In a certain country, there are \(n\) cities. Some of them are connected by
roads, all of which go in both directions. It is possible to get from
any city to any other city using only roads. However, for any pair of
cities, there is always only one way to get from one of them to the
other and there are no alternative routes.
Show that there are exactly \(n-1\)
roads in this country.
If \(x\) is any positive real number and \(n \ge 2\) is a natural number, show that \[(1+x)^n > 1+nx\]
Anna and Bob play a game with the following rules: they both receive
a positive integer number. They do not know each other’s numbers, but
they do know that their numbers come one after another – they do not
know which one is larger. (If Anna gets \(n\), Bob gets either \(n-1\) or \(n+1\)). Anna then asks Bob – “do you know
what number I have?” If Bob does know, he has to say Anna’s number and
he wins the game. If he does not, he has to say that he does not. Then,
he asks Anna if she knows his number. If Anna does not know, she asks
Bob. This continues until one of them finds out what is the other’s
number. Assuming that both Anna and Bob know mathematics sufficiently
well to be able to solve this problem, find out who wins the game and
how.
For simplicity let’s assume Bob always gets the odd number and Anna
always gets the even number - two consecutive numbers have opposite
parity!
A real number \(y\) is such that \(y+\frac1{y}\) happens to be an integer. Show that for any natural \(n\), it is also true that \(y^n + \frac1{y^n}\) is an integer.
We have a very large chessboard, consisting of white and black squares. We would like to place a stain of a specific shape on this chessboard and we know that the area of this stain is less than the area of one square of the chessboard. Show that it is always possible to place the stain in such a way that it does not cover a vertex of any square.
Some Star Trek fans and some Doctor Who fans met at a science fiction convention. It turned out that everyone knew exactly three people at the convention. However, none of the Star Trek fans knew each other and none of the Doctor Who fans knew each other. Show that there are the same number of Star Trek fans as the number of Doctor Who fans at the convention.
Multiply an odd number by the two numbers either side of it. Prove that the final product is divisible by \(24\).
Mattia is thinking of a big positive integer. He tells you what this number to the power of \(4\) is. Unfortunately it’s so large that you tune out, and only hear that the final digit is \(4\). How do you know that he’s lying?
You might want to know what day of the week your birthday is this
year. Mathematician John Conway invented an algorithm called the
‘Doomsday Rule’ to determine which day of the week a particular date
falls on. It works by finding the ‘anchor day’ for the year that you’re
working in. For \(2025\), the anchor
day is Friday. Certain days in the calendar always fall on the anchor
day. Some memorable ones are the following:
‘\(0\)’ of March - which is \(29\)th February in a leap year, and \(28\)th February otherwise.
\(4\)th April, \(6\)th June, \(8\)th August, \(10\)th October and \(12\)th December. These are easier to remember as \(4/4\), \(6/6\), \(8/8\), \(10/10\) and \(12/12\).
\(9\)th May, \(11\)th July, \(5\)th September and \(7\)th November. These are easier to see as
\(9/5\), \(11/7\), \(5/9\) and \(7/11\). A mnemonic for them is “9-5 at the
7-11".
Then find the nearest one of these dates to the date that you’re looking
for and find remainders.
For example, \(\pi\) day, (\(14\)th March, which is written \(3/14\) in American date notation. It’s also Albert Einstein’s birthday) is exactly \(14\) days after ‘\(0\)’th March, so is the same day of the week - Friday in \(2025\).
What day of the week will \(25\)th December be in \(2025\)?
\(6\) friends get together for a game of three versus three basketball. In how many ways can they be split into two teams? The order of the two teams doesn’t matter, and the order within the teams doesn’t matter.
That is, we count A,B,C vs. D,E,F as the same splitting as F,D,E vs A,C,B.