For many of us, the first introduction to the famous Fibonacci series (1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89…), in which every term is the sum of the preceding two terms, comes through an engaging story about rabbits. Fibonacci placed a hypothetical pair of newborn rabbits in a pen and proposed the following rules: (1) each pair matures at age one month, (2) each pair gives birth to a male and a female at the end of two months, (3) each reproductive-age pair gives birth to a new male-female pair in every subsequent month, (4) no rabbit dies.
In real life, that’s wishful thinking. Rabbits are prolific breeders, but not as fast as Fibonacci proposed. Different breeds of rabbits start giving birth at different ages, which ranges from a few months to 10, but none starts reproducing at two months as far as I can make out.
Here is a practical, if fictional, situation that may arise if you try to replicate Fibonacci’s sequence in your backyard.
Puzzle #219.1
A young student of Fibonacci wants to observe his master’s series unfold in real life. He goes to a pet shop. “Rabbits, please.”
“How many do you need?”
“How many do you have?”
“Almost a thousand, I think,” the shopkeeper replies, and then checks his records. “Yes, we have 960 rabbits, 480 males and 480 females. That’s a lot more than you could possibly need.”
“That’s what you think,” Fibonacci’s student says, and buys the entire lot.
{{/usCountry}}“That’s what you think,” Fibonacci’s student says, and buys the entire lot.
{{/usCountry}}At home, he builds two rabbit batteries, one in his courtyard and the other in his garden, with each battery split into a number of colony pens. Our young mathematician divides the rabbits into 480 male-female pairs, and places an equal number of pairs in each colony pen. He then waits for the rabbits to breed.
Alas, the rabbits have no knowledge of Fibonacci’s prescription of “first litter after two months, followed by one litter every subsequent month”. Being of various ages, they reproduce rather randomly. The student, however, ensures one aspect of Fibonacci’s idealised situation is replicated: there are no deaths.
After six months, he makes a fresh checklist. He first goes to the battery in the courtyard at the front of his house. “A round 300 pairs,” he observes. “Every colony pen now has four pairs of rabbits more than I had placed in it six months ago.”
The young man proceeds to the garden at the back of his house. “Ah,” he notes. “Here too, the growth is uniform across all 35 colony pens. Each pen has four pairs of rabbits more than I had placed six months ago.”
How many pairs of rabbits did the student originally place in each colony pen, and how many pairs are in each pen now?
Puzzle #219.2
[3(280 + x)]² = 72y904
In the above equation, the letter y stands for a digit in the right-hand number, while x is an integer. It is much easier to solve than it may appear at first sight.
What are x and y?
MAILBOX: LAST WEEK’S SOLVERS
#Puzzle 218.1
Hi Kabir,
The number of moves to transfer 5 coins is 31. In general, the number of moves required for n coins is (2ⁿ – 1). This can be explained as follows.
To transfer a pile of n coins from the green mat (say A) to the red mat (say B), using the amber mat (say C) as an intermediate transit point, we can use the following sequence: (1) Move the top (n – 1) coins from A to C. (2) Move the largest of the n coins (which is now exposed) from A to B. (3) Move (n – 1) coins from mat C to mat B. This brings (n – 1) coins to their original positions.
Steps (1) and (3) require the same number of moves, while step (2) is a single move. Thus the number of moves required to transfer n coins = 2 x (number of moves required to transfer (n – 1) coins) + 1.
When n = 1, the number of moves = 1
When n = 2, the number of moves = 2 x 1 + 1 = 3
When n = 3, the number of moves = 2 x 3 + 1 = 7
When n = 4, the number of moves = 2 x 7 + 1 = 15
When n = 5, the number of moves = 2 x 15 + 1 = 31
and so on.
The sequence 1, 3, 7, 15, 31... is of the form (2ⁿ – 1).
— Professor Anshul Kumar, New Delhi
#Puzzle 218.2
Hi,
The minimum is 8 marks, placed at:
1, 3, 6, 13, 20, 27, 31, 35 inches
The ruler’s endpoints, at 0 and 36 inches, are already available, so they do not count as additional marks. These 10 marks will enable any integer measurement between 1 and 36 inches.
— Ajay Ashok, New Delhi
Solved both puzzles: Professor Anshul Kumar (Delhi)
Solved #Puzzle 218.1: Dr Sunita Gupta (Delhi), Kanwarjit Singh (Chief commissioner of income-tax, retired), Vinod Mahajan (Delhi), Yadvendra Somra (Sonipat)
Solved #Puzzle 218.2: Ajay Ashok (Delhi)
Problematics will be back next week. Please send in your replies by Friday noon to problematics@hindustantimes.com