In the expansion of \((x + y)^n\), using the Newton binomial formula, the second term was 240, the third – 720, and the fourth – 1080. Find \(x\), \(y\) and \(n\).
Show that any natural number \(n\) can be uniquely represented in the form \(n = \binom{x}{1} + \binom{y}{2} + \binom{z}{3}\) where \(x, y, z\) are integers such that \(0 \leq x < y < z\), or \(0 = x = y < z\).
Here is a fragment of the table, which is called the Leibniz triangle. Its properties are “analogous in the sense of the opposite” to the properties of Pascal’s triangle. The numbers on the boundary of the triangle are the inverses of consecutive natural numbers. Each number is equal to the sum of two numbers below it. Find the formula that connects the numbers from Pascal’s and Leibniz triangles.
Numbers \(a, b, c\) are integers with \(a\) and \(b\) being coprime. Let us assume that integers \(x_0\) and \(y_0\) are a solution for the equation \(ax + by = c\).
Prove that every solution for this equation has the same form \(x = x_0 + kb\), \(y = y_0 - ka\), with \(k\) being a random integer.
Solve the equations in integers:
a) \(3x^2 + 5y^2 = 345\);
b) \(1 + x + x^2 + x^3 = 2^y\).
Draw all of the stairs made from four bricks in descending order, starting with the steepest \((4, 0, 0, 0)\) and ending with the shallowest \((1, 1, 1, 1)\).
A frog jumps over the vertices of the triangle \(ABC\), moving each time to one of the neighbouring vertices.
How many ways can it get from \(A\) to \(A\) in \(n\) jumps?
Valentina added a number (not equal to 0) taken to the power of four and the same number to the power two and reported the result to Peter. Can Peter determine the unique number that Valentina chose?
Is it possible to:
a) load two coins so that the probability of “heads” and “tails” were different, and the probability of getting any of the combinations “tails, tails,” “heads, tails”, “heads, heads” be the same?
b) load two dice so that the probability of getting any amount from 2 to 12 would be the same?
The segment \(OA\) is given. From the end of the segment \(A\) there are 5 segments \(AB_1, AB_2, AB_3, AB_4, AB_5\). From each point \(B_i\) there can be five more new segments or not a single new segment, etc. Can the number of free ends of the constructed segments be 1001? By the free end of a segment we mean a point belonging to only one segment (except point \(O\)).