Monthly Archives: March 2020

Palindrome Numbers

By Chandrahas M. Halai

1 x 1 = 1

11 x 11 = 121

111 x 111 = 12321

1111 x 1111 = 1234321

11111 x 11111 = 123454321

111111 x 111111 = 12345654321

1111111 x 1111111 = 1234567654321

11111111 x 11111111 = 123456787654321

111111111 x 111111111 = 12345678987654321

I have received this as Whatsapp forward from many of my contacts with the title Number Magic. All the squares calculated here are Palindrome Numbers. Palindrome Numbers are the numbers which when read from either left-to-right or right-to-left come out to be the same.

You might be wondering who must have been the person to come out with this fascinating thing. It was Digambara Jain monk Mahavira (815 – 877 CE). In his book Ganitasarasangraha written around 850 CE he had calculated the value for

111111 x 111111 = 12345654321

At that time the rest of the world didn’t even know how to count using decimal numbers.

 

Romance and Indian Mathematics

By Chandrahas M. Halai

In my earlier article, Mathematics, Everywhere! https://www.indictoday.com/quick-reads/mathematics-everywhere-solving-culinary-problem-with-mathematics/ , I had quoted a verse from ancient Indian mathematician and Jain monk Mahaviracharya’s book Ganitasarasangraha (written around 850 AD):

In worldly life, in Vedic learning, in religious practice, in business, in everything, Mathematics is useful. In romance, economics, in music, dance and drama, in cooking, medicine and in architecture, in prosody, poetry, logic and grammar, and in all the arts, Mathematics reigns supreme. It is used in calculating movements of the Sun and other heavenly bodies. Whatever there is in all the three worlds, which are possessed by moving and non-moving beings- all that indeed cannot exist as apart from Mathematics?

You might be wondering whether the reference to romance in the above verse is a mistake. No! It is not. It is very much there.

Actually some examples in ancient Indian mathematics work alluded to romance and love making, one of them is in Ganitasarasangraha itself. Ancient Indian mathematicians had a sense of humour and were fun-loving. They used to make learning and doing mathematics interesting and fun-filled activity. Let me present here two examples from Bhaskaracharya’s Lilavati (written in 1150 AD) which refer to romance.

लोलाक्ष्या प्रियनिर्मिता वसुलवा भूषा ललाटेऽर्पिता

यच्छिष्टास्त्रिगुणाद्रिभागकलिता न्यस्ता कुचान्त: स्त्रजि ।

शिष्टार्धं भुजनालयोर्मणिगण: शेषाब्धिकस्त्र्याहत:

कांच्यात्मा मणिराशिमाशु वद मे वेण्यां हि यत्षोडश ॥

A lover gave his fiancée some jewels for making ornaments. She used one-eighth of them for an ornament for the forehead. She used 3/7th of the remaining for a necklace. Half of the remainder were used to make armlets. Three quarters of the remaining jewels, along with little tinkling bells, were used to make a belt. Finally, she put 16 jewels in her wreathed hair. Find quickly the total number of jewels?

(Which lady wouldn’t like to have such a fiancé?)

Solution:

Let the total number of jewels be x.

The lady used 1/8th of the jewels for an ornament for the forehead, i.e. x/8.

Therefore, x – (x/8) = 7x/8 are the remaining jewels.

She used 3/7th of the remaining jewels for making a necklace. i.e. (3/7) * (7x/8) = 3x/8

Therefore, now (7x/8) – (3x/8) = x/2 jewels remain.

Of these, half were used to make armlets. i.e. x/4

Therefore, now x/4 jewels remain.

Of these, 3/4th was used to make a belt. i.e. (3/4)(x/4) = 3x/16

Therefore, now (x/4) – (3x/16) = x/16 jewels remain, which the lady had put into her hair.

Hence, we have

x/16 = 16

x = 256

Therefore, the total numbers of jewels that the lover had gifted his fiancée were 256.

 

Let us now look at another problem from Lilavati :

हारस्तारस्तरुण्या निधुवनकलहे मौक्तिकानां विशीर्णो

भूमौ यातस्त्रिभागः शयनतलगतः पंचमांशोऽस्य दृष्टः ।

प्राप्तः षष्ठः सुकेश्या गणक दशमकः संगृहीतः प्रियेण

दृष्टं षट्कं च सूत्रे कथय कतिपयैमौंक्तिकैरेष हारः ॥

In a coital sport of a couple, the lady’s pearl necklace was broken. One-third of the pearls fell on the ground; one-fifth went under the bed. The lady collected one-sixth and her lover collected one-tenth. Six pearls remained on the original thread. Find the total number of pearls in the necklace.

Solution:

Let the total number of pearls in the necklace be x.

Hence, we now have

(x/3) + (x/5) + (x/6) + (x/10) + 6 = x

Therefore, the total numbers of pearls in the necklace were 30.

From this example, it is clear that talking about love making was not a taboo during Bhaskaracharya’s times.

 

Here, let me present a similar example from Ganitasarasangraha :

One night, in a month of the spring season, a certain young lady was lovingly happy along with her husband on the floor of a big mansion, white like the moon, and situated in a pleasure-garden with trees bent down with the load of the bunches of flowers and fruits, and resonant with the sweet sounds of parrots, cuckoos and bees which were all intoxicated with the honey obtained from the flowers therein. Then on a love-quarrel arising between the husband and the wife, that lady’s necklace made of pearls broke and fell on the floor. One-third of the pearls reached the maid-servant there; 1/6 fell on the bed; then ½ of what remained (and one-half of what remained thereafter and again ½ of what remained thereafter) and so on, counting six times (in all), fell everywhere; and there were found to remain (unscattered) 1,161 pearls, and if you know how to work on fractions, give the total number of pearls in that necklace.

It is quite possible that Bhaskaracharya might have been inspired by this and other examples from Ganitasarasangraha.

Also, note that in the 9th century AD, a Jain monk, Mahavira was not averse to talk about love and romance.

Three Digit Kaprekar’s Constant

By Chandrahas Halai

 

This is in continuation to my earlier article on Kaprekar’s constant, 6174.

https://www.indictoday.com/quick-reads/kaprekar-constant/

 

Let’s carry out Kaprekar’s operations on some three digit numbers where all the digits are not the same.

 

Let us apply Kaprekar’s operations on 628, the year Brahmagupta wrote Brahmasphutasiddhanta.

1) 862 – 268 = 594

2) 954 – 459 = 495

 

Now, let us apply Kaprekar’s operations on 850, the year Mahaviracharya wrote Ganitasarasangraha.

1) 850 – 058 = 792

2) 972 – 279 = 693

3) 963 – 369 = 594

4) 954 – 459 = 495

 

Now, let us try Kaprekar’s operations on 443.

1) 443 – 344 = 099

2) 990 – 099 = 891

3) 981 – 189 = 792

4) 972 – 279 = 693

5) 963 – 369 = 594

6) 954 – 459 = 495

 

We always get the number 495 when we carry out Kaprekar’s operations on any three digit number where all the digits are not the same in at most 6 steps. The number 495 is the three digits Kaprekar’s constant.

 

As a tribute to Ganitanand Kaprekar, let’s carry out Kaprekar’s operations on 357.

1) 753 – 357 = 396

2) 963 – 369 = 594

3) 954 – 459 = 495

A cyclist and Trains

By Chandrahas M. Halai

train by trees against blue sky

Photo by RAJAT JAIN on Pexels.com

This article is inspired by a problem “Tramcars and a Pedestrian” from the book “Algebra can be Fun” by Yakov Perelman. I am presenting here a modern day adaptation of the problem and an easy to understand solution of the same.

While cycling along train tracks I noticed that a train caught up with me every 9 minutes, and every four and a half minutes a train coming in the opposite direction passed me. Trains and my cycle are moving at a uniform rate.

Can you figure out what the time interval is between trains leaving their terminal stations?

Also find the speed of my cycling as compared to that of the train?

Solution:

Challenging problem, isn’t it? But at the same time it is pretty interesting.

I live in the lively city of Mumbai. In the early hours of the morning, everyday, local suburban trains start their journey from CST terminal station towards another terminal station Kalyan at regular intervals. Similarly the trains at Kalyan ply towards CST station at the same interval.

In Mumbai I stay at the suburb of Ghatkopar. As a part of my exercise routine every morning I cycle on the road along the railway tracks in the direction of CST.

The best strategy to solve difficult problems is to break them into simple units.

We will split the current problem into two simple cases.

In the first case we will analyse the motion of the cycle and the trains moving in the direction of CST.

And in the second case we will analyse the motion of the cycle and the trains coming from CST and going towards Kalyan.

Firstly, let the regular interval at which the trains are leaving the terminal stations be x minutes.

Case I

Myself and the trains moving in the direction of CST

Refer to Figure 1.

Let us assume that at some time t = 0 minutes, I and train T1 are at some point A on the railway tracks. Then, at time t = x minutes the next train T2 will arrive at the same point A. This train T2 will catch up with me at time t = 9 minutes at point B.

I have covered the distance between points A and B (distance AB) in 9 minutes. The same distance AB will be covered by train T2 in (9 – x) minutes.

Therefore, the distance that I travel in a minute will be covered by train T2 (or any other train) in = (9 – x) / 9 minutes.

 

Figure 1

Figure1
 

Case II

Myself and the trains moving in the direction of Kalyan

Figure 2

Figure2

Refer to Figure 2.

Let us assume that at some time t = 0 minutes, I and train T3 are at some point P on the railway tracks. At the same time t = 0 minutes, train T4 is at some point Q on the railway tracks. This train T4 will meet me at time t = 9/2 minutes at point R. Also, at time t = x minutes, train T4 will reach point P.

I have covered the distance between points P and R (distance PR) in 9/2 minutes. The same distance PR will be covered by train T4 in (x – (9/2)) minutes.

Therefore, the distance that I travel in a minute will be covered by train T4 (or any other train) in = (x – (9/2)) / (9/2) minutes.

Now equal distances are covered in equal time intervals. Hence, from case I and II, we have

(9 – x) / 9 = (x – (9/2)) / (9/2)

Solving this, we get

x = 6 minutes

The trains are leaving the terminal stations at a regular interval of 6 minutes.

 

Now, to solve the second part of the problem:

From case I, we know that the distance I cover on cycle in 9 minutes is covered by the train in 9 – 6 = 3 minutes.

This means that my cycling speed is one-third that of the trains.