Problems

Age
Difficulty
Found: 274

Louise is confident that all her classmates have different number of friends. Is she right?

There are 100 cities all connected by roads. Each city has 6 roads coming in (or going out). How many roads do connect those cities?

There are 15 cities in a country named The Country of Fifteen Cities. The king ordered his main architect to build roads in such a way that each city was connected with other cities by exactly 5 roads, otherwise he would hang the architect. Do you think that the architect can accomplish the task or should he flee that country immediately?

The architect decided to flee The Country of 15 Cities and began to travel around the world. He arrived to a country, where every city had exactly 3 roads going to and from it. Can there be all together 100 roads in that country?

There are 9 cities named City 1, City 2, City 3, …, and City 9 in a country named The Country of the Nine Cities. Two cities are connected by a road only if the sum of the numbers made up by their names is divisible by 3. Can our travelling architect reach City 9 by starting his journey from City 1 and travelling along those roads?

Show that among any 6 people there are always either 3 people who all know each other or 3 total strangers.

Show that the number of people who ever lived and made an odd number of handshakes is even.

Is it possible to trace the lines in the figures below in such a way that you trace each line only once?

Can you draw 9 line segments in such a way that each segment crosses exactly 3 other segments?

Take any two non-equal numbers \(a\) and \(b\), then we can write \[a^2 - 2ab + b^2 = b^2 - 2ab + a^2\] Using the formula \((x-y)^2 = x^2 - 2xy + y^2\), we complete the squares and rewrite the equality above as \[(a-b)^2 = (b-a)^2.\] As we take a square root from the both sides of the equality, we get \[a-b = b-a.\] Finally, adding to both sides \(a+b\) we get \[\begin{aligned} a-b + (a+b) &= b-a + (a+ b)\\ 2a&= 2b\\ a&=b. \end{aligned}\] Therefore, All NON-EQUAL NUMBERS ARE EQUAL! (This is gibberish, isn’t it?)