PUZZLES
1) A little girl can take either one or two steps at a time. She can therefore climb a three-step staircase in three different ways: 1+1+1, 1+2, and 2+1. In how many ways can she climb an eleven-step staircase.?
2) Another simple classic: a man realizes he is dying. He calls for his children and asks them to divide his gold coins in the following manner. To the eldest child, he gives one gold coin and a 7th of the remaining coins. To the second child, he gives two gold coins and a 7th of what remains. To the third child, he gives three gold coins and a 7th of what remains. And so on, giving coins and a 7th of the dwindling remainder to each successive child. The last child receives all that is left. The children find, to their surprise, that each receives the same amount from their father’s estate. So…how many children were there, and how large was the estate?
The answers to April’s puzzles were supplied in the April issue.
Here are the answers to this month’s puzzles:
1) 144. We don’t have to count, of course, because the child’s possibilities form a Fibonacci sequence. As the sequence increases, the ratio of successive numbers approaches closer and closer to 1.6180339887…
2) The question appeared in Fibonacci’s Liber Abaci of 1202, and the answer is straightforward. If the estate is E and the first child receives x, the first child in fact receives 1+1/7 (E – 1). The second child receives 2+1/7 (E – 2 – x). The two shares are equal, so the two sums can be equated. The result allows us to cancel out E, showing that x = 6. Each child therefore receives six gold coins. If we equate 6 to 1+1/7 (E – 1), we find that the estate was 36 gold coins. The number of children is therefore 36/6 or six.
