answer:To determine whether the five-digit number is odd, we need to check the digit placed in the unit’s position. The number is odd if its unit's digit is either 3, 5, 7, or 9. - Total possible digits for the unit's position that result in an odd number: 3, 5, 7, 9 (4 possibilities). - Total digits available for the unit's position: 2, 3, 5, 7, 9 (5 possibilities). Probability calculation: [ text{Probability} = frac{text{Number of favorable outcomes}}{text{Total outcomes}} = frac{4}{5} ] Thus, the probability that the number is odd is boxed{frac{4}{5}}.

answer:1. To find the third derivative of ( y = x^5 - 7x^3 + 2 ), we proceed as follows: 1.1. Differentiate ( y ) with respect to ( x ): [ y^{prime} = frac{d}{dx}(x^5 - 7x^3 + 2) = 5x^4 - 21x^2 ] 1.2. Differentiate ( y^{prime} ) to find ( y^{primeprime} ): [ y^{primeprime} = frac{d}{dx}(5x^4 - 21x^2) = 20x^3 - 42x ] 1.3. Differentiate ( y^{primeprime} ) to find ( y^{primeprimeprime} ): [ y^{primeprimeprime} = frac{d}{dx}(20x^3 - 42x) = 60x^2 - 42 ] [ boxed{y^{primeprimeprime} = 60x^2 - 42} ] 2. To find the fifth derivative of ( y = ln x ): 2.1. Start with the first derivative: [ y^{prime} = frac{d}{dx} (ln x) = frac{1}{x} = x^{-1} ] 2.2. Differentiate successively: [ y^{primeprime} = frac{d}{dx}left(x^{-1}right) = -x^{-2} ] [ y^{primeprimeprime} = frac{d}{dx}left(-x^{-2}right) = 2x^{-3} ] [ y^{(4)} = frac{d}{dx}left(2x^{-3}right) = -6x^{-4} ] [ y^{(5)} = frac{d}{dx}left(-6x^{-4}right) = 24x^{-5} = frac{24}{x^5} ] [ boxed{y^{(5)} = frac{24}{x^6}} ] 3. To find the second derivative of ( s = arctan(2x) ) at ( x = -1 ): 3.1. First derivative: [ s^{prime} = frac{d}{dx} (arctan(2x)) = frac{2}{1 + (2x)^2} = frac{2}{1 + 4x^2} ] 3.2. Second derivative: [ s^{primeprime} = frac{d}{dx} left(frac{2}{1 + 4x^2}right) = frac{d}{dx} left(2(1 + 4x^2)^{-1}right) ] Using the chain rule: [ s^{primesecond} = 2 cdot (-1) cdot frac{d}{dx}(1 + 4x^2)^{-2} cdot 4x = -8x(1 + 4x^2)^{-2} = frac{-8x}{(1 + 4x^2)^2} ] Substitute ( x = -1 ): [ s^{primesecond}(-1) = frac{-8(-1)}{(1 + 4(-1)^2)^2} = frac{8}{25} ] [ boxed{s^{primesecond}(-1) = frac{8}{25}} ] 4. To show that ( y = e^{-varphi} sin(varphi) ) satisfies the differential equation ( y^{second} + 2y^{prime} + 2y = 0 ): 4.1. First derivative: Using the product rule: [ y^{prime} = frac{d}{dvarphi} (e^{-varphi} sin(varphi)) = (e^{-varphi})^{prime} sin(varphi) + e^{-varphi} (sin(varphi))^{superprime} ] [ y^{superprime} = e^{-varphi} cos(varphi) - e^{-varphi} sin(varphi) = e^{-varphi} (cos(varphi) - sin(varphi)) ] 4.2. Second derivative: [ y^{second} = frac{d}{dvarphi} left( e^{-varphi} (cos(varphi) - sin(varphi)) right) ] [ y^{second} = e^{-varphi} (-sin(varphi) - cos(varphi)) - e^{-varphi} (cos(varphi) - sin(varphi)) = -2 e^{-varphi} cos(varphi) ] 4.3. Verify the differential equation: [ y^{second} + 2y^{superprime} + 2y = -2 e^{-varphi} cos(varphi) + 2 e^{-varphi} (cos(varphi) - sin(varphi)) + 2 e^{-varphi} sin(varphi) ] [ = -2 e^{-varphi} cos(varphi) + 2 e^{-varphi} cos(varphi) - 2 e^{-varphi} sin(varphi) + 2 e^{-varphi} sin(varphi) = 0 ] Therefore, [ boxed{0 = 0} ] 5. To find the 24th derivative of ( y = e^{x}(x^{2} - 1) ): Using the Leibniz formula for derivatives: [ y^{(24)} = left( e^x (x^2 - 1) right)^{(24)} = sum_{k=0}^{24} binom{24}{k} left( e^x right)^{(24-k)} left( x^2 - 1 right)^{(k)} ] Since any derivative of ( x^2 - 1 ) of order greater than 2 is zero: [ y^{(24)} = binom{24}{0} e^x (x^2 - 1) + binom{24}{1} e^x frac{d}{dx} (x^2 - 1) + binom{24}{2} e^x frac{d^2}{dx^2} (x^2 - 1) ] [ y^{(24)} = e^x (x^2 - 1) + 24 e^x (2x) + 276 e^x (2) ] [ y^{(24)} = e^x (x^2 - 1) + 48xe^x + 552e^x ] [ boxed{y^{(24)} = e^x (x^2 + 48x + 551)} ] 6. To find the ( k )-th derivative of ( y = x^m ): 6.1. Using the general formula for the ( k )-th derivative of ( x^m ): [ y^{(k)} = frac{d^k}{dx^k} (x^m) = begin{cases} m(m-1)cdots(m-k+1) x^{m-k} & text{for } k le m 0 & text{for } k > m end{cases} ] 6.2. If ( m ) is a positive integer, then [ y^{(k)} = boxed{begin{cases} frac{m!}{(m-k)!} x^{m-k} & text{if } k le m 0 & text{if } k > m end{cases}} ] In summary: 1. (boxed{y^{primeprimeprime} = 60x^2 - 42}) 2. (boxed{y^{(5)} = frac{24}{x^6}}) 3. (boxed{s^{primesecond}(-1) = frac{8}{25}}) 4. (boxed{0 = 0}) 5. (boxed{y^{(24)} = e^x (x^2 + 48x + 551)}) 6. (boxed{begin{cases} frac{m!}{(m-k)!} x^{m-k} & text{if } k le m 0 & text{if } k > m end{cases}})

answer:1. **Define the sequences and sums:** Let ( P_i ) be the sequence formed by the ordered weights of the four stones colored ( i ): ( P_i = (a_i, b_i, c_i, d_i) ) with ( a_i < b_i < c_i < d_i ). Define the differences ( x_i = b_i - a_i ), ( y_i = c_i - b_i ), and ( z_i = d_i - c_i ). Let ( S ) represent the sums of these groups: [ S_a = sum_{i=1}^n a_i, quad S_b = sum_{i=1}^n b_i, quad S_c = sum_{i=1}^n c_i, quad S_d = sum_{i=1}^n d_i ] [ S_x = sum_{i=1}^n x_i, quad S_y = sum_{i=1}^n y_i, quad S_z = sum_{i=1}^n z_i ] 2. **Establish relationships between sums:** It is trivial that: [ S_x = S_b - S_a, quad S_y = S_c - S_b, quad S_z = S_d - S_c ] Adding these, we get: [ S_x + S_y = S_c - S_a, quad S_x + S_y + S_z = S_d - S_a ] Summing the first and last equations, we obtain: [ 3S_x + 2S_y + S_z = S_a + S_b + S_c + S_d - 4S_a ] Since the total sum of weights is ( 2n(4n+1) ), we have: [ 3S_x + 2S_y + S_z = 2n(4n+1) - 4S_a ] Simplifying, we get: [ 2(S_x + S_y) + S_x + S_z = 2(n(4n+1) - 2S_a) ] Therefore: [ S_x + S_z equiv 0 pmod{2} ] 3. **Check parity conditions:** We also have: [ S_a + S_d equiv S_b + S_c pmod{2}, quad S_x equiv S_z pmod{2} ] If ( S_a + S_d = S_b + S_c ), the problem is solved by taking: [ A = bigcup {a_i, d_i}, quad B = bigcup {b_i, c_i} ] Both sums are equal. 4. **Assume ( S_a + S_d neq S_b + S_c ):** Let ( alpha_i = x_i - z_i = (b_i - a_i) - (d_i - c_i) = (b_i + c_i) - (a_i + d_i) ). We need to find a subset ( T_{half} ) of ( {1 le i le n mid alpha_i neq 0} ) such that: [ sum_{i in T_{half}} alpha_i = frac{S_z - S_x}{2} ] 5. **Proof by contradiction:** Suppose ( T_{half} ) does not exist. For any ( T ), let ( S_T = sum_{i in T} alpha_i ). Order the sequence ( (S_{T_0}, S_{T_1}, ldots, S_{T_N}) ) with the sums of all subsets ( T ) of ( V = {1 le i le n mid alpha_i neq 0} ). Take ( k < l in V ) such that ( S_{T_k} ) and ( S_{T_l} ) are consecutive elements of the ordered sequence and: [ S_{T_k} < frac{S_z - S_x}{2} < S_{T_l} ] Consider ( alpha_j > 0 ) the least natural such that ( j in T_l ) and ( j notin T_k ). If ( T_m = T_k cup {j} ), then: [ S_{T_m} = S_{T_k} + alpha_j > S_{T_k} ] Since ( j in T_l ), ( S_{T_m} ge S_{T_l} ). If ( S_{T_m} > S_{T_l} ), there would be ( j' in V ) such that ( S_{T_k} + alpha_{j'} = S_{T_l} ), leading to a contradiction. Thus, ( T_l = T_m = T_k cup {j} ) and ( S_{T_l} = S_{T_k} + alpha_j ). 6. **Complement and contradiction:** Let ( bar{T_i} ) be the complement of ( T_i ) in ( V ) and ( bar{S_{T_i}} = sum_{i in bar{T_i}} alpha_i ). We have: [ S_{T_i} + bar{S_{T_i}} = S_z - S_x ] By the absurdity hypothesis, ( S_{T_l} > frac{S_z - S_x}{2} ), so: [ 2S_{T_l} > S_{T_l} + bar{S_{T_l}} Rightarrow S_{T_l} > bar{S_{T_l}} ] Similarly, ( S_{T_k} < frac{S_z - S_x}{2} ), so: [ 2S_{T_k} < S_{T_k} + bar{S_{T_k}} Rightarrow S_{T_k} < bar{S_{T_k}} ] This implies ( bar{S_{T_k}} ge S_{T_l} ). 7. **Final contradiction:** Substituting, we get: [ bar{S_{T_k}} ge S_{T_k} + alpha_j Rightarrow bar{S_{T_k}} - S_{T_k} ge alpha_j ] Also: [ bar{S_{T_l}} le S_{T_k} Rightarrow bar{S_{T_k}} - alpha_j le S_{T_k} Rightarrow bar{S_{T_k}} - S_{T_k} le alpha_j ] Thus: [ bar{S_{T_k}} - S_{T_k} = alpha_j ] and: [ T_l = bar{T_k} ] Since ( bar{S_{T_k}} > S_{T_k} ), we have: [ bar{S_{T_k}} ge S_{T_l} Rightarrow bar{S_{T_k}} ge S_{T_k} + alpha_j Rightarrow bar{S_{T_l}} + alpha_j ge S_{T_k} + alpha_j Rightarrow bar{S_{T_l}} ge S_{T_k} ] By the absurdity hypothesis: [ S_{T_l} le bar{S_{T_l}} Rightarrow S_{T_l} - bar{S_{T_l}} le 0 Rightarrow S_{T_k} + alpha_j - (bar{S_{T_k}} - alpha_j) le 0 Rightarrow bar{S_{T_k}} - S_{T_k} ge 2alpha_j Rightarrow alpha_j ge 2alpha_j Rightarrow alpha_j le 0 ] This is a contradiction. 8. **Construct the partition:** Take the subset ( T_{half} ) of ( V ) such that: [ sum_{i in T_{half}} alpha_i = frac{S_z - S_x}{2} ] Partition the stones into two groups ( A_1 ) and ( B_1 ): [ A_1 = {a_i, d_i mid i in T_{half}} cup {b_i, c_i mid i notin T_{half}} ] [ B_1 = {b_i, c_i mid i in T_{half}} cup {a_i, d_i mid i notin T_{half}} ] Each group has exactly two stones of each color ( i ). The weight difference is: [ S_{A_1} - S_{B_1} = sum_{i in T_{half}} (a_i + d_i) + sum_{i notin T_{half}} (b_i + c_i) - sum_{i in T_{half}} (b_i + c_i) - sum_{i notin T_{half}} (a_i + d_i) = 0 ] 9. **Handle ( alpha_i = 0 ):** For stones such that ( alpha_i = 0 ), the solution is similar. Let ( V' = {1 le i le n mid alpha_i = 0} ), ( S_x' = sum_{i in V'} x_i ), ( S_z' = sum_{i in V'} z_i ), and ( S_{xz}' = sum_{i in V'} (x_i + z_i) ). There is a subset ( T_{half}' ) of ( V' ) such that: [ sum_{i in T_{half}'} x_i = frac{S_{xz}'}{2} ] Partition the remaining stones into two groups ( A_2 ) and ( B_2 ): [ A_2 = {a_i, c_i mid i in T_{half}'} cup {b_i, d_i mid i notin T_{half}'} ] [ B_2 = {b_i, d_i mid i in T_{half}'} cup {a_i, c_i mid i notin T_{half}'} ] Each group has exactly two stones of each color ( i ). The weight difference is: [ S_{A_2} - S_{B_2} = sum_{i in T_{half}'} (a_i + c_i) + sum_{i notin T_{half}'} (b_i + d_i) - sum_{i in T_{half}'} (b_i + d_i) - sum_{i notin T_{half}'} (a_i + c_i) = 0 ] 10. **Combine the partitions:** Let ( A = A_1 cup A_2 ) and ( B = B_1 cup B_2 ). This solves the problem. (blacksquare)

answer:The dealer claims to sell the goods at the cost price but actually sells less than a kilogram by using a false weight. This means that for every kilogram he claims to sell, he is actually selling 100% - 53.84615384615387% = 46.15384615384613% of that kilogram. To find out the weight he uses per kilogram, we can simply take the percentage he actually sells and convert it to weight: Weight per kg = 46.15384615384613% of 1 kg = 0.4615384615384613 kg So, the dishonest dealer uses a weight of boxed{0.4615384615384613} kg for every kilogram he claims to sell.