Physical Address

304 North Cardinal St.
Dorchester Center, MA 02124

Problematics | Nobel Prizes in other people’s subjects

John Nash, the mathematician, was unique in many respects. If you have watched the Hollywood film ‘A Beautiful Mind’, with Nash portrayed by Russell Crowe, you will have got a sense of his extraordinary personal life, besides his genius. Some aspects of his work will also delight puzzlers, and I certainly plan to bring them to Problematics sometime in the future.
Nash also remains the only mathematician to have won a Nobel Prize. There is, of course, no Nobel for mathematics, but Nash won for economics, which brings us to the context. To win a Nobel in any discipline, one does not need to be a specialist in that discipline. This year’s Nobel Prizes begin on Monday, October 7, and all the winners in the following puzzle, as you will see, will get their awards in disciplines that are outside of their own fields of research.
#Puzzle 111.1

It’s Nobel Prize week and all hopefuls are mindful of the schedule, which remains unchanged year after year. There’s medicine on the first Monday of October, followed by physics on Tuesday, chemistry on Wednesday and literature on Thursday. Every contender is dressed in a suit of their favourite colour, ready for the TV cameras if and when as the big moment arrives.
As it turns out, the winners are a medicine researcher, a physicist, a chemist and a professor of literature, but none of them wins in their own subject. The countries they hail from are (in alphabetical order): Chile, Lithuania, Moldova and Peru. We cannot divulge the winners’ actual names at this point, so let us identify them with codenames (in alphabetical order): Curie, Linnaeus, Mendel and Pascal. Other useful information below:
1. For every single award, the codename of the winning scientist, the field for which they have won the Nobel, their actual subject of specialisation, and their country of origin each begins with a different letter.
2. The physicist is wearing a white suit when his award is announced, a day after Mendel’s award.
3. The awards of the chemist and the winner in the blue suit are announced on consecutive days, in whichever order.
4. The Moldovan’s award is announced a day before the Chilean’s award.
5. The Lithuanian wins in physics.
6. The medical researcher’s suit is red.
7. Another winner is wearing a grey suit.
8. Pascal’s award is the last to be announced among these four.
#Puzzle 111.2
A motorist wants to average 80km per hour from Delhi to Agra and back. But his car breaks down along the way. After fixing it and reaching Agra, he finds that his average has been only 40km/h. On the return journey, he wants to rush it so that his overall speed across both legs becomes 80km/h.
MAILBOX: Last week’s solvers
#Puzzle 110.1

Dear Kabir,
Solving these problems was an interesting experience. Being a maths guy with little interest in English poetry, I was totally disinclined to even attempt at decoding the poem, but then thought I would try. First, the single letter P had to be A or I. Then W and R in two-letter words were likely to be S, T or N.
I turned to the author’s name and realised it could be the only one I cherish and always remember as the poem that was part of things kept close by Nehru, our first Prime Minister. The guess turned out to be right and once I had the poem (‘Stopping by Woods on a Snowy Evening’ by Robert Frost) before me, it was a cakewalk.
— Kanwarjit Singh, Chief Commissioner of Income-Tax (retired)
Sabornee Jana and Sanjay Gupta found it easy, taking a hint from the fact that the last two verses are the same. Professor Anshul Kumar took Robert Frost as a potential candidate, and beginning from there ultimately revealed the rest.
#Puzzle 110.2
Hi Kabir,
Let the age of the father be f and the ages of the nine children be (a – 4d), (a – 3d), (a – 2d), (a –d), (a), (a + d), (a + 2d), (a + 3d) and (a + 4d). Then, the sum of squares of the children’s ages is
9a² + 60d² = f²
Since the left-hand side is a multiple of 3, the right had side also must be a multiple of 3. Therefore, f must be a multiple of 3 and consequently, both sides are multiples of 9. This means that d is a multiple of 3. Now, we can also make some reasonable assumptions about the age difference between the father and the eldest child and between the father and the youngest child, as follows.
f – 3 (a + 4d) > 20, and f – (a – 4d) < 60
From this we get,
a + 4d + f – 60 < a – 4d + f – 20
=> 8d < 40, or, d < 5
Since d is a multiple of 3, we can conclude that d = 3. Now if we define f = 3g, the equation (9a² + 60d² = f²)becomes,
a² + 60 = g² => (g + a) (g – a) = 60
We can express 60 as a product of two factors and equate these factors to g + a and g – a. The sum of the two factors would be 2g and the difference would be 2a. This means that either both the factors are even or both are odd. There are only two options:
(i) 60 = 30 x 2 => g = 16, a = 14, and
(ii) 60 = 10 x 6 => g = 8, a = 2
Option (ii) can be ruled out because a must be greater than 4d. Thus, option (i) is the only solution. Accordingly, the age of the father is 16 x 3 = 48. And the ages of the children are 2, 5, 8, 11, 14, 17, 20, 23, 26.
— Professor Anshul Kumar, Delhi
 
Solved both puzzles: Kanwarjit Singh (Chief Commissioner of Income-Tax, retired), Prof Anshul Kumar (Delhi), Sabornee Jana (Mumbai), Dr Sunita Gupta (Delhi), Shishir Gupta (Indore), Ajay Ashok (Delhi), Aditya Krishnan (Coimbatore), Sanjay Gupta (Delhi), Shri Ram Aggarwal (Delhi)
Solved #Puzzle 110.2: Anil Khanna (Ghaziabad), Yadvendra Somra (Sonipat), YK Munjal (Delhi)
Postscript: No one has solved Sanjay Gupta’s bonus puzzle correctly. To repeat: “I have five apples to distribute and there are five little girls. I give an equal share of apples to each of the girls. Each girl I give an apple to eats it up instantly. Yet, there is one whole apple left with me. What share does each girl get?” The answer: “Each girl gets one whole apple. I am one of the little girls and I did NOT eat my share.”

en_USEnglish