Problematics | Five Oscar nominees, five viewers

If you look at old issues of this paper or previous online episodes of Problematics, you may notice a kind of hierarchy among mathematical, logic and word puzzles. When deciding which one of the two puzzles in any week comes first, I have tended to prioritise mathematical > logic > word puzzles on most occasions. This is actually quite inadvertent. I try to begin with the tougher puzzle and close with the easier one, and usually maths is tougher than logic is tougher than wordplay.

The reason for bringing this up is that we are flipping the order this time. We begin with a logic puzzle before moving on to mathematics. I think the two are equal on the scale of hardness. There are two reasons for flipping the order. One, just for variety. Two, the logic puzzle is about movies, which means it is colourful.

#Puzzle 79.1

There are 10 movies in contention for Best Picture at the Oscars this year. The number of contenders has varied since 1927-28, but 10 in any case is more than this puzzle needs. We narrow it down to five: Oppenheimer, Barbie, Killers of the Flower Moon, Maestro and The Holdovers. We now move to a college canteen where students, being students, are chatting about movies rather than worrying about their studies.

1. Margot and Cillian think the award will go to Oppenheimer, which both have seen.

2. The two keep discussing Oppenheimer until Emily joins them. The conversation now moves on to Barbie, which is the only common movie that all three have seen.

3. Eventually, Emily and Cillian leave for class. Margot, who remains in the canteen, is joined by Ryan and Florence. They find out that among the five movies named above, the only one seen by all three is Killers of the Flower Moon.

4. Later, Cillian returns, but it’s time for Margot and Ryan to go. That leaves Cillian and Florence together, and they discuss the only movie that both have seen, which is Maestro.

5. Barbie has the highest number of viewers among these five students.

6. Of the five students, three have seen The Holdovers.

7. One student has seen all five films, another has seen four, and yet another has seen three. Of the remaining two students, one has seen two films while the other has seen only one.

Which student has seen which films?

#Puzzle 79.2

Over the last year and a half, there have been several occasions when we have met a smart alec who tries to trick you at every party you go to. So, here he is again.

He shows you three dice. “Each one has six faces, and each face has a number of spots between 1 and 6 inclusive,” he says, as if you didn’t know already. “Choose any number of spots from 1 to 6,” he adds.

“Six,” you choose, before he sets down the rules of the game.

“We roll all three dice together. We do that for 216 rounds. In each round, for every 6 that appears, I pay you ten gold coins or the value thereof. For every 1, 2, 3, 4 or 5, you pay me four gold coins or an equivalent value.”

“Hang on,” you reply, “I am not sure that’s fair. It’s 5 times more probable that any die will not show 6 than show 6. But 10 gold coins is only = 2.5 x 4 gold coins.”

“Okay then,” the smart alec offers, “here’s an alternative. Same rules, but 6 is mine, while 1, 2, 3, 4 and 5 are yours.”

Should you accept the alternative or stick to the original plan?

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 78.1

Hi Kabir,

While the 8th conditions spells out exact numbers, the last condition does not. My interpretation is that it specifies a theoretical lower limit on the size of overlap between the three subsets (families that sent husbands, sons and daughters) of a set (all families).

Let F be the number of families in the LoseLose tribe. Then,

the number of husbands sent for recruitment = H = 2F/3,

the number of sons sent = S = 3F/4, and

the number of daughters sent = D = 4F/5.

The situation in which there is minimum overlap among these three subsets is when a portion of one subset fully covers the non-overlapping part of the other two subsets, and the remaining portion of the first subset overlaps with the overlapping part of the other two subsets. This remaining portion can be shown to have a size of H + S + D – 2F.

Therefore, the minimum number of families that sent all three = H + S + D - 2F = 13F/60.

This is given to be 52. Therefore, F = 240 and the numbers of husbands, sons and daughters sent are 160, 180 and 192, respectively.

The number of bags of rice paid = 160x750 + 180x1000 + 192x1000 = 4,92,000.

— Professor Anshul Kumar, Delhi

#Puzzle 78.2

Hi Kabir,

The anagram solutions are:

A ROLE IN THE MATCH = MICHAEL ATHERTON

A SPARKLING DRIVE = DILIP VENGSARKAR

FINAL DELIVERIES = FANIE DE VILLIERS

SOURS CRICKET MATCH = MARCUS TRESCOTHICK

DRAVID ALIAS EVAN = ARAVINDA DE SILVA

— Akshay Bakhai, Mumbai

Notes on #78.1

In their attempts to solve the numerical puzzle, a number of readers start by observing that the number of families is a multiple of 60 (this part is right). These readers then go to reason like this: if at least 52 families sent all three members, the total number of families cannot be 60 because of the other conditions. But the next multiple, 120, will work, and so 80 families sent their husbands. And so on.

Agreed, it is possible to have a set 120 families with 52 of them in all three subsets. So why is their answer incorrect? I discussed this with the first reader to send the correct answer, Professor Anshul Kumar. As he notes, with 120 families, the size of the overlap (families in all three subsets) can be anything in the range 26 to 80. This means that 52 families could have sent all three but we cannot assert that “at least 52 families sent all three”.

Solved both puzzles: Professor Anshul Kumar (Delhi), Akshay Bakhai (Mumbai), Sampath Kumar V (Coimbatore)

Solved #Puzzle 78.1: Kanwarjit Singh (Chief Commissioner of Income Tax, retired)

Solved #Puzzle 78.2: Dr Sunita Gupta (Delhi), Sundarraj C (Bengaluru), Group Captain RK Shrivastava (retired, Delhi), Shishir Gupta (Indore), Jaikumar Inder Bhatia & Disha Bhatia (Ulhasnagar, Thane), Ajay Ashok (Mumbai), Amarpreet (Delhi), Sanjay S (Coimbatore)