The product of 1986 natural numbers has exactly 1985 different prime factors. Prove that either one of these natural numbers, or the product of several of them, is the square of a natural number.
The product of a group of 48 natural numbers has exactly 10 prime factors. Prove that the product of some four of the numbers in the group will always give a square number.
In March 2015 a teacher ran 11 sessions of a maths club. Prove that if no sessions were run on Saturdays or Sundays there must have been three days in a row where a session of the club did not take place. The 1st March 2015 was a Sunday.
Prove that from any 27 different natural numbers less than 100, two numbers that are not coprime can be chosen.
7 different digits are given. Prove that for any natural number \(n\) there is a pair of these digits, the sum of which ends in the same digit as the number.
In a group of seven boys, everyone has at least three brothers who are in that group. Prove that all seven are brothers.
A council of 2,000 deputies decided to approve a state budget containing 200 items of expenditure. Each deputy prepared his draft budget, which indicated for each item the maximum allowable, in his opinion, amount of expenditure, ensuring that the total amount of expenditure did not exceed the set value of \(S\). For each item, the board approves the largest amount of expenditure that is agreed to be allocated by no fewer than \(k\) deputies. What is the smallest value of \(k\) for which we can ensure that the total amount of approved expenditures does not exceed \(S\)?
Izzy wrote a correct equality on the board: \(35 + 10 - 41 = 42 + 12 - 50\), and then subtracted 4 from both parts: \(35 + 10 - 45 = 42 + 12 - 54\). She noticed that on the left hand side of the equation all of the numbers are divisible by 5, and on the right hand side by 6. Then she took 5 outside of the brackets on the left hand side and 6 on the right hand side and got \(5(7 + 2 - 9)4 = 6(7 + 2 - 9)\). Having simplified both sides by a common multiplier, Izzy found that \(5 = 6\). Where did she go wrong?
A carpet of size 4 m by 4 m has had 15 holes made in it by a moth. Is it always possible to cut out a 1 m \(\times\) 1 m area of carpet that doesn’t contain any holes? The holes are considered to be points.
Prove that in any group of 2001 whole numbers there will be two whose difference is divisible by 2000.