Mark has 1000 identical cubes, each of which has one pair of opposite faces which are coloured white, another pair which are blue and a third pair that are red. He made a large \(10 \times 10 \times 10\) cube from them, joining cubes to one another which have the same coloured faces. Prove that the large cube has a face which is solidly one colour.
In six baskets there are pears, plums and apples. The number of plums in each basket is equal to the total number of apples in the other baskets. The number of apples in each basket is equal to the total number of pears in the other baskets. Prove that the total number of fruits is divisible by 31.
Solve the following inequality: \(x+y^2 +\sqrt{x-y^2-1} \leq 1\).
Is it true that, if \(b>a+c>0\), then the quadratic equation \(ax^2 +bx+c=0\) has two roots?
Suppose that: \[\frac{x+y}{x-y}+\frac{x-y}{x+y} =3.\] Find the value of the following expression: \[\frac{x^2 +y^2}{x^2-y^2} + \frac{x^2 -y^2}{x^2+y^2}.\]
Pinocchio correctly solved a problem, but stained his notebook. \[(\bullet \bullet + \bullet \bullet+1)\times \bullet= \bullet \bullet \bullet\]
Under each blot lies the same number, which is not equal to zero. Find this number.
Seven coins are arranged in a circle. It is known that some four of them, lying in succession, are fake and that every counterfeit coin is lighter than a real one. Explain how to find two counterfeit coins from one weighing on scales without any weights. (All counterfeit coins weigh the same.)
Four people discussed the answer to a task.
Harry said: “This is the number 9”.
Ben: “This is a prime number.”
Katie: “This is an even number.”
And Natasha said that this number is divisible by 15.
One boy and one girl answered correctly, and the other two made a mistake. What is the actual answer to the question?
Peter recorded an example of an addition on a board, after which he replaced some digits with letters, with the same figures being replaced with the same letters, and different figures with different letters. He did it such that he was left with the sum: \(CROSS + 2011 = START\). Prove that Peter made a mistake.
Compute the following: \[\frac{(2001\times 2021 +100)(1991\times 2031 +400)}{2011^4}.\]