Japanese vs Vedic Methods for Multiplication

International Journal of Mathematics Trends and Technology (IJMTT) – Volume 54 Issue 3- February 2018

Japanese vs Vedic Methods for Multiplication D.N. Garain#1 and Sanjeev Kumar*2 #

Head, University Department of Mathematics, S.K.M.University, Dumka, Jharkhand, India *Research Scholar, S.K.M.University, Dumka, Jharkhand, India

Abstract -Japanesemultiplicationmethod is very usefulfor a person who has only skill of counting ability without having knowledgeof addition and multiplication. Vertically and crosswise Vedic multiplication method is useful as aspeedy calculator. In thepresent study, we havedescribedbothmethodsin our own manner by keeping Sole idea unchanged. Also it has been given the logic for the procedures.Comparative Study of both methods has been given which reflects the merit and demerit of both Methods. Keywords: Vedic method, Japanese method, Multiplication methods. 1. Introduction Professor Fujisawa Rikitaro(1900)of imperialUniversity of Tokyo created the Japanese multiplication method. His methodis geometric based Multiplication method. After deep study and analysis JagadguruSankaracarya Sri BharatiKrsnaTirthaji Maharaja (1884–1960) of GovardhanaMatha, Puri, India obtained sixteensimpleMathematical Sutras (formulae) from veda. V. S. Agrawala(1965) edited the book ‘Vedic Mathematics’ by taking ideas and concept of sixteen formulae of Sri BharatiKrsnaTirthaji Maharaja ji. The thirdsutra (formula) is known asUrdhva – Triyagbhyam(Verticallyand crosswise).With thisformula,Multiplication of two numbers maybeobtainedvery easilyandquickly. Here, weshall describebothmethodsin ourown mannerkeepingsoleidea Unchangedbygiving thelogicforprocedure andthencomparetheir meritsand demerits. 2. Geometrical Construction of Fujisawa’s Concept of multiplication Fujisawa gavea concept formultiplyingtwo numbers in his own way whichisknown asJapanesemultiplication method.For this,he took help oftwo kinds ofparallellines – Horizontal andVertical.Letmn and pqrare two numberssuch that m, n, p, q, r  N  {0},Nis the set of natural number,mn = 1 × n + 10 × m and pqr = 1 × r + 10 × q + 100 × p. Let us drawm number of closed parallellines inhorizontal directionandafter leavingsomespacebelow, letus drawn number of closed parallellines along thesamedirection (Figure – 1). Now, from the left side, letus draw p numberofclosed parallel lines along the verticaldirectionsuch thatthese parallellinesintersect horizontal parallellines. Similartoabove process, leaving somespaces on the right, let us draw q and rnumbersclosed parallellinesrespectively alongvertical direction.

Fig.–1 Here,m* means m numberof parallellines. 3. Process of Calculation of multiplication

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Letus draw thesamefigureas mentioned in figure – 1.We startour work by counting the number of points of intersection at different position accordingto the orderof different arrow lines.

Fig.–2 Now,we count number of point of intersection made by two kindsof parallellinesat different position A, B, C, D, EandF by following arrowlinesin order. At the position F, The number of pointsof intersectionis ab (let), a,b N  {0} , Where ab = 1 × b + 10 ×a. Here, b is writtenin positionF anda is written below b as carry number andb isincludedin the result. Let, total numberof points of intersectionof parallel linesat E and C whichis equal to cd (let), c,d  N  {0}.Let cd and the carry numbera make thenumberef, f is writtenatthe position E and carry number e is written belowf. f is included in the result left to b. Lettotalnumberof points of intersection of parallellines at D andB whichisequal togh (let).Letgh and e make number ij, j is writtenat the position D and carry number i is written below j. j is includedinthe resultleft to f. The number of pointsof intersection at A is kl (let). kl and i make number st.Let us write st at the positionA. st is included intheresultleft to j. Then, mn ×pqr = stjfb Example – 1. Evaluate: 45 ×123

Fig.–3 Number of points of intersectionatposition F is 15. Here, carrynumber 1 is written below 5. 5isincluded in theresult.

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Total number of points of intersectionat position E and C is 22 which makes 23 withthe carry number 1 and hence 3 is writtenat E and 2 is written below 3 ascarry number. 3 isincluded in the result leftto 5. Totalnumber of pointsof intersection at position D and B is13 whichmakes 15 with the carry number 2 and hence 5iswritten at D and 1 is written below 5 ascarry number. 5isincludedinthe resultleft to 3. Number of points of intersection atposition A is 4 which make 5 with the carry number 1. 5 isincluded in the result leftto 5. Then,  45 × 123 = 5535. Example – 2.Evaluate:

23 × 14

Fig.–4 Then,

 23 × 14 = 322.

Example – 3.

Evaluate:

132 × 141

Fig.-5 Then ,

 132 ×141 = 18612.

4. Geometry and process of counting when 0 is one of the digits When 0 isone of thedigits inanumber then wedraw adotted line as parallel line for zero digit and wedon’t count the points ofintersection of aparallelline with this dotted line. Example – 4.

Evaluate:

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102 × 141


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Fig.–6 Then,

 102 × 141 = 14382.

Example – 5.

Evaluate:

34 × 203

Fig.–7 Then,

 34 × 203 = 6902

5. Logic for the Procedure The followingfigure shows the positionof differentpoints.

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Fig.–8 Place value at the position F is Unit × Unit = Unit Place value at the position E is Unit × Tens = Tens Place value at the position C is Tens × Unit = Tens Place value at the position D is Unit × Hundreds = Hundreds Place value at the position B is Tens × Tens = Hundreds Place value at the position A is Tens × Hundreds = Thousands The number of points of intersectionattheposition E and C areobtainedbyadding them fortheirsame place values. Similarly, points of intersection at the position D and B are added for their same place values. At theposition F, Number of points of intersection = n × r = ab (let), a,b N  {0}, Where, ab = 1 × b + 10 × a. Here, a be the carry number and b isincludedin the result. Total number of points of intersection at the position E and C be n × q + m × r = cd (let); c, d N  {0}. The carry numbera is added with cd whichgives cd + a = ef, Where eisthe carry number. f is included inthe result left tob. Totalnumber of points of intersectionatthe position D and B be n × p + m × q =gh (let); g, h  N  {0}.The carry number eisadded with gh which gives gh + e=ij,Where i is the carry number. jisincludedin theresult left to f. Numberof points of intersectionat A bem × p = kl (let). The carry number iis added with kl which gives kl + i = st.stisincluded inthe resultleft to j. Finally, we getthe resultof the multiplicationby arranging the numbers st, j, f and b at theirplace value.  mn × pqr = stjfb We may describeexample – 1 withthislogic as follow Number of points of intersection at F = 5 × 3 = 15; Here, 1 isthe carry number. Total numberof pointsof intersection at E and C be5 × 2 + 4 × 3 = 22. Adding the carry number 1, we get 22 + 1 = 23;here, 2 isthe carrynumber. Total number of pointsof intersection at D andBbe 5 × 1 + 4 × 2 = 13. Adding the carry number2, we get 13 + 2 = 15;here, 1 is thecarrynumber. Number of points ofintersection at A be 4 × 1 = 4.Addingthe carrynumber 1, we get 4+1=5 Hence,  45 × 125 = 5535. 6. Commutativitycharacterof Japanese multiplication method As thenumber of points of intersectionatdifferentpositionsis obtained by multiplying number of horizontal parallel lines with number of vertical parallellineswhich mustbe equal to the multiplication of numberofverticalparallellineswith number of horizontal parallel lines. So,it shows the commutativitycharacter. 7. Verticallyand CrosswiseVedicmethod formultiplication

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For vertically and crosswise product of mn and pqr, we have to makenumber of digits of two number to be same. For this we include 0 (Zero) at hundreds place of mn and mn is written as 0mn. Multiplication is written as -

Multiplication process is as follow (i) Step – 1.

p q r    = r × n = ab 0 m  n  a, b  N  {0}, ab = 1 × b + 10 × a

Fig.–9

Here, a be the carry number and b is included in the result.

p q  0  m

(ii) Step – 2.

r  = n × q + r × m = cd n 

andcd + a = ef Here, e be the carry number and f is included in the result left to b of step – 1.

Fig.–10

p q r   = n × p + r × 0 + q × n = gh 0 m n

(iii) Step – 3.

andgh + e = ij Here, i be the carry number and j is included in the result left to f of step – 2.

Fig.–11

p qr  = m × p + q × 0 = kl   0 m n

(iv) Step – 4.

andkl + I =st

Fig.–12

Here, s be the carry number and t is included in the result left to j of step – 3.

 p q r   =p×0=0 0  m n

(v) Step – 5.

ando + s = s Here, s is included in the result left to t of step – 4. Hence,  mn ×pqr = stifb Example – 7. Evaluate: We write

(i) Step – 1.

45 × 123

1 2 3   = 3 × 5 = 15 0 4  5 

Here, 1 be the carry number and 5 is included in the result.

(ii) Step – 2.

Fig.–13

1 0

Fig.–14

 2 3   = 5 × 2 + 3 × 4 = 22  4 5

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and22 + 1 = 23

Fig.–15

Here, 2 be the carry number and 3 is included in the result left to 5 of step – 1.

(iii) Step – 3.

1 2  0 4

3  = 5 × 1 + 3 × 0 + 2 × 4 = 13 5  and13 + 2 = 15

Fig.–16

Here, 1 be the carry number and 5 is included in the result left to 3 of step – 2.

(iv) Step – 4.

1  0

2 3  =4×1+2×0=4 4  5

and4 + 1 = 5 Here, 5 is included in the result left to 5 of step – 3.

(v) Step – 5.

Fig.–17

1  2 3 =1×0=0   0 4 5

Here, 0 is included in the result left to 5 of step – 4. Hence,  45 × 123 = 5535. In short we may calculate this multiplication with the help of above process as follow

0 5152315

Fig.–18

 Answer = 5535.

Example – 8. Evaluate: 23 × 14 Multiplication process is done by following steps -

Fig.–19

3 12 12

 Answer = 322. Example – 9. Evaluate:

132 × 141

1 8 16 11 2

 Answer = 18612. Example – 10. Evaluate:

102 × 141

1438 2

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 Answer = 14382. 8. Logic for the Procedure Sincethemultiplicationprocess isdone by following steps –

Fig.–20 Place values of dots are shown in table-1 Table-1 (Place value at different position of dots)

Where, U = Unit, T = Tens, H = Hundreds. Step – (i) is Unit × Unit = Unit Step – (ii) is Unit × Tens + Unit × Tens = Tens Step – (iii) is Unit × Hundreds + Unit × Hundreds + Tens × Tens = Hundreds Step – (iv) is Tens × Hundreds + Tens × Hundreds = Thousands Step – (v) is Hundreds × Hundreds = Ten Thousands

9. Comparision of two methods (i)Japanese method of multiplicationis performed with thehelp of geometrical figure where as in Vedicmethod it is evaluate mentally. So, Japanese methodis moreeffectivefora personhaving lessknowledge. (ii) Without knowledgeof additionand multiplicationtablebut with the help of counting ability,theJapanesemethod of multiplicationcan be performed where as during the course of Vedic method, the knowledge of addition and multiplicationis necessary. (iii) Japanese methodismore time consumingand Vedic methodis lesstime consuming.So, Vedic method actsas a speedycalculator. (iv) In Japanesemethod, more spaceis necessary but in Vedic methodvery less space is required i.e.,in a singlelineit can beevaluated. (v) Numberscontainingsmall digitsgivesquick resultinJapanesemethod sincenumber of lines drawn isless.In Vedicmethod,it willtake less timethan the same caseof Japanese method. (vi) Numbercontaininglarge digits willtake moretimeto draw more number of parallel lines. So, inthiscase Japanesemethod ismoretime consuming. But Vedic methodis easyand lesstimeconsuming with a little practice. References [1]FugisawaRikitarou,SugakuKyoujyuhouKougiHikki, DainihonnTosyo, 1900. [2]JagatguruSankaracarya Sri BharatiKrsnaTirtha Maharaja and, V.S.Agrawala,Vedic mathematics, MotilalBanarsidass Publishers Private Limited, Delhi, pp. 33 – 44, 1965. [3] A.P. Nicholas, K.R.William, and j.Pickles, Vertically and Crosswise,MotilalBanarsidassPublishersPrivate Limited, Delhi, pp.1 – 3, 1999.

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