Problematics | Walk, don’t run

A chestnut of a puzzle that is based on a chestnut among themes — speed and distance — describes an object racing ahead of a slow-moving object and then returning. It appears in many versions, some over a century old, and now demands a reincarnation. I have tried to mix multiple elements in my version, besides packaging it in an entirely different situation.

#Puzzle 80.1

The woods are lovely, dark and deep, just as the great poet once described them. Unfortunately, however, the characters in this puzzle have no promises to keep. All they want to do is cross the forest in the middle of the night, for whatever reason.

Because wild animals may be on the prowl, they send a runner ahead to check out. Everyone starts at the same instant. The walkers proceed briskly at 6 kilometres per hour, while the brave runner, putting his own life at risk, jogs ahead at 9 km/hr. After some distance into the forest, the runner turns and runs back. Maintaining his original speed, he reaches the rest of the team exactly 3 hours after they had all started off.

“All safe up to the distance I travelled,” he announces.

“What do you mean, up to the distance you travelled?” one of his teammates berate him. “Go to the edge of the forest and back, and don’t come back until you are sure the entire forest is safe.”

“And be faster this time,” another walker warns the runner. “Three hours is too long.”

Once again, they start together from that point, the walkers maintaining their 6 km/hr speed, and the runner raising his pace to 12 km/hr. He runs for 12 km more before reaching the edge of the forest, where he turns and runs back (still @12km/hr) to tell his team that the entire forest is safe.

The easy question, which I shall not ask, is the total distance across the forest. Instead, take two questions that are only slightly tougher.

(a) On the first leg of his reconnaissance run, which took 3 hours, how much time does the braveheart travel running ahead and how much time running back? (b) On the second leg, what is the total time he spends running (forward and back) before meeting his teammates?

#Puzzle 80.2

There is a clock on a wall, and a mouse runs up it in true Hickory Dickory Dock tradition. However, the clock is stuck so loosely on the wall that before it can strike one and the mouse can be gone, it collapses under the weight of the intruder. Falling on the ground, its face breaks into three parts. An observer with a mathematical bent of mind notes that the numbers on each part adds up to the same total.

Can you draw the three broken parts? Extra points for neatness.

MAILBOX: LAST WEEK’S SOLVERS

#Puzzle 79.1

Hi Kabir,

The solution to the puzzle is as shown in the table. I have added colour to it. Yellow means the movie has seen by the student concerned, red means it has not seen by that student.

— Sundarraj C, Bengaluru

#Puzzle 79.2

Dear Kabir,

I can read the question in two different meanings and I have different answers to both versions.

Version 1: Getting an outcome of 6, 4, 3 means one six and the smart alec pays me 10 coins. One 4 and one 3 means I have to pay 4 gold coins for each of them. So, a total 8 gold coins for this one throw. If this is the way the question was intended, then out of the 216 rounds, there will be 216*3 = 648 numbers. Out of which 6 would have come 108 times and 1/2/3/4/5 540 times. The smart alec would pay me 108*10=1080 coins but I would have to pay 540*4 = 2160 coins. So, I am at a loss. In this scenario if he offers me the option to switch, I would switch and would avoid the loss.

Version 2: Getting an outcome of 6,4,3 means one six and Alec pays me 10 coins. One 4 and One 3 means I have to pay just 4 gold coins altogether for that round. If this is the case, then out of the total 216 rounds, 6 will not appear in 5*5*5 rounds = 125 rounds. For the remaining 91 rounds, the smart alec has to pay me 10 coins and so he gives me 910 coins in total. But for me 1/2/3/4/5 does not come only in 1 case (6,6,6). For the remaining 215 cases, I have to pay 4 coins to Alec. So, that is 215*4 = 860 coins. Here, I am at a gain of 50 coins. In this scenario, if the smart alec offers me the option to switch, I would not switch, as I am already at a gain.

— Sampath Kumar V, Coimbatore

Sundarraj C and Akshay Bakhai too felt the puzzle could be interpreted in two different ways. In fact, the language was clear about “every 6” and “every 1, 2, 3, 4 or 5”, so the version Sampath describes first was what was intended. A few other readers, meanwhile, have interpreted the puzzle either this way or that. Although I think there was no ambiguity in the language, the fact that there have been so many dual interpretations calls for some leniency. Anyone who has given either of the two solutions, therefore, counts as correct — provided their mathematical reasoning is sound.

Solved both puzzles: Sundrarraj C (Bengaluru), Sampath Kumar V (Coimbatore), Akshay Bakhai (Mumbai), Dr Sunita Gupta (Delhi), Amarpreet (Delhi), Shruti Sethi (Ludhiana), Professor Anshul Kumar (Delhi). Kanwarjit Singh (Chief Commissioner of Income Tax, retired), Sanjay S (Coimbatore)

Solved #Puzzle 79.1: Vishnuvardhan (University of Technology Sydney), Ajay Ashok (Mumbai), Aafiya (Coimbatore)

Solved #Puzzle 79.2: Group Captain RK Shrivastava (retired, Delhi), Yadvendra Somra (Sonipat), Dr Vivek Jain (Baroda), Shri Ram Aggarwal (Delhi), YK Munjal (Delhi)

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