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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 dark room on a shelf there are 4 pairs of socks of two different sizes and two different colours that are not arranged in pairs. What is the minimum number of socks necessary to move from the drawer to the suitcase, without leaving the room, so that there are two pairs of socks of different sizes and colours in the suitcase?

In a group of seven boys, everyone has at least three brothers who are in that group. Prove that all seven are brothers.

The function \(f (x)\) for each real value of \(x\in (-\infty, + \infty)\) satisfies the equality \(f (x) + (x + 1/2) \times f (1 - x) = 1\).

a) Find \(f (0)\) and \(f (1)\). b) Find all such functions \(f (x)\).

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 \(S\)?

Aladdin visited all of the points on the equator, moving to the east, then to the west, and sometimes instantly moving to the diametrically opposite point on Earth. Prove that there was a period of time during which the difference in distances traversed by Aladdin to the east and to the west was not less than half the length of the equator.

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?