We solve problems

Problem #PRU-35379

Problems Calculus Functions of one variable. Continuity Certain properties of a function and recurrence relations. Algebra Algebraic equations and systems of equations Systems of linear equations

14–15 3.0

Find all functions $f (x)$ defined for all real values of $x$ and satisfying the equation $2 f (x) + f (1 - x) = x^{2}$ .

Problem #PRU-35430

Problems Discrete Mathematics Algorithm Theory Game theory Game theory (other) Number Theory Divisibility Odd and even numbers

12–13 3.0

Given a board (divided into squares) of the size: a) $10 \times 12$ , b) $9 \times 10$ , c) $9 \times 11$ , consider the game with two players where: in one turn a player is allowed to cross out any row or any column if there is at least one square not crossed out. The loser is the one who cannot make a move. Is there a winning strategy for one of the players?

Problem #PRU-35457

Problems Calculus Real numbers Integer and fractional parts. Archimedean property Methods Pigeonhole principle Pigeonhole principle (angles and lengths)

13–15 3.0

Prove that the equation $a_{1} \sin x + b_{1} \cos x + a_{2} \sin 2 x + b_{2} \cos 2 x + \dots + a_{n} \sin n x + b_{n} \cos n x = 0$ has at least one root for any values of $a_{1}, b_{1}, a_{2}, b_{2}, \dots, a_{n}, b_{n}$ .

Problem #PRU-35476

Problems Calculus Functions of one variable. Continuity Methods Algebraic methods Iterations

15–16 3.0

Let $f (x)$ be a polynomial about which it is known that the equation $f (x) = x$ has no roots. Prove that then the equation $f (f (x)) = x$ does not have any roots.

Problem #PRU-35512

Problems Methods Pigeonhole principle Pigeonhole principle (other)

12–14 3.0

Prove that in any group of friends there will be two people who have the same number of friends.

Problem #PRU-35541

Problems Calculus Real numbers Integer and fractional parts. Archimedean property Methods Pigeonhole principle Pigeonhole principle (angles and lengths)

13–15 3.0

Solve the equation $2 x^{x} = \sqrt{2}$ for positive numbers.

Problem #PRU-35599

Problems Algebra and arithmetic Algebraic geometry Cryptography Number Theory Divisibility Division with remainders. Arithmetic of remainders Division with remainder Methods Pigeonhole principle Pigeonhole principle (other)

14–15 3.0

Upon the installation of a keypad lock, each of the 26 letters located on the lock’s keypad is assigned an arbitrary natural number known only to the owner of the lock. Different letters do not necessarily have different numbers assigned to them. After a combination of different letters, where each letter is typed once at most, is entered into the lock a summation is carried out of the corresponding numbers to the letters typed in. The lock opens only if the result of the summation is divisible by 26. Prove that for any set of numbers assigned to the 26 letters, there exists a combination that will open the lock.

Problem #PRU-35606

Problems Methods Pigeonhole principle Pigeonhole principle (other)

13–14 3.0

In an $n$ by $n$ grid, $2 n$ of the squares are marked. Prove that there will always be a parallelogram whose vertices are the centres of four of the squares somewhere in the grid.

Problem #PRU-35627

Problems Methods Pigeonhole principle Pigeonhole principle (other) Discrete Mathematics Set theory and logic Set theory Set theory (other) Algebra Fractions Simple fractions

12–15 3.0

A hostess bakes a cake for some guests. Either 10 or 11 people can come to her house. What is the smallest number of pieces she needs to cut the cake into (in advance) so that it can be divided equally between 10 and 11 guests?

Problem #PRU-35635

Problems Calculus Real numbers Integer and fractional parts. Archimedean property Methods Pigeonhole principle Pigeonhole principle (angles and lengths)

13–15 3.0

Two players play the following game. They take turns. One names two numbers that are at the ends of a line segment. The next then names two other numbers, which are at the ends of a segment nested in the previous one. The game goes on indefinitely. The first aims to have at least one rational number within the intersection of all of these segments, and the second aims to prevent such occurring. Who wins in this game?