=0.9975

Thenewareais0.9975lbandhencetheareadecreasesby(1.0000-0.9975)whichis0.25%.Or,thedecreasecanalsobereadoffdirectlyastherightsideoftheproductviz100/-25.

3)Compute896×896-204×204

Solution:Theproblembelongstothetypea2-b2wherethefactorsare(a-b)and(a+b).Hencethesolutionis(896-204)(896+204)=692*1100

=761200(ReferChapter3)4)Ifallthedigitsinnumbers1to24arewrittendown,wouldthenumbersoformedbedivisibleby9?

Solution:Thenumberwouldbe

123456789101112131415161718192021222324.

123456789101112131415161718192021222324.Thesumofthefirst‘n’naturalnumbersisn(n+1)/2.

Hence,thesumwouldbe24*25/2=300andtheremainderis3.Thereforethenumberisnotdivisibleby9.(ReferChapter2)5)Whatshouldbethevalueof*inthefollowingnumbersothatitisdivisibleby11?

8287*845*381

Solution:Forthefirstnumber,thesumsofthealternatesetsare22and(20+).Ifthevalueofthe''is2,thedifferencewillbezeroandthenumberwillbedivisibleby11.Hence,'*'=2.

Forthesecondnumber,thesumsare4and(8+'*').Ifthevalueofthe''is7,thedifferencewillbe11andthenumberwillbedivisibleby11.Hence,''=7.

6)Whichofthefollowingnumbersisdivisibleby99?

135792,114345,3572404,913464

Solution:Thenumbershouldbedivisiblebothby9and11.

Navaseshofeachofthenumbersis0,0,7,0

So,3rdnumberisrejected.

Ofthebalance,only2ndnumberisdivisibleby11.(Referchapter3).

Hence,114345isdivisibleby99.

7)Compute781×819.

Solution:Usethebaseas800inNikhilamandwriteas781–19819+19

Hence,theproductwillbe800/-381.

Multiplytheleftpartby8toconverttotheprimarybaseof100,toget6400/–381

381

So,borrow4fromlefttogetthefinalansweras6396/19.

8)A gardenerplanted 103041 trees in such away that the number of rowswereasmanyastreesinarow.Findthenumberofrows.

Solution:Computethesquarerootof103041

Hence,thenumberoftreesineachrowis321.

9)Findthevalueof5112.

Solution:Mentally,using500asthebase,Weget5112=522/121.

Now,converttothebaseofthousandbydividingtheleftpartoftheanswerby2.


10) Packets of chalk are kept in a box in the form of a cube. If the totalnumber of packets is 2197, find the number of rows of packets in eachlayer.

Solution:RefertoChapter10,takethecuberootof2197whichis13.Hence,eachrowcontains13packets.

11)Givena+b=10,ab=1.Computethevalueofa4+b4.

Solution:Sincea+b=10andab=1,

(a+b)2 = a2+b2+2ab

a2+b2 = 100-2=98

=

(a2+b2)2=a4+b4+2a2b2

a4+b4 = (a2+b2)2-2a2b2

= 982-2

= 9604-2=9602.(UseNikhilamtoget982orally)

12)The population in a village increases at a rate of 5% every year. If thepopulation in a given year is 80000, what will be the population after 3years.

Solution:Finalpopulation = 80000(1+5/100)3

= 80000(21/20)3

= 10*213

Therefore,thepopulationwillbe92610.

13)Whatisthelengthofthegreatestrodwhichcanfitintoaroomoflength=24mt,breadth=8mtandheight=6mt.

Solution:LetthelengthofthegreatestrodbeL.

Then,L2 = 242+82+62

= 576+64+36=676

Therefore,L = 26mt.

UsetechniquesinChapters8and9togetthesquareof24andthesquarerootof676.

14)Thedifferenceonincreasinganumberby8%anddecreasingitby3%is407.Whatistheoriginalnumber?

Solution:LetthenumberbeN.

Now,(1.08N-0.97N)=407

Hence,0.11N=407(407isdivisibleby11)

GivingN=3700

Therefore,theoriginalnumberis3700.

15)Rs.500growstoRs.583.20in2yearscompoundedannually.Whatistherateofinterest?

Solution:583.20=500(1+r)2

116.64=(1+r)2

ByNikhilam,weknowthat116.64=10.82

Hence,requiredrateofinterestis8%.

16)Anumber 19107 is divided into 2 parts in the ratio 2 : 7. Find the twonumbers.

Solution:Thefirstnumberwouldbe(2/9)*19107.

Thisisadivisionby9,refertoChapter2,wegettheansweras4246.

Thesecondnumberis19107-4246,usemishranktogetthesecondas14861.

17)A sumofRs 21436 is divided among 92boys.Howmuchmoneywouldeachboyget?

Solution:Thisisadivisionproblem,refertoChapter6.

18)Evaluate13992.

Solution:Usemishranktoconvertthegivennumberto140

Now,140 2=1/9/16/ / /0/1

=1957201

19)Evaluate397*397+397104+104104+397*104

Solution:Theexpressionis

(397+104)2 = 5012

= 502/001(byNikhilam)

= 251/001

20)Ifthesurfaceareaofthesideofacubeis225cm2,whatisthevolumeofthecube?

Solution:Theareaofanyonesideis225.

Therefore,eachsideis15cm.(Sincethenumberendswith25,squarerootendswith5,andtogetthefirstdigit,1×2=2)Thereforethevolumeis153whichis

3375cm3.

ProblemsforPractice

1)Ashopkeeperknows that13%orangeswithhimarebad.Hesells75%of thebalanceandhas261orangesleftwithhim.Howmanyorangesdidhestartwith?

2)Whatshouldbethevalueof*inthefollowingnumberssothattheyaredivisibleby9?

38*1451*603

3)Theareaofasquarefieldis180625mt2. If thecostoffencingit isRs10permetre,howmuchwoulditcosttofenceit?

4)Find thenumbernearest to99547which isdivisibleby687 :100166,99615,99579,98928

5) Inaparade,soldiersarearranged in rowsandcolumns insuchaway that thenumbersofrowsisthesameasthenumberofcolumns.Ifeachrowconsistsof333soldiers,howmanysoldiersarethereintheparade?

6)Onesideofa square is increasedby15%and theother isdecreasedby11%.Whatisthe%changeinthearea?

7)Findthesquarerootof

530.38090.7

8)If1000isinvestedinbankandearns5%interest,compoundedyearly,whatistheamountattheendofthreeyears?

9) The difference in compound interest compounded half-yearly and simpleinterestat10%onasumforayearisRs25.Whatisthesum?

10)Computethefollowing:1/35,3/14

11)‘A’sellsaradioto‘B’atagainof10%and‘B’sellsitto‘C’atagainof5%.If‘C’paysRs924forit,whatdiditcost‘A’?

12)Findthenearestnumberto77993whichisexactlydivisibleby718.

13)‘A’received15%higherthan‘B’and‘C’received15%higherthan‘A’.If‘A’received9462,whichofthefollowingdid‘B’and‘C’receive?

(8218,10032or8042,10881or1419,10881or8228,10881)

14)Findthegreatestandtheleast5digitnumberexactlydivisibleby654.

15)Anilwalks120metresnorth,70metreswest,175metressouthand22metreseast.Findhisdistancefromhisstartingpoint.

Answers

1)1200oranges

2)6,8

3)Side=425mt,cost=Rs.17000

4)99615

5)110889soldiers

6)Increaseof2.35%

7)Squarerootsare23.03,0.8366

8)Rs.1157.625

9)Rs.10000

10)0.0285714,0.2142857

11)CostisRs.800

12)78262

13)‘B’received8228,‘C’received10881

14)10464,99408

15)Anilis73metresaway.

The Power of Vedic Maths with Trigonometry

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