answer:Given a prime number p, we need to find all natural numbers n such that p mid varphi(n) and for all a with (a, n) = 1, it follows that n mid left( a^{frac{varphi(n)}{p}} - 1 right). 1. **Prime Factorization of ( n ):** Let n = prod_{i=1}^m p_i^{a_i}, where p_i are the prime factors of n and a_i are their respective exponents. The Euler's totient function varphi(n) is given by: [ varphi(n) = prod_{i=1}^m varphi(p_i^{a_i}) ] Since varphi(p_i^{a_i}) = p_i^{a_i - 1}(p_i - 1), we have: [ varphi(n) = prod_{i=1}^m left[ p_i^{a_i - 1} (p_i - 1) right] ] 2. **Condition on ( p ) dividing ( varphi(n) ):** For p mid varphi(n), there are two main cases to consider: - There exists some p_i = p. - There exists some p_i such that p mid (p_i - 1). 3. **For all ( a ) such that ( (a, n) = 1 ):** We need the condition: [ n mid left( a^{frac{varphi(n)}{p}} - 1 right) ] This implies: [ p_i^{a_i} mid left( a^{frac{varphi(n)}{p}} - 1 right) ] for each i = 1, 2, ldots, m. 4. **Case Discussions:** **Case 1:** At least two p_k, p_l satisfy p mid (p_k - 1) and p mid (p_l - 1). - Here, (frac{varphi(p_k^{a_k})}{p} in mathbb{N}_+) and (frac{varphi(p_l^{a_l})}{p} in mathbb{N}_+). - Hence, since (varphi(p_i^{a_i}) mid frac{varphi(n)}{p}) for all (i = 1, 2, ldots, m), condition holds for n. **Case 2:** Exactly one p_k satisfies p mid (p_k - 1). Let p^alpha | varphi(p_k^{alpha_k}), where alpha in mathbb{N}_+. - (i) If p^2 mid n, then p^{alpha-1} | frac{varphi(n)}{p}, leading to a contradiction that (varphi(p_k^{a}) mid frac{varphi(n)}{p}). So, n cannot meet the condition. - (ii) If p mid n, assume p_1 = p. If alpha_1 geq 2, ( frac{varphi(p_1^{a_1})}{p} in mathbb{N}_+) ensures (varphi(p_i^{a_i}) mid frac{varphi(n)}{p} ) for all (i). If alpha_1 = 1, contradiction is found ( p^{a-1} | frac{varphi(n)}{p} ). **Case 3:** No p_k satisfies p mid (p_k - 1), thus p mid n. Assume p_1 = p, then p^{alpha_1 - 1} | varphi(p_1^{a_1}) and ( p^{alpha_1 - 2} | frac{varphi(n)}{p} ), leading to contradiction. Thus, the necessary and sufficient conditions given, the modified conditions tell us that no such n satisfies both (p mid varphi(n)) and ( n mid left( a^{frac{varphi(n)}{p}} - 1 right)) for all a where (a, n) = 1. (blacksquare)

answer:1. Identify the relevant property: Numbers ending in 25 and their powers both end in 25. This can be observed through modular arithmetic. 2. Consider the structure of the problem. The term we need to evaluate is 33 cdot 92025^{1989}. 3. Focus on the powers of 92025 modulo 100. Specifically, find what 92025^{1989} mod 100 equals: [ 92025 equiv 25 mod 100 ] 4. Use the property that numbers ending in 25, when raised to any power, will also end in 25: [ (25 mod 100)^{1989} equiv 25^{1989} mod 100 equiv 25 mod 100 ] 5. Consequently, 92025^{1989} ends with the digits "25". 6. Now, consider the product 33 cdot 25. Compute this product explicitly to find its last two digits: [ 33 cdot 25 = 825 ] The last two digits of 825 are "25". 7. Therefore, the last two digits of 33 cdot 92025^{1989} are 25. # Conclusion: The correct answer is: [ boxed{25} ]

answer:Consider the following problem where we must show that among the areas of all possible triangles formed by eight given points in the plane, we can assign (pm) signs such that the sum equals zero. Let's break down the solution step-by-step. 1. **Observation for Four Points:** To begin with, let's consider any four points in the plane. Without loss of generality, label them (A, B, C, D). 2. **Case 1: Point Forms a Convex Quadrilateral:** - When the points form a convex quadrilateral (ABCD), the following holds: [ S_{ABC} + S_{ACD} - S_{BCD} - S_{ABD} = 0 ] Here, (S_{XYZ}) represents the area of triangle (XYZ). 3. **Case 2: One Point Lies Inside the Triangle:** - If one of the points (say (D)) lies inside the triangle (ABC), the following equality holds: [ S_{ABC} - S_{ABD} - S_{ACD} - S_{BCD} = 0 ] These two cases demonstrate that for any four points, the areas of triangles can be grouped and assigned signs so that their total sum is zero. 4. **Extending to Eight Points:** Next, assign the vertices of a cube (ABCDA'B'C'D') to represent our eight points. Consider 14 sets of four points each: six faces of the cube, six sectional planes passing through pairs of opposite edges, and two inscribed tetrahedra (AB'CD') and (A'BC'D). 5. **Non-Redundant Triangles:** By construction, any triangle formed by these eight points will appear exactly once in one of these fourteen sets of four points—ensuring no redundancy. 6. **Application of Initial Insight:** Each set of four points, as we showed initially, can have their triangular areas combined to zero by appropriately assigning (pm) signs. Therefore, by employing the strategy proven for any four points to the systematically crafted sets from our eight points, we confidently achieve the desired result. To summarize: [ text{Each triangle's area can be assigned a sign such that their sum is zero.} ] Using this insight, the required conclusion to the problem is attained. Thus, [boxed{text{The given problem statement is proven true.}}]

answer:To find out how long it will take for the train to pass a telegraph post, we need to convert the speed of the train from kilometers per hour (kmph) to meters per second (m/s), because the length of the train is given in meters. The conversion factor between kmph and m/s is: 1 kmph = 1000 meters / 3600 seconds So, to convert 72 kmph to m/s: 72 kmph * (1000 m / 3600 s) = 20 m/s Now that we have the speed of the train in meters per second, we can calculate the time it takes to pass the telegraph post. Since the train is 200 meters long, and it needs to cover its own length to completely pass the post, we can use the formula: Time = Distance / Speed Plugging in the values we have: Time = 200 m / 20 m/s = 10 seconds Therefore, it will take the train boxed{10} seconds to pass the telegraph post.