Problems
Downtown MathHattan has a grid pattern, with \(4\) streets going east-west and \(6\) streets south-north. You take a taxi from School (A) to cinema (point B), but you would like to stop by an ice cream shop first. In how many ways can a taxi get you there if you don’t want to take a route that is longer than necessary?
Gabby the Gnome has \(3\) cloaks of different colours: blue, green, and brown. He also has \(5\) different hats: \(3\) yellow and \(2\) red. Finally, he owns \(6\) different pairs of shoes: \(2\) yellow, and \(4\) red. Gabby is selecting an outfit: a cloak, a hat, and a pair of shoes. In how many ways can he do it if he wants the colour of his shoes to match the colour of the hat?
An airplane is flying from Prague to Tokyo, which are cities in the northern hemisphere with different latitudes. Suppose that the airplane must touch the equator. Could you help the pilot find the shortest path that the airplane can take, assuming that the Earth is a perfect sphere?
Suppose that we have symbols \(a,b,c,d,e\) and an operation \(\clubsuit\) on the symbols satisfying the following rules:
\(x\;\clubsuit\;e = x\), where \(x\) can be any of \(a,b,c,d,e\).
\(a\;\clubsuit\;c = c\;\clubsuit\;a = b\;\clubsuit\;d = d\;\clubsuit\;b = e\).
any bracketing of the same string of symbols are the same; for example, \(((a\;\clubsuit\;c)\;\clubsuit\;d)\;\clubsuit\;(a\;\clubsuit\;d) = (a\;\clubsuit\;(c\;\clubsuit\;(d\;\clubsuit\;(a\;\clubsuit\;d))))\).
\((a\;\clubsuit\;b)\clubsuit\;c = d\).
We use the power notation. If \(n\geq 1\) is a natural number, we write \(a^n\) for \((\dots(a\;\clubsuit\;a)\;\clubsuit\dots)\;\clubsuit\; a\), where \(a\) appears \(n\) times. Similarly for other symbols. Let \(p,q,r,s\geq 1\) be natural numbers. Express \(a^p\;\clubsuit\;b^q\;\clubsuit\;a^r\;\clubsuit\;b^s\) using the symbols \(a,b,c,d\) no more than once (power notation allowed).
A goofy robot named Zippity only speaks using \(0\)s and \(1\)s. Every message Zippity sends is made of \(10\) digits. How many different \(10\)-digit messages can Zippity send if each message must include exactly one run of five zeros in a row? For example, \(0011000001\) would count as a valid message, but not \(1001010001\).
In the jungle, \(100\) monkey families live on tree houses, which we think of the little squares arranged on a \(10\times 10\) square grid. Today it turns out that \(9\) of these houses have been infected with Bananavid-19, which spreads every day! Thankfully, a healthy home can only be infected if has at least two infected neighbours (we don’t count diagonals on the grid as neighbours), prove that the infection will not spread to the whole monkey village.
Consider the two following grids. We consider two of the little squares to be a neighbor if they share a side (so that diagonals don’t count as neighbours). You are allowed to choose any two neighbouring little squares and add the same whole number to both squares. Show that by repeating this step, you cannot turn the first grid into the second one.
We build a sequence of numbers as follows: the first six terms are \(1,0,1,0,1,0\) and from then onwards, the next term of the sequence is equal to the last digit of the sum of the last \(6\) terms of the sequence. For example: if we had at some point the numbers \(1,0,3,5,10,9\) the next term would be the last digit of \(1+0+3+5+10+9=28\), i.e: \(8\). Will the terms \(0,1,0,1,0,1\) ever appear in the sequence?
On a board there are the following \(+\) and \(-\) signs drawn:
\[+\; + \; + \; - \; - \; + \; + \; - \; + \; - \; +\]
You can choose any two signs, erase them, and then draw a \(+\) sign if the signs you erased were both equal, and a \(-\) sign if the signs you erased were different. Show that regardless of the order you perform these erasures, the sign that is left at the end is always the same.
\(2025\) lilly pads are placed in a row. Some number of frogs are on top of the pads. Each minute, if there are two frogs on the same lily pad, and this lily pad is not at one of the ends of the row, some two of the frogs will jump: one to the left lily pad, and one to the right lily pad (they will jump in opposite directions) show that this process cannot repeat forever.