;Fkk f’k[kk
e;wjk.kka ukxkuka e.k;ks ;Fkk A
.
r}}snk³~x
‘kkL=k.kka xf.kra ew/kZfu fLFkre~ AA
Like the crest on the heads of peacocks,
like the gems on the hoods of the cobras, mathematics is at the top of Vedānga
śāstras.(Vedāńga Jyaotişa)
We all boast on our present
knowledge of mathematics but the foundation of it is some how related to the
ancient mathematics. The most applauded Pythagoras problem was much earlier
known to the Indians. One of the greatest problems squiring the circle,
which remained unsolved for centuries, was solved by the writers of Sulva-Sutras. The well known formula nCr=n!/n-r!r! used in
the combination was given by the Indian mathematician Mahavir(9th century) much earlier, but was wrongly credited
to Herigone (1634AD). Though Prof.Smith in his History of mathematics
vol.2 (page527)give credit to Bhaskara
for giving the formula of nCr,and nPr to
find the combination and permutation of n
things taking r of them at a time.
The
Bakshali Manuscript was discovered at
a village called Bakshali about 70
miles away from Taxila in 1881 in the
course of excavation. The manuscript consisting of 70 birch-bark leaves is rich
with the mathematical content.
 |
| The Bakshali Manuscript |
Ahmes papyrus or the Rhind papyrus, the main source of information of
Egyptian mathematics which is now kept in the British Museum
2200 BC. |
| Ahmes Papyrus |
 |
| Moscow Papyrus |
This papyrus
is called the Ahmes Papyrus in the honour of Ahmes who was responsible for copying it
from an ancient mathematical book. It has 80 problems on algebra, geometry,
mensuration etc. The other oldest book in the mathematics is known as Moscow Papyrus has its origin in 1700 BC contains 25 problems. It is now kept in Moscow
Museum of fine Art.
Euclid's Element
Aryabhatta 1(476AD) was one of the remarkable mathematicians and
astronomer of India
 |
| A page from Arybhata's book Arybhatiam |
The present chapter will give the
reader the depth of ancient mathematical problems taken from the Sulbha sutras(there are 7 sulbha sutras known by the name of Bodhayana,Apasthambha Katyayana,Manava, Maitryana ,Varaha and Vadhula after the names of rishis or sages who wrote them.),the ancient Chinese text book Jiuzhang
Suanshu or the Nine Chapters on the
Mathematical Arts(a practical hand book of mathematics consisting of 246
problems.), Vedic and Jain
Mathematics book, Aryabhattiam, Lilavati, Ahmes papyrus, Moscow Papyrus and many other books. All add
the beauty of ancient mathematics and exhibit the marvelous knowledge of theirs
in mathematics.
 |
| Nine Chapters on the Mathematical Arts |
Problem 1:
Given Kou = 3cĥih , Ku =4 cĥih.
Find the length of hsien? (Chinese Problem)
Solution:
Here kou (base)=3 cĥih
Ku (perpendicular AB)=4 cĥih
hence, (hsien)2 =32+42=
52 (applying Pythagoras theorem)
So hsien =5 cĥih
Problem 2:
There is a square town of unknown
dimensions. There is a gate in the middle of each side. Twenty paces outside
the north gate is a tree. If one leaves the town by the south gate, walks 14
paces due south, and then walk due west for 1775 paces, tree will just come
into view. What are the dimensions of the town?
(Chinese Problem)
Solution:- 250
Some more problems for practice
from Chinese mathematical book
a) Suppose a field has width 1+ 1/2
+ 1/3 + ... + 1/n. What must its
length be if its area is 1?
b) A good runner can go 100 paces while a poor runner covers 60 paces. The poor runner has covered a
distance of 100 paces before
the good runner sets off in pursuit. How many paces does it take the good
runner before he catches up the poor runner.
c) A cistern is filled through five canals.
Open the first canal and the cistern fills in 1/3 day; with the second, it fills in 1 day; with the third, in 21/2 days; with the fourth, in 3 days, and with the fifth in 5 days. If all the canals are opened, how
long will it take to fill the cistern?
d) Certain items are purchased jointly. If each
person pays 8 coins, the
surplus is 3 coins, and if each
person gives 7 coins, the
deficiency is 4 coins. Find the
number of people and the total cost of the items.
e) There are two piles, one containing 9 gold coins and the other 11 silver coins. The two piles of coins weigh
the same. One coin is taken from each pile and put into the other. It is now
found that the pile of mainly gold coins weighs 13 units less than the pile of mainly silver
coins. Find the weight of a silver coin and of a gold coin.
Problem 3:
The two parallel faces of a figure resembling
a panava( a drum shaped musical instrument) are each 8 units, the central width
2 units, and the length between the faces is 16 units. Say O mathematician,
what is the area of this figure resembling the panava?
Solution:
Ar(ABCD)= Ar(AFED)+ Ar(EFBC)
=
½.AF(EF+AD)+1/2. FB(EF+BC)
=1/2 AF.EF+1/2AF.AD+1/2FB.EF+1/2FB.BC
=1/2EF.AB+1/2AD.AB (since AD=BC)
=1/2.AB.(EF+AD)
= ½
.16.(2+8)
= 80 sq.units.
Problem4:
There is a reservoir of water of
dimension 6x12. At the corner of the reservoir there is a fish and at the north west corner there is
a baka(crane). For the fear of the crane the fish, crossing the reservoir
hurriedly went towards the south in an oblique direction but was killed by the
crane who came along the sides of the reservoir. Give out the distances traveled
by them (assuming their speeds are same)?
Solution:-
Let LBQP is a reservoir in which
BQ=LP=12
LB=PQ=6
LB
= eastern side, PQ= western side.
LP=northern
and BQ= southern side.
The fish is at L and crane at P.
Distance traveled by them = LO=PQ+QO
Produce OQ to C such that OC=OL
and draw a circle passing through L.
BC=BQ+PQ=12+6=18
Let CB produced intersect the
circle at A.
Then AB.BC=LB2
AB.18= 62=36
So AB =2
Hence AC=18+2=20 and OC=OL=20/2=10
(The problem 3and 4 are due to Bhaskara I, the well-known Indian
mathematicians. He wrote three famous books Maha
Bhaskariyam,Laghu Bhaskariyam and Aryabhattiya Bhasya.)
Some more problems for practice
from Bhaskara I
 |
| Lilavati of Bhaskaracharya |
a)
O friend, what is the area of the scalene ∆ in which
one lateral side is 13, the other 15 and the base 14?
b)
A hawk is sitting on the pole whose height is 18
cubits. A rat which had gone out of its dwelling, at the foot of the pole, to a
distance of 81 cubits, while returning to its dwelling, is killed by the cruel
hawk on the way. Say how far has it gone towards the hole and horizontal motion
of the hawk (the speeds of the rat and the hawk being the same).
c)
If 8 inch out of 32 inch of the angular diameter of the
moon is eclipsed (at a lunar eclipse) by the shadow of the diameter 80 inch, I
want to know what are the arrows (sara) of the interceptor arc of the shadow
and the full moon.
Problem5:
In a love quarrel arising between
the loving couple, the lady’s necklace made up of pearl became sundered and the
pearls fell helter-skelter. A third of the pearls in the necklace reached the
maid-servant there, a sixth fell on the bed, then a half of the last fraction,
again a half of the later and so on, counting this way six times in all, of the
pearls fell on the floor scattered(in groups); and there were found 1161 pearls
still in necklace. If you know how to work find the total number of pearls.
Solution:
Let the no. of pearls in the
necklace be x
Hence by the question,
x/3+x/6+x/12+x/24+x/48+x/96+x/192+x/384+1161
=x
(128+64+32+16+8+4+2+1)x +1161 =x
384
or, x-255x/384=1161
or, 129x/384=1161
or, x=3456
Problem 6:
A wizard having magical power sees a cock
fight going on and speaks privately to both the owners of the cocks. To one he says, “If your bird wins then you
give me your stake money, but if you don’t win I shall pay you 2/3 of its.”
Then going to other owner he promises in the same way to give ¾ if he loses.
Tell me, O ornament of the first rate mathematicians the stake of money of each
of the cock owners if from either of them the wizard would earn a profit of 12
gold coin.
Solution:
Let A and B have x and y stake of
money.
If A wins, hence B loses
x-3y/4=12-----------(1)
If B wins and A loses
y-2x/3 =12---------(2)
on solving we get, x=42,y=40
Problem7:
A snake’s hole is at the feet of
a pillar, nine feet high and a peacock is perched on its summit. Seeing a snake
at a distance of thrice the height of the pillar moving towards its hole, the peacock
pounces obliquely upon the snake. Say at what distance from the snake’s hole they
meet if both of them move and at the same speed.
Solution:-
Let B be the position of the
snake hole and A be the position of Peacock, E the position of snake.
Since AB =9, and BE =27
If BC =x, CE=AC =27-x (since they
cover the equal distance in equal time.
)
AC2 =AB2
+BC2 (by Pythagoras theorem)
(27-x)2 =92
+x2
On solving we get x =12
Hence at a distance of 12 feet
from the hole they will meet.
(Since due to influence of
earth’s gravitation the path of the bird will not be straight but here the path
is assumed to be straight.)
Problem 8:
The mixed price of 9 citrons and
7 fragrant wood apples is 107; again the mixed price of 7 citrons and 9
fragrant wood apples is 101.O arithmetician, tell me quickly the price of a
citron and a wood apple here.
Solution:-
If the price of a citron is Rs.x and a wood
apple cost is Rs. y
The equation will be 9x+7y
=107and 7x+9y =101
On solving we get x =8 and y =5
Problem 9:
A powerful un -vanquished excellent
black snake which is 80 angulas in length, enters into a hole at the rate of 7
½ angulas in 5/14 of a day, and in the course of ¼ th of a day, its tail grows
¼ of a angula. O ornament of mathematicians, tell me at what time this snake
will enters fully into the hole.
Solution:
Let x be the number of days snake
takes to enter the hole.
By question,
80+(x+x/4) =15/2x
¼ 5/14
or, 80+5x =21x
or, 16x=80
or, x=5
Problem10:
Four different sums of money
equal 40, 30,20 and 50 respectively are lent out at the same rate of interest for
5, 4, 3 and 6 units of time respectively. The total interest is 34. What is the
interest that each amount fetches?
Solution:
Let the interest received be x1,
x2 ,x3 and x4
x1 = x2 =
x3 = x4
40*5 30*4 20*3 50*6
by the method of proportion
Each ratio = sum of antecedent/
sum of consequent
x1+x2+ x3 + x4 = 34 = 1
200+120+60 +300 680 20
(Since the sum of interest is 34)
Equating each ratio
x1/200 =1/20 ,x2/120
=1/20 ,x3/60 =1/20 ,x4/300 =1/20 gives
x1 =10 , x2 =6 ,x3 = 3 and x4= 15
(Problem 5 to 10 are due to Mahavira
one of the greatest Indian mathematician. The celebrated Jain mathematician was
one of the gems of the royal court of the famous king Amoghavarsa. Mahavir
wrote his famous book Ganita Sara Sangraha in 850 AD. The text of Ganita Sara
Sangraha contains topics in arithmetic, algebra, geometry and mensuration
divided into nine chapters of about 1100 slokas. )
Some more problems for practice
from Mahivircahrya
a)
Of a collection of mango fruit, the king took 1/6 th
, the queen 1/5th of the
remainder, the first prince 1/4th of the remainder, the second
prince 1/3rd of the remainder and the third prince ½ of the
remainder and three mangoes remained, O you who are clever in miscellaneous
problem in fractions, give out the measure of
the collections of the mangoes?
b)
Out of a certain number of birds, one fourth moves out
to lotus plant, one fourth and seven times the square root of the number move
to a hill and 56 birds remain on the tree. What is the total number of the
birds?
c)
Three merchants found a purse on the way. One of them
said, “ if I get this purse I shall become twice as rich as both of you.” Then
the second said, ‘I shall become as rich’ the third man said, ‘I shall become
five times as rich.’ What is the value of the money in the purse as well as with
each of them
Problem 11:
700 loaves are to be divided among recipients
where the amounts they are to receive are in the continued proportion 2/3 : ½:
:1/3 : ¼ .
Solution:
If ‘a’ is the base number then the fractions will be x1 =2a/3
,x2=a/2, x3=a/3 and x4=a/4
Now, on adding,
8a+6a+4a+3a =
700
12
on solving
a=400 the respective value of x1,
x2 ,x3 and x4 can be found by putting the value of a.
Problem12:
This is the 14th problem of Moscow
Papyrus which was considered the most difficult problem of that time which states
that “A pyramid has been divided in such a way that the top area is a square of
length 2 units and bottom area is a square of length 4 units and height 6
units. Find the volume of the frustrum?
Here the word hau represents an unknown
quantity
x
+ x/7 =19
On solving x=133/8
Problem14:
As I was going to St. Ives, I met a man with seven wives; every
wife had seven sacks, every sack had seven cats, every cat had seven kits.
Kits, cats, sacks, and wives, how many were going to St. Ives?
Solution:
This sum requires the knowledge
of geometrical progression 7+49+343+2401=2800
Problem 15:
Multiply 41 x 59
Solution:
In Rhind papyrus there was a
unique method for multiplication and division. Suppose we are to multiply 41 x
59 then does the following activities:
Take 59 and keep multiplying with
1, 2, 4, 8, 16….as shown in the table.
41
|
59
|
1
|
59
þ
|
2
|
118
|
4
|
236
|
8
|
472
þ
|
16
|
944
|
32
|
1888
þ
|
Since 32 x 2=64>41 so we need
not go beyond 32, moreover 1+8+32=41 so only these with correct sign was shown
in the table. Now add the value respective to 1,8,32 and the result is then
59+472+1888=2419, i.e. 41 x 59 =2419
Problem 16:
Divide 1465 by 65
Solution:
The division work in the Rhind
papyrus is also used by doubling the number as done in multiplication.
1
|
65 þ
|
2
|
130 þ
|
4
|
260 þ
|
8
|
520
|
16
|
1040 þ
|
We stop here as 65+130+260+1040
=1465, so the quotient is the sum of number of respective columns, i.e.
1+2+4+16=23
Problem 17:
Divide 10 measures among 10
persons so that each person shall receive 1/8 less than the preceding ones.
Solution:
Here n =10,Sn =10 and
d = -1/8
By applying the sum formula of
AP; Sn= n/2[2a+ (n-1) d] we get a = 25/16 =1 9/16
Problem 18:
Divide 100 square measures into 2
squares such that the side of one of the squares shall be three fourths of side
of others. (This problem is taken from Berlin Papyrus of middle kingdom
dated 2160-1700BC)
Solution:
If the sides of squares are x and
y then the equation becomes
x2+y2=100
and y =3/4 x
on substitution, we get x =8 and
y =6
(Problem 11 to 18 are due to the
oldest mathematical book Ahmes and Moscow
papyrus)
Some more problems for practice
from papyrus
a)
A quantity, its 2/3, its ½ and its 1/7 added together,
becomes 33. What is the quantity?
b)
Given that 13
hekat of upper Egyptian grain is made into 18 des of besha date substitute
beer, and that I des of this makes 2 1/6 des of barley
beer, what is the strength of the barley beer?
c)
A sandal maker
works for 15 days receiving wages every 5 days. If he does the work in 10 days
after what period should he be paid?
d)
Divide 100
loaves among 5 men in such a way that the share received shall be in the
arithmetic progression and that one- sevenths of the sum of the largest three shares
shall be equal to the sum of the smallest two.
Problem19:
Five merchants together buy a jewel. Its price
is equal to half the money possessed by the first together with the money
possessed by the others or one third the money possessed by the second together
with the moneys of the others or one fourth the money possessed by the third
together with the money of the others etc. Find the price of the jewel and he
money possessed by the merchant.
Solution:-
We have the
following system of equations
x1/2+x2+x3+x4+x5
= x1+x2/3+x3+x4+x5
= x1+x2+x3/4+x4+x5 =
x1+x2+x3+x4/5+x5 = x1+x2+x3+x4+x5/6=p-------(1)
if x1/2+x2/3+x3/4+x4/5+x5/6
=q
Then on adding the equation (1) becomes,
377q/60 = p
A number of possible answers can
be obtained. The Bakshali Manuscript has taken the value 60.For instance if q =
60 then we have p =377 and x1=
120,x2 =90, x3=80, x4=75,x5=72
Problem 20:
If a man earns 50 dinars in 8
days, how much will he earn in 12 days?
Solution:-
The Bakshali describes the rule
or three where the three numbers are written as
8 50 12
where 8 = Pramana , 50 = Phala , 12=
Iccha
Answer = Phala x Iccha = 50 x12
Pramana 8
Problem 21:
Find the square root of 41?
Solution:- The Bakshali manuscript has the following statement for finding the square root
‘In the case of a non square number subtract the nearest square number ,
divide the remainder by twice the nearest square; half of the square of this is then divided by the sum
of the approximate root and the fraction, this is subtracted and will give the
corrected root.’
The mathematical formula can be
written as:
√Q = √A2+b =A+b/2A –(b/2A)2
2(A+b/2A)
√41= √36+5
Here Q =41,
A=36 and b=5(41-36 =5 as stated in the method). Hence by the above formula
√41= 6+5/2*6 –(5/12)2
2(6+5/12)
= 6+0.416666666-0.17361111/12.83333333
=6+0.416666666-0.013528138
= 6.403138527
The modern value is 6.403124237
which is correct to four decimal places. The Bakshali manuscript formula for approximating the square root is
faster than the Newton ’s
method in giving √41 to ten decimal places.
Problem 22:
Gold swarnas 1,2,3,4 are subjected to wastage
of mashaksa numbering1, 2, 3, 4 respectively. What is the average number?
Solution:
Average = 1.1+2.2+3.3+4.4
1+2+3+4
=3
Problem23:
A merchant pays duty in certain goods at three
different places. At the first he gives one third of the goods, at the second
one fourth of the reminder and at the third one fifth of the remainder. The total
duty is 24. What was the original amount?
Solution
Let the original amount be x
Then by question,
x/3+1/4(x-x/3)+1/5(x-x/3-x/6)
= 24
so x =40
(Problem 19 to 23 is taken from Bakshali manuscript. The Bakshali
manuscript is a hand book which contains rule and illustration with solved
examples. An inspector of police Mian An-Wan-Udin whose tenant had
probably found it while digging a stone in a village named Bakshashali near
Peshawar in 1881AD took this manuscript to the Assistant Commissioner of
Peshawar who forwarded it to Lahore museum, and this was published in the
Indian Antiquary in 1883.This
contains the example on algebra,
geometry, mensuration and arithmetic.)
Some more problems from Bakshali
manuscript from practice
a)
One person
possesses seven superior horses, another nine ordinary horses, and another ten
camels. Each gives one animal that he possesses to each of the others. They are
then equally well off. Find the price of each animal and the total value of the
animals possesses by each person?
b)
Two Rajputs
are the attendants of a king. For their services one gets 13/6 dinaras a day and the other 3/2. The first owes the second 10 dinaras. Calculate and tell me quickly, in
what time they will have equal amount?
c)
A number added to 5 is a square. When 7 is
subtracted from it, it is also a square. What is the number?
Problem24:
Inside a forest a number of apes
equal to the square of 1/8 of the total number re playing noisy games. The
remaining 12 apes, who are of a more serious disposition, are on a nearby hill
and irritated by the shrieks coming from the forest. What is the total number
of apes?
Solution:
If x be the number of apes then
the following question will change into the equation, (x/8)2+12=x
which on solving will give x=16 and x=48
Problem25:
How many are the variations in the god Shambhu
by the exchange of his ten attributes, held reciprocally in his several hands;
namely the rope, the elephants hook, the serpent, the tabor, the skull, the
trident, the bedstead, the dagger, the arrow, the row; as those of Hari by the
exchange of the mace, the discuss, the lotus and the conch.
Solution:-
Since Shambhu has 10 attributes
those can be exchanged among themselves in 10! ways and 10! =3628800 and similarly Hari has 4 items
and that can be exchanged into 4! =24 ways.
Problem 26:
Find the sum
of all the number obtained by permuting digits of 23456789?
Solution:
Total number
of permutation is 8! =40320
Sum of every
column is (2+3+4+5+6+7+8+9) x40320/8 =221760
Sum of all the
number = 221760(1+10+102+103+….+107)
= 221760 x (108-1)
10-1
= 2463999975360
(Problem 24 to 26 is taken from the texts of great Indian
mathematician Bhaskara who wrote Siddhanta Siromni in the year 1150AD when he
was 36 years old. This book consists of four parts namely, Lilavati,
Bijaganitam, Grahaganitam and Goladhyaya. The first two treated as independent
texts dealing exclusively with arithmetic, geometry and algebra and the last
two deals with astronomy.)
Some more problem from Bhaskara
a) The
square root of half of a swarm of bees went to a rose bush to suck honey
followed by 89 of entire swarm. One lady bee was caught in a lotus flower which
closed at night. She was humming in response to the humming call of a male bee.
O lady, tell me the number of bees in the swarm ?
b) Partha
with raise shot a round of arrow to kill Karna in the war. With half of those
arrows he destroyed (Karna’s) arrows, then killed his horses with four times
the square root, hit Salya with 6 arrows, destroyed Karna’s umbrella, flag and
bow with three arrows and finally beheaded Karna with one arrow. How many arrow did Arjuna shoot?
c) There are two bamboos of height 15 and 10 feet
standing upon the ground and tight strings are tied from the summit of either
bamboo to the foot of the others. Find the perpendicular distance of the points
of intersection of the two strings from the ground?
d) In
an expedition to seize his enemy’s elephants the king marched two yojnas on the
first day. Say, intelligent calculator with what increasing rate of daily march
did he proceed, since he reached his foe’s city, a distance of 80 yojnas in a
week?
Problem 27:
Ass and Mule problem of Euclid
The mule says to the ass, “if you give me on of your sacks,
I would have as many as you.” The ass replied, ‘if you give one of your sacks I
would have twice as many as you.’ How many sacks do they have?
Solution:
If x= number of sacks of mule and y = number of sacks of ass
then the equation will be
x+1 =y+1 and 2(x-1) =y+1
on solving we get x
=5 and y =7
Problem 28:
One third of a number of apples is to be given to one man,
one eight to a second man, one fourth to a third man, one fifth to a fourth
man, a fifth man is to get 10 of them and a sixth man only 1 of them. How many
apples will be required? (From Greek Anthology written around 500AD)
Solution:
Let the number of
apples be x
Then x = x/3 +x/8+x/4+x/5+10+1
x-109x/120 =11
x =120
Problem 29:
His boyhood lasted one sixth of his life; his beard grew
after one twelfth more; he married one seventh more; his son was born 5 years
later, the son lived to half his father’s age and the father died 4 years after
his son. How old was he (Diophantus) when he died?
Solution:
Let x represent the number of years Diophantus lived.
x/3+x/12+x/7+5+x/2+4 = x
on solving we get x = 84
(The above problem was written about 250 AD about an
Alexandrian mathematician Diophantus. There is no much known about his personal
life. Though this problem from Greek anthology tells he died at the age of 84)
Problem 30:
If the cost of a rectangular stone of 9 units x5 units x1unit
is Rs.8, what will be the cost of 2 such stone where measure is 10 units x7
units x2 units?(From Sridhar’s book)
Solution:
Answer = Phala x Iccha = 10x7x2x2x8
Pramana 9x5x1x1
= Rs 49 7/9
Problem 31:
If 2/3 of 2/3 of a certain quantity of a barley is taken and
100 units of barley is added to it the originally quantity is recovered. What
is the quantity?
Solution:
If the quantity is x
Then, 2/3
2/3 x
+100 = x
2/3 x
+100 = x
or, (1- 4/9 )x =100
or,5/9 x=100
or, x=180
Answers of some of the unsolved problem:
Chinese problem:
Answer (b):- 250
Answer (c):-15/74 of a days
Answer (d):- 7 people and 53 coins
Bhaskara I problem:
Answer (a):-84 square units
Answer (b):-distance traveled by the rat is 42 ½ cubits and
horizontal distance is 38 ½ cubits
Answer (c):- 2inches and 6 inches
Mahaviracharya’s
problem:-
Answer (a):- Third prince= 6 mangoes, second prince = 9
mangoes, first prince = 12 mangoes,Queen = 15 mangoes and the king gets 18
mangoes.
Answer (b):-576
Answer (c):-1:3:5:15
Bakshali Manuscript:
Answer (a):-21:14:12
Answer (b):-30days
Answer (c):-11
Bhaskara problem:
Answer (a):-72
Answer (b):-100
Answer (c):- 6ft
Answer (d):-22/7
(Courtesy:-Pictures taken from the website of Mac Tutor Archives with
due permission)
Rajesh Kumar Thakur
rkthakur1974@gmail.com
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