We solve problems

Problem #PRU-103787

Problems Methods Algebraic methods Counting in two ways Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (other)

12–14 3.0

One term a school ran 20 sessions of an after-school Astronomy Club. Exactly five pupils attended each session and no two students encountered one another over all of the sessions more than once. Prove that no fewer than 20 pupils attended the Astronomy Club at some point during the term.

Problem #PRU-107762

Problems Methods Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (other)

13–15 4.0

Each of the 1994 deputies in parliament slapped exactly one of his colleagues. Prove that it is possible to draw up a parliamentary commission of 665 people whose members did not clarify the relationship between themselves in the manner indicated above.

Problem #PRU-116560

Problems Methods Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (other) Proof by contradiction

14–15 4.0

2011 numbers are written on a blackboard. It turns out that the sum of any of these written numbers is also one of the written numbers. What is the minimum number of zeroes within this set of 2011 numbers?

Problem #PRU-116873

Problems Methods Extremal principle Extremal principle (other) Mathematical induction Mathematical induction (other) Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other)

14–15 4.0

We are given \(n+1\) different natural numbers, which are less than \(2n\) (\(n>1\)). Prove that among them there will always be three numbers, where the sum of two of them is equal to the third.

Problem #PRU-32132

Problems Methods Extremal principle Extremal principle (other) Discrete Mathematics Set theory and logic Order relations Algebraic methods Proof by exhaustion

13–15 3.0

The order of books on a shelf is called wrong if no three adjacent books are arranged in order of height (either increasing or decreasing). How many wrong orders is it possible to construct from \(n\) books of different heights, if: a) \(n = 4\); b) \(n = 5\)?

Problem #PRU-34991

Problems Methods Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (other) Discrete Mathematics Combinatorics Systems of points and line segments Systems of points, lines, and line segments

13–15 4.0

On a line, there are 50 segments. Prove that either it is possible to find some 8 segments all of which have a shared intersection, or there can be found 8 segments, no two of which intersect.

Problem #PRU-35578

Problems Methods Extremal principle Extremal principle (other)

13–14 2.0

In how many ways can you rearrange the numbers from 1 to 100 so that the neighbouring numbers differ by no more than 1?

Problem #PRU-60489

Problems Number Theory Divisibility Equations in integer numbers Methods Extremal principle Extremal principle (other) Methods from geometry Geometric interpretations in algebra Pigeonhole principle Pigeonhole principle (other) The greatest common divisor (GCD) and the least common multiplier (LCM). Mutually prime numbers

13–15 4.0

Let \(a\), \(b\), \(c\) be integers; where \(a\) and \(b\) are not equal to zero.

Prove that the equation \(ax + by = c\) has integer solutions if and only if \(c\) is divisible by \(d = \mathrm{GCD} (a, b)\).

Problem #PRU-65204

Problems Geometry Plane geometry Triangles Similar triangles Auxiliary similar triangles Methods Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (area and volume)

14–16 4.0

A unit square is divided into \(n\) triangles. Prove that one of the triangles can be used to completely cover a square with side length \(\frac{1}{n}\).

Problem #PRU-78557

Problems Discrete Mathematics Combinatorics Dissections, partitions, covers and tilings Covers Methods Extremal principle Extremal principle (other) Pigeonhole principle Pigeonhole principle (other)

13–15 4.0

Several pieces of carpet are laid along a corridor. Pieces cover the entire corridor from end to end without omissions and even overlap one another, so that over some parts of the floor lie several layers of carpet. Prove that you can remove a few pieces, perhaps by taking them out from under others and leaving the rest exactly in the same places they used to be, so that the corridor will still be completely covered and the total length of the pieces left will be less than twice the length corridor.