Problems
Is \(F_{100}\) a multiple of \(3\)?
This week we’re looking at Fibonacci numbers, and other sequences of numbers.
We say that the ‘zeroth’ Fibonacci number is \(0\) and the first Fibonacci number is \(1\). Then, from that point, every Fibonacci number is found by adding the two previous Fibonacci numbers. This means that the sequence begins \(0,1,1,2,3,5,8,13,21,34,55,89,144,...\)
The Fibonacci numbers hide lots of patterns which we’ll explore today, for example snail’s spiral.
We have a sequence where the first term (\(x_1\)) is equal to \(2\), and each term is \(1\) minus the reciprocal of the previous term (which we can write as \(x_{n+1}=1-\frac{1}{x_n}\)).
What’s \(x_{57}\)?
Let \(n\) be a positive integer. Can \(n^7-77\) ever be a Fibonacci number?
There is a very, very fast way of computing the greatest common divisor of two positive integers. It was in fact known even to the Greeks two thousand years ago. This procedure is called the Euclidean algorithm, named after Euclid, a famous ancient Greek mathematician.
The algorithm works as follows. Take two positive integers \(a,b\). Let’s say \(a\geq b\).
Calculate the remainder of \(a\) when divided by \(b\). Call it \(r_1\).
Calculate the remainder of \(b\) when divided by \(r_1\). Call it \(r_2\).
Calculate the remainder of \(r_1\) when divided by \(r_2\). Call it \(r_3\).
Continue to divide the remainder from two steps prior by the remainder from the last step, until...
The remainder \(r_n\) is divisible by \(r_{n+1}\). The Euclidean algorithm stops now and \(r_{n+1}\) is \(\gcd(a,b)\).
Show that there is indeed some natural number \(n\) such that \(r_n\) is divisible by \(r_{n+1}\), so that the Euclidean algorithm must stop eventually. Furthermore, show that \(r_{n+1}\) is actually \(\gcd(a,b)\) (otherwise it is all in vain!).
Prime numbers are like atoms that build every integer number. That is, a prime decomposition of a number is unique and then we can use it to find the number’s factors. Today we will explore this idea a bit more.
We will introduce a couple of new terms. First, a common divisor of two numbers is simply a number that both of these numbers can be divided by. Two numbers which have no common divisors (except from 1) are called relatively prime. We can establish if two numbers are relatively prime by looking at their prime factorizations - if they share no common primes, then they cannot share a common divisor!
Out of all the common divisors two numbers have, one must be the largest. This is an important number and is called the Greatest Common Divisor (GCD). You can find it by looking at the prime factorizations of the two numbers. For every prime number appearing in both factorizations, we take the smaller power. Then we multiply all our choices together. If we divide both numbers by their GCD, the resulting numbers will have no common divisors left and so will be relatively prime.
Similar to the notion of a common divisor is the one of common multiple. It is simply a number that is divisible by both numbers. Among the common multiples, one must be the smallest – this is called the Least Common Multiple (LCM). Again, the LCM can be found by looking at the prime factorizations of the two numbers. For every prime number appearing in any of the two factorizations, we take the larger power. Then we multiply all our choices together.
Cut a packet of 4 cards. Is any of the cards in the same place as it was before?
We have a packet of 13 cards from Ace to King. Let Ace be the first card, 2 the second card and so on with King being the thirteenth card. How can you interchange 4 and 7 (and leave all other cards where they are) by only switching adjacent pairs of cards?
How many permutations of 13 cards leaves the third card where it started?