Problematics | Speed test while walking and running
One person runs and another one walks on a circular road, moving in opposite directions, before they switch roles. Can you determine where and when they meet?
Any set of puzzles that aims to be diverse will always have its fair share of speed-and-distance problems. Regular readers of Problematics have solved several of these over the last one year and a little, but it has been a while since we had the last one of these. So why not return to one of them this week?
Such puzzles are never extremely difficult, but many of them do require some thought. You should find the following one enjoyable, although it is nowhere near as difficult as the matchstick puzzle last week.
A ring road circles an area of a town measuring roughly 2 square kilometres, but the area is not of any consequence to our puzzle. What does matter is the length of the road, which is exactly 5 km.
At the crack of dawn one day, when there is hardly any traffic to speak of, two students set out to create and solve a puzzle athletically.
“I am faster than you, as we have clocked and established several times in the past,” one of them said.
“That’s nothing to brag about,” said the other; “you run only a little faster than my 6 km/hr.”
“I don’t think the difference is all that little,” the first one argued. “And I also walk faster than your 2 km/hr.”
“Have you noticed,” the slower one mused, “that the difference between your running speed and mine is the same as the difference between your walking speed and mine?”
“Interesting observation,” agreed the faster one, as the two marked the starting point of their race — if it could be called a race, given that it was not really a competition.
“You run clockwise and I’ll walk anticlockwise,” the faster one continued, setting down the rules. “When we meet, you start walking and I shall start running.”
They met in due course. As agreed, the runner began to walk from this point while the walker now ran.
“Let’s continue this until we meet at the starting point again,” the slower one called back over his shoulder.
When they met at that point, the faster one had circuited the road two times overall while the slower one had done one lap.
For how long did the two walk and run, and which points did they meet?
The town where the two students live is a small one, with a population that runs into a few lakhs. A rich old billionaire dies and wills some of his money to other elderly townspeople who, he figures, are nearing the end of their lives, like he was when he wrote his will. What he leaves them is not much, though: Any woman of the town over the age of 90 is entitled to collect ₹100, should she be interested in that amount. Likewise, any man aged over 90 can collect ₹75, if interested.
Needless to say, when the time for payments came, a number of elderly people showed no interest in the paltry amount and declined the offer. A number of others, however, did accept the money. To get down to exact numbers, out of 30,000 people over 90, two in every three women and eight in every nine men accepted.
How much money was given to these elders in total?
MAILBOX: LAST WEEK’S SOLVERS
The strategy adopted by the first player is that for each move, he expresses the number of sticks N as a sum of numbers in decreasing order n₁ + n₂ + n₃… nₖ, as per the following method and then removes nₖ sticks.
n₁ is the largest Fibonacci number less than or equal to N,
n₂ is the largest Fibonacci number less than or equal to N – n₁,
n₃ the largest Fibonacci number less than or equal to N – n₁ – n₂,
and so on.
The first player also needs to ensure that the initial number of sticks is not a Fibonacci number itself.
It should be noted that nᵢ and nᵢ₊₁ cannot be consecutive terms in the Fibonacci series. Therefore, after the removal of nₖ sticks by the first player, the number of sticks that can be removed by the second player in the subsequent move has to be less than nₖ₋₁. Suppose the second player removes s sticks. The first player will again express the remaining number of sticks as a sum of Fibonacci numbers. It can be shown that the last term will not be more than 2s and hence those many sticks can be removed by the first player.
When the first player makes a move, the number of terms always decreases by 1 and when the second player makes a move, the number of terms remains the same or increases. Therefore, it will always be the first player who removes the last set of sticks and wins.
— Professor Anshul Kumar, Delhi
Apart from Prof Kumar, only two other readers (see list below) have got the Fibonacci connection here. Some readers have given other strategies, but I don’t think those will work. Those require the first player to collect larger numbers of matches on various moves, but the second player can preempt that by picking up 1 or 2, say, thereby restricting the number that the first player can pick up on the subsequent move.
The anagrams are
PYTHAGORAS AND MARIE CURIE
ALBERT EINSTEIN AND GALILEO
NEWTON AND ARCHIMEDES
DARWIN AND COPERNICUS
FARADAY & DA VINCI.
— Sundarraj C, Bengaluru
For those who might have missed the anagrams last week, I have repeated them in the image, along with the answers this time.
Solved both puzzles: Professor Anshul Kumar (Delhi), Ajay Ashok (Mumbai), Group Captain RK Shrivastava (retired; Delhi)
Solved #Puzzle 67.2: Dr Sunita Gupta (Delhi), Sundarraj C (Bengaluru), Geetansha Gera (Faridabad), Shishir Gupta (Indore)
Problematics will be back next week. Please send in your replies by Friday noon to firstname.lastname@example.org