answer:We are given the system of equations: [ begin{array}{l} a^{2} + b^{2} = n operatorname{lcm}(a, b) + n^{2} b^{2} + c^{2} = n operatorname{lcm}(b, c) + n^{2} c^{2} + a^{2} = n operatorname{lcm}(c, a) + n^{2} end{array} ] We seek to find all positive integer triples ((a, b, c)) such that these equations hold for some positive integer (n). 1. **Claim**: The only solutions are of the form ((k, k, k)). 2. **Verification**: If ((a, b, c) = (k, k, k)), substitute into the equations to verify: - [ k^2 + k^2 = n operatorname{lcm}(k, k) + n^2 implies 2k^2 = nk + n^2 implies n = k ] - Hence, ((k, k, k)) is indeed a valid solution with (n = k). 3. **Opposite Direction**: Suppose ((a, b, c)) is a solution. We need to show (a = b = c). 4. **Suppose**: There exists an integer (d > 1) such that (d) divides (a), (b), and (c). Then from any of the given equations, (d) divides (n). 5. Replace ((a,b,c)) with ((frac{a}{d}, frac{b}{d}, frac{c}{d})) and (n) with (frac{n}{d}). Then ((frac{a}{d}, frac{b}{d}, frac{c}{d})) is a new solution with (frac{a}{d}, frac{b}{d}, frac{c}{d}), and (frac{n}{d}). 6. **Without Loss of Generality (WLOG)**: We assume that (a), (b), (c), and (n) share no common divisor other than 1. 7. To solve for (2a^2), consider the first equation: [ 2a^2 = n (operatorname{lcm}(a, b) - operatorname{lcm}(b, c) + operatorname{lcm}(c, a) + n) ] Hence, (n mid 2a^2). 8. Similarly, we get: - (n mid 2b^2) - (n mid 2c^2) Since (a), (b), (c) and (n) share no common divisor other than 1, it follows that (n) must be either 1 or 2. 9. **Case (n = 1)**: [ a^2 + b^2 = operatorname{lcm}(a, b) + 1 ] This implies ((a - b)^2 leq 2a b leq ab + 1), which is possible only if (a = b = c = 1). Thus, we have the family of solutions ((k, k, k)). 10. **Case (n = 2)**: [ a^2 + b^2 = 2operatorname{lcm}(a, b) + 4 leq 2ab + 4 ] This implies ((a - b)^2 leq 4 implies |a - b| leq 2). Similarly, (|b - c| leq 2) and (|c - a| leq 2). Suppose without loss of generality (a = b - 1). Substituting back, it yields contradictions. 11. Therefore, no two of (a), (b), and (c) can be consecutive if they differ by more than 1 or 2. At least two of (a, b), and (c) must be equal. 12. **Conclusion**: Exclude any contradiction, the only possible solution after assumptions and simplifications are the family ((k, k, k)). Therefore, the solution is: [ boxed{(k, k, k)} ]

answer:Let's solve each part of the problem separately and in a detailed manner. 1. **Part (a): Probability that the two drawn cards are of the same suit.** 1.1. Consider a Hungarian card pack which consists of 32 cards with four suits, each containing 8 cards. 1.2. When drawing the first card from the first pack, the second card drawn from the second pack must match the suit of the first card. 1.3. There are 32 possible outcomes when drawing a card from the second pack. 1.4. Since only 8 cards in the second pack will have the same suit as the first card drawn from the first pack, the probability is calculated as follows: [ v_{1} = frac{8}{32} = frac{1}{4} = 0.25 ] **Conclusion:** [ boxed{0.25} ] 2. **Part (b): Probability that the two drawn cards are of the same rank.** 2.1. The rank of the card drawn from the second pack must match the rank of the card drawn from the first pack. 2.2. There are 8 possible ranks in a pack of Hungarian cards. 2.3. For each rank of the first card drawn, there are 4 cards with the same rank in the second pack out of the 32 available cards. 2.4. Thus, the probability of drawing a card from the second deck that matches the rank of the card from the first deck is: [ v_{2} = frac{4}{32} = frac{1}{8} = 0.125 ] **Conclusion:** [ boxed{0.125} ] 3. **Part (c): Probability that the two drawn cards are either of the same suit or the same rank.** 3.1. We need to count the favorable cases where the cards are either of the same suit or the same rank. 3.2. From previous parts, there are 8 favorable outcomes (same suit) and 4 favorable outcomes (same rank) with 1 outcome counted twice where the card is both of the same suit and rank. 3.3. Using the principle of inclusion-exclusion, the total number of favorable outcomes is: [ 8 + 4 - 1 = 11 ] 3.4. The probability is given by the ratio of the number of favorable outcomes to the total number of outcomes (32): [ v_{3} = frac{11}{32} approx 0.344 ] **Conclusion:** [ boxed{0.344} ] 4. **Part (d): Probability that the second card drawn has a higher rank than the first card drawn.** 4.1. As there are 8 ranks, each combination of ranks from the two packs (total of (8 times 8 = 64) combinations) is possible. 4.2. Out of these, 8 combinations result in the same rank for both cards. 4.3. Among the remaining 56 combinations, half of them will have the first card’s rank higher than the second, and the other half will have the second card’s rank higher: [ frac{56}{2} = 28 ] 4.4. Thus, the probability that the second card drawn has a higher rank is: [ v_{4} = frac{28}{64} = frac{7}{16} approx 0.438 ] **Conclusion:** [ boxed{0.438} ]

answer:If Jinsu can complete the entire task in 4 hours, then in one hour, he can complete 1/4 of the task. Therefore, the amount of work Jinsu can do in one hour is boxed{1/4} of the task.

answer:1. **Assign Coordinates to Points**: - Let J = (0, 0), K = (5, 0), L = (5, 6), M = (0, 6). - Given MJ = JK = KL = 2: - J = (2, 6) (since MJ = 2 and M = (0, 6)) - K = (5, 4) (since KL = 2 and L = (5, 6)) 2. **Find Equations of Lines**: - **Line MK**: Points M = (0, 6) and K = (5, 0). - Slope of MK = frac{0 - 6}{5 - 0} = -frac{6}{5} - Equation using point-slope form: y - 6 = -frac{6}{5}(x - 0) Rightarrow y = -frac{6}{5}x + 6 - **Line JL**: Points J = (0, 0) and L = (5, 6). - Slope of JL = frac{6 - 0}{5 - 0} = frac{6}{5} - Equation using point-slope form: y - 0 = frac{6}{5}(x - 0) Rightarrow y = frac{6}{5}x 3. **Find Intersection of Lines MK and JL**: - Set y = -frac{6}{5}x + 6 equal to y = frac{6}{5}x: [ frac{6}{5}x = -frac{6}{5}x + 6 Rightarrow frac{12}{5}x = 6 Rightarrow x = 2.5 ] [ y = frac{6}{5}(2.5) = 3 ] - Intersection point P = (2.5, 3). 4. **Calculate Area of Shaded Triangles Using Shoelace Theorem**: - **Triangle MJP**: Points M = (0, 6), J = (0, 0), P = (2.5, 3). - Using Shoelace Theorem: [ text{Area} = frac{1}{2} left| (0 cdot 0 + 0 cdot 3 + 2.5 cdot 6) - (6 cdot 0 + 0 cdot 2.5 + 3 cdot 0) right| ] [ = frac{1}{2} left| 0 + 0 + 15 - 0 - 0 - 0 right| = 7.5 ] - **Triangle KLP**: Points K = (5, 0), L = (5, 6), P = (2.5, 3). - Using Shoelace Theorem: [ text{Area} = frac{1}{2} left| (5 cdot 6 + 5 cdot 3 + 2.5 cdot 0) - (0 cdot 5 + 6 cdot 2.5 + 3 cdot 5) right| ] [ = frac{1}{2} left| 30 + 15 + 0 - 0 - 15 - 15 right| = 7.5 ] 5. **Total Area of Shaded Regions**: - Total area = Area of MJP + Area of KLP = 7.5 + 7.5 = 15 Conclusion: The total shaded area is 15 square units. The final answer is boxed{C}