💡To determine the exponent of (3) in the prime factorization of (30!), we need to count how many times the prime number (3) appears in the factors of the numbers from (1) to (30).First, we count the multiples of (3) between (1) and (30). These are:[3, 6, 9, 12, 15, 18, 21, 24, 27, 30]There are (10) such multiples.Next, we count the multiples of (3^2 = 9) between (1) and (30). These are:[9, 18, 27]There are (3) such multiples.Finally, we count the multiples of (3^3 = 27) between (1) and (30). There is only one such multiple:[27]Adding these counts together, the total exponent of (3) in the prime factorization of (30!) is:[10 + 3 + 1 = 14]

💡First, calculate the total cost of the umbrellas and raincoats before any discounts or refunds.1. **Cost of umbrellas:** [ 2 text{ umbrellas} times 8/text{umbrella} = 16 ]2. **Cost of raincoats:** [ 3 text{ raincoats} times 15/text{raincoat} = 45 ]3. **Total cost before discount:** [ 16 + 45 = 61 ]Next, add the cost of the waterproof bag.4. **Cost of waterproof bag:** [ 25 ]5. **Total cost before discount (including bag):** [ 61 + 25 = 86 ]Apply the 10% discount to the total cost.6. **Discount amount:** [ 10% text{ of } 86 = 0.10 times 86 = 8.60 ]7. **Total cost after discount:** [ 86 - 8.60 = 77.40 ]Calculate the refund for the defective raincoat, which is 80% of its cost.8. **Refund for defective raincoat:** [ 80% text{ of } 15 = 0.80 times 15 = 12 ]Subtract the refund from the total cost after discount to find the final amount spent.9. **Total amount spent:** [ 77.40 - 12 = 65.40 ]

💡First, I need to determine the number of faces, edges, and vertices of a pentagonal prism.A pentagonal prism has two pentagonal bases and five rectangular lateral faces. This gives a total of 7 faces.Next, I'll count the edges. Each pentagonal base has 5 edges, and there are 5 vertical edges connecting the corresponding vertices of the two bases. Adding these together, there are 15 edges in total.Finally, I'll count the vertices. Each pentagonal base has 5 vertices, resulting in 10 vertices overall.Adding the number of faces, edges, and vertices together: 7 + 15 + 10 equals 32.

💡Okay, so I'm trying to solve this problem where in a carriage, any m passengers (where m is at least 3) have a unique common friend. I need to figure out how many friends the person with the most friends has. Hmm, let me break this down step by step.First, let me make sure I understand the problem correctly. We have a group of people in a carriage, and the friendship is mutual—meaning if A is friends with B, then B is friends with A. Also, no one is their own friend. The key condition here is that any m passengers have exactly one common friend. So, if I pick any m people, there's exactly one person who is friends with all of them.I need to find the maximum number of friends that any single person can have in this setup. Let's denote this maximum number as k. So, I need to find the value of k.Let me start by considering small values of m to get an intuition. Let's say m = 3. So, any 3 passengers have exactly one common friend. What does this imply?Suppose there is a person, let's call them Alice, who has k friends. Let's denote her friends as Bob, Charlie, Dave, etc. Now, if I pick any 3 of Alice's friends, they must have a unique common friend. Since Alice is friends with all of them, she is a common friend. But the condition says that any 3 passengers have exactly one common friend. So, if I pick 3 friends of Alice, their only common friend must be Alice. That means none of these 3 friends can have another common friend besides Alice.Wait, but if Alice has k friends, then each of these friends must have their own set of friends. If any 3 of Alice's friends have only Alice as their common friend, that restricts how many friends each of them can have. They can't be friends with anyone else outside of Alice, otherwise, those three could have another common friend.Hmm, so does that mean that all of Alice's friends can only have Alice as their friend? But that would mean each of them has only one friend, which is Alice. But if that's the case, then any 3 of them would have only Alice as their common friend, which satisfies the condition. But then, what about Alice herself? How many friends does she have?If each of her friends only has her as their friend, then she can have as many friends as she wants, right? But wait, the problem is asking for the maximum number of friends any person can have. So, if Alice can have an unlimited number of friends, each of whom only has her as their friend, then k could be very large. But that doesn't seem right because the problem is asking for a specific number, not an arbitrary one.Wait, maybe I'm missing something. The problem says any m passengers have a unique common friend. So, if I pick any m passengers, they have exactly one common friend. If Alice has k friends, and each of those friends only has Alice as their friend, then any m passengers chosen from Alice's friends would have Alice as their only common friend. That works. But what about passengers who are not Alice's friends?Oh, right! There might be other people in the carriage who are not friends with Alice. Let's denote the set of all passengers as P. So, P includes Alice, her friends, and possibly others. If I pick m passengers from P, they must have exactly one common friend.Suppose there's someone else in the carriage, let's call them Eve, who is not friends with Alice. Then, if I pick Eve and m-1 of Alice's friends, their common friend must be someone. But Eve isn't friends with Alice, so the common friend can't be Alice. Therefore, Eve must have another common friend with those m-1 of Alice's friends. But since those m-1 friends only have Alice as their friend, Eve can't have a common friend with them unless Eve is friends with someone else.Wait, this is getting a bit tangled. Let me try to formalize this.Let me denote the set of all passengers as P, and the friendship relation as a graph where each person is a node, and edges represent friendships. The problem states that any m nodes have exactly one common neighbor. So, in graph theory terms, this is a graph where every m-vertex subset has exactly one common neighbor.This kind of graph is known in combinatorics. It's similar to a projective plane or a block design, but I'm not entirely sure. Maybe it's a type of strongly regular graph? I'm not too familiar, but perhaps I can reason it out.Let me consider the properties of such a graph. If every m passengers have exactly one common friend, then the graph is highly symmetric and structured. It must have certain regularity properties.Suppose that each person has the same number of friends, say k. Then, the graph is k-regular. But the problem doesn't specify that everyone has the same number of friends, only that any m passengers have exactly one common friend. So, it might not be regular.However, the question is about the person with the most friends, so we need to find the maximum degree in such a graph.Let me think about the implications of the condition. If I fix a person, say Alice, and look at her friends, then any m-1 of her friends must have exactly one common friend, which must be Alice. Because if they had another common friend, say Bob, then Alice and those m-1 friends would have two common friends: Alice and Bob, which violates the condition.Wait, that seems important. So, if I take any m-1 friends of Alice, their only common friend is Alice. Therefore, no m-1 friends of Alice can have another common friend besides Alice. That implies that the subgraph induced by Alice's friends has the property that any m-1 nodes have no common neighbors outside of Alice.But in the overall graph, any m nodes have exactly one common neighbor. So, if I take m nodes that include Alice and m-1 of her friends, their common neighbor is Alice. But if I take m nodes that are all friends of Alice, their common neighbor is also Alice. Wait, but the problem says any m passengers have exactly one common friend. So, if I take m friends of Alice, their common friend is Alice. But if I take m passengers that include some friends of Alice and some non-friends, their common friend must be someone else.This is getting a bit complex. Maybe I should use some combinatorial arguments.Let me denote the total number of passengers as n. So, n is the number of nodes in the graph. Each person has some number of friends, and we're looking for the maximum degree k.Given that any m passengers have exactly one common friend, this imposes a structure on the graph. Let me consider how many common friends a set of m passengers can have.If I fix a set S of m passengers, they have exactly one common friend, say f(S). So, each m-set S has a unique f(S) such that f(S) is friends with everyone in S.Now, let's fix a person, say Alice, and count the number of m-sets that include her. Wait, no, maybe it's better to fix a person and count how many m-sets have that person as their common friend.Suppose Alice is the common friend for some m-sets. How many m-sets can have Alice as their common friend? Well, if Alice has k friends, then the number of m-sets that include Alice and m-1 of her friends is C(k, m-1), since we need to choose m-1 friends from her k friends to form an m-set with her.But each m-set can only have one common friend, so the number of m-sets for which Alice is the common friend is C(k, m-1). Similarly, every other person can be the common friend for some m-sets.Now, the total number of m-sets in the entire graph is C(n, m). Each m-set has exactly one common friend, so the sum over all persons of C(k_i, m-1) must equal C(n, m), where k_i is the number of friends of person i.But we're interested in the maximum k, so let's denote k_max as the maximum degree. Then, the sum over all persons of C(k_i, m-1) = C(n, m).But without knowing n, this seems difficult. Maybe there's another way.Wait, perhaps we can use double counting. Let's count the number of pairs (S, f(S)), where S is an m-set and f(S) is its unique common friend.On one hand, there are C(n, m) such pairs, since each m-set has exactly one common friend.On the other hand, for each person, the number of m-sets for which they are the common friend is C(k_i, m-1), as we discussed earlier. So, the total number of such pairs is the sum over all persons of C(k_i, m-1).Therefore, we have:Sum_{i=1 to n} C(k_i, m-1) = C(n, m)Now, we want to maximize k_max, the maximum degree. To maximize k_max, we might suspect that the graph is as "unbalanced" as possible, with one person having as many friends as possible, and others having fewer.But we need to ensure that the sum of C(k_i, m-1) equals C(n, m). Let's see.Suppose that one person, Alice, has k friends, and all other n-1 people have degrees such that C(k_i, m-1) is as small as possible. The minimal degree for a person is 0, but if someone has 0 friends, then any m-set including them would need to have a common friend, which would have to be someone else. But if someone has 0 friends, then any m-set including them and m-1 others would need to have a common friend among the others, which might complicate things.Alternatively, maybe all other people have degree 1, but that might not work either.Wait, perhaps the graph is such that there is one person with k friends, and all others have degree m-1. Let me test this idea.If Alice has k friends, and each of the other n-1 people has degree m-1, then the sum becomes C(k, m-1) + (n-1)*C(m-1, m-1) = C(k, m-1) + (n-1)*1 = C(k, m-1) + n - 1.This must equal C(n, m). So,C(k, m-1) + n - 1 = C(n, m)But C(n, m) = n! / (m! (n - m)!) and C(k, m-1) = k! / ((m-1)! (k - m + 1)! )This seems complicated, but maybe for specific values of m, we can find a pattern.Wait, let's consider m=3. Then, the equation becomes:C(k, 2) + n - 1 = C(n, 3)So,k(k - 1)/2 + n - 1 = n(n - 1)(n - 2)/6Hmm, this is a cubic equation in n, which might not be easy to solve. But maybe there's a specific structure here.Alternatively, perhaps the graph is such that each person has exactly m friends, making it m-regular. Then, the sum would be n*C(m, m-1) = n*m = C(n, m).But C(n, m) = n(n - 1)(n - 2)...(n - m + 1)/m!So, n*m = n(n - 1)(n - 2)...(n - m + 1)/m!Simplifying, m = (n - 1)(n - 2)...(n - m + 1)/m!This seems unlikely unless n is small.Wait, maybe for m=3, n=7. Let's test that.If m=3, n=7, then C(7,3)=35. If the graph is 3-regular, then the sum would be 7*C(3,2)=7*3=21, which is less than 35. So, that doesn't work.Alternatively, if one person has k friends, and the rest have some other degrees.Wait, maybe the graph is a projective plane graph. In projective planes, each pair of lines intersects in exactly one point, and each pair of points lies on exactly one line. This seems similar to our condition where any m passengers have exactly one common friend.In projective planes, the number of points is n = q^2 + q + 1 for some prime power q, and each line contains q + 1 points. So, if we set m = q + 1, then the number of common friends would be 1, as required.But I'm not sure if this directly applies here. Maybe the maximum degree k is equal to m.Wait, in the projective plane, each point is on the same number of lines, which is q + 1. So, if we set m = q + 1, then each line corresponds to a common friend for m points. So, the maximum degree would be q + 1, which is m.Hmm, that suggests that the maximum number of friends is m.But wait, in our problem, the condition is that any m passengers have exactly one common friend. In the projective plane, any m points lie on exactly one line, which would correspond to having exactly one common friend. So, the maximum degree would be the number of lines through a point, which is q + 1 = m.Therefore, the maximum number of friends is m.But I need to verify this.Suppose that the graph is such that each person has exactly m friends, and any m passengers have exactly one common friend. Then, the graph would satisfy the conditions, and the maximum degree would be m.Alternatively, if someone had more than m friends, say m + 1, then we could have a situation where m passengers have more than one common friend, which would violate the condition.Wait, let me think about that. If someone, say Alice, has m + 1 friends, then any m of her friends would have Alice as a common friend. But if any m passengers have exactly one common friend, then those m friends can't have another common friend besides Alice. So, that's fine.But what about passengers who are not friends with Alice? If I take m passengers that include some friends of Alice and some non-friends, their common friend must be someone else. But if Alice has m + 1 friends, then the number of non-friends is n - 1 - (m + 1) = n - m - 2.Wait, but if n is not specified, this might not hold. Maybe the structure of the graph forces n to be a specific number.Wait, going back to the projective plane idea, if n = m^2 - m + 1, then each person would have m friends, and any m passengers would have exactly one common friend. This is similar to a finite projective plane of order m - 1.In that case, the maximum degree would indeed be m.But I'm not entirely sure if this is the only possibility. Maybe there are other graphs that satisfy the condition with a higher maximum degree.Wait, let's consider the case where m=3. If m=3, then any 3 passengers have exactly one common friend. Let's see what the maximum degree could be.Suppose we have a graph where one person, Alice, is friends with everyone else. So, if there are n passengers, Alice has n - 1 friends. Now, any 3 passengers that include Alice and two others would have Alice as their common friend. But what about any 3 passengers that don't include Alice? They must have exactly one common friend, which would have to be someone else.But if Alice is friends with everyone, then any other person has at least one friend (Alice). But if we take 3 passengers who are all friends of Alice, their common friend is Alice. However, if we take 3 passengers who are not friends with each other, their common friend must be someone else.Wait, but if Alice is friends with everyone, then any other person has at least Alice as a friend. So, if I take 3 passengers who are not friends with each other, their common friend must be someone who is friends with all three. But if Alice is friends with everyone, then she is the common friend for any 3 passengers, regardless of whether they are friends with each other or not.But that contradicts the condition because if I take 3 passengers who are not friends with each other, their common friend would still be Alice, but the problem states that any 3 passengers have exactly one common friend. So, if Alice is friends with everyone, then any 3 passengers would have Alice as their common friend, which is fine. But wait, if I take 3 passengers who are all friends with each other, their common friend is Alice, but they might also have another common friend among themselves.Wait, no, because if they are all friends with each other, then they form a triangle. But the problem doesn't restrict the number of friends a person can have, only that any m passengers have exactly one common friend.Wait, but if three passengers are all friends with each other, then their common friend is Alice, but they also have each other as friends. However, the problem states that any m passengers have exactly one common friend, meaning that the number of common friends is exactly one. So, if three passengers are all friends with each other, their common friends would be each other and Alice, which would mean more than one common friend, violating the condition.Therefore, in this case, if Alice is friends with everyone, then any three passengers who are mutual friends would have multiple common friends, which is not allowed. Therefore, Alice cannot be friends with everyone, which means that the maximum degree k must be less than n - 1.So, in the case of m=3, the maximum degree k cannot be n - 1. Therefore, the maximum degree is less than n - 1.But how much less? Let's try to find a structure where the maximum degree is m.Suppose that each person has exactly m friends, and the graph is such that any m passengers have exactly one common friend. This would satisfy the condition, and the maximum degree would be m.But is this possible? Let's think about it.If each person has exactly m friends, then the graph is m-regular. Now, in an m-regular graph, the number of edges is (n * m)/2. But how does this relate to the condition that any m passengers have exactly one common friend?Let me consider the number of common friends for any m passengers. If the graph is m-regular, then the number of common friends for any m passengers would depend on the structure.Wait, perhaps this is similar to a strongly regular graph, but with parameters adjusted for m.Alternatively, maybe it's a type of design where each m-set has exactly one common neighbor.Wait, in combinatorial design theory, a projective plane has the property that any two lines intersect in exactly one point, and any two points lie on exactly one line. This seems analogous to our problem where any m passengers have exactly one common friend.In a projective plane of order q, each line contains q + 1 points, and each point is on q + 1 lines. The total number of points is q^2 + q + 1.If we set m = q + 1, then the number of common friends for any m passengers would be 1, as required. Therefore, in this case, the maximum degree would be q + 1 = m.So, this suggests that the maximum number of friends is m.But let me check if this holds for m=3. If m=3, then q=2, and the projective plane has 7 points, each line has 3 points, and each point is on 3 lines. So, in this case, the maximum degree is 3, which matches m.Therefore, it seems that in general, the maximum number of friends is m.But wait, earlier I considered the case where one person is friends with everyone, which led to a contradiction. So, in the projective plane setup, no single person is friends with everyone, but each person has exactly m friends.Therefore, the maximum number of friends any person can have is m.So, putting it all together, the person with the most friends has m friends.

💡Okay, so I have this problem: Can a natural number, which is expressed in the decimal system using (6k - 1) ones, be prime? Hmm, interesting. Let me try to break this down step by step.First, I need to understand what it means for a number to be expressed using (6k - 1) ones. So, if (k) is some positive integer, then (6k - 1) gives the number of ones in the decimal representation of the number. For example, if (k = 1), then (6(1) - 1 = 5), so the number would be 11111. If (k = 2), then (6(2) - 1 = 11), so the number would be 11111111111 (eleven ones). Got it.So, the number in question is a string of (6k - 1) ones. Let's denote this number as (N). So, (N) is a number like 111...1 with (6k - 1) digits, all ones.Now, the question is whether such a number can be prime. That is, can (N) be a prime number?I know that numbers consisting of all ones are called repunit numbers. A repunit is a number like 11, 111, 1111, etc., consisting entirely of ones. Repunit numbers have interesting properties, especially in number theory.I recall that repunit numbers can sometimes be prime, but it's not always the case. For example, 11 is a prime repunit, but 111 is not, since it factors into 3 and 37. Similarly, 11111 is 41 times 271, so that's also composite. So, it's not guaranteed that a repunit is prime, but some are.Given that, I need to check if a repunit with (6k - 1) digits can be prime. Let's think about the structure of such numbers.First, let's express (N) mathematically. A repunit with (q) digits can be written as:[N = frac{10^q - 1}{9}]So, in our case, (q = 6k - 1). Therefore,[N = frac{10^{6k - 1} - 1}{9}]Now, for (N) to be prime, (frac{10^{6k - 1} - 1}{9}) must be a prime number. Let's think about the properties of such numbers.I remember that if (q) is composite, then the repunit (N) is also composite. This is because if (q = ab), then (10^q - 1) can be factored as ((10^a - 1)(10^{a(b-1)} + 10^{a(b-2)} + dots + 10^a + 1)), so (N) would be composite.Therefore, for (N) to be prime, (q) must be prime. So, (6k - 1) must be a prime number. Let's denote (p = 6k - 1), so (p) is prime.So, now we have (N = frac{10^p - 1}{9}), where (p) is a prime number of the form (6k - 1).Now, I need to check if such (N) can be prime. I know that not all repunit primes are known, but some are. For example, the repunit with 2 digits (11) is prime, with 19 digits is prime, 23 digits is prime, etc.Wait, 23 is a prime number, and 23 can be written as (6k - 1). Let's check: (6k - 1 = 23) implies (6k = 24), so (k = 4). So, yes, 23 is of the form (6k - 1).Therefore, the repunit with 23 ones is prime. So, that would be an example where (k = 4), and the number (N) is prime.But wait, is that the only case? Or are there others?I think there are more repunit primes, but they are rare. For example, the repunit with 313 digits is also prime, and 313 is a prime number. Let's check if 313 is of the form (6k - 1). (6k - 1 = 313) implies (6k = 314), so (k = 314/6 ≈ 52.333). Hmm, not an integer. So, 313 is not of the form (6k - 1). Therefore, that repunit prime doesn't fit our case.Similarly, the repunit with 103 digits is prime, and 103 is a prime. Let's see: (6k - 1 = 103) implies (6k = 104), so (k = 104/6 ≈ 17.333). Again, not an integer. So, 103 is not of the form (6k - 1).Wait, so maybe 23 is the only known repunit prime of the form (6k - 1)? Or are there others?I think the known repunit primes are for primes (p) where (p) is 2, 19, 23, 317, 1031, etc. Let me check if any of these are of the form (6k - 1).- 2: (6k - 1 = 2) implies (6k = 3), (k = 0.5). Not integer.- 19: (6k - 1 = 19) implies (6k = 20), (k = 20/6 ≈ 3.333). Not integer.- 23: As before, (k = 4). Integer.- 317: (6k - 1 = 317) implies (6k = 318), (k = 53). Integer.- 1031: (6k - 1 = 1031) implies (6k = 1032), (k = 172). Integer.Wait, so 317 and 1031 are also of the form (6k - 1). So, the repunits with 317 and 1031 ones are also prime. Therefore, these are examples where (N) is prime.Therefore, it seems that yes, there are natural numbers expressed with (6k - 1) ones that are prime. For example, when (k = 4), (k = 53), and (k = 172), the corresponding repunits are prime.But wait, are these known? I think 23 is known, but I'm not sure about 317 and 1031. Let me check.Upon checking, I recall that repunit primes are extremely rare, and only a few are known. The known repunit primes correspond to primes (p) where (p) is 2, 19, 23, 317, 1031, 49081, 86453, 109297, 270343, etc. These are all primes, and some of them are of the form (6k - 1).For example:- 23: (6*4 - 1 = 23), so yes.- 317: (6*53 - 1 = 317), yes.- 1031: (6*172 - 1 = 1031), yes.- 49081: Let's see, (6k - 1 = 49081) implies (6k = 49082), (k = 49082/6 ≈ 8180.333). Not integer, so 49081 is not of the form (6k - 1).- 86453: (6k - 1 = 86453) implies (6k = 86454), (k = 14409). Integer.- 109297: (6k - 1 = 109297) implies (6k = 109298), (k = 18216.333). Not integer.- 270343: (6k - 1 = 270343) implies (6k = 270344), (k = 45057.333). Not integer.So, among the known repunit primes, 23, 317, 86453 are of the form (6k - 1). Therefore, these are examples where the number (N) is prime.Therefore, the answer is yes, such numbers can be prime.But wait, I should also consider whether all such numbers of the form (6k - 1) are composite or not. For example, is there a case where (6k - 1) is prime, but the corresponding repunit is composite?Yes, certainly. For example, take (k = 2), so (6*2 - 1 = 11), which is prime. The repunit with 11 ones is 11111111111. Is this prime? I think it's composite. Let me check.I recall that 11111111111 factors into 21 and 5291, but wait, 21 is 3*7, so 11111111111 = 3 * 7 * 11 * 13 * 37 * 41 * 271. So, it's composite. Therefore, even though (6k - 1 = 11) is prime, the repunit is composite.Therefore, not all numbers of the form (6k - 1) ones are prime, but some are. Specifically, when (6k - 1) is a prime number and the corresponding repunit is also prime.So, in conclusion, yes, there exist natural numbers expressed with (6k - 1) ones that are prime, such as when (k = 4) (23 ones), (k = 53) (317 ones), and (k = 172) (1031 ones), among others.Therefore, the answer is yes, such numbers can be prime.

💡Okay, so I have this question here with three statements about geometric bodies, and I need to figure out which one is correct. Let me go through each statement one by one and see if I can understand them properly.First, statement ① says: "A geometric body that has one polygonal face and all other faces are triangles is a pyramid." Hmm, okay. I know that a pyramid typically has a base that's a polygon, and the other faces are triangles that meet at a common point called the apex. So, if there's one polygonal face and the rest are triangles, that does sound like a pyramid. But wait, is that always true? Let me think. If I have a base that's a polygon, say a pentagon, and then five triangular faces connecting each side of the pentagon to a single apex, that's definitely a pyramid. But what if the triangular faces don't all meet at a single point? Is it still a pyramid? I don't think so. So, maybe the statement is missing something about the triangles needing to share a common vertex. So, statement ① might not be entirely correct because it doesn't specify that the triangular faces must converge at a single apex.Moving on to statement ②: "Cutting a pyramid with a plane, the geometric body formed between the base and the cutting plane is called a frustum." Alright, I remember that a frustum is like a portion of a solid, usually a cone or pyramid, that lies between two parallel planes cutting it. So, if you cut a pyramid with a plane parallel to the base, the part between the base and the cutting plane is indeed a frustum. But the statement doesn't specify that the cutting plane has to be parallel to the base. If the plane isn't parallel, then the resulting figure isn't a frustum—it's just a truncated pyramid, but not a frustum. So, statement ② is only correct if the cutting plane is parallel, which isn't mentioned here. Therefore, statement ② might be incorrect because it doesn't specify the parallel condition.Lastly, statement ③: "Any face of a triangular pyramid can be considered as the base." A triangular pyramid is another name for a tetrahedron, right? A tetrahedron has four triangular faces, and in theory, any of those faces can serve as the base. Unlike a pyramid with a polygonal base, where the base is specifically the polygon, in a tetrahedron, all faces are triangles, so you can choose any one of them as the base. That makes sense because all faces are equivalent in a regular tetrahedron, and even in an irregular one, any face can be the base depending on how you orient it. So, statement ③ seems correct.Let me recap:- Statement ①: It describes a pyramid but misses the crucial point that the triangular faces must meet at a common apex. Without that, it's not necessarily a pyramid. So, I think this is incorrect.- Statement ②: It talks about cutting a pyramid with a plane to form a frustum but doesn't specify that the plane must be parallel to the base. Since a frustum specifically requires parallel planes, this statement is incomplete and thus incorrect.- Statement ③: It correctly states that any face of a triangular pyramid (tetrahedron) can be considered the base. Since all faces are triangles and interchangeable in this role, this statement is correct.So, after analyzing each statement, I'm pretty confident that statement ③ is the correct one.