💡Okay, so I have this problem where a marksman is shooting at a target four times, and each shot is independent. I know that the probability of hitting the target at least once is 80/81. I need to find the marksman's hit rate, which I assume is the probability of hitting the target on a single shot.Hmm, let me think. If the probability of hitting the target at least once is 80/81, that means the probability of missing all four shots is 1 minus that, right? So, 1 - 80/81 equals 1/81. That makes sense because the only way to not hit the target at least once is to miss every single shot.Now, since each shot is independent, the probability of missing all four shots should be the product of missing each shot individually. Let's denote the probability of missing a single shot as (1 - p), where p is the hit rate. So, missing all four shots would be (1 - p)^4.We already found that the probability of missing all four shots is 1/81. So, setting up the equation: (1 - p)^4 = 1/81.To solve for p, I need to take the fourth root of both sides. The fourth root of 1/81 is... let me think. 81 is 3^4, right? Because 3*3*3*3 is 81. So, the fourth root of 1/81 is 1/3. Therefore, 1 - p = 1/3.Solving for p, I subtract 1/3 from 1, which gives p = 2/3. So, the hit rate is 2/3.Wait, let me double-check that. If p is 2/3, then the probability of missing is 1/3. Raising that to the fourth power gives (1/3)^4 = 1/81, which matches the probability of missing all four shots. Then, 1 - 1/81 is indeed 80/81, which is the probability of hitting at least once. So, that checks out.I think that's it. The hit rate is 2/3.

💡Okay, so I have this question about the digital Earth, and I need to figure out which statements are correct. Let me try to break it down step by step.First, let's understand what the digital Earth is. From what I remember, the digital Earth refers to the concept of creating a digital representation of Earth using various technologies like GIS (Geographic Information Systems), remote sensing, and the internet. It's about integrating all kinds of Earth data into a digital format that can be accessed and analyzed globally.Now, looking at the statements:1. **Geographic information technology is the foundation for building a digital Earth.** Hmm, I think this is true because GIS is a key tool in managing and analyzing spatial data, which is essential for creating a digital Earth. Without GIS, it would be hard to integrate different types of Earth data.2. **The most significant feature of the digital Earth is its ability to simulate reality.** This one is a bit tricky. I know that digital Earth aims to represent Earth in a digital form, but does it simulate reality? I think simulation might be more about creating models or scenarios, whereas digital Earth is more about data integration and visualization. Maybe this statement is stretching it a bit.3. **The digital Earth adopts a centralized method for information storage.** I'm not sure about this. I thought digital Earth relies on distributed systems because it involves data from various sources around the world. Centralized storage would mean all data is kept in one place, which doesn't seem efficient or practical for a global system.4. **The digital Earth is a technological system that manages Earth's information digitally through computer networks.** This sounds accurate. It describes digital Earth as a system that uses computer networks to manage Earth's information digitally, which aligns with what I know about its purpose.So, based on my understanding:- Statement ① is correct.- Statement ② might not be entirely accurate.- Statement ③ seems incorrect.- Statement ④ is correct.Therefore, the correct statements are ① and ④. Looking at the options, that would be option D: ①②④. Wait, but I thought statement ② wasn't correct. Maybe I need to reconsider.Perhaps statement ② is correct because digital Earth does simulate reality by creating a digital model of Earth. If that's the case, then statements ①, ②, and ④ are correct, which would make option D the right choice.I'm a bit confused about statement ②, but I'll go with option D since it includes the most accurate statements.

💡Okay, so I'm trying to solve this problem where I have to find all m-tuples of non-negative integers (a₁, a₂, ..., aₘ) such that p divides f(a₁, a₂, ..., aₘ). Here, f is defined as the sum over all m-tuples v in S of the product of v_i raised to the power a_i. First, let me make sure I understand the setup correctly. We have a positive integer m and a prime number p. The set S consists of all m-tuples where each component v_i is between 1 and p, inclusive. So, each v is like a vector with m components, each ranging from 1 to p. The function f takes an m-tuple of non-negative integers (a₁, a₂, ..., aₘ) and maps it to the sum over all such vectors v in S of the product of v_i raised to the respective a_i. So, for each vector v, we compute the product of each component raised to the corresponding a_i, and then sum all these products together.Our goal is to find all m-tuples (a₁, a₂, ..., aₘ) such that p divides this sum f(a₁, a₂, ..., aₘ). Let me think about how to approach this. Maybe I can start by looking at the function f more closely. Since f is a sum over all possible vectors v in S, and each term in the sum is a product of powers of the components of v, perhaps I can factor this sum into a product of sums. Wait, that might be a good idea. If I consider each component separately, then f(a₁, a₂, ..., aₘ) can be written as the product of sums for each coordinate. Specifically, for each i from 1 to m, we have a sum over v_i from 1 to p of v_i^{a_i}, and then the total f is the product of these individual sums. So, f(a₁, a₂, ..., aₘ) = (1^{a₁} + 2^{a₁} + ... + p^{a₁}) × (1^{a₂} + 2^{a₂} + ... + p^{a₂}) × ... × (1^{aₘ} + 2^{aₘ} + ... + p^{aₘ}).That makes sense because when you expand the product, each term corresponds to choosing one term from each sum and multiplying them together, which is exactly the same as summing over all possible m-tuples v.So, if I denote S_i = 1^{a_i} + 2^{a_i} + ... + p^{a_i}, then f(a₁, ..., aₘ) = S₁ × S₂ × ... × Sₘ.Therefore, the problem reduces to finding all m-tuples (a₁, ..., aₘ) such that p divides the product S₁ × S₂ × ... × Sₘ.Now, since p is a prime, by Euclid's lemma, if p divides a product, it must divide at least one of the factors. So, p divides f(a₁, ..., aₘ) if and only if p divides at least one of the S_i.Therefore, the problem now is to find all m-tuples (a₁, ..., aₘ) such that for at least one i, p divides S_i = 1^{a_i} + 2^{a_i} + ... + p^{a_i}.So, I need to characterize for each i, the exponents a_i such that p divides the sum S_i.Let me focus on a single S_i. Let's fix an index i and consider S_i = 1^{a} + 2^{a} + ... + p^{a}, where a is a non-negative integer.I need to find all a such that p divides S_i.First, let's note that if a = 0, then each term in S_i is 1, so S_i = p. Therefore, p divides S_i when a = 0.Similarly, if a = 1, then S_i = 1 + 2 + ... + p. The sum of the first p integers is p(p + 1)/2. Since p is prime, p divides p(p + 1)/2, so p divides S_i when a = 1.Wait, but hold on, when a = 1, S_i = p(p + 1)/2. Since p is prime, p divides p(p + 1)/2 because p divides p. So, yes, p divides S_i when a = 1.But wait, let me check for a specific prime, say p = 3. Then S_i = 1 + 2 + 3 = 6, which is divisible by 3. Similarly, for p = 5, S_i = 15, which is divisible by 5. So, yes, when a = 1, S_i is divisible by p.What about a = 2? Let's compute S_i for a = 2. For p = 3, S_i = 1 + 4 + 9 = 14, which is not divisible by 3. For p = 5, S_i = 1 + 4 + 9 + 16 + 25 = 55, which is divisible by 5. Hmm, interesting.Wait, so for p = 3, a = 2, S_i = 14, which is not divisible by 3. For p = 5, a = 2, S_i = 55, which is divisible by 5. So, it seems that sometimes S_i is divisible by p, sometimes not, depending on p and a.Wait, maybe I need a general approach. Let me recall some number theory. There is a theorem called Fermat's Little Theorem which says that for any integer a not divisible by p, a^{p-1} ≡ 1 mod p. Also, if a is divisible by p, then a^k ≡ 0 mod p for any k ≥ 1.So, maybe I can use this to compute S_i modulo p.Let me consider S_i = 1^a + 2^a + ... + p^a. Let's compute this modulo p.Note that p^a ≡ 0 mod p, so S_i ≡ 1^a + 2^a + ... + (p-1)^a mod p.So, S_i ≡ Σ_{k=1}^{p-1} k^a mod p.Now, I need to compute this sum modulo p.I remember that the sum of k^a modulo p has some properties depending on a.In particular, if a is not divisible by p-1, then the sum Σ_{k=1}^{p-1} k^a ≡ 0 mod p.Wait, is that true? Let me think.Yes, in fact, there is a result that says that if a is not divisible by p-1, then Σ_{k=1}^{p-1} k^a ≡ 0 mod p.And if a is divisible by p-1, then Σ_{k=1}^{p-1} k^a ≡ -1 mod p.Wait, let me verify this.Suppose that a ≡ 0 mod (p-1). Then, for each k from 1 to p-1, k^{a} ≡ 1 mod p, because of Fermat's Little Theorem. Therefore, Σ_{k=1}^{p-1} k^a ≡ (p-1) × 1 ≡ -1 mod p.On the other hand, if a is not divisible by p-1, then the sum Σ_{k=1}^{p-1} k^a ≡ 0 mod p.So, putting it all together, we have:If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p.If a ≡ 0 mod (p-1), then S_i = Σ_{k=1}^{p} k^a ≡ Σ_{k=1}^{p-1} k^a + 0 ≡ -1 mod p.If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p, so p does not divide S_i.If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p, so p does not divide S_i.Wait, but wait, earlier when a = 0, S_i = p, which is divisible by p. So, when a = 0, S_i ≡ 0 mod p.Similarly, when a = 1, S_i = p(p + 1)/2, which is divisible by p.But according to the above, when a is not divisible by p-1, S_i ≡ 0 mod p, which includes a = 1, 2, ..., p-2, etc.Wait, but when a = p-1, S_i ≡ -1 mod p, so p does not divide S_i.Wait, so let me clarify:If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p, so p does not divide S_i.If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p, so p does not divide S_i.Wait, but when a = 0, S_i = p, which is 0 mod p. So, a = 0 is a special case.Wait, maybe I need to adjust the statement.Let me rephrase:For a non-negative integer a, S_i = Σ_{k=1}^{p} k^a.We can write S_i = Σ_{k=1}^{p-1} k^a + p^a.Since p^a ≡ 0 mod p, S_i ≡ Σ_{k=1}^{p-1} k^a mod p.Now, if a = 0, then Σ_{k=1}^{p-1} k^0 = Σ_{k=1}^{p-1} 1 = p - 1 ≡ -1 mod p. But S_i = p, which is 0 mod p. So, in this case, S_i ≡ 0 mod p.Wait, so when a = 0, S_i = p, which is 0 mod p.When a > 0, then:If a ≡ 0 mod (p-1), then Σ_{k=1}^{p-1} k^a ≡ -1 mod p, so S_i ≡ -1 mod p.If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p.If a ≡ 0 mod (p-1), then S_i ≡ -1 mod p.Wait, but when a = 1, which is not divisible by p-1 (unless p = 2, but let's take p > 2 for now), then Σ_{k=1}^{p-1} k^1 = (p-1)p/2 ≡ 0 mod p, because p divides (p-1)p/2.Wait, but for p > 2, (p-1)/2 is an integer, so (p-1)p/2 is divisible by p, hence S_i ≡ 0 mod p.Similarly, for a = 2, when p = 5, S_i = 55, which is 0 mod 5.But for p = 3, a = 2, S_i = 14, which is 2 mod 3, so not 0.Wait, so maybe my earlier statement is not entirely accurate.Let me think again.I think the correct statement is:For a positive integer a, Σ_{k=1}^{p-1} k^a ≡ 0 mod p if p-1 does not divide a, and ≡ -1 mod p if p-1 divides a.But when a = 0, Σ_{k=1}^{p-1} k^0 = p - 1 ≡ -1 mod p, but S_i = p, which is 0 mod p.So, the key is that when a = 0, S_i is 0 mod p, regardless of p.When a > 0:- If p-1 divides a, then Σ_{k=1}^{p-1} k^a ≡ -1 mod p, so S_i ≡ -1 mod p.- If p-1 does not divide a, then Σ_{k=1}^{p-1} k^a ≡ 0 mod p, so S_i ≡ 0 mod p.Therefore, for each S_i:- If a_i = 0, then S_i ≡ 0 mod p.- If a_i > 0 and p-1 does not divide a_i, then S_i ≡ 0 mod p.- If a_i > 0 and p-1 divides a_i, then S_i ≡ -1 mod p.Therefore, S_i is divisible by p if and only if either a_i = 0 or p-1 does not divide a_i.Wait, that seems to be the case.So, putting it all together, for each S_i:- If a_i = 0, then S_i is divisible by p.- If a_i > 0 and p-1 does not divide a_i, then S_i is divisible by p.- If a_i > 0 and p-1 divides a_i, then S_i is not divisible by p.Therefore, for the product S₁ × S₂ × ... × Sₘ to be divisible by p, it suffices that at least one of the S_i is divisible by p.Which, as per above, happens if for that i, either a_i = 0 or p-1 does not divide a_i.Therefore, the m-tuples (a₁, a₂, ..., aₘ) such that p divides f(a₁, ..., aₘ) are exactly those for which at least one a_i is either 0 or not divisible by p-1.In other words, the m-tuples where not all a_i are positive integers divisible by p-1.Wait, let me rephrase that.The m-tuples (a₁, ..., aₘ) such that p divides f(a₁, ..., aₘ) are those where at least one a_i is either 0 or not a multiple of p-1.So, the complement would be the m-tuples where all a_i are positive integers divisible by p-1, in which case none of the S_i would be divisible by p, hence the product would not be divisible by p.Therefore, the answer is all m-tuples (a₁, ..., aₘ) where at least one a_i is 0 or not divisible by p-1.Let me check this with some examples.Take m = 1, p = 3.Then, S consists of vectors (1), (2), (3).f(a) = 1^a + 2^a + 3^a.We need to find a such that 3 divides f(a).From our earlier analysis:- If a = 0, f(a) = 3, which is divisible by 3.- If a = 1, f(a) = 6, which is divisible by 3.- If a = 2, f(a) = 14, which is not divisible by 3.- If a = 3, which is divisible by p-1 = 2, f(a) = 1 + 8 + 27 = 36, which is divisible by 3.Wait, but according to our earlier conclusion, if a is divisible by p-1, then S_i ≡ -1 mod p, so f(a) ≡ -1 mod p, which is not 0. But in this case, a = 3, which is divisible by p-1 = 2, but f(a) = 36, which is 0 mod 3.Wait, that contradicts our earlier conclusion.Wait, maybe I made a mistake.Wait, for p = 3, p-1 = 2.If a = 2, which is divisible by p-1, then S_i = 1^2 + 2^2 + 3^2 = 1 + 4 + 9 = 14, which is 2 mod 3, so not divisible by 3.But when a = 3, which is not divisible by p-1 = 2, S_i = 1^3 + 2^3 + 3^3 = 1 + 8 + 27 = 36, which is 0 mod 3.Wait, so when a is divisible by p-1, S_i is not divisible by p, and when a is not divisible by p-1, S_i is divisible by p.But in the case when a = p-1, which is 2 for p = 3, S_i = 14, which is 2 mod 3, not 0.But when a = 3, which is not divisible by p-1 = 2, S_i = 36, which is 0 mod 3.Similarly, for p = 5, p-1 = 4.Take a = 4, which is divisible by p-1. Then S_i = 1^4 + 2^4 + 3^4 + 4^4 + 5^4.Compute this modulo 5:1^4 ≡ 1, 2^4 ≡ 1, 3^4 ≡ 1, 4^4 ≡ 1, 5^4 ≡ 0.So, S_i ≡ 1 + 1 + 1 + 1 + 0 = 4 ≡ -1 mod 5.So, S_i ≡ -1 mod 5, which is not 0.If a = 5, which is not divisible by p-1 = 4, then S_i = 1^5 + 2^5 + 3^5 + 4^5 + 5^5.Compute modulo 5:1^5 ≡ 1, 2^5 ≡ 2, 3^5 ≡ 3, 4^5 ≡ 4, 5^5 ≡ 0.So, S_i ≡ 1 + 2 + 3 + 4 + 0 = 10 ≡ 0 mod 5.So, indeed, when a is divisible by p-1, S_i ≡ -1 mod p, and when a is not divisible by p-1, S_i ≡ 0 mod p.But when a = 0, S_i = p, which is 0 mod p.So, in the case when a = 0, regardless of p, S_i is 0 mod p.Therefore, the conclusion is:For each i, S_i is divisible by p if and only if a_i = 0 or a_i is not divisible by p-1.Therefore, the product S₁ × S₂ × ... × Sₘ is divisible by p if and only if at least one of the S_i is divisible by p, which happens if for that i, a_i = 0 or a_i is not divisible by p-1.Hence, the m-tuples (a₁, a₂, ..., aₘ) such that p divides f(a₁, ..., aₘ) are those where at least one a_i is either 0 or not divisible by p-1.In other words, the m-tuples where not all a_i are positive integers divisible by p-1.Therefore, the answer is all m-tuples (a₁, a₂, ..., aₘ) where at least one a_i is 0 or not a multiple of p-1.I think that's the conclusion.

💡Okay, so I have this problem where I need to find all strictly monotone functions ( f: mathbb{N} rightarrow mathbb{N} ) such that for every natural number ( n ), the equation ( f(f(n)) = 3n ) holds. Hmm, interesting. Let me try to unpack this step by step.First, let's understand what a strictly monotone function is. It means that the function is either strictly increasing or strictly decreasing. Since the function maps natural numbers to natural numbers, and given that ( f(f(n)) = 3n ) which is clearly increasing, I think ( f ) must be strictly increasing as well. If ( f ) were decreasing, composing it twice would give an increasing function, but I need to confirm that.Now, the equation ( f(f(n)) = 3n ) suggests that applying ( f ) twice scales the input by 3. So, ( f ) is some kind of square root of the function that multiplies by 3. That's an abstract way to think about it, but maybe I can find a concrete form for ( f ).Let me start by testing small values of ( n ) to see if I can spot a pattern.For ( n = 1 ):( f(f(1)) = 3 times 1 = 3 ). So, ( f(1) ) must be a number such that when I apply ( f ) again, I get 3. Let's denote ( f(1) = a ), then ( f(a) = 3 ).Similarly, for ( n = 2 ):( f(f(2)) = 3 times 2 = 6 ). Let ( f(2) = b ), so ( f(b) = 6 ).For ( n = 3 ):( f(f(3)) = 9 ). Let ( f(3) = c ), so ( f(c) = 9 ).Hmm, so each ( f(n) ) is mapped to another number such that applying ( f ) again gives ( 3n ). This seems recursive. Maybe I can find a sequence or a pattern here.Let me assume that ( f ) is linear, like ( f(n) = kn ) for some constant ( k ). Then, ( f(f(n)) = f(kn) = k(kn) = k^2n ). According to the equation, this should equal ( 3n ), so ( k^2 = 3 ). But ( k ) must be a natural number, and ( sqrt{3} ) is not an integer. So, ( f ) can't be linear.Alright, scratch that idea. Maybe ( f ) is piecewise defined or has some multiplicative properties. Let me think about the properties of ( f ).Since ( f ) is strictly increasing, it must be injective (one-to-one). Also, because ( f(f(n)) = 3n ), which is surjective onto multiples of 3, ( f ) must map natural numbers in such a way that every multiple of 3 is hit exactly once.Let me try to construct ( f ) step by step.Starting with ( n = 1 ):( f(f(1)) = 3 ). Let's say ( f(1) = 2 ). Then ( f(2) = 3 ).Now, for ( n = 2 ):( f(f(2)) = 6 ). Since ( f(2) = 3 ), then ( f(3) = 6 ).For ( n = 3 ):( f(f(3)) = 9 ). So, ( f(6) = 9 ).For ( n = 4 ):( f(f(4)) = 12 ). Let me denote ( f(4) = d ), so ( f(d) = 12 ).Similarly, for ( n = 5 ):( f(f(5)) = 15 ). Let ( f(5) = e ), so ( f(e) = 15 ).For ( n = 6 ):( f(f(6)) = 18 ). Since ( f(6) = 9 ), then ( f(9) = 18 ).Hmm, I'm starting to see a pattern here. It seems like ( f ) is mapping numbers in a way that alternates between multiplying by 2 and 1.5 or something like that. Wait, let me see:From ( f(1) = 2 ), ( f(2) = 3 ), ( f(3) = 6 ), ( f(6) = 9 ), ( f(9) = 18 ), and so on. It looks like every time we apply ( f ), we're either doubling or tripling or something in between.But this might not be the right way to think about it. Maybe I should consider the function in terms of base 3 or some other base since the equation involves multiplying by 3.Let me try to represent numbers in base 3 and see if that helps. For example:- ( 1 ) in base 3 is ( 1 ).- ( 2 ) in base 3 is ( 2 ).- ( 3 ) in base 3 is ( 10 ).- ( 4 ) in base 3 is ( 11 ).- ( 5 ) in base 3 is ( 12 ).- ( 6 ) in base 3 is ( 20 ).- ( 7 ) in base 3 is ( 21 ).- ( 8 ) in base 3 is ( 22 ).- ( 9 ) in base 3 is ( 100 ).Looking at the values we have so far:- ( f(1) = 2 ) (which is ( 2 ) in base 3)- ( f(2) = 3 ) (which is ( 10 ) in base 3)- ( f(3) = 6 ) (which is ( 20 ) in base 3)- ( f(6) = 9 ) (which is ( 100 ) in base 3)- ( f(9) = 18 ) (which is ( 200 ) in base 3)Hmm, interesting. It seems like ( f ) is taking a number, and if it starts with a '1' in base 3, it changes it to a '2', and if it starts with a '2', it changes it to '1' followed by a zero. Let me test this hypothesis.For example:- ( f(1) = 2 ) (changing '1' to '2')- ( f(2) = 3 ) (changing '2' to '10')- ( f(3) = 6 ) (changing '10' to '20')- ( f(6) = 9 ) (changing '20' to '100')- ( f(9) = 18 ) (changing '100' to '200')This seems consistent. So, maybe the function ( f ) works by manipulating the base 3 representation of the number. Specifically, if the number starts with a '1' in base 3, ( f ) changes that '1' to a '2', and if it starts with a '2', ( f ) changes that '2' to a '1' and appends a '0' at the end.Let me try to formalize this:- If a number ( n ) in base 3 is written as ( 1abc...d ), then ( f(n) = 2abc...d ).- If a number ( n ) in base 3 is written as ( 2abc...d ), then ( f(n) = 1abc...d0 ).This seems to fit the pattern we've observed. Let me test this with another number, say ( n = 4 ).In base 3, ( 4 ) is ( 11 ). According to the rule, since it starts with '1', ( f(4) = 21 ) in base 3, which is ( 2 times 3 + 1 = 7 ). Then, ( f(7) ) should be ( 3 times 4 = 12 ). Let's see:( f(7) ) in base 3 is ( 21 ). Since it starts with '2', ( f(21_3) = 110_3 ), which is ( 1 times 9 + 1 times 3 + 0 = 12 ). Perfect, that works.Another test: ( n = 5 ). In base 3, ( 5 ) is ( 12 ). Applying the rule, ( f(5) = 22 ) in base 3, which is ( 2 times 3 + 2 = 8 ). Then, ( f(8) ) should be ( 3 times 5 = 15 ). Let's check:( f(8) ) in base 3 is ( 22 ). Since it starts with '2', ( f(22_3) = 120_3 ), which is ( 1 times 9 + 2 times 3 + 0 = 15 ). Great, that works too.So, it seems like this function ( f ) defined by manipulating the base 3 digits is satisfying the condition ( f(f(n)) = 3n ). But I need to make sure that this function is strictly monotone.Let's check the order:- ( f(1) = 2 )- ( f(2) = 3 )- ( f(3) = 6 )- ( f(4) = 7 )- ( f(5) = 8 )- ( f(6) = 9 )- ( f(7) = 12 )- ( f(8) = 15 )- ( f(9) = 18 )- ( f(10) = 19 ) (since 10 in base 3 is 101, so f(101) = 201 which is 2*9 + 0*3 +1=19)- ( f(11) = 20 ) (11 in base 3 is 102, so f(102)=202=2*9 +0*3 +2=20)- ( f(12) = 21 ) (12 in base 3 is 110, so f(110)=210=2*9 +1*3 +0=21)- And so on.Looking at these values, ( f(n) ) is indeed strictly increasing. Each subsequent ( f(n) ) is larger than the previous one. So, the function is strictly monotone.But wait, is this the only possible function? Or are there other functions that satisfy ( f(f(n)) = 3n )?Let me think. Suppose there's another function ( g ) that also satisfies ( g(g(n)) = 3n ). Would it necessarily have the same base 3 behavior as ( f )?Given that ( f ) is strictly increasing and the way it manipulates the base 3 digits ensures that each application of ( f ) either increases the number by a certain factor or shifts digits in a way that maintains the strict monotonicity, it seems plausible that this might be the unique solution.But to be thorough, let's consider if there could be another function. Suppose ( f ) doesn't follow the base 3 rule. For example, maybe ( f(1) = 3 ) instead of 2. Then ( f(f(1)) = f(3) ). According to the equation, this should be 3, so ( f(3) = 3 ). But then ( f(3) = 3 ), which would mean ( f(f(3)) = f(3) = 3 ), but according to the equation, it should be ( 3 times 3 = 9 ). Contradiction. So, ( f(1) ) can't be 3.Similarly, if ( f(1) = 1 ), then ( f(f(1)) = f(1) = 1 ), but it should be 3. So, ( f(1) ) can't be 1. Therefore, ( f(1) ) must be 2.This suggests that the function is uniquely determined by the condition ( f(f(n)) = 3n ) and the requirement of being strictly monotone.Another way to think about it is that ( f ) pairs numbers in such a way that each pair ( (n, f(n)) ) satisfies ( f(f(n)) = 3n ). Since ( f ) is strictly increasing, these pairs must be ordered in a way that maintains the increasing sequence.For example, the pairs are:- ( 1 leftrightarrow 2 leftrightarrow 3 leftrightarrow 6 leftrightarrow 9 leftrightarrow 18 leftrightarrow ... )- ( 4 leftrightarrow 7 leftrightarrow 12 leftrightarrow 21 leftrightarrow ... )- ( 5 leftrightarrow 8 leftrightarrow 15 leftrightarrow 24 leftrightarrow ... )- And so on.Each of these sequences is generated by applying ( f ) repeatedly, and they all follow the pattern of multiplying by 3 every two steps.This structure is similar to a binary tree where each node has two children, but in this case, it's more like a linked list where each element points to the next in the sequence.Given this structure, it's clear that the function ( f ) must map each number to its "partner" in the sequence, ensuring that applying ( f ) twice results in multiplying by 3.Therefore, the function ( f ) defined by manipulating the base 3 digits as described earlier is indeed the unique solution that satisfies both the functional equation and the strict monotonicity condition.To summarize, the function ( f ) works by treating numbers in base 3. If a number starts with a '1', ( f ) changes that '1' to a '2'. If it starts with a '2', ( f ) changes that '2' to a '1' and appends a '0' at the end. This ensures that applying ( f ) twice results in multiplying the original number by 3, and the function remains strictly increasing.

💡First, I need to determine how much Vince earns per customer. Let's denote this amount as ( E ).Vince serves 80 customers a month, so his total earnings for the month are ( 80 times E ).He spends 280 on rent and electricity and saves 872. Therefore, the amount left after these expenses is:[80E - 280 - 872]Vince also spends 20% of his earnings on recreation and relaxation, which means he keeps 80% of his earnings. This amount should equal the remaining money after paying for rent, electricity, and savings:[0.80 times (80E) = 80E - 280 - 872]Simplifying the equation:[64E = 80E - 1152]Subtracting ( 80E ) from both sides:[-16E = -1152]Dividing both sides by -16:[E = frac{1152}{16} = 72]Therefore, Vince earns 72 per customer.

💡First, determine the amount of paint remaining after the first day. Since one third of the paint is used, two thirds remain.Next, calculate the amount of paint used on the second day, which is one third of the remaining paint from the first day. Subtract this amount from the remaining paint after the first day to find the paint available for the third day.Finally, express the remaining paint as a fraction of the original amount.