THEPOWEROFVEDICMATHS2ndEDITION
ATULGUPTA
PublishedbyJaicoPublishingHouseA-2JashChambers,7-ASirPhirozshahMehtaRoadFort,Mumbai-400001
©AtulGupta
THEPOWEROFVEDICMATHSISBN81-7992-357-6
FirstJaicoImpression:2004ThirteenthJaicoImpression(Revised&Updated):2010
FifteenthJaicoImpression:2011
Nopartofthisbookmaybereproducedorutilizedinanyformorbyanymeans,electronicormechanicalincludingphotocopying,recordingorbyany
informationstorageandretrievalsystem,withoutpermissioninwritingfromthepublishers.
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
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
3-Digitmultiplication
Multiplying3-digitand2-digitnumbers
4-Digitmultiplication
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’
Divisorsendingwith‘1’
Divisorsendingwith‘8’
Divisorsendingwith‘7’
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.
Example2-Considerthemultiplicationof52×47
C)Case1-Primarybase10,secondarybase50for52×47
Secondarybaseis50Scalingoperation:Factor=5=50/10
(secondarybase÷primarybase)Onedigittoberetainedintherighthandpartoftheanswer
Consider:52×47(subtract50fromeachnumber)
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.
Sothefinalanswerbecomes24/44=2444.
Since 24.5 hundreds is 2450, this can be interpreted also as 2450 / (–6) =2444
Wegetthesameanswerirrespectiveoftheprimarybase.Thefractionalpartcanbeavoidedifwechooseascalingoperationinvolvingmultiplicationinsteadofdivision.
II.Secondarybaseof250
Example2-Considerthemultiplicationof266×235
E)Primarybase1000andsecondarybase250for235×266
Secondarybaseis250Scalingoperation:Factor=1/4=250/1000
(secondarybase÷primarybase)Threedigitstoberetainedintherighthandpartoftheanswer
Consider:235×266(subtract250fromeachnumber)
Considerthesteps
Wewillscaledownthisintermediateresultbydividing251by4.
Since251/4=62¾,theanswerbecomes62¾/(–240).
Sincetheprimarybaseis1000,wewillcarry3/4fromtheleftwhichisequaltoacarryof750(primarybase1000*¾)andsubtract240from750giving510.
Sothefinalanswerbecomes62/510=62510.
Or,itcanbeconsideredas62750-240=62510.
III.Secondarybaseof500
Example2-Considerthemultiplicationof532×485
F)Primarybase100andsecondarybase500for532×485
Secondarybaseis500Scalingoperation:Factor=5=500/100
(secondarybase÷primarybase)Twodigitstoberetainedintherighthandpartoftheanswer
Consider:532×485(Subtract500fromeachnumber)
Considerthesteps
Wewillscaleupthisintermediateresultbymultiplying517by5.
Since517×5=2585
Sotheanswerbecomes2585/(–480).
Sincetheprimarybaseis100,wewillcarry5fromtheleftwhichisequaltoacarryof500(primarybase100*5)andsubtract480from500giving20.
Sothefinalanswerbecomes2580/020=258020.
Or,itcanbeconsideredas258500-480giving258020.
G)Primarybase1000andsecondarybase500for532×485
Secondarybaseis500Scalingoperation:Factor=1/2=500/1000
(secondarybase÷primarybase)Threedigitstoberetainedintherighthandpartoftheanswer
Considerthesteps
Wewillscaledownthisintermediateresultbydividing517by2.
Since517/2=258½,theanswerbecomes258½/(–480).
Sincetheprimarybaseis1000,wewillcarry½fromtheleftwhichisequaltoacarryof500(primarybase1000*½)andsubtract480from500giving20.
Sothefinalanswerbecomes258/020=258020.
Theleftpartoftheansweris258.5hundredswhichisequalto258500.
So,itcanbeconsideredas258500-480giving258020.
IV.Secondarybaseslike40,60
Example3-Considerthemultiplicationof32×48
H)Primarybase10andsecondarybase40for32×48
Secondarybaseis40Scalingoperation:Factor=4=40/10
(secondarybase÷primarybase)Onedigittoberetainedintherighthandpartoftheanswer
Consider:32×48(subtract40fromeachnumber)
Wewillscaleupthisintermediateresultbymultiplying40by4.
Since40×4=160,theanswerbecomes160/(–64).
Sincetheprimarybaseis10,wewillcarry7fromtheleftwhichisequaltoacarryof70(primarybase10*7)andsubtract64from70giving6.
Thisisalsoequivalentto1600-64.
So,thefinalansweris1536.
Example4-Considerthemultiplicationof64×57
I)Primarybase10andsecondarybase60for64×57
Secondarybase=60Scalingoperation:Factor=6=60/10
(secondarybase÷primarybase)Onedigittoberetainedintherighthandpartoftheanswer
Consider:64×57(subtract60fromeachnumber)
Wewillscaleupthisintermediateresultbymultiplying61by6.
Since61×6=366,theanswerbecomes366/(–12)=364/8.
Sincetheprimarybaseis10,wewillcarry2fromtheleftwhichisequaltoacarryof20(primarybase10*2)andsubtract12from20giving8.
Thisisalsoequivalentto3660-12.
So,thefinalansweris3648.
V.Secondarybaseof300
Example-Considerthemultiplicationof285×315
J)Primarybase100andsecondarybase300for285×315
Secondarybase=300Scalingoperation:Factor=3=300/100
(secondarybase÷primarybase)Twodigitstoberetainedintherighthandpartoftheanswer
Consider:285×315(subtract300fromeachnumber)
Wewillscaleupthisintermediateresultbymultiplying300by3.
Since300×3=900,theanswerbecomes900/(–225).
Sincetheprimarybaseis100,wewillcarry3fromtheleftwhichisequaltoacarryof300(primarybase100*3)andsubtract225from300giving75.
Sothefinalanswerbecomes897/75=89775.
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
Case2:3-digitmultiplication
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
Case4:4-digitmultiplication
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)
Divide31by8togetthequotientas3andremainderas7
Carrytheremainder7tothenextdigit0andget70.
Multiplythequotient3bytheflag4toget12andsubtractitfrom70toget58
Divide58by8togetthequotientas7andremainderas2
Atthispoint,theintegerpartofthefinalquotienthasbeenobtainedsincewehavereachedthecolon.Allthebalancedigitswouldbethedecimalpart
Carrytheremainder2tothenextdigit8andget28
Multiplythequotient7bytheflag4toget28andsubtractitfrom28toget0
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:
Placeacolononedigitfromtheright(=numberofdigitsofflag)
Divide43by8togetquotientas5andremainderas3
Carrytheremainder3tothenextdigit4andget34
Multiplythequotient5bytheflag3toget15andsubtractitfrom34toget19
Divide19by8togetthequotientas2andremainderas3
Carrytheremainder3tothenextdigit0andget30
Multiplythequotient2bytheflag3toget6andsubtractitfrom30toget24
Divide24by8togetthequotientas3andremainderas0
Carry the remainder 0 to the next digit 9 and get 09. At this point, theintegerpartof thefinalquotienthasbeenobtainedsincewehavereachedthecolon.Allthebalancedigitswouldbethedecimalpart.
Multiplythequotient3bytheflag3toget9andsubtractitfrom9toget0
Since the difference is now 0, the division is completed and we get theansweras523
Trya)4341283andb)4340373
Case2:Divisionbyanumberwithaflagofonedigit(withremainder)
Considerthedivisionof26382by77
Thequotientis342andtheremainderis48,whichappearsunderthecolumnjustafterthecolon.Thefinalanswerindecimalformis342.623.
Steps:
Thestepsasexplainedabovearecarriedouttillwereachthecolumnafterthecolonwhichshowstheremainderas48.Ifwewanttogettheanswerindecimal form, we have to carry out division further by using the sameprocedure.
Divide48by7togetquotientas6andremainderas6
Now,weattachazero to thedividendandwrite the remainder to its left,givingthenextnumberas60.
Multiplythequotient6bytheflag7toget42andsubtractitfrom60toget18
Divide18by7togetquotientas2andremainderas4
Carrytheremainder4tothenextdigit0(attachedasbefore)andget40
Multiplythequotient2bytheflag7toget14andsubtractitfrom40toget26
Divide26by7togetthequotientas3andremainderas5
Thisprocesscanbecarriedoutuptoanydesiredlevelofaccuracy.
Case3:Divisionwithadjustments
Considerthedivisionof25382by77
Steps:
Placeacolononedigitfromtheright(=numberofdigitsofflag)
Divide25by7togetquotientas3andremainderas4
Carrytheremainder4tothenextdigit3andget43(GD)
Multiplythequotient3bytheflag7toget21andsubtractitfrom43toget22(ND)
Divide22by7togetthequotientas3andremainderas1
Carrytheremainder1tothenextdigit8andget18(GD)
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)
Carrytheremainder8tothenextdigit8andget88(GD)
Multiplythequotient2bytheflag7toget14andsubtractitfrom88toget74(ND)whichisnowpositive
Carry out the process in the remaining columns, making an adjustmentwheneverthenetdividendbecomesnegative.
Thefinalresultis329withremainder49orindecimalform,itis329.6363.
Case4:Divisionbyanumberwithaflagof2digits
Letusnowconsiderdivisionbya3digitdivisorwheretheflagwouldhavetwodigits.
Considerthedivisionof2329989by514
Steps:
Placeacolontwodigitsfromtherightsincewenowhavetwodigitsintheflag
Divide23by5togetquotientas4andremainderas3
Carrytheremainder3tothenextdigit2andget32(GD)
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)
Divide28by5togetthequotientas5andremainderas3
Carrytheremainder3tothenextdigit9andget39(GD)
Now,theprevioustwoquotientdigitsare4and5whicharemultipliedby
theflag14asfollows:
Wewillsubtract21fromtheGD39toget18(ND).
Repeattheprocessfortheremainingcolumns.
Case5:Divisionbyanumberwithaflagof3digits
Letusnowconsiderdivisionbya4-digitdivisorwheretheflagwouldhavethreedigits.
Considerthedivisionof6851235by65245246
Steps:
Placeacolonthreedigitsfromtherightsincenowtherearethreedigitsintheflag
Divide6by6togetquotientas1andremainderas0
Carrytheremainder0tothenextdigit8andget08(GD)
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)
Divide3by6togetthequotientas0andremainderas3
Carrytheremainder3tothenextdigit5andget35(GD)
Now,theprevioustwoquotientdigitsare0and1.Wepadthemwithone
zeroandmultiplybytheflag524asfollows:
Wewillsubtract2fromtheGD35toget33(ND).
Divide33by6togetthequotientas5andremainderas3
Carrytheremainder3tothenextdigit1andget31(GD)
Now,thepreviousthreequotientdigitsare1,0and5.Wemultiplybythe
flag524asfollows:
Wewillsubtract29fromtheGD31toget2(ND).
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
33.Examplesforsquaresof4-digitnumbers
34.Examplesforsquaresof5-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
Letusconsiderasimpleexamplei.e.124
Herethefirstdigitis1anditsfourthpowerisalso1.Theratioofthe2ndtothe1stdigitis2.
Theremaining4termscanbeobtainedbymultiplyingeachoftheprevioustermsby2asshown:Case2:Ratioisequalto1
Letusconsiderasimpleexamplei.e.114
Herethefirstdigitis1anditsfourthpowerisalso1.Theratioofthe2ndtothe1stdigitis1.
Theremaining4termscanbeobtainedbymultiplyingeachoftheprevious
termsby1asshown:
Case3:Ratioislessthan1
Letusconsiderasimpleexamplei.e.214
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
On dividing 12 by 3, we get q = 4 and r = 0. 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
Remove the zero from the divisor and place a decimal point in thenumeratorattheappropriateposition
Carryoutastep-by-stepdivisionbyusingthenewdivisor
Everydivisionshouldreturnaquotientandaremainderandshouldnotbeadecimaldivision
Subtractthequotientfrom9ateachsteptoformthenextdividend
Example1:
Letusnowseethedetailsfor4/21
Steps
Subtract1fromthenumeratortoget3
Subtract1fromthedivisortoget20
Themodifieddivisionnowis3/20
Remove thezero from thedenominatorbyplacingadecimalpoint in thenumerator
Themodifieddivisionis0.3/2
Wewillnotwritethefinalansweras0.15but
Wewillcarryoutastep-by-stepdivisionasexplainedbelow
Stepsforstep-by-stepdivision
Divide3by2,togetaquotient(q)=1andremainder(r)=1
Writedownthequotientintheanswerlineandtheremainderjustbelowit
toit’sleftasshown
Subtractthequotient(1)from9andwriteitdown
Now,thenextnumberfordivisionis18andnot11.
Ondividing18by2,weget theq=9andr=0.Theresultnowlooksas
Subtractthequotient(9)from9andwriteitdown
Now,thenextnumberfordivisionis0andnot9.
Repeatthedivisionasexplainedbefore.
On dividing 0 by 2, we get q = 0 and r = 0. The result now looks as
The final result upto 8 digits of accuracy is
andtheresultof4/21=0.19047619
Example2
Letusnowseethedetailsfor7/111
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.
E.g.ifthedivisoris13,wecanconvertitto91bymultiplyingby7.
Hence,3/13=21/915/23=35/161
Similary,fornumbersendingwith7,wecanmultiplyby3togetanumberendingwith1.
E.g.ifthedivisoris17,wecanconvertitto51bymultiplyingby3.
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
Add2tothedenominatorandusethatasthedivisor
Remove the zero from the divisor and place a decimal point in thenumeratorattheappropriateposition
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:
Letusnowseethedetailsfor5/28
Steps
Add2tothedivisortoget30
Themodifieddivisionisnow5/30
Remove thezero from thedenominatorbyplacingadecimalpoint in thenumeratorgivingthemodifieddivisionas0.5/3
Wewillnotwritethefinalansweras0.16666but
Wewillcarryoutastep-by-stepdivisionasexplainedbelow
Stepsforstep-by-stepdivision
Divide5by3,togetaquotient(q)=1andremainder(r)=2
Writedownthequotientintheanswerlineandtheremainderjustbelowittoitsleftasshown
Now,thebasedividendis21
Addthequotientdigit(1)totheinitialdividendtoget22.
Thenextnumberfordivisionis22whichondivisionby3givesq=7andr
=1.Theresultnowlooksas
Now,thebasedividendis17
Addthequotientdigit(7)tothebasedividendtoget24
Thenextnumberfordivisionis24whichondivisionby3givesq=8andr
=0.Theresultnowlooksas
Repeatthedivisiontogetthefinalanswerasshownbelow.
Thefinalresultupto5digitsofaccuracyis
andtheresultof5/28=0.17856.
42.Examplesforauxilliaryfractions-divisorsendingwith8
IV)Divisorsendingwith‘7’
Letusconsidercaseswherethedivisorendswith7.
E.g.17,27,37,57etc.
GeneralSteps
Add3tothedenominatorandusethatasthedivisor
Remove the zero from the divisor and place a decimal point in thenumeratorattheappropriateposition
Carryoutastep-by-stepdivisionasexplainedbefore
Everydivisionshouldreturnaquotientandaremainderandshouldnotbeadecimaldivision
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:
Letusnowseethedetailsfor5/27
Steps
Add3tothedivisortoget30
Themodifieddivisionisnow5/30
Remove thezero from thedenominatorbyplacingadecimalpoint in thenumeratorgivingthemodifieddivisionas0.5/3
Wewillnotwritethefinalansweras0.16666but
Wewillcarryoutastep-by-stepdivisionexplainedbelow
Stepsforstep-by-stepdivision
Divide5by3,togetaquotient(q)=1andremainder(r)=2
Writedownthequotientintheanswerlineandtheremainderjustbelowit
toitsleftasshown
Now,theinitialbasedividendis21
Multiplythequotientdigit(1)by2andaddtothebasedividend21toget23
Thenextnumberfordivisionis23whichondivisionby3givesq=7andr
=2.Theresultnowlooksas
Now,thebasedividendis27
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’
Letusconsidercaseswherethedivisorendswith6.
E.g.26,36,56,86etc.
GeneralSteps
Add4tothedenominatorandusethatasthedivisor
Remove the zero from the divisor and place a decimal point in thenumeratorattheappropriateposition
Carryoutastep-by-stepdivisionasexplainedbefore
Everydivisionshouldreturnaquotientandaremainderandshouldnotbeadecimaldivision
The quotient and the remainder is taken as the next base dividend as
explainedbefore
Sincethelastdigitofthedivisoris6whichhasadifferenceof3from9,ateachstage,wemultiply thequotientby3andadd to thebasedividend tocomputethenextgrossdividend.
Example1:
Letusnowseethedetailsfor6/76
Steps
Add4tothedivisortoget80
Themodifieddivisionisnow6/80
Remove thezero from thedenominatorbyplacingadecimalpoint in thenumeratorgivingthemodifieddivisionas0.6/8
Wewillnotwritethefinalansweras0.075but
Wewillnowcarryoutastep-by-stepdivisionasexplainedbelow
Stepsforstep-by-stepdivision
Divide6by8,togetaquotient(q)=0andremainder(r)=6
Writedownthequotientintheanswerlineandtheremainderjustbelowit
toit’sleftasshown
Now,theinitialbasedividendis60
Multiplythequotientdigit(0)by3andaddtothebasedividend60toget60.This is a redundant stephere since thequotientdigit is zero,but it isshownforuniformity.
Thenextnumberfordivisionis60whichondivisionby8givesq=7andr
=4.Theresultnowlooksas
Now,thebasedividendis47
Multiplythequotientdigit(7)by3toget21andaddittothebasedividend(47)toget68
Thenextnumberfordivisionis68whichondivisionby8givesq=8andr
=4.Theresultnowlooksas
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)
Multiplyitbytheosculatortoget36(18×2)
Subtractmaximumpossiblemultiplesofthedivisorfromit.Hence,19canbesubtractedfromittoget17.
Digit3
Addittothedigittoitslefttoget19(17+2)
Since19isdivisibleby19,thegivennumberisalsodivisibleby19.
Computationofpositiveosculatorsfordifferentnumbers
48.Examplesforcheckingthedivisibilityofthefollowingby19
49.Examplesforcheckingthedivisibilityofthefollowingby29
50.Examplesforcheckingthedivisibilityofthefollowingby13
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
Multiplyitbytheosculatortoget6(3×2)
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.
So,thefinalansweris261121.
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.