answer:Let's denote the percentage of material A in the second solution as x%. Since the first solution is 80% of the mixture, the second solution must be 20% of the mixture (because the total must add up to 100%). The amount of material A in the mixture from the first solution is 20% of material A times 80% of the mixture, which is 0.20 * 0.80. The amount of material A in the mixture from the second solution is x% of material A times 20% of the mixture, which is x/100 * 0.20. The total amount of material A in the mixture is 22%, so we can set up the equation: 0.20 * 0.80 + x/100 * 0.20 = 0.22 Solving for x: 0.16 + 0.002x = 0.22 0.002x = 0.22 - 0.16 0.002x = 0.06 x = 0.06 / 0.002 x = 30 Therefore, the second solution contains boxed{30%} of material A.

answer:1. **Identify roots of ( P(X) ):** Given the equation: [ (X + 2019)P(X) = X P(X + 1), ] we first inspect it at specific values of (X = 0) and (X = -2018). - Evaluating the original equation at (X = 0): [ (0 + 2019)P(0) = 0 cdot P(1), ] which simplifies to: [ 2019P(0) = 0 implies P(0) = 0. ] Therefore, (0) is a root of (P(X)). - Evaluating the original equation at (X = -2018): [ (-2018 + 2019)P(-2018) = (-2018) P(-2017), ] which simplifies to: [ P(-2018) = -2018 P(-2017). ] From the above, for consistency in integral domain, we must have (P(-2018) = 0). Thus, (-2018) is also a root of (P(X)). 2. **Factorization and substitution:** Given (P(0) = 0) and (P(-2018) = 0), we know that (X) and ((X + 2018)) are factors of (P(X)). Therefore, we can write: [ P(X) = X (X + 2018) Q(X), ] for some polynomial (Q(X)). 3. **Substitute into original equation:** Substitute (P(X) = X(X+2018)Q(X)) into the original equation, so it becomes: [ (X + 2019) X (X + 2018) Q(X) = X (X + 1)(X + 2018 + 1) Q(X + 1). ] Simplify the equation: [ (X + 2019) X (X + 2018) Q(X) = X (X + 1)(X + 2019) Q(X + 1). ] Since (X neq 0) and (X + 2019 neq 0), we can divide both sides by (X(X + 2019)): [ (X + 2018) Q(X) = (X + 1) Q(X + 1). ] 4. **Recursive property:** Consider: [ (X + 2018) Q(X) = (X + 1) Q(X + 1). ] Evaluate at (X = 0): [ 2018Q(0) = Q(1). ] Evaluate at (X = 1): [ 2017Q(1) = 2Q(2). ] Notice the pattern here. For any (X = k), we have: [ (2018 - k) Q(k) = (k + 1) Q(k + 1). ] 5. **Form of Q(X):** For (Q(X)) to satisfy the equation and given the recursive relation, (Q(X)) must be a constant because the relationship (Q(X) = Q(X+1)) implies (Q(X)) is the same for all (X). Consequently, let (Q(X) = C), where (C) is a constant. Therefore, [ P(X) = X (X + 2018) C. ] 6. **Final form of the polynomial:** Accordingly, the polynomial (P(X)) can be expressed as: [ P(X) = C X (X + 2018), ] where (C) is any real constant. # Conclusion: The required polynomial is: [ boxed{P(X) = C X (X + 2018)}, ] where (C) is a constant.

answer:We will prove the statement using induction on the dimension ( d ). # Base Case: ( d = 2 ) Consider a 2-dimensional grid of size ( n times n ). We need to show that ( n ) lines can be chosen to cover all ( n^2 ) points of the grid. 1. Let ( A ) and ( B ) be finite sets such that ( |A| = a ) and ( |B| = b ) with ( a leq b ). 2. The grid ( C = A times B ) has ( a times b ) points. 3. Suppose ( a + b - 3 ) lines cover ( C ). We need to show that ( n ) lines can be chosen to cover all points of the grid. We proceed by contradiction: 4. Assume that ( C' ) is the set of points obtained from ( C ) by omitting all points lying on the perpendicular lines. 5. Let ( k_1 ) and ( k_2 ) be the number of lines perpendicular to the basis vectors of ( mathbb{R} ). Then ( C' ) has ((a - k_1)(b - k_2)) points. 6. If ( C' = emptyset ), then ( a = k_1 ) or ( b = k_2 ), and a coverage of ( C' ) can be chosen. 7. Suppose ( C' neq emptyset ). Then ( C' ) is covered by ( a + b - 3 - k_1 - k_2 ) lines not left out of the original coverage. 8. If ( a - k_1 = 1 ), then the point ( b - k_2 ) in ( C' ) is on a section parallel to the vectors, so to cover them, you need a line ( b - k_2 ). However, ( b - k_2 > b - k_2 - 2 = a + b - 3 - k_1 - k_2 ), which is a contradiction. 9. The same holds if ( b - k_2 = 1 ). 10. Suppose ( a - k_1 geq 2 ) and ( b - k_2 geq 2 ). Then the convex hull of ( C' ) is a rectangle with boundary ( 2(a - k_1) + 2(b - k_2) - 4 ) points. 11. Any non-axial line can contain at most two points on the boundary of the rectangle, so the line ( a + b - 3 - k_1 - k_2 ) has at most ( 2(a + b - 3 - k_1 - k_2) < 2(a - k_1) + 2(b - k_2) - 4 ) points on the boundary of the rectangle, which means it cannot cover ( C' ), leading to a contradiction. Thus, the case ( d = 2 ) is proven. # Inductive Step: Assume the statement is true for ( d-1 ) Assume that for a ( (d-1) )-dimensional grid, ( n ) hyperplanes can be chosen to cover all points of the grid. 1. Consider a ( d )-dimensional grid. If for some ( i ) where ( 1 leq i leq n ), we have ( V_i = {x_1 = i} subset mathbb{R} ) is not between the intercepting hyperplanes, then we can apply the induction hypothesis to the intersection of the lattice ( mathbb{R}_i ) with ( V_i ). 2. There are ( L_1^{(i)}, L_2^{(i)}, ldots, L_n^{(i)} ) hyperplanes such that the ( L_k^{(i)} cap V_i ) ( (d-2) )-dimensional planes are perpendicular to the ( x_2 )-axis and together cover ( mathbb{R}_i ). 3. If ( V_j ) (( j neq i )) is also not between the overlapping hyperplanes, then the ( (d-2) )-dimensional planes ( L_1^{(j)} cap V_j, L_2^{(j)} cap V_j, ldots, L_n^{(i)} ) defined for ( V_j ) are also perpendicular to the ( x_2 )-axis. 4. Otherwise, ( L_1^{(i)}, L_2^{(i)}, ldots, L_n^{(i)}, L_1^{(j)}, L_2^{(j)}, ldots, L_n^{(j)} ) would be ( 2n ) different hyperplanes, which is impossible as ( 2n > 2n - 3 ). Suppose ( L_k^{(i)} = L_s^{(j)} ): 5. Then ( L_k^{(i)} cap V_i ) is parallel to ( L_s^{(j)} cap V_j ), since ( V_i ) is parallel to ( V_j ), so ( L_s^{(j)} cap V_j ) is perpendicular to the ( x_2 )-axis. 6. If ( L_s^{(j)} cap V_j ) is also perpendicular to an axis other than the ( x_2 )-axis, then the ( (d-2) )-dimensional plane ( L_s^{(j)} cap V_j ) has a planar dimension of at least 2 for the ( (d-1) )-dimensional plane ( V_j ), which is impossible. Now notice: 7. If ( L_k^{(i)} cap V_i ) is perpendicular to the ( x_2 )-axis, then the hyperplane ( L_k^{(i)} ) is perpendicular to the plane ([x_1, x_2]). 8. Indeed, the 2-codimensional plane ( L_k^{(i)} cap V_i ) in ( mathbb{R} ) is perpendicular to the ( x_1 )-axis, the ( x_2 )-axis, and the normal vector ( n_i ) of the plane ( L_k^{(i)} ), so ( n_i ) must be parallel to the plane ([x_1, x_2]). Thus, if we consider only the intercept hyperplanes perpendicular to the plane ([x_1, x_2]), then they already cover the grid. Therefore, the projection of everything on the plane ([x_1, x_2]) reduces the problem to the case for ( d = 2 ). # Conclusion For ( d geq 2 ), the cube grid can be covered with ( n ) hyperplanes so that none of the ( 2n - 2 ) overlapping hyperplanes can be omitted. (blacksquare)

answer:We are given an arithmetic sequence of polynomials ((P_n)_{n in mathbb{N}}) with a common difference (Q(x)) and the first term (P(x)). Both (P(x)) and (Q(x)) are monic polynomials with integer coefficients and do not share any integer roots. Each term of the sequence has at least one integer root. We need to prove two statements: a) (P(x)) is divisible by (Q(x)). b) (text{deg}left(frac{P(x)}{Q(x)}right) = 1). # Part (a) 1. **Define the GCD**: Let (G = gcd(P, Q)). Since (P) and (Q) are monic polynomials, (G) is also a monic polynomial with integer coefficients. 2. **Express (P) and (Q) in terms of (G)**: Write (P = GA) and (Q = GB), where (A) and (B) are monic polynomials with integer coefficients. 3. **Use the fact that (P) and (Q) do not share integer roots**: Since (P) and (Q) do not share any integer roots, (G) must not have any integer roots. 4. **Form the sequence (P_n)**: The sequence is given by (P_n = P + (n-1)Q). Substituting the expressions for (P) and (Q), we get: [ P_n = GA + (n-1)GB = G(A + (n-1)B) ] 5. **Existence of integer roots**: Each (P_n) has at least one integer root. Let (x_n) be an integer root of (P_n), i.e., (P_n(x_n) = 0). This implies: [ G(A(x_n) + (n-1)B(x_n)) = 0 ] Since (G) has no integer roots, it must be that: [ A(x_n) + (n-1)B(x_n) = 0 ] 6. **Simplify the equation**: Rearrange the equation to get: [ A(x_n) = -(n-1)B(x_n) ] 7. **Analyze the values of (B(x_n))**: Since (A(x_n)) and ((n-1)B(x_n)) are integers, and this holds for infinitely many (n), (B(x_n)) must be either (1) or (-1) for all (n). Because (B) is monic, it must be constant. Therefore, (B = 1) or (B = -1). 8. **Determine the value of (B)**: Since (B) is monic, it must be (1). Thus, (Q = G). 9. **Conclusion for part (a)**: Since (Q = G), (P(x)) is divisible by (Q(x)). # Part (b) 1. **Express (P(x)) in terms of (Q(x))**: Since (Q = G), we have (P = GA) and (Q = G). Therefore: [ frac{P(x)}{Q(x)} = A(x) ] 2. **Determine the degree of (A(x))**: From the equation (A(x_n) = -(n-1)), we see that (A(x)) must be a linear polynomial because it maps (x_n) to (-(n-1)), which is a linear function of (n). 3. **Conclusion for part (b)**: Therefore, (text{deg}(A(x)) = 1), which implies: [ text{deg}left(frac{P(x)}{Q(x)}right) = 1 ] (blacksquare)