Find the mistake in the sequence of equalities: \(-1=(-1)^{\frac{2}{2}}=((-1)^2)^{\frac{1}{2}}=1^{\frac{1}{2}}=1\).
Theorem: If we mark \(n\) points on a circle and connect each point to every other point by a straight line, the lines divide the interior of the circle is into is \(2n-1\) regions.
"Proof": First, let’s have a look at the smallest natural numbers.
When \(n=1\) there is one region (the whole disc).
When \(n=2\) there are two regions (two half-discs).
When \(n=3\) there are \(4\) regions (three lune-like regions and one triangle in the middle).
When \(n=4\) there are \(8\) regions, and if you’re still not convinced then try \(n=5\) and you’ll find \(16\) regions if you count carefully.
Our proof in general will be by induction on \(n\). Assuming the theorem is true for \(n\) points, consider a circle with \(n+1\) points on it. Connecting \(n\) of them together in pairs produces \(2n-1\) regions in the disc, and then connecting the remaining point to all the others will divide the previous regions into two parts, thereby giving us \(2\times (2n-1)=2n\) regions.
Let’s "prove" that the number \(1\) is a multiple of \(3\). We will use the symbol \(\equiv\) to denote "congruent modulo \(3\)". Thus, what we need to prove is that \(1\equiv 0\) modulo \(3\). Let’s see: \(1\equiv 4\) modulo \(3\) means that \(2^1\equiv 2^4\) modulo \(3\), thus \(2\equiv 16\) modulo \(3\), however \(16\) gives the remainder \(1\) after division by \(3\), thus we get \(2\equiv 1\) modulo \(3\), next \(2-1\equiv 1-1\) modulo \(3\), and thus \(1\equiv 0\) modulo \(3\). Which means that \(1\) is divisible by \(3\).
There are real numbers written on each field of a \(m \times n\) chessboard. Some of them are negative, some are positive. In one move we can multiply all the numbers in one column or row by \(-1\). Is that always possible to obtain a chessboard where sums of numbers in each row and column are non-negative?
Tom found a large, old clock face and put \(12\) sweets on the number \(12\). Then he started to play a game: in each move he moves one sweet to the next number clockwise, and some other to the next number anticlockwise. Is it possible that after finite number of steps there is exactly \(1\) of the sweets on each number?
There are numbers \(1,2,3,4,5,6,7,8,9\) and \(10\) written on a board. Each time you make a "move" you can erase three of the remaining numbers, \(a,b\) and \(c\), and replace them with the numbers \(2a+b,2b+c\) and \(2c+a\). Is it possible to make all the \(10\) numbers left on the board equal?
On a certain island there are \(17\) grey, \(15\) brown and \(13\) crimson chameleons. If two chameleons of different colours meet, both of them change to the third colour. No other colour changes are allowed. Is it possible that after a few such colour transitions all the chameleons have the same colour?
Sixteen lightbulbs are arranged in a \(4 \times 4\) grid. Some of them are on, the other ones are off. You are allowed to change the state of all the bulbs in a column, in a row, or along any diagonal (note: there are \(14\) diagonals in total!). Is it possible to go from the arrangement on the left to the one on the right by repeating this operation?
The numbers \(1\) to \(2025\) are written on a board. In one move we can erase two numbers and replace them with the absolute value of their difference. Can we achieve a sequence consisting of only \(0\) after some number of moves?
A magic square is a square filled with numbers, one in each cell, in such a way that the sums of the numbers in each row, each column and along each of the two main diagonals are the same. The value of this sum is known as the magic constant of the square. Show that in any \(4 \times 4\) magic square (which contains \(16\) numbers) the sum of all the numbers in the \(4\) central squares is also equal to the magic constant of the square.