E-Paper
Problematics | Celebrating the loser
Five candidates suffer badly in the elections. Who is the grocer, who contests from Puducherry, and who plays the santoor?
Checking poll results on the website of the Election Commission of India usually means looking at who has defeated whom by how many votes. Our attention is largely focused on the top two candidates, and sometimes on the top three if the contest in a seat is three-cornered. What we tend to ignore is those who follow lower down, and there are often many of them.
The following puzzle is in honour of those who contest just because they feel like it or because someone has bluffed them that they could win, and who end up losing their security deposits.
#Puzzle 191.1
Our puzzle has five protagonists who are contesting as independent candidates in Assam, West Bengal, Tamil Nadu, Kerala and Puducherry (each candidate in a different state/UT). Although they do not know one another, they have a few things in common. Each one likes a different sport and plays a different musical instrument, but all of them belong to professions (a different one for each) that have nothing to do with sports or music. Also, each one performs abysmally in the elections.
The following clues should tell us everything we need to know about them.
(1) The five contestants are aged 31, 32, 33, 34 and 35
(2) The 34-year-old contestant wins 1501 votes
(3) The contestant who likes football wins 1372 votes
(4) The realtor is older than the Puducherry contestant
(5) The builder is one year younger than the transporter
(6) The grocer is one year older than the contestant who plays the santoor
(7) The 35-year-old likes badminton; another contestant likes cricket
(8) The contestant who likes hockey is older than the Puducherry contestant but younger than the West Bengal contestant
(9) The Tamil Nadu contestant is younger than the one who wins 1223 votes
(10) The contestant who likes football wins 1372 votes
(11) The contestant who likes tennis is 1 year younger than the one who wins 1496 votes
(12) The Kerala contestant is younger than the grocer but older than the one who wins 1125 votes
(13) The Kerala contestant is also younger than the contestant who plays the santoor
(14) The Tamil Nadu contestant is younger than the one who plays the guitar
(15) The realtor plays the tabla; another contestant plays the violin
(16) The contestant who plays the sitar wins 1372 votes
(17) The salesman is not the one contesting from Assam
(18) The West Bengal contestant wins 1372 votes
From the youngest to the eldest, arrange the candidates in tabular form giving their states, professions, musical instruments, favourite sports and votes won.
#Puzzle 191.2
Consider the following:
1 + 6 + 7 + 17 + 18 + 23 = 72 = a + b + c + x + y + z
1² + 6² + 7² + 17² + 18² + 23² = 1228 = a² + b² + c² + x² + y² + z²
1³ + 6³ + 7³ + 17³ + 18³ + 23³ = 23472 = a³ + b³ + c³ + x³ + y³ + z³
Remarkably, a, b, c, x, y, and z are the same integers in each equation (they are unequal, and none of them is equal to 1, 6, 7, 17, 18, or 23 either). Even more remarkably, the elegant relationships above will hold true even if you use fourth powers or fifth powers on both sides of the equation.
It was only a few days ago that I learnt there is a formula to derive such sets of 12 numbers. It holds for exponential powers of 1 to 5, but no higher. Finding the formula, however, is not the puzzle, because that would be more luck than skill.
What are the values of the integers a, b, c, x, y, and z? The order is immaterial.
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 190.1
In the fight between 13 honeybees and 4 giant wasps, the bees can kill the wasps in the shortest time if they split into groups of (3, 3, 3, 4), then (4, 4, 5) after one wasp is dead, then (6, 7) after the second wasp is dead, and finally all 13 together after the third wasp is dead, leaving only one at the mercy of the entire team. All readers have begun with (3, 3, 3, 4) honeybees, but some have differed in the way subsequent groups are formed, thus leading to longer durations. I have ignored minor differences in calculations, and credited every reader who has followed a (3, 3, 3, 4)—(4, 4, 5)—(6, 7)—(13) sequence. See the correct answer below:
Hi Kabir,
The 13 honeybees will kill the 4 wasps in a little under 5 minutes and 47 seconds. The optimal war strategy is as follows :
Round 1 (3 minutes): The honeybees split into four groups (3, 3, 3, 4) to take on each of the four giant wasps. The group of 4 honeybees finish off one wasp in 3 minutes
Round 2 (2 minutes and 24 seconds): The group of 4 splits and its members join the other three groups, forming three new groups of (4, 4, 5). The group of 5 honeybees finishes off the second wasp in 3 x 4/5 = 12/5 minutes (or 2 minutes and 24 seconds).
In these 12/5 minutes the other two groups (four honeybees each) have been fighting one giant wasp each. Since 4 honeybees require 3 minutes to kill one giant wasp, each of the two surviving giant wasps are down to [1 – (12/5)/3] = 1/5 of their initial strength.
Round 3 (20 and 4/7 seconds): The honeybees now split into two groups (6, 7), with each group fighting a wasp at 1/5 of its original strength. The group of 7 finishes off the third wasp in 1/5 x (4 x 3) x 1/7 = 12/35 minutes (or 20 and 4/7 seconds).
Meanwhile, the other group of 6 honeybees would normally require 1/5 x 4/6 x 3 = 2/5 minutes to kill the fourth wasp. During the 12/35 minutes taken to kill the third wasp, the strength of the fourth wasp is down to 1/5 [1 – (12/35)/(2/5)] = 1/35 of its original strength.
Round 4 (1 and 53/91 seconds): After the third wasp is killed, all 13 honeybees gather together against the fourth wasp, which is at 1/35 of its original strength. They kill it in 1/35 x (4/13 x 3) = 12/455 minutes (or 1 and 53/91 seconds)
Total time taken
= 3 minutes + (2 minutes and 24 seconds) + (20 and 4/7 seconds) + (1 and 53/91 seconds)
= 5 minutes and (46 and 34/91 seconds)
— Sabornee Jana, Mumbai
#Puzzle 190.2
Hi Kabir,
After midnight, the hands of a clock will meet a little after 01:05. At T hours after midnight, where T is between 1 and 2, the angle between the hour hand and the vertical is 30T degrees and the angle between the minute hand and the vertical is 360(T – 1) degrees. The two hands meet when these two are equal, that is, when 30T = 360(T – 1) = 12/11. Therefore, starting from midnight, the two hands will meet after intervals of 12/11 hours. Thus, the total number of times the two hands meet in between two midnights = 24 x (11/12) – 1 = 21.
— Professor Anshul Kumar, New Delhi
Solved both puzzles: Sabornee Jana (Mumbai), Professor Anshul Kumar (Delhi), Dr Sunita Gupta (Delhi), Yadvendra Somra (Sonipat), Kanwarjit Singh (Chief Commissioner of Income-tax, retired), Shishir Gupta (Delhi)
Solved #Puzzle 190.1: YK Munjal (Delhi)
Solved #Puzzle 190.2: Vinod Mahajan (Delhi), Ajay Ashok (Delhi), Dr Vivek Jain (Baroda)
