Some Equalization Problems from the Bakhshali Manuscripts

R. C. Gupta

Department of Mathematics
Birla Institute of Technology
P.O. Mesra, Ranchi-835 215

(Received 15 January 1985)

Introduction

The Bakhshali Manuscript (= BM) is the name given to the mathematical work written on birch-bark and found in 1881 near the village Bakhshali (or Bakhshalai), in the Peshawar district, situated about 70 miles from the famous Taxila (now in Pakistan). It was passed on for study and publication to Dr. Rudolf Hcernle of the Calcutta Madrasa. After working on the manuscript for about 20 years, he presented it to the Bodleian Library, Oxford in 1902 where it is still there (Shelf mark: MS. Sansk. d.14).

The BM consists of 70 leaves arranged in the present order by Hoernle from the mass of sheets which reached his hands. The language of the work is Gatha dialect or the Western Prakrit which was used in the then N. W. India till about 300 a.d. (according to Hoernle). It is an admixture of Sanskrit and Prakrit. The script of the manuscript is the Sarada which is said to be developed from the Siddha-matrka about the 8th century A.D. in the N. W. Indian subcontinent. According to Datta, the work is not a treatise on mathematics in the true sense but a running commentary on some earlier original text. Thus we must distinguish between

(i) the date of the original treatise consisting of the sutras (rules) and udaharanas (examples) only,

(ii) the date of the commentary which the present BM work is and which consists of rules, examples, solution of the examples, verification, etc. and

(iii) the date of the present copy of the manuscript which may be quite late and involves many scribes.

Datta places the BM work in the early centuries of the Christian era while the present manuscript is placed in the 9th century by Hoernle. Anyway, there is a. need to make a fresh and thorough study of the BM including the arrangement of the text especially in the light of new findings.

Equalization Problems

The basic rule for solving simple equalization problems is given explicitly in the Aryabhatiya, II, 30 of Aryabhata I (b 476 A.D.) as follows
    "Divide the difference between the rupakas (or rupees) with the two persons by the difference between their gulikas (precious things, animate or inanimate). The quotient is the value (in terms of rupees) of one gulika if the possessions (of the persons) are equal."

That is, if one person has a gulikas (such as horses, gems, etc.) and b rupees, and another has c gulikas and d rupees, then for the equality of their wealths
        ax + b = cx + d,                     .. (1)
where x is the value of each gulika, then
        x = (d - b)/(a - c)                       .. (2)

Instead of equalization of wealth, we may equalize the coordinates (or position vectors) of two travellers which will give us, e.g. the time of their meeting. In fact, the very next verse given by Aryabhata is in this connection. If s1 and s2 are the initial distances (from origin) of two travellers who start simultaneously and move along a line (passing through the origin) with constant speeds v1 and v2, then their meeting time will be given by
$\qquad \qquad \qquad \qquad \qquad s_1 \pm v_1t = s_2 + v_2 t \\ or \qquad \qquad \qquad t = (s_1 - s_2)/(v_2 \pm v_1)$                       .. (3)
Here (s1 — s2) represents their initial distance apart, and (v2 ± v1) is their relative velocity.

We have quoted the Aryabhatiya because it is the earliest extant work whose date is certain, otherwise similar rules and examples are found in the BM (which may be earlier (?) than the time of Aryabhata). For example, BM, folio 3r, rule 15 (Kaye III, p. 171) states:
Sutram: gatisyaiva visesanca vibhaktam purva gantunah, tenaiva kalam bhavati......

"Rule: Divide the distance already covered by the previous traveller by the difference of (their) speeds; that gives the time (of their meeting)...."
      That is, t = s1/(v2 - v1)                       .. (4)
by taking the origin at the starting point of the other traveller. An example on this rule is found in BM, folio 4r (Kaye III, p. 175) which has been restored by Kaye as

'One goes at the rate of 5 yojanas for 7 days, and then a second starts at the rate of 9 yojanas a day. When will they have travelled equal distance?'

This is fully worked out in the BM getting the answer 35/4 days (after the 9yojana - traveller starts). Then the text proceeds to verify this answer by the Rule of Three (trairasikena) but details are missing. Then follows another example with v1 = 18, v2 = 25, s1 = 8v1 (Kaye III, pp. 175-176).

In (4), s1 is the initial distance apart which is covered by the relative velocity (v2 — v1) to give the time t. Similarly if s1 is considered the initial stock (bhandagaram) of any quantity (including wealth), and v1 and v2 are rates of earning/income and expenditure/consumption, then t will be the time in which the stock will be fully consumed. The BM, folio 60r, rule 52 (Kaye III, p. 216) states this as:
Sutram: ayavyaya visesam tu vibhajya drsya samgunam/yallabdham sa bhavet kalam......//

"The known quantity is divided by the difference of earning and expenditure. The quotient becomes the time (when the quantity will be consumed)."

The accompanying example reads (Kaye III, p. 216):

'In two days one earns five; in three days he consumes nine. His store is 30. In what time will the whole stock be consumed?'

The above principle is applied to another class cf equalization (samadhana) problems of which the following is an example (Kaye III, p. 217):

Udaharana: "In three days one pandit earns a wage of five and a second wise man earns six (rasa) in five days. The second is given by the first 7 from his earnings. By this gift, they became equally rich (samadhana). Tell in what time (this happened)?" (BM, folio 60v).

By making a gift g, the gap or difference in their wealth will be 2g which is to be covered by the difference of the rates of their earnings. So the BM, rule 53 for this gives:

"The difference of the daily earnings being divisor of the amount given. Twice the quotient (obtained) is the lime when the gift makes their wealth equal." (Kaye, III, p. 216). That is,
        t = 2g/(e1 - e?2)                       .. (5)
For the above example,
      $t = 14/\left(\frac{5}{3} - \frac{6}{5}\right) = 30 \textrm{ days,}$
as has been worked out (but the karanam is missing) and verified in the BM, folio 61r (Kaye III, p. 217). There is another example of the same type in BM, folio 61 (Kaye III, p. 218):

'Two rajaputras are servants of a king. The wages of one are 2 plus 1/6 a day, of second 1 plus 1/2. The second is given 10 dinaras by the first. Calculate and tell me quickly in what time there will be equality (samatam)'.

The answer found and verified is 30 days. Still another example (g = 7, e1 = 7/4, e2 = 5/6) is there in BM, folio 31r (Kaye III, p. 186).

Samadhana Problems Involving Simultaneous Equations

The BM, folio 3 (Kaye III, pp. 170-171) contains an interesting problem involving mutual gifts by three merchants. The text is not fully preserved but can be restored, from the workingout details given, as follows:

Example: "One possesses 7 horses (asvas), another 9 hayas, and the third 10 camels. Each gives one of his animals to both the others (by which they become equally wealthy). It is required to find the capital of each merchant or the price of each animal. If thou art able, solve me this riddle."

Let x, y, z be the price of a horse, haya, and camel respectively, and let the merchants possess a, b, c of these animals (instead of 7, 9, 10). After the mutual gifts are made, let S be the wealth of each merchant. So that we must have
            (a - 2)x + y + z = S,
            x + (b — 2)y + z = S,
            x + y + (c — 2)z = S.                   .. (6)

These equations can be reduced to the type
  rx + k = sy + k = tz + k = S,                         .. (7)
where       k = x + y + z,
                 r = a - 3, s = b - 3, t = c - 3.

Since S is not known, the set (7) is indeterminate and will have infinite solutions given by
            rx = sy = tz = S - k                   .. (8)
One set of simple integral solutions is obviously
            x = st, y = tr, z = rs                  .. (9)
or         x = P/r, y = P/s, z = P/t,                 .. (10)
where   P = S - k = rst.                       .. (11)

The BM solution, as is clear from the given working details (Kaye III, p. 171) is equivalent to (10). BM first obtains r = 4, s = 6, and t = 7 from given a = 7, b = 9, and c = 10. Then is obtained P = rst = 168, whence
          x = 42, y = 28, z = 24.                   .. (12)

Finally, the capitals of the merchants are found to be 7 X 42, 9 X 28, 10 X 24, or 294, 252, 240 respectively and the samadhana wealth of each to be 262.

It must be noted that the above solution (10) is not the least or lowest integral for which we should take the L.C.M. of r, s, t, instead of P in (10). Then the solution will be x = 21, y = 14, z = 12, and not (12). However, the BM solution (10) is same as said to be given by Sridhara (c. 800 A.D.) whose rule is found quoted in the Kriya-kramakari (c. 1534), a commentary on the famous Lilavati (1150 A.D.), as follows.7
    Purusasamasena hatam datavyam tadvisodhya panyebhyah/
    Sesam parasparahatam svasesabhaktam manermaulyam//

"Multiply the number (of gems) to be given by the total number of men (taking part in the exchange) and subtract the product from the number (of gems) for sale (owed by each). The continued product of the remainders divided by any one's own remainder gives the price of his gem."

Of course this rule is more general being applicable to the case of n merchants possessing n types of gems (instead of animals) and making mutual gift of g gems (instead of one). Here we shall have the system
          r1x1 + k = r2x2 + k = .... = rnxn + k = S         .. (13)
where x1, x2,..., xn are the prices of the gems and
          r1 = a1 - ng, r2 = a2 - ng,....
with           k = (x1 + x2 + .... + xn)g.

It is clear that the lost (or untraced) BM sutra must have been similar to die above rule. In fact, the BM phrase sesam paraspara krtam gunita..svasesena tu vibhaktam is quite comparable to Sridhara's phrase in the above sutra. It must also be noted, however, that the above sutra is found (with only slight variation) in the Ganitasara-samgraha (= GSS), VI, 163 of Mahavira (850 A.D.). Mahavira's example (n = 3, g = 1,a1 = 6, a2 = 7, a3 = 8) has been also reproduced (along with some other examples of his) in the Kriyakramakari but without mentioning the source (may be because he was a Jaina). I have also not been yet able to locate the quoted Sridhara's sutra in any of his extant works.

The BM, folios 1—2 (Kaye III, pp. 168-170) contains another sutra (No. 11) with an elaborate example involving similar theory. The text is mutilated and wrongly interpreted by Kaye. Correct interpretation of the example is given by Gurjar but the BM sutra has not been explained so far (which we attempt to do here). The example is about five merchants (with capitals x1 to x5, say) who together want to purchase a jewel of price S, say. The given conditions lead to the system
        (x1/2) + x2 + x3 + x4 + x5 = S
        x1 + (x2/3) + x3 + x4 + x5 = S
        x1 + x2 + (x3/4) + x4 + x5 = S
        x1 + x2 + x3 + (x4/5) + x5 = S
        x1 + x2 + x3 + x4 + (x5/6) = 5             .. (14)

These easily reduce to the system (13) with g = 1, r1 = -1/2, r2 = -2/3, etc. which are not positive integers, and so solution (10) is not given. However, the values of the unknowns are inversely proportional to r1, r2, etc. So we can take them proportional to 2/1, 3/2, 4/3, 5/4 and 6/5. The BM has reduced these to common denominator (sadrsam kriyate) as
    120/60, 90/60, 80/60, 75/60, 72/60.

So that 120, 90, 80, 75, 72 are taken to be the values of the unknowns as an integral solution and the price of the jewel is found to be 377 (esa mani mulyam). The rest of the working is done to verify all the equations of (14).

Suppose (p/q) is the fractional coefficient of any unknown along the leading diagonal in (14). The value of the corresponding r-coefficient in (13) will be
    (p/q) - 1 = -(q- p)/q, q > p.

Since the value of the corresponding unknown is inversely proportional to r (and the minus sign can be ignored), the prescribed BM sutra (No. 11) rightly asks
    'Amsam visoddhya - cchedebhya kuryat tatparivartanam'
That is, "Subtract the numerator-parts (p) from the denominator (q) and invert it (the fraction)".

Then we reduce the resulting fractions to a common denominator etc. to get a suitable integral solution. This explains the sutra and the working out of the example with it. Interestingly there follows another example in three unknowns for which fractional coefficients along the diagonal in (14) are all negative, namely
        -7/12, -3/4 (with correction), -5/6
so that, in this case by the above sutra and theory, the unknowns will be proportional, 12/19, 4/7 and 6/11. The values of the unknowns are then correctly found to be 924, 836, and 798 respectively and that of jewel's price to be 1095 (Kaye III, p. 170).

Equalization of Uniform and Accelerated Growths

Let
          S1 = a + (a + d) + (a + 2d) + ... to n terms, and
          S2 = b + b + b + ... to n terms
represent the accelerated and uniform growths of any types of quantities. For equalization or samadhana type of problems S1 is equal to S2.

Hence
          [(n - 1) d/2 + a]. n = b.n                 .. (15)
or         (n - 1)d + 2a = 2b
which is comparable to (1) in (n - 1), and hence by (2)
          n = (2b - 2a)/d + 1                                     .. (16)
          = 2(b - a)/d + 1 = (b - a)/(d/2) + 1.                 .. (17)
For the purpose of ancient rhetoric mathematics, it is often necessary and useful to distinguish between the above three forms for n. The form (16) is contained in the BM, folio 8r (Kaye III, p. 172) as: Sutram: Dvigunam prabhavam suddha dvigunam niyatam tatha.
    Uttarena bhajecchesam labdham rupam vinirdiset.

That is, "Subtract twice the first term from twice the constant rate, and divide the remainder by the common difference; then add one (to get n)."

The accompanying example is somewhat: 'One servant does work at fixed rate equivalent to 10 masakam, and another works 2 units on first day increasing by 3 on each subsequent days. In what time will there be equality of their work?'

So that here a = 2, d = 3, and b = 10. But full working is not available, the answer will be found to be n = 19/3.

An interesting point in the working is that (2b — 2a) is correctly found to be 16. Then a wrong phrase 'uttarardhena bhajayet' (instead of 'uttarena bhajayet') was quoted and was afterwards cancelled. It may be that the phrase 'uttarardhena bhajayet' was quoted from another sutra which gave the last form (17) of the rule for which it is correct. In fact, an example (a = 3, d = 4, b = 7) on BM, folio 7v (Kaye III, p.174) has been indeed worked out by this very form of the sutra. Consequently, it is believed that BM folio 7v did contain such a sutra which is lost.11 This folio has another example (a =1, d = 2, b = 5) which began to be solved by the same sutra but the details are lost.

The example in BM, folio 9r has been restored by Hoernle as (Kaye III, p. 173):

"For a certain feast one Brahmana is invited on the first day, and on every succeeding day one more Brahmana is invited. For another feast 10 Brahmanas are invited every day. In how many days will their numbers be equal, and how many Brahmanas were invited."
This example (a = 1, d = 1, b = 10) is fully worked out in the text getting n = 19 days, and then by 'rupona karana', i.e. L.H.S. of eq. (15), the number of Brahmanas invited in the first feast
              = [(19 - 1).(1/2) + 1]. 19 = 190
which will obviously be also the number for the other feast.

As a word-numeral rupa means one (in algebra rupa stands for the absolute term) and BM's frequently used phrase 'rupona karana' refers to the formula
          [(n - 1).d/2+a].n = Sn

The GSS, II, 61 (p. 20) which contains this formula also starts with the phrase 'rupenono', This work, VI, 319 (p. 173) contains the last form (17), while the other form in (17) is found in the Patiganita rule 96, of Sridhara.

Equality of Two Uniformly Accelerated Growths

Let,
        S1 = a + (a + d) + (a + 2d) + ... to n terms,
        S2 = b + (b + e) + (b + 2e) + ... to n terms,

If these two are equal, we must have
        (n - 1)d + 2a = (n - 1)e + 2b
which is again comparable to (1), and hence by (2)
        n = 2(b - a)/(d — e) + 1                         .. (18)

This formula is contained in BM, folio 4v, rule 17 (Kaye III, p. 176) as follows:
        Adyor visesa dvigunam cayasuddhir-vibhdjitam
        Rupadhikam tatha kalam gati samyam tada bhavet.

"Twice the difference of the initial terms divided by the difference of the common differences is increased by one. That will be time (represented by n, cf. kala iha padasyopalaksanam) when the distances moved (by the two travellers) will be same."

The accompanying example reads :

"The initial speed (of a traveller) is 2 and subsequent daily increment is 3. That of another, these are 3 initially and 2 as increment. Find in what time will their distances covered attain equality."

The working is lost, but the answer, by (18),
        = 2(3 - 2)/(3 - 2) + 1 = 3 days.

Another example is found in BM, folio 5r (Kaye III, p. 177). It is on attaining samadhana (equality of wealth), the given rates of earning are a = 5, d = 6, and b = 10, e = 3. The example has been fully worked out by (18) and verified, by the usual 'ruponakarana'. The answer obtained is n equal to 13/3 (days) when each would have pooled 65 units of wealth. It must be noted that the value of n found is fractional which the ancient Indians accepted even as number of terms of a series. Of course here there is no difficulty as it represents time.

Now, we discuss the interesting situation presented in one portion of the BM, folio 4 (Kaye III, p. 176). The available part of a sutra (No. 16) in the recto side of the folio, and the partially extant working of an example in the beginning of the available verso side shows that the intended rule was the following form of (18):
        n = [(b - a)/(d - e)]x2 +                 .. (19)

Since the beginning part of the worked out example is not available, we cannot be sure of given data which is only to be guessed or constructed. Here I shall discuss Kaye's conjectures and then give my own restoration with explanation. All that follows from the available working is that, in the said example, difference of first terms =2 = b - a, say and difference of the common differences = 2 = d - e say.

Hence by (19), n = 3. Also then the sum of first series is found to be 21 (which should also be the sum of the other series). So that we must have (assuming no other lacuna) (7 — d), 7, (7 + d) and (7 - e), 7, (7 + e) as the terms of the series. Now for finding d and e we get only one relation, namely, d = 2 + e. So that there can be many possibilities even of simple type namely (i) 4, 7, 10 and 6, 7, 8; (ii) 3, 7, 11 and 5, 7, 9; (iii) 2, 7, 12 and 4, 7, 10; (iv) 1, 7, 13 and 3, 7, 11.

Kaye took (i) and says it was the (lost) example. He gave no reason (as it was only a guess). At the end of the rule there is a statement that 'sutre bhrantimasti' which means that the commentator or scribe found some confusion or mistake in the rule. Kaye explains this away simply by saying that the above remark is about the wrong numbering of the rule, because the numbering phrase (also) happens to be wrong being 'sodasam sutram 17'. Actually the figure should be 16 (sodasa) instead of 17, but this may be considered a minor scribal slip.14

I think that the remark that 'there is confusion or mistake in the sutra' was made because some different type of trouble was faced while working out the example by applying the rule. I advance the following possibility.

When any new or more general rule is given, it is quite natural to illustrate it with the help of examples already given earlier. Here also it happened so. In order to illustrate that the rule corresponds to (19), the commentator took the same two series which were used to illustrate the rule for (17). These are (see under uniform and accelerated growth):

    1st series, a = 3, d = 4; 2nd series a = 1, d = 2.

Now ancient rules are verbal (not symbolic) and working is done rhetorically (and not by substitution in a formula as we do now). The original sutra needed the "difference" of the first terms and of common-differences. These were found tc be (3 - 1) = 2, and (4 - 2) =2 for the above series. Then the first difference is to be divided by the second difference just obtained thereby getting 2/2 = 1 which is to be doubled and then increased by unity. So that the BM solver got n = 3 finally. He then found the sum of three terms of the first series to be 21. But for the second series the sum cf three terms will be 9 (as he might have found) but he knew that it should also be 21. And here is the trouble which he faced.

The BM solver might have again checked his working steps, but due to lack of symbology and rhetoric way of working, he could not detect the mistake he committed or the fallacy in which he was trapped. But since both the sums should be same (and not 21 and 9 as he got), he was compelled to state that there is some confusion or error in the rule itself. There is also another remark at the end (but the plate is not clear)
            kim prabhutepi likhite
which might mean

"How the (equal sums) are obtained? Even then I am writing them (what they should be)."

Of course, some emendations may also be suggested to make the matters more clear.

The confusion started with the choice of bad example for illustration here because the a and d of one series are both greater than those of the other (so that their sums will never be equal for positive n). Then the BM solver fell prey to verbal "difference" which he took as positive or numerical one, thereby getting wrong n = 3. The correct value, by (19), will be n = —1. And then S1 and S2 will be both equal to 1. But such theoretical discussions are difficult at rhetoric stage.

Our above interpretations are not impossible. The resulting restorations, if plausible, will have some historical value.


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