Problematics | When Einstein and Galileo watched Oppenheimer - Hindustan Times

# Problematics | When Einstein and Galileo watched Oppenheimer

Dec 18, 2023 08:00 AM IST

## Archimedes, Bohr, Curie, Darwin, Einstein, Faraday and Galileo watch a movie after distributing their money. How much money did each have in the beginning?

Just a couple of weeks ago, we had a set of five anagrams, which you unscrambled to get the names of 10 scientists. Arrange them in alphabetical order and you find, by a remarkable coincidence, that six of the first seven letters are represented, the exception being B. That is not really a problem, for we can easily fill the missing link with, say, Bohr. Six other names come from the anagrams you solved: Archimedes, Curie, Darwin, Einstein, Faraday and Galileo.

We need these names because there are seven characters in the story unfolding below. The puzzle is inspired by one from Shakuntala Devi. I have invested quite heavily into it, changing not only the names but also the actions of the dramatis personae. The mathematical principles, however, remain the same as described by the late genius.

## #Puzzle 69.1

Our seven friends want to watch Oppenheimer, for which the ticket price is 350. To cover the cost of travel and popcorn, followed by dinner, they figure they need about 600 each. They count their money, which includes coins, and find that they have a total of 4,480. Edited excerpts from their conversation:

Archimedes: That’s 640 per viewer, which is more than we need.

Bohr: Yes, but not each of us has 640 or even 600 individually.

Curie: Then let’s pool our resources so that everyone can pick up 640 from the pool.

Darwin: No Madame, let’s distribute the money more ingeniously. Archie, you being the one with the most funds, give each of us some money.

Archimedes: How much?

Einstein: Give each of us as much money as we already have.

Bohr: How much do I give you all?

Galileo: Give each of us as much as we already have now, just like Archie did.

This continues in alphabetical order. After Archimedes and Bohr, it is Curie, Darwin, Einstein, Faraday and finally Galileo who, in that order, distribute their money in the pattern described.

Chorus: Bingo. Each one of us now has exactly 640.

## #Puzzle 69.2

You play two separate Wordle games, each with a different answer, but you use the same four words trying to uncover that answer. Do you have enough information from both games to determine the respective hidden words?

## #Puzzle 68.1

The illustration with the answers below is by Professor Anshul Kumar. I have included parts of his solution as well as parts from Shri Ram Aggarwal, but I have included only Professor Kumar’s illustration to cover both.

Hi Kabir,

Suppose the difference between the two walking speeds and the two running speeds is v km/hr. Accordingly, the speeds of the faster friend for walking and running are (2 + v) and (6 + v) respectively.

In the illustration below, A is the starting point of the race and B is the point where the two racers meet first. The slower one (S) runs from A to B and the faster one (F) walks from A to B. Then S walks from B to A, whereas F runs, crosses A again, continues running and completes his two laps at A.

Their effective combined speed, interestingly, turns out to be the same in both phases: 6 + (2 + v) = 8 + v in the first phase and 2 + (6 + v) = (8 + v) in the second phase. They collectively covered one lap in the first phase and two laps in the second phase. Clearly, the duration of the second phase was twice the duration of the first phase.

Now, S covered 5 km by running at the speed of 6 km/hr for some time and then walking for twice that duration at 2km/hr. Thus, the distance covered in the two phases to complete a lap of 5 km is in the ratio of 6:2x2 or 3:2. Which means that the clockwise distance from A to B is 3 km and B to A is 2 km. With the distances known, now we can get the durations of the two phases as follows.

First phase duration = (3km)/(6km/hr) = ½ hours

Second phase duration = (2km)/(2km/hr) = 1 hour

Therefore, S ran for half an hour and walked for 1 hour, while F walked for half an hour and ran for 1 hour.

## — Professor Anshul Kumar, Delhi

Dear Mr Kabir,

By their first meeting, the slower one had covered 3 km by running at 6km/hr while the faster one had covered 2km/hr by walking. They meet at point B in 30 minutes. From this, the faster one’s walking speed can be shown to be 4km/hr and his running speed to be 8km/hr.

They meet again at point C, say. This happens after 30 more minutes, by when the slower student has covered 1 km @ 2km/hr and the faster one has covered 4 km @ 8km/hr. Finally, they meet at A again, covering another 1km and 4km respectively.

## — Shri Ram Aggarwal, Delhi

[Note: In total, as Yadvendra Somra points out, the slower one ran 3km and walked 2km, while the faster one walked 2km and ran 8km.]

## #Puzzle 68.2

Sir,

68.2

Let there be w women and m men 90 years of age.

Given, w + m = 30,000

[(2/3)w x 100] + [(8/9)m x 75] = (200/3)(m + w) = (200/3) x (30000)

= 20,00,000

Solved both puzzles: Prof Anshul Kumar (Delhi), Shri Ram Aggarwal (Delhi), Yadvendra Somra (Sonipat), Vivek Aggarwal (Bangalore)

Solved #Puzzle 68.2: Dr Sunita Gupta (Delhi), Dr Vivek Jain (Baroda), Abhishek Garg (Chandigarh), Kanwarjit Singh (Chief Commissioner of Income Tax, retired), Shishir Gupta (Indore), YK Munjal (Delhi), Group Captain RK Shrivastava (retired; Delhi)

Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com

Unlock a world of Benefits with HT! From insightful newsletters to real-time news alerts and a personalized news feed – it's all here, just a click away! -Login Now!