We solve problems

Problem #PRU-103780

Problems Methods Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other) Proof by exhaustion

11–13 3.0

In any group of 10 children, out of a total of 60 pupils, there will be three who are in the same class. Will it always be the case that amongst the 60 pupils there will be: 1) 15 classmates? 2) 16 classmates?

Problem #PRU-116539

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

13–15 3.0

There are a thousand tickets with numbers 000, 001, ..., 999 and a hundred boxes with the numbers 00, 01, ..., 99. A ticket is allowed to be dropped into a box if the number of the box can be obtained from the ticket number by erasing one of the digits. Is it possible to arrange all of the tickets into 50 boxes?

Problem #PRU-21985

Problems Methods Algebraic methods Counting in two ways Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other)

12–13 3.0

The total age of a group of 7 people is 332 years. Prove that it is possible to choose three members of this group so that the sum of their ages is no less than 142 years.

Problem #PRU-22000

Problems Number Theory Divisibility Divisibility of a number. General properties Methods Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other)

12–13 3.0

You are given 11 different natural numbers that are less than or equal to 20. Prove that it is always possible to choose two numbers where one is divisible by the other.

Problem #PRU-31363

Problems Methods Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other) Proof by contradiction

12–14 3.0

a) In a group of 4 people, who speak different languages, any three of them can communicate with one another; perhaps by one translating for two others. Prove that it is always possible to split them into pairs so that the two members of every pair have a common language.

b) The same, but for a group of 100 people.

c) The same, but for a group of 102 people.

Problem #PRU-32046

Problems Methods Algebraic methods Counting in two ways Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other)

11–13 3.0

You are given 25 numbers. The sum of any 4 of these numbers is positive. Prove that the sum of all 25 numbers is also positive.

Problem #PRU-32787

Problems Number Theory Numeral systems Decimal number system Methods Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (other)

12–13 3.0

Is it possible to find 57 different two digit numbers, such that no sum of any two of them was equal to 100?

Problem #PRU-35149

Problems Geometry Plane geometry Geometrical inequalities Area inequalities Methods Algebraic methods Partitions into pairs and groups; bijections Pigeonhole principle Pigeonhole principle (area and volume)

13–15 3.0

In a square which has sides of length 1 there are 100 figures, the total area of which sums to more than 99. Prove that in the square there is a point which belongs to all of these figures.

Problem #PRU-35719

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

12–14 3.0

26 numbers are chosen from the numbers 1, 2, 3, ..., 49, 50. Will there always be two numbers chosen whose difference is 1?

Problem #PRU-64568

Problems Methods Examples and counterexamples. Constructive proofs Calculus Number sequences Number sequences (other) Algebraic methods Partitions into pairs and groups; bijections

13–14 3.0

Hannah placed 101 counters in a row which had values of 1, 2 and 3 points. It turned out that there was at least one counter between every two one point counters, at least two counters lie between every two two point counters, and at least three counters lie between every two three point counters. How many three point counters could Hannah have?