🧠:Okay, let's try to tackle this problem. So, we have a finite increasing sequence of natural numbers a₁, a₂, ..., aₙ where n is at least 3. The recurrence relation given is that for all k ≤ n-2, a_{k+2} = 3a_{k+1} - 2a_k - 1. Also, the sequence must include the term 2021. We need to find the maximum number of three-digit numbers divisible by 25 that this sequence can contain.First, let me understand the recurrence relation. The relation is a_{k+2} = 3a_{k+1} - 2a_k -1. Hmm. So, this is a linear recurrence with constant coefficients, but it's nonhomogeneous because of the -1 term. Maybe I can solve the recurrence relation to find a general formula for the terms of the sequence. Once I have a general formula, I can analyze which terms are three-digit numbers divisible by 25 and how to maximize their count while ensuring 2021 is in the sequence.Let me start by solving the recurrence relation. The given recurrence is linear, so I can find the homogeneous solution and then a particular solution.The homogeneous part is a_{k+2} - 3a_{k+1} + 2a_k = 0. The characteristic equation would be r² - 3r + 2 = 0. Solving that: r = [3 ± sqrt(9 - 8)] / 2 = [3 ± 1]/2, so r = 2 or r = 1. Therefore, the homogeneous solution is a_k^h = A(2^k) + B(1^k) = A*2^k + B.Now, for the particular solution. The nonhomogeneous term is -1, which is a constant. So, let's assume a particular solution is a constant, say C. Plugging into the recurrence:C = 3C - 2C -1Simplify: C = (3C - 2C) -1 => C = C -1 => 0 = -1. Wait, that's a contradiction. So, our guess was wrong. Since the nonhomogeneous term is a constant and 1 is a root of the characteristic equation (r=1), we need to multiply our guess by k. Let's try a particular solution of the form C*k.So, set a_p(k) = C*k. Then:a_p(k+2) = C*(k+2)3a_p(k+1) - 2a_p(k) -1 = 3C*(k+1) - 2C*k -1 = (3C(k+1) - 2Ck) -1 = (3Ck + 3C - 2Ck) -1 = (Ck + 3C) -1Set equal to a_p(k+2):C*(k + 2) = Ck + 3C -1Left side: Ck + 2CRight side: Ck + 3C -1Subtract Ck from both sides:2C = 3C -1 => 2C -3C = -1 => -C = -1 => C = 1So, the particular solution is a_p(k) = k. Therefore, the general solution is:a_k = A*2^k + B + kNow, the general term of the sequence is a combination of the homogeneous and particular solutions. So, a_k = A*2^k + B + k.Since the sequence is increasing, we need to ensure that each term is larger than the previous. Let's note that the sequence is defined for k from 1 to n, and the recurrence applies for k ≤ n-2. So, given a₁, a₂, the rest of the terms are determined by the recurrence.But the general solution is a_k = A*2^k + B + k. Let's express this formula. To find constants A and B, we need initial conditions. Since the problem doesn't specify the starting terms, but just says the sequence is increasing and contains 2021, we need to figure out possible sequences that include 2021 and have as many three-digit multiples of 25 as possible.Our goal is to maximize the number of terms in the sequence that are three-digit numbers divisible by 25. Three-digit numbers divisible by 25 are 100, 125, 150, ..., 975, 1000. Wait, 1000 is a four-digit number, so the last three-digit multiple of 25 is 975. So, they start at 100 and go up by 25 each time: 100, 125, 150, ..., 975. So, the three-digit numbers divisible by 25 are from 100 to 975 inclusive, stepping by 25. Let's compute how many there are: (975 - 100)/25 + 1 = (875)/25 +1 = 35 +1 = 36. So, there are 36 three-digit numbers divisible by 25.But the question is not how many there are in total, but how many can be in the sequence. So, we need to arrange the sequence so that as many of these 36 numbers as possible are included, while the sequence is increasing, follows the recurrence, and contains 2021.First, perhaps we need to analyze the structure of the sequence given by the recurrence relation a_{k} = A*2^k + B + k.Given that, let's see how the terms behave. The term has an exponential component (A*2^k), a linear component (k), and a constant term B. Depending on the value of A, the exponential term can dominate or not. Since the sequence is increasing, we need to ensure that each term is larger than the previous one.Let's compute a_{k+1} - a_k:a_{k+1} - a_k = [A*2^{k+1} + B + (k+1)] - [A*2^k + B + k] = A*2^{k+1} - A*2^k + (k+1 - k) = A*2^k(2 -1) +1 = A*2^k +1Since the sequence is increasing, a_{k+1} - a_k > 0. So, A*2^k +1 > 0. Since 2^k is positive for all k, and we are dealing with natural numbers, A must be such that A*2^k +1 is positive. If A is non-negative, then A*2^k is non-negative, so A*2^k +1 ≥1 >0. If A is negative, then we need to ensure that A*2^k +1 >0, so A*2^k > -1. Since 2^k is at least 2 (for k ≥1), then if A is negative, A*2^k ≥ A*2. So, A*2 > -1 => A > -1/2. Since A must be an integer? Wait, the problem states the sequence consists of natural numbers. So, each term a_k must be a natural number.Given that a_k = A*2^k + B + k must be natural for all k. Since 2^k and k are integers, A and B must be integers. Because if A and B are integers, then each term is an integer. Moreover, since the sequence is increasing, a_{k+1} - a_k = A*2^k +1 must be positive. So, A*2^k +1 >0. Since 2^k is positive, even if A is negative, for large k, A*2^k could be negative and large in magnitude. However, the sequence must be increasing for all k up to n. So, perhaps A must be non-negative? Let's check.Suppose A is negative. Let A = -m where m is a positive integer. Then, a_k = -m*2^k + B +k. The difference a_{k+1} - a_k = -m*2^k +1. For this to be positive, we need -m*2^k +1 >0 => m*2^k <1. But m is at least 1, and 2^k is at least 2 for k ≥1. So, m*2^k ≥2, which would imply that -m*2^k +1 ≤ -2 +1 = -1 <0. Therefore, the difference would be negative, which contradicts the sequence being increasing. Therefore, A cannot be negative. Therefore, A must be non-negative.So, A is a non-negative integer. Therefore, the difference a_{k+1} - a_k = A*2^k +1, which is positive because A*2^k is non-negative and we add 1. Therefore, the sequence is always increasing as long as A is non-negative. So, A must be a non-negative integer, and B must be an integer such that all terms a_k are natural numbers.So, given that, we have a general formula a_k = A*2^k + B +k. Now, the problem states that the sequence must contain the term 2021. So, there exists some index m such that a_m = 2021. So, 2021 = A*2^m + B + m.Our goal is to choose A and B such that as many three-digit numbers divisible by 25 are in the sequence. To maximize the count, we need to align the terms of the sequence to hit as many multiples of 25 in the three-digit range as possible. But also, since the sequence is determined by the recurrence, which in turn is determined by the choices of A and B, and the starting terms a₁ and a₂.Wait, but the general solution is a_k = A*2^k + B +k, so once A and B are fixed, the entire sequence is determined. However, the problem says "a finite increasing sequence... such that for all k ≤ n-2, the equality a_{k+2} = 3a_{k+1} - 2a_k -1 holds". So, the sequence can start at any a₁ and a₂, as long as the rest of the terms follow the recurrence. Therefore, the sequence is determined by the first two terms a₁ and a₂. Then, a₃ = 3a₂ -2a₁ -1, a₄ = 3a₃ -2a₂ -1, etc.But the general solution is a_k = A*2^k + B +k, so the closed-form expression can be written in terms of a₁ and a₂. Alternatively, maybe we can relate A and B to a₁ and a₂.Let me verify that. Suppose we have a₁ = A*2^1 + B +1 = 2A + B +1a₂ = A*2^2 + B +2 = 4A + B +2Then, solving for A and B in terms of a₁ and a₂:From a₁ = 2A + B +1,From a₂ = 4A + B +2,Subtract the first equation from the second:a₂ - a₁ = (4A + B +2) - (2A + B +1) = 2A +1Therefore, 2A = (a₂ - a₁) -1 => A = (a₂ - a₁ -1)/2Then, substitute back into the first equation:a₁ = 2A + B +1 => B = a₁ -2A -1So, A must be (a₂ - a₁ -1)/2, which must be an integer because in the general solution, A is an integer (since a_k must be natural numbers). Therefore, (a₂ - a₁ -1) must be even. So, a₂ - a₁ must be odd. Therefore, the difference between a₂ and a₁ is odd.Also, since the sequence is increasing, a₂ > a₁. So, a₂ - a₁ ≥1, and since a₂ - a₁ is odd, the minimal difference is 1 (if a₂ - a₁ =1, then A = (1 -1)/2=0). If a₂ - a₁ is 3, then A = (3 -1)/2=1, etc.Therefore, A is determined by the difference between a₂ and a₁. So, given that, A is a non-negative integer (as we saw before), so (a₂ - a₁ -1) must be non-negative and even. Therefore, a₂ - a₁ -1 ≥0 and even. Therefore, a₂ - a₁ ≥1, and (a₂ - a₁ -1) is even. So, a₂ - a₁ is odd and at least 1, which is consistent with the sequence being increasing.So, given that, we can model the entire sequence based on a₁ and a₂, with a₂ = a₁ + 1 + 2A, where A is a non-negative integer.Alternatively, since the general term is a_k = A*2^k + B +k, and we can express A and B in terms of a₁ and a₂, but perhaps this might complicate things. Maybe it's better to work with the closed-form formula.Given that a_k = A*2^k + B +k, and we need to have a term equal to 2021. Let's denote that for some m, a_m = A*2^m + B + m = 2021.Our goal is to choose A and B such that as many terms as possible in the sequence a_k are three-digit numbers divisible by 25. So, the terms a_k must satisfy 100 ≤ a_k ≤ 999 and a_k ≡0 mod25.Additionally, the sequence is increasing, so once a term exceeds 999, the subsequent terms will be larger and thus not three-digit numbers. Similarly, terms before the first three-digit number would be less than 100. Therefore, the three-digit numbers in the sequence are a contiguous block from some index p to q where 100 ≤ a_p ≤ a_q ≤999, and outside of that, terms are either below 100 or above 999.But since the sequence must contain 2021, which is a four-digit number, the sequence must extend beyond the three-digit numbers up to at least 2021. Therefore, the three-digit terms in the sequence must all be before the term 2021, or after? Wait, the sequence is increasing, so once it reaches 2021, all subsequent terms will be larger than 2021, hence four-digit numbers. Therefore, the three-digit terms must all be before the term 2021. Therefore, the term 2021 is somewhere in the sequence, and before it, there are terms increasing up to 2021, with some of them being three-digit numbers divisible by 25.But since the sequence is finite, it's a finite increasing sequence containing 2021, so 2021 could be the last term, or somewhere in the middle. Wait, but the problem says the sequence is given, and it's finite. So, 2021 is one of the terms, and the rest of the terms are determined by the recurrence. However, since the sequence is finite, we can't have terms after 2021 unless they are part of the sequence, but since the sequence is increasing, once it passes 2021, subsequent terms are larger. However, the problem doesn't specify whether the sequence must terminate at 2021 or can continue beyond. It just says it's a finite sequence containing 2021. So, 2021 can be anywhere in the sequence, but since the sequence is finite, we can choose the starting terms such that 2021 is included, and as many three-digit multiples of 25 as possible are before it.But to maximize the number of three-digit multiples of 25, we need the part of the sequence that lies in the three-digit range to include as many multiples of 25 as possible. Since the sequence is determined by the recurrence, which is linear with an exponential component, the spacing between consecutive terms increases exponentially if A >0. If A=0, the recurrence becomes a_{k+2} = 3a_{k+1} -2a_k -1. Let's see what happens when A=0.If A=0, then the general term becomes a_k = B +k. So, the sequence is linear: each term increases by 1. Because then the difference a_{k+1} -a_k = (B +k +1) - (B +k) =1. So, the sequence would be B +1, B +2, B +3, ..., B +n. But in this case, the recurrence must hold. Let's check with A=0:If a_k = B +k, then a_{k+2} = B +k +2According to the recurrence: 3a_{k+1} -2a_k -1 = 3(B +k +1) -2(B +k) -1 = 3B +3k +3 -2B -2k -1 = B +k +2, which matches a_{k+2}. So, that works.Therefore, when A=0, the sequence is linear, increasing by 1 each time. In this case, the sequence would be an arithmetic progression with common difference 1. So, if A=0, then the sequence is simply starting from some B+1, B+2, etc. But in this case, since we need to include 2021, the term 2021 would be B +m =2021, so B=2021 -m. But since the sequence is finite, m can be any index from 1 to n.But if the sequence is an arithmetic progression with difference 1, then the three-digit numbers divisible by 25 would be 100, 125, 150, ..., 975. Since the difference is 1, the sequence would pass through each number once, so it would include each multiple of 25 exactly once when it reaches that number. Therefore, the number of three-digit multiples of 25 in the sequence would be 36 (from 100 to 975 inclusive, step 25). However, in this case, the sequence would need to start before 100 and end at 2021. So, the total length of the sequence would be from, say, B +1 up to 2021, with each term increasing by 1. But if the sequence starts before 100, say at 1, then it would include all numbers from 1 to 2021, which would include all three-digit numbers, hence 900 three-digit numbers, but among them, 36 divisible by 25. However, since the problem asks for the maximum number of three-digit numbers divisible by 25 that the sequence can contain, in this case, it's 36.But wait, if the sequence is allowed to start anywhere, perhaps we can start the sequence such that the first term is 100, then the next is 101, ..., up to 2021. But then the number of three-digit numbers would be 2021 -100 +1 = 1922 terms, but only 36 of them are divisible by 25. But if we start the sequence at 100, the terms are 100,101,...,2021. But in this case, the number of three-digit numbers is 900 (from 100 to 999), but only 36 are divisible by 25. However, in this case, the number of three-digit numbers divisible by 25 is still 36. So, regardless of where the sequence starts (as long as it's an arithmetic progression with difference 1), the number of three-digit multiples of 25 would be 36. But is that the case?Wait, no. If the sequence is an arithmetic progression with difference 1, then it includes every number between its starting point and ending point. So, if the starting point is ≤100 and the ending point is ≥999, then it includes all three-digit numbers, hence 900 terms, of which 36 are divisible by 25. However, if the sequence starts above 100 or ends before 999, the number would be less. But since we need to include 2021, which is above 999, the sequence must continue beyond 999. Therefore, to include 2021, the sequence must start at some point before 2021 and go up to 2021. If the sequence is an arithmetic progression with difference 1, starting from some number a₁ ≤100 and ending at 2021, then the number of three-digit numbers in the sequence would be 900 (from 100 to 999), hence 36 divisible by 25. But if we can arrange the sequence to have larger steps, perhaps skipping some numbers, but still hitting multiple multiples of 25, maybe we can have more than 36? Wait, no. Because multiples of 25 are spaced 25 apart. If the step is larger than 1, say step is 25, then we can hit each multiple of 25. But in our case, the sequence is determined by the recurrence relation. If A is positive, then the differences between terms increase exponentially. Let's explore that.Suppose A >0. Then the general term is a_k = A*2^k + B +k. Let's see how this behaves. The difference between consecutive terms is a_{k+1} -a_k = A*2^k +1. So, the differences themselves form a sequence that grows exponentially. Therefore, the sequence starts with a certain difference, then each subsequent difference doubles (plus 1). Wait, no. The difference is A*2^k +1. So, if A is positive, then the differences increase exponentially. For example, if A=1:Difference between a₁ and a₂: 1*2^1 +1=3Between a₂ and a₃:1*2^2 +1=5Between a₃ and a₄:1*2^3 +1=9And so on, each difference is doubling minus 1? Wait, no. The difference at step k is A*2^{k} +1. So, for A=1, starting at k=1:Difference[1] = 1*2^1 +1=3Difference[2] =1*2^2 +1=5Difference[3] =1*2^3 +1=9Difference[4] =1*2^4 +1=17And so on. Each difference is doubling the previous exponent's 2^k term plus 1. So, the differences are increasing exponentially. Therefore, the sequence's terms grow exponentially.Therefore, in such a case, the sequence would start with small terms, then rapidly increase. If we can make the sequence pass through several three-digit multiples of 25 before skyrocketing to 2021, perhaps we can have more such terms. However, because the differences are increasing exponentially, the number of terms in the three-digit range might be limited.Alternatively, if A is very large, the sequence might jump over the three-digit range entirely. For example, if A is large enough that a₁ and a₂ are already above 999, but then we need to include 2021, which is four digits. But since the sequence is increasing, it cannot go back down. However, the problem states the sequence must contain 2021, but it doesn't specify that it has to have terms before or after. Wait, but a sequence must have at least three terms (n≥3). So, if A is very large, the sequence could start at some high number, but it must have three terms. However, since the sequence is increasing, the terms must be increasing. But if A is large, the terms would be huge, but since 2021 is required, perhaps A cannot be too large. Hmm, this is getting complicated.Wait, maybe we can approach this by trying to model the sequence. Let's consider that the general term is a_k = A*2^k + B +k. We need a_m =2021 for some m. So, 2021 =A*2^m + B +m. We need to choose A and B such that as many a_k in the three-digit range are divisible by 25.Alternatively, since a_k =A*2^k + B +k, for a_k to be divisible by 25, we need A*2^k + B +k ≡0 mod25. This is a congruence condition. To maximize the number of k's for which this holds, we need to choose A and B such that A*2^k +k ≡ -B mod25. Since B is a constant, -B mod25 is also a constant. Therefore, for each k in the three-digit range (which corresponds to a_k between 100 and 999), we need A*2^k +k ≡C mod25, where C is fixed. The maximum number of such k's would depend on how many times A*2^k +k can hit the same residue C mod25 within the range of k where a_k is three-digit.But this seems quite abstract. Maybe we need another approach.Since the problem is about maximizing the number of terms divisible by 25 in the three-digit range, let's note that 25 divides a_k, so a_k ≡0 mod25. Given that a_k =A*2^k + B +k, then:A*2^k + B +k ≡0 mod25.If we can set B ≡ -A*2^k -k mod25 for each k where a_k is a three-digit multiple of 25. However, B is a constant, so this congruence must hold for all such k. This is only possible if -A*2^k -k is constant mod25 for all k where a_k is a three-digit multiple of 25. But this seems restrictive. Alternatively, if we can choose A and B such that for as many k as possible, A*2^k +k ≡ -B mod25.Since B is fixed, we need to choose A such that A*2^k ≡ -B -k mod25 for multiple k's. The maximum number of k's would correspond to how many times the term A*2^k +k can hit the same residue mod25.This seems related to finding a value of A such that A*2^k ≡C -k mod25 for some constant C, across multiple k's. The number of solutions depends on the multiplicative inverses of 2^k mod25 and the linear equations.Alternatively, perhaps we can fix B such that for a particular A, multiple k's satisfy A*2^k +k ≡ -B mod25. However, this seems complicated.Alternatively, note that if we can make the term a_k =25*m for some integer m, where 100 ≤25*m ≤999, so 4 ≤m ≤39 (since 25*4=100 and 25*39=975). So, m ranges from 4 to39, giving 36 terms.If we can make the sequence a_k pass through as many of these 25*m as possible. Given that the sequence is determined by a linear recurrence with an exponential component, the spacing between terms increases rapidly if A>0. Therefore, to maximize the number of terms in the three-digit range, we need the sequence to pass through as many multiples of 25 as possible before increasing beyond 999. If A is small, the growth is slower, allowing more terms in the three-digit range.Wait, but when A=0, the sequence grows linearly, which would allow the maximum number of three-digit terms (900 terms), but only 36 of them divisible by25. However, with A>0, the sequence grows exponentially, so it would pass through the three-digit range more quickly, possibly hitting fewer multiples of25. However, maybe by carefully choosing A and B, we can align the terms to hit more multiples of25.Alternatively, maybe the optimal solution occurs when A=0, giving 36 terms. But the problem asks for the maximum number, so perhaps there's a way to have more than 36? That seems impossible since there are only 36 three-digit multiples of25. However, if the sequence can pass through some multiples of25 more than once, but since the sequence is strictly increasing, each multiple of25 can be included at most once. Therefore, the maximum possible is36. However, the sequence might not pass through all of them if it's increasing too quickly. Therefore, the question is whether we can arrange the sequence to pass through all36 multiples of25. If the sequence is an arithmetic progression with difference25, then it would hit each multiple of25 exactly once. But our recurrence relation does not allow for that unless A=0 and the difference is1, but with difference1, you pass through all numbers. But the problem's sequence has a difference that depends on the recurrence.Wait, no. If the sequence is determined by the recurrence a_{k+2}=3a_{k+1}-2a_k -1, then even with A=0, which gives a linear sequence with difference1, then yes, you pass through all numbers. But if A>0, the sequence grows faster. So, the maximum number of three-digit multiples of25 would be36, achievable only when the sequence is the linear one (A=0). However, the problem states that the sequence must contain2021, which is a four-digit number. If the sequence is linear (A=0), then it would need to go from some B +1 up to2021, increasing by1 each time. Therefore, in this case, the three-digit numbers would be from100 to999, which are all included if B +1 ≤100 and the sequence continues past999 up to2021. Therefore, the number of three-digit multiples of25 would indeed be36. However, is this possible?Wait, if A=0, then the general term is a_k = B +k. To have 2021 in the sequence, there exists some m such that B +m =2021. Therefore, B=2021 -m. The sequence starts at a₁ =B +1=2021 -m +1, and a₂ =B +2=2021 -m +2, etc., up to a_m =2021. Therefore, if we choose m to be at least 2021 -100 +1? Wait, no. If we want the sequence to include all three-digit numbers from100 to999, then the sequence must start at a₁ ≤100 and end at a_m=2021. Therefore, the starting term a₁ =2021 -m +1 ≤100. So, 2021 -m +1 ≤100 => m ≥2021 -100 +1=1922. Therefore, m=1922, meaning the sequence has 1922 terms, starting at a₁=2021 -1922 +1=100, and each subsequent term increases by1 up to a_{1922}=2021. Therefore, this sequence would include all numbers from100 to2021, which includes all three-digit numbers from100 to999, hence36 multiples of25. Therefore, in this case, the maximum number of three-digit multiples of25 is36.But wait, the problem says "a finite increasing sequence... of natural numbers". Natural numbers usually start at1, but in this case, the sequence can start at100 if we choose B=2021 -m and m=1922, so B=2021 -1922=99. Therefore, a₁=99 +1=100, a₂=101, ..., a_{1922}=2021. Therefore, this is a valid sequence, and it contains all three-digit numbers from100 to999, hence36 multiples of25. However, in this case, the sequence would have1922 terms, which is allowed as n≥3.But the problem asks for the maximum number of three-digit numbers divisible by25 that this sequence can contain. If the sequence can contain all36, then36 is the maximum. However, the problem might be designed such that the answer is less than36 due to the constraints of the recurrence relation. Wait, but when A=0, the recurrence is satisfied, and the sequence is just a linear sequence with difference1, which does satisfy the recurrence relation as we checked earlier. Therefore, in this case,36 is achievable. However, let's verify this.If A=0, then the general term is a_k =B +k. The recurrence is a_{k+2}=3a_{k+1}-2a_k -1. Let's check with a_k =B +k:Left side: a_{k+2}=B +k +2Right side:3a_{k+1} -2a_k -1=3(B +k +1) -2(B +k) -1=3B +3k +3 -2B -2k -1=B +k +2They are equal, so yes, the linear sequence satisfies the recurrence. Therefore, choosing A=0 and B=99 (as in the example above), we get a sequence starting at100, increasing by1 each time up to2021. Therefore, it contains all three-digit numbers from100 to999, inclusive. Hence, the number of three-digit numbers divisible by25 is(975 -100)/25 +1= (875)/25 +1=35 +1=36. Therefore,36 is achievable.But the question is asking for the maximum number. Is it possible to get more than36? Since there are only36 three-digit numbers divisible by25, the answer cannot exceed36. Therefore, the maximum is36.However, the problem might have a trick. If the sequence is allowed to loop or something, but no, it's a finite increasing sequence, so no repeats. Therefore, the maximum is36. However, let's check if the initial terms can be set such that A=0 is the only way to get36, or if there are other sequences with A>0 that can also achieve36.Suppose A>0. Then the general term is a_k =A*2^k +B +k. The differences between terms are increasing exponentially, so the sequence would accelerate upwards. Therefore, it would spend fewer terms in the three-digit range compared to the linear case. Hence, it would have fewer opportunities to hit multiples of25. Therefore,36 is only achievable when A=0, and for A>0, the number would be less.Therefore, the maximum number is36. However, let's check with a small example.Suppose A=1, B=0. Then the general term is a_k=2^k +0 +k=2^k +k. Let's compute some terms:k=1:2+1=3k=2:4+2=6k=3:8+3=11k=4:16+4=20k=5:32+5=37k=6:64+6=70k=7:128+7=135k=8:256+8=264k=9:512+9=521k=10:1024+10=1034So, starting from k=1, the terms are 3,6,11,20,37,70,135,264,521,1034,...In this sequence, the three-digit terms are135,264,521,1034. Wait,1034 is four-digit. So, three-digit terms are135,264,521. Among these, which are divisible by25?135:135/25=5.4→no;264:264/25=10.56→no;521/25=20.84→no. So, none of the three-digit terms in this sequence are divisible by25. Therefore, in this case, the count is0.If we choose A=1 and B such that the terms align better. Let's say we want a_k divisible by25. Suppose we want a_7=135 to be divisible by25. Then 135 + x ≡0 mod25. But since a_k=2^k +k +B, changing B would shift all terms by B. Wait, no. Wait, the general term is a_k =A*2^k +B +k. So, B is a constant. If we set B such that a_m=25*c for some m, then B=25*c -A*2^m -m. For example, if we want a_7=135 to be 25*5=125, then B=125 -128 -7= -10. Let's check:a_k=2^k + (-10) +k. Then:k=1:2 -10 +1= -7 (not natural number)So, invalid. Therefore, we can't have negative terms. Therefore, B must be chosen such that all terms are natural numbers. So, for a_k= A*2^k + B +k ≥1 for all k from1 ton.Therefore, if A=1, B must be at least such that a₁=2 +B +1 ≥1 → B≥-2. But since we need all terms to be natural numbers, B must be chosen such that even the first term is natural. So, B must be an integer such that A*2^k +B +k ≥1 for all k in the sequence. If we set B= -2, then a₁=2 +(-2) +1=1, which is okay, a₂=4 +(-2) +2=4, a₃=8 +(-2)+3=9, etc. So, let's try B= -2:a_k=2^k -2 +kk=1:2 -2 +1=1k=2:4 -2 +2=4k=3:8 -2 +3=9k=4:16 -2 +4=18k=5:32 -2 +5=35k=6:64 -2 +6=68k=7:128 -2 +7=133k=8:256 -2 +8=262k=9:512 -2 +9=519k=10:1024 -2 +10=1032So, the three-digit terms are133,262,519,1032. Again, none are divisible by25.Alternatively, if we set B such that one of the terms is divisible by25. Let's say we want a_7=128 +B +7=135 +B. Let's set135 +B=150 → B=15. Then:a_k=2^k +15 +kk=1:2 +15 +1=18k=2:4 +15 +2=21k=3:8 +15 +3=26k=4:16 +15 +4=35k=5:32 +15 +5=52k=6:64 +15 +6=85k=7:128 +15 +7=150 (divisible by25)k=8:256 +15 +8=279k=9:512 +15 +9=536k=10:1024 +15 +10=1049So, in this case, a_7=150 is divisible by25. Then, the three-digit terms are150,279,536,1049. Among these,150 is divisible by25, but the rest are not. So, only one three-digit term divisible by25. Even though we forced one term to be150, the others are not aligned. Therefore, even with A=1 and B chosen to make one term divisible by25, we only get one such term.If we try to make two terms divisible by25. Let's say we want a_7=150 and a_8=175. Then:For a_7=150:2^7 +B +7=150 →128 +B +7=150 →B=15For a_8=175:2^8 +B +8=256 +15 +8=279≠175. Doesn't work.Therefore, it's not possible to have two terms divisible by25 in this case.Therefore, when A>0, the number of three-digit terms divisible by25 is very limited, often just one or none. Whereas when A=0, we can get all36. Therefore, the maximum number is36.But wait, the problem states that the sequence is finite and must contain2021. In the case where A=0, the sequence can be made to include2021 by having a sufficiently large n. However, does such a sequence satisfy the original recurrence relation for all k ≤n-2?Yes, because when A=0, the sequence is a linear progression with difference1, which satisfies the recurrence a_{k+2}=3a_{k+1} -2a_k -1. As we checked earlier, substituting a_k =B +k into the recurrence gives equality on both sides.Therefore, the sequence starting at100 and increasing by1 up to2021 includes all three-digit numbers from100 to999, hence36 multiples of25. This sequence satisfies the recurrence relation, is increasing, and contains2021. Therefore, the answer should be36.But let's check if there's any catch. The problem says "a finite increasing sequence... of natural numbers". If we start at100, then the earlier terms (a₁=100, a₂=101, etc.) are natural numbers, so that's fine. The sequence can start at100 and go up to2021, which is allowed. Therefore, the maximum number is36.But I recall that in some problems, especially Olympiad-level, the obvious answer might not be the correct one due to hidden constraints. Let me think again.The key point is that when A=0, the general term is a_k =B +k, which is an arithmetic sequence with difference1. Therefore, it will include every natural number from B+1 to B+n. To include2021, we need B +m =2021 for some m. If we choose B=99, then the sequence starts at100 and goes up to2021, which requires m=2021 -99=1922 terms. In this sequence, the three-digit numbers are from100 to999, which are 900 terms, of which36 are divisible by25. Therefore, the answer is36.But let's confirm that such a sequence indeed satisfies the recurrence relation for all k ≤n-2.Take any three consecutive terms in the sequence: a_k=100 +k -1, a_{k+1}=100 +k, a_{k+2}=100 +k +1.According to the recurrence, a_{k+2}=3a_{k+1} -2a_k -1.Let's check:3a_{k+1} -2a_k -1=3*(100 +k) -2*(100 +k -1) -1=300 +3k -200 -2k +2 -1=100 +k +1= a_{k+2}.Yes, it works. Therefore, the sequence is valid.Therefore, the maximum number of three-digit numbers divisible by25 is36.However, the problem is presented in a context where the recurrence might restrict the answer. However, our analysis shows that when A=0, the sequence can indeed cover all three-digit multiples of25. Therefore, the answer should be36.But let me check another example where A is not zero but still allows multiple terms divisible by25.Suppose we set A=1 and B= such that several terms hit multiples of25. Let's try.Suppose we want a_7=150. Then, as before, B=150 -128 -7=15.Then, the terms are:k=1:2 +15 +1=18k=2:4 +15 +2=21k=3:8 +15 +3=26k=4:16 +15 +4=35k=5:32 +15 +5=52k=6:64 +15 +6=85k=7:128 +15 +7=150k=8:256 +15 +8=279k=9:512 +15 +9=536k=10:1024 +15 +10=1049Now, let's see if any other terms are divisible by25:k=5:52 → nok=6:85 → nok=7:150 → yesk=8:279 → nok=9:536 →536/25=21.44→nok=10:1049→noSo, only one term divisible by25.Alternatively, choose B such that two terms are divisible by25. Let's attempt:Suppose we want a_7=150 and a_9=175.For a_7=150:128 + B +7=150 ⇒B=15For a_9=175:512 + B +9=175 ⇒B=175 -512 -9= -346Contradiction, since B cannot be both15 and-346. Therefore, impossible.Alternatively, set B such that a_k=100. Let's say a_6=100.a_6=64 +B +6=100 ⇒B=30Then, the terms are:k=1:2 +30 +1=33k=2:4 +30 +2=36k=3:8 +30 +3=41k=4:16 +30 +4=50k=5:32 +30 +5=67k=6:64 +30 +6=100k=7:128 +30 +7=165k=8:256 +30 +8=294k=9:512 +30 +9=551k=10:1024 +30 +10=1064So, three-digit terms:100,165,294,551,1064. Among these,100 and165:165/25=6.6→no;294/25=11.76→no;551/25=22.04→no. Only100 is divisible by25. But100 is a three-digit number, so that's one term.But when A=1, even if we set one term to100, we only get one three-digit multiple of25. Compared to A=0 where we get36.Therefore, it's clear that when A=0, we can achieve the maximum number of36, which is the total number of three-digit multiples of25. Therefore, the answer should be36.But to confirm, let's consider whether the sequence with A=0 and including2021 can indeed exist.Yes, as shown earlier, setting B=99 and starting from100, the sequence goes up to2021 with difference1, satisfying the recurrence and including all three-digit multiples of25. Therefore,36 is achievable.Therefore, the maximum number is36.But wait, the problem says "the sequence must contain the term a_{k}=2021". In the case where the sequence is100,101,...,2021, then2021 is the last term. The problem doesn't restrict the sequence from having terms beyond2021, but since it's finite, it must end at some term. However, to include2021, it can end at2021. Therefore, the sequence can be100,101,...,2021, which has2021 -100 +1=1922 terms. This sequence contains all three-digit numbers from100 to999, hence36 multiples of25. Therefore, the answer is36.However, I need to check the problem statement once again to make sure I didn't miss anything."A finite increasing sequence a₁, a₂, ..., aₙ (n ≥3) of natural numbers is given, such that for all k ≤n-2, the equality a_{k+2}=3a_{k+1} -2a_k -1 holds. The sequence must contain the term a_k=2021. Determine the maximum number of three-digit numbers divisible by25 that this sequence can contain."Yes, so the sequence must be finite, start at some a₁, end at some aₙ, contain2021 somewhere, and satisfy the recurrence for all k up ton-2. We constructed such a sequence starting at100, ending at2021, which satisfies the recurrence, contains2021, and includes all three-digit numbers from100 to999, hence36 multiples of25. Therefore, the answer is36.But I want to make sure that no other sequence with A>0 can include more than36 such terms. Since when A>0, the sequence grows exponentially, the number of terms in the three-digit range is logarithmic in nature. For example, with A=1, the three-digit terms might be around k=7 tok=9 or something, giving only a few terms. Therefore, it's impossible for A>0 sequences to include more than a few three-digit multiples of25. Whereas the A=0 sequence includes all36. Therefore, the maximum is indeed36.Therefore, the final answer is36, which in boxed form isboxed{36}.

✅:To determine the maximum number of three-digit numbers divisible by 25 that the sequence can contain, we start by analyzing the given recurrence relation (a_{k+2} = 3a_{k+1} - 2a_k - 1). Solving this recurrence relation, we find the general solution to be (a_k = A cdot 2^k + B + k), where (A) and (B) are constants.Given that the sequence must contain the term 2021, we know there exists some (m) such that (a_m = 2021). This translates to (2021 = A cdot 2^m + B + m).To maximize the number of three-digit numbers divisible by 25, we consider the case where (A = 0). In this scenario, the sequence simplifies to an arithmetic progression with a common difference of 1, i.e., (a_k = B + k). For the sequence to include 2021, we set (B = 2021 - m), ensuring the sequence starts at (a_1 = B + 1) and ends at (a_m = 2021).By choosing (B = 99), the sequence starts at 100 and ends at 2021, covering all three-digit numbers from 100 to 999. There are 36 three-digit numbers divisible by 25 in this range (100, 125, 150, ..., 975).Verifying the recurrence relation for this arithmetic progression, we confirm that it satisfies (a_{k+2} = 3a_{k+1} - 2a_k - 1). Each term increases by 1, ensuring the sequence is strictly increasing and valid.Thus, the maximum number of three-digit numbers divisible by 25 that this sequence can contain is (boxed{36}).

🔑:Given a finite increasing sequence (a_1, a_2, ldots, a_n) of natural numbers ((n geq 3)) satisfying the recurrence relation:[a_{k+2} = 3a_{k+1} - 2a_{k} - 1]for all (k leq n-2). We need to determine the largest number of three-digit numbers that are multiples of 25 in this sequence, which must include the term (a_k = 2021).1. We establish that (a_k) is in arithmetic progression: We will demonstrate that an arithmetic progression (a_k = k) (where (a_k = k)) satisfies the given recurrence relation: Given: [ a_{k+2} = 3a_{k+1} - 2a_{k} - 1 ] Substituting (a_k = k), (a_{k+1} = k + 1), and (a_{k+2} = k + 2): [ a_{k+2} = k + 2 ] Evaluate the right-hand side of the recurrence relation with the substitutions: [ 3a_{k+1} - 2a_{k} - 1 = 3(k + 1) - 2k - 1 ] [ = 3k + 3 - 2k - 1 ] [ = k + 2 ] Since both sides are equal, the arithmetic progression (a_k = k) satisfies the recurrence relation. 2. Sequence containing 2021: We can have a sequence containing the term 2021. For instance, consider the sequence starting close to 3 and ending at 2021: [ 3, 4, 5, ..., 2018, 2019, 2020, 2021 ]3. Finding the count of three-digit numbers divisible by 25: We need to count the three-digit numbers from 100 to 999 that are divisible by 25. These numbers form an arithmetic progression: First term: (100) Common difference: (25) Last term: (975) The general term for an arithmetic progression is given by: [ a_n = a + (n-1)d ] Here: [ 975 = 100 + (n-1) times 25 ] Solving for (n): [ 975 = 100 + 25(n - 1) ] [ 975 - 100 = 25(n - 1) ] [ 875 = 25(n - 1) ] [ 35 = n - 1 ] [ n = 36 ] Therefore, there are 36 three-digit numbers between 100 and 999 that are divisible by 25.4. Conclusion: The largest number of three-digit numbers that are multiples of 25 in the sequence is: [ boxed{36} ]

🧠:Okay, let's see. I need to figure out how many cows would eat all the grass in 96 days, given that 70 cows can finish it in 24 days and 30 cows in 60 days. Hmm, this sounds like a problem involving rates of consumption and growth. I remember these types of problems sometimes being called "grass growing problems" where you have to consider both the cows eating the grass and the grass growing simultaneously. First, let me try to model the problem. The grass is growing at a constant rate, right? So, every day, a certain amount of grass grows. At the same time, the cows are eating the grass. The key here is that the amount of grass isn't constant because it's being replenished. So, the total grass at any time is the initial amount plus the growth over time minus the amount eaten by the cows. Let me define some variables to make this clearer. Let's say:- Let G be the initial amount of grass in the field.- Let r be the rate at which grass grows per day (in units equivalent to the amount a cow eats, maybe).- Let c be the number of cows.- Let d be the number of days.The total grass consumed by the cows in d days would be c * d (assuming each cow eats 1 unit per day). The total grass available is the initial grass plus the grass grown over d days, which is G + r * d. Since the grass is all eaten by the cows in each scenario, we can set up equations. For the first case with 70 cows in 24 days:G + r * 24 = 70 * 24And for the second case with 30 cows in 60 days:G + r * 60 = 30 * 60So now we have two equations:1. G + 24r = 1680 (since 70*24=1680)2. G + 60r = 1800 (since 30*60=1800)Now, we can solve these two equations to find G and r. Subtract equation 1 from equation 2:(G + 60r) - (G + 24r) = 1800 - 1680G - G + 60r -24r = 12036r = 120r = 120 / 36r = 10/3 ≈ 3.333...So the grass grows at a rate of 10/3 units per day. Now plug r back into equation 1 to find G:G + 24*(10/3) = 1680G + 80 = 1680G = 1680 - 80G = 1600So the initial grass is 1600 units. Let me check with equation 2 to make sure:G + 60*(10/3) = 1600 + 200 = 1800, which matches 30*60=1800. Okay, that seems correct.Now, the question is asking for the number of cows, let's call it x, that would eat all the grass in 96 days. Using the same logic:Total grass after 96 days is G + r*96 = 1600 + (10/3)*96Calculate that:(10/3)*96 = 10*32 = 320So total grass = 1600 + 320 = 1920This total grass should be equal to the amount eaten by x cows in 96 days, which is x*96.Therefore:x*96 = 1920Solve for x:x = 1920 / 96 = 20Wait, so 20 cows? That seems too straightforward. Let me verify again.Wait, but hold on. Let me check all steps once more.First, the equations were set up as:Initial grass + growth over days = cows * days.So:1. G + 24r = 70*24 = 16802. G + 60r = 30*60 = 1800Subtracting equation1 from equation2:36r = 120 => r = 10/3 ≈3.333.Then G = 1680 -24*(10/3)=1680 -80=1600. Correct.Then, for 96 days:Total grass = 1600 + 96*(10/3) =1600 + 320=1920.Total consumption by x cows in 96 days: 96x.Set equal: 96x=1920 =>x=20.Hmm, so according to this, 20 cows. But wait, the answer seems counterintuitive. If 70 cows take 24 days, and 30 cows take 60 days, then more days should require fewer cows. Since 96 is more than 60, and 30 cows take 60 days, then 96 days would require even fewer cows. So 20 is less than 30, which makes sense. So maybe 20 is correct.But wait, let's check another way. Maybe there's a different approach where you consider the rate of consumption versus the rate of growth.Let me think. Let's model the problem with differential equations. Let’s denote:- Let y(t) be the amount of grass at time t.- The grass grows at a rate of r per day.- The cows consume grass at a rate of c per day (c being the number of cows, assuming each cow eats 1 unit per day).So, the rate of change of grass is:dy/dt = r - cWe want to find c such that y(96) = 0, given that y(0) = G.But wait, integrating this:y(t) = G + (r - c) * tBut if this is the case, then setting y(t) = 0:G + (r - c) * t = 0 => c = (G + r * t)/t = G/t + rBut this seems different from the previous approach. Wait, but in the previous approach, we set G + r * t = c * t. Which is the same as this: G = (c - r) * tSo, rearranged, c = (G + r * t)/t = G/t + r. Wait, so according to this formula, c = G/t + r. So in the first case:c = 1600/24 + 10/3 ≈ 66.666 + 3.333 ≈ 70. So that works.Similarly, in the second case:c = 1600/60 +10/3 ≈26.666 +3.333≈30. Correct.Therefore, for 96 days:c =1600/96 +10/3 ≈16.666 +3.333≈20. So that gives 20. Therefore, this method also gives 20. Therefore, 20 cows. So that seems correct.But wait, this seems too straightforward. However, in the problem statement, it says "grass grows equally fast and dense". Maybe I need to consider that the grass might have a growth rate that depends on the current amount? But the problem says "grows equally fast and dense", which probably means the growth is constant, not exponential. So it's a constant rate r per day, independent of the current grass. Therefore, the model is linear.Therefore, the answer is 20.But let me check again with the two initial conditions.First case: 70 cows, 24 days.Total grass consumed:70*24=1680Total grass available: G +24r=1600 +24*(10/3)=1600 +80=1680. Correct.Second case:30 cows,60 days.30*60=1800.G +60r=1600 +60*(10/3)=1600 +200=1800. Correct.Third case: x cows,96 days.x*96=1600 +96*(10/3)=1600 +320=1920.Thus x=1920/96=20. So that's right.Therefore, the answer should be 20 cows.But let me check another way. Let's think about the rate of consumption versus the rate of growth.The idea is that the cows are eating the grass, but the grass is growing. So the net rate at which the grass is being depleted is (number of cows - growth rate). Wait, no: if the grass is growing at r per day, and the cows are eating c per day, then the net rate of change is (r - c). So if c > r, the grass is decreasing; if c < r, the grass is increasing.But in our problem, the grass is being eaten completely in a finite time. That requires that the cows eat the grass faster than it grows. Otherwise, if c <= r, the grass would never be depleted. So in our problem, since with 30 cows, it takes 60 days, which is finite, so 30 cows must be eating faster than the grass grows. But according to the calculation, r =10/3 ≈3.333. So 30 cows eat 30 units per day, which is way higher than the growth rate. But in our model, even if the cows eat more than the growth rate, the grass will be depleted in finite time. Wait, but in the model above, the grass is G +rt, and the consumption is c*t. So even if c < r, but G is positive, then you can have G + rt = c*t => G = (c - r)*t. But if c < r, then (c - r) is negative, so t would be negative, which is not possible. So actually, in reality, if c <= r, then the grass would never be depleted, because the growth is more than the consumption. Therefore, for the grass to be depleted, we must have c > r. So in our problem, since 30 cows can deplete the grass in 60 days, 30 cows must be greater than the growth rate. But according to our calculation, r=10/3≈3.333, so 30 cows is way higher. Therefore, our model is okay. So as long as c > r, the grass can be depleted in time t= G/(c - r). Wait, but according to our previous equations, G + rt = ct => t= G/(c - r). But in our case, we have t=24 when c=70, so 24=1600/(70 -10/3). Let's check:70 -10/3 ≈70 -3.333≈66.6661600 /66.666≈24. So correct.Similarly, t=60 when c=30: 1600/(30 -10/3)=1600/(80/3)=1600*3/80=60. Correct.Therefore, the formula t= G/(c - r). So solving for c: c= (G + rt)/t. Wait, but we can also express c= G/t + r. So c= G/t + r. Which is the same as the previous result.Therefore, in the case of 96 days, c=1600/96 +10/3≈16.666 +3.333≈20. Therefore, 20 cows.Thus, all checks out. So the answer is 20.Final AnswerThe number of cows needed to eat all the grass in 96 days is boxed{20}.

✅:Given that 70 cows can eat all the grass in 24 days and 30 cows can eat it in 60 days, we need to determine how many cows would eat all the grass in 96 days. We start by defining the variables:- ( G ) as the initial amount of grass.- ( r ) as the rate at which grass grows per day.- ( c ) as the number of cows.- ( d ) as the number of days.The total grass consumed by ( c ) cows in ( d ) days is ( c times d ). The total grass available is the initial grass plus the grass grown over ( d ) days, which is ( G + r times d ).Setting up the equations for the given scenarios:1. For 70 cows in 24 days: [ G + 24r = 70 times 24 ] [ G + 24r = 1680 ]2. For 30 cows in 60 days: [ G + 60r = 30 times 60 ] [ G + 60r = 1800 ]Subtracting the first equation from the second:[(G + 60r) - (G + 24r) = 1800 - 1680][36r = 120][r = frac{120}{36} = frac{10}{3} approx 3.333]Substituting ( r ) back into the first equation to find ( G ):[G + 24 left(frac{10}{3}right) = 1680][G + 80 = 1680][G = 1680 - 80 = 1600]For 96 days, the total grass available is:[G + 96r = 1600 + 96 left(frac{10}{3}right)][1600 + 320 = 1920]The number of cows ( x ) needed to consume this grass in 96 days is:[x times 96 = 1920][x = frac{1920}{96} = 20]Thus, the number of cows needed to eat all the grass in 96 days is (boxed{20}).

🔑:To solve this problem, we need to determine how many cows will consume all the grass in 96 days, given that the grass grows uniformly and steadily over time. Let ( x ) be the amount of grass that grows per day.# Step-by-Step Solution:1. Initial Setup: We start by setting the initial amount of grass to 1 unit. Let's denote the total grass eaten by the cows over a certain number of days taking into account the grass growth within those days.2. Formula for Grass Consumption: After ( t ) days, the total amount of grass available will be ( 1 + t cdot x ).3. Calculations Based on Provided Data: - For 70 cows in 24 days: [ text{Total grass eaten} = 1 + 24x ] Each cow consumes: [ frac{1 + 24x}{24 cdot 70} text{ units of grass per day} ] - For 30 cows in 60 days: [ text{Total grass eaten} = 1 + 60x ] Each cow consumes: [ frac{1 + 60x}{60 cdot 30} text{ units of grass per day} ]4. Equating Per Day Consumption for Both Situations: [ frac{1 + 24x}{24 cdot 70} = frac{1 + 60x}{60 cdot 30} ]5. Solving for ( x ): To solve for ( x ), we equate the two expressions: [ frac{1 + 24x}{1680} = frac{1 + 60x}{1800} ] Cross-multiplying to solve for ( x ): [ 1800(1 + 24x) = 1680(1 + 60x) ] Expanding both sides, we get: [ 1800 + 43200x = 1680 + 100800x ] Rearranging terms to isolate ( x ): [ 43200x - 100800x = 1680 - 1800 ] Simplify: [ -57600x = -120 ] Solving for ( x ): [ x = frac{1}{480} ]6. Calculating Grass Consumption per Day per Cow: Knowing ( x = frac{1}{480} ), we now calculate the grass consumed by one cow per day in terms of the initial amount of grass: [ text{Grass eaten per day by one cow} = frac{1 + 24 cdot frac{1}{480}}{1680} = frac{1 + frac{1}{20}}{1680} = frac{21}{20 cdot 1680} = frac{1}{1600} ]7. Determine the Number of Cows for 96 Days: We need to find how many cows will consume all the grass in 96 days. Let ( n ) be the number of cows: [ n cdot 96 cdot frac{1}{1600} = 1 + 96 cdot frac{1}{480} ] Simplifying: [ n cdot 96 cdot frac{1}{1600} = 1 + 0.2 = 1.2 ] [ n cdot frac{96}{1600} = 1.2 ] [ n cdot frac{3}{50} = 1.2 ] [ n = 1.2 cdot frac{50}{3} = 20 ] Conclusion:The number of cows that would consume all the grass in 96 days is:[boxed{20}]

🧠:Alright, let's try to tackle this soccer tournament problem step by step. It's a bit complex, but if I break it down, maybe I can figure it out. So, there are five teams: A, B, C, D, E. Each plays exactly once against every other team. That means each team plays 4 matches. The scoring is 3 points for a win, 1 for a draw, 0 for a loss. First, let's note the key points given:1. All teams have different points.2. Team A has the most points, even though they lost to Team B.3. Teams B and C didn't lose any games.4. Team C has fewer points than Team D.Our goal is to find the points for each team.Since each team plays 4 matches, the maximum points a team can get is 12 (4 wins). But given that there are 5 teams, the total number of matches is C(5,2) = 10 matches. Each match contributes either 3 points (if one team wins) or 2 points (if it's a draw) to the total points pool. So, total points can range from 20 (all draws) to 30 (all wins). But since B and C didn't lose any games, their matches must have all been wins or draws. Wait, but if B didn't lose any games, then in their 4 matches, they either won or drew. Similarly for C. But Team A lost to Team B. So in the match between A and B, B won, so B gets 3 points, A gets 0. But Team A still ended up with the highest points overall. That suggests that Team A won all their other matches except the one against B. Let's check.Wait, Team A has to have the most points. They lost to B, so they have at most 3 wins (since they can't win against B). If they lost only to B, then they have 3 wins and 1 loss. But wait, each team plays 4 matches. If they lost one (to B), then they have three other matches (against C, D, E). If they won all three, they would get 3*3=9 points. But if B didn't lose any games, B must have either won or drew all their matches. Since B already beat A, they have 3 points from that match. Then, in their other three matches (against C, D, E), B could have won or drawn. But since B didn't lose any matches, all their matches are wins or draws. So B's total points would be 3 (for beating A) plus points from matches against C, D, E. Let's think about B's maximum possible points. If B wins all their matches, they would have 4 wins: 4*3=12 points. But then they would have the highest points, but the problem says Team A has the most points. So that's a contradiction. Therefore, B must have drawn some matches. Since B can't lose, but they can draw. So B's total points would be 3 (from beating A) plus the results of their matches against C, D, E.Similarly, Team C didn't lose any games. So in all their matches (against A, B, D, E), they either won or drew. But Team C's points are less than D's. So Team D has more points than C.Also, all teams have different points. So the points must be unique integers from some range.Let me start by figuring out possible points for each team. Let's list the possible total points each team can have. Since each team plays 4 matches, possible points are 0,1,2,...,12, but with all different, and certain constraints.First, since B and C didn't lose any games, they have at least some points. Let's think about B first. B has at least the 3 points from beating A. Then, in their other three matches (against C, D, E), since they didn't lose any, they could have wins or draws. Let's denote B's results against C, D, E. Each of those matches can be either a win (3 points) or a draw (1 point). So B's total points would be 3 + sum of points from those three matches.Similarly, Team C didn't lose any games, so in their four matches (A, B, D, E), all are either wins or draws. Wait, but C's match against B must have been either a draw or a win for C. But B didn't lose any games, so if C played against B, the result must be a draw. Because B can't lose, so C can't win against B. Therefore, the match between B and C must have been a draw. Therefore, both B and C get 1 point from that match.So, in B's matches: B beat A (3 points), drew with C (1 point), and has two other matches against D and E. Since B didn't lose any, those could be wins or draws.Similarly, Team C's matches: drew with B (1 point), and their other three matches (A, D, E) are either wins or draws.But Team A lost only to B, so in their other three matches (C, D, E), they might have won all. If A won all three, that's 9 points. But let's check if that's possible.But wait, Team C didn't lose any games. So if A played against C, then the result must be either a win for A (which would mean C lost, but C can't lose) or a draw. Wait, contradiction! Team C didn't lose any games, so if A played against C, A couldn't have beaten C. Therefore, the match between A and C must have been a draw. Therefore, A's match against C was a draw, giving A 1 point and C 1 point. So Team A's results: lost to B (0 points), drew with C (1 point), and then played against D and E. If A won those two matches, they would get 3*2=6 points, plus the 1 from the draw with C, totaling 7 points. Wait, but that seems low. Wait, but wait, if A lost to B, drew with C, and won against D and E, that's two wins (6 points), one draw (1 point), one loss (0), total 7 points. But then B might have more points. Let's check B's possible points.B has beaten A (3 points), drew with C (1 point), and then has matches against D and E. If B won both of those, that's 3*2=6 points, plus 3 +1 = total 10 points. Then B would have 10 points, which is more than A's 7. But the problem says Team A has the most points. So this can't be. Therefore, there must be a mistake in my reasoning.Wait, so if A can't beat C (because C didn't lose any games), then A vs C must be a draw. Therefore, A has 1 point from that match. Then A's other matches: D and E. If A won both, that's 6 points, plus 1 from C, total 7. But B could have 3 (from beating A) +1 (draw with C) + results against D and E. If B drew both D and E, that's 2 points, so total 3+1+2=6. Then B would have 6, A has 7. Then A is higher. That works. But then C, who drew with B and drew with A, and their other matches against D and E. Since C didn't lose any games, those could be wins or draws. If C drew all matches, they would have 1 (vs A) +1 (vs B) + two draws (vs D and E) = 4 points. But D could have more points.Wait, but Team C has fewer points than D. So D must have more than C. Let's see. Let's try to map this out step by step.First, the matches:Each team plays 4 matches. Let's list all matches for each team:Team A: plays B, C, D, E.Result: lost to B, drew with C, and maybe won against D and E.Team B: plays A, C, D, E.Result: beat A, drew with C, and two matches against D and E.Team C: plays A, B, D, E.Result: drew with A and B, and two matches against D and E (either wins or draws).Team D: plays A, B, C, E.Team E: plays A, B, C, D.Now, let's consider Team C. Since they didn't lose any games, their matches against D and E must be either wins or draws. Let's assume they drew both. Then C's total points would be 1 (A) +1 (B) +1 (D) +1 (E) = 4 points. But D could have beaten E and maybe others. Wait, but D's points have to be higher than C's.Alternatively, if C won both against D and E, they would get 3+3=6 points from those, plus the two draws with A and B: total 6+2=8. Then C has 8, D must have more than 8. But D's points would depend on their results. Let's see.But let's start with possible points for each team, considering all constraints.First, since all teams have different points, and there are 5 teams, the points must be 5 distinct integers. The maximum possible points any team can have is 12, but given the constraints, it's probably lower.Let me list possible point distributions. Let's consider that Team A has the highest, then D > C, B and C didn't lose any games, and all have unique points.First, let's think about Team B. Since B didn't lose any games, their points come from wins and draws. B has at least 3 points (from beating A) and 1 point from drawing with C, so total so far 4. Then their remaining two matches (against D and E) could be wins or draws. Let's denote B's points as 3 (A) +1 (C) + x (D) + y (E), where x and y are either 3 or 1 (since no losses). So B's total points: 4 + x + y. Possible values:If B wins both D and E: 4 + 3 +3 =10 points.If B wins one and draws one: 4 +3 +1=8.If B draws both: 4 +1 +1=6.Similarly, Team A's points: they lost to B, drew with C, and need to have the highest total. So A's points come from matches against D and E. If A won both, that's 6 points, plus the draw with C (1), total 7. But if B has 6,8,10, then A's 7 would be higher than B's 6 but lower than 8 or 10. But the problem states Team A has the highest points. So if B can have 10, then A must have more than 10, which isn't possible because A's maximum is 3*3=9 (but they lost to B, so they can only have 3 wins). Wait, wait, A lost to B, drew with C, so they have two remaining matches (against D and E). If they won both, that's 6 points, plus the draw with C (1), total 7. So maximum possible for A is 7? But that can't be right because 7 is lower than B's possible 10. Contradiction. So there must be an error here.Wait, maybe I made a mistake in assuming that A drew with C. Let me double-check. Team C didn't lose any games. So when C plays A, the result must be a win for C or a draw. But if C won against A, then A lost to both B and C. But the problem only mentions that A lost to B. So does that mean A didn't lose to anyone else? The problem says "Team A scored the most points overall, although they lost to team B." It doesn't say they only lost to B, but given that they have the most points, it's likely they lost only to B. Otherwise, if they lost to others, their points would be lower. So maybe A only lost to B and won the rest. But wait, if A lost to B, and also drew with C (since C can't lose), then A's remaining matches against D and E must be wins. But if C didn't lose any games, then A's match against C must be a draw. Therefore, A's record is: L vs B, D vs C, W vs D, W vs E. Total points: 0 +1 +3 +3 =7.But then B's points: W vs A (3), D vs C (1), and then matches against D and E. If B wins both, B has 3+1+3+3=10. That would make B higher than A's 7, which contradicts the condition that A is highest. Therefore, this can't be. So there's a problem here. Therefore, my assumption that A drew with C must be wrong. Wait, but C cannot lose any games. So if A didn't draw with C, then C must have beaten A. But then A would have two losses (to B and C), which might make their points too low. Let's check.If A lost to B and C, then A's matches: 0 vs B, 0 vs C, and then against D and E. If A wins both, that's 6 points. Total 6. Then B's points: 3 vs A, and their other matches. If C beat A, then C has 3 points from A, and since C didn't lose any games, their other matches (B, D, E) must be draws or wins. But B didn't lose any games either, so C vs B must be a draw. Therefore, C has 3 (from A) +1 (from B), and then their matches against D and E. If C wins those, they get 3+3=6, total 3+1+6=10. But then C would have 10 points, which might be higher than D, but the problem says D has more than C. So if C has 10, D must have more than 10, which is impossible since maximum is 12. But if D also has high points, maybe. But this is getting complicated. Let me try to structure this.Alternative approach: Let's consider all matches and deduce results step by step.Total teams: A, B, C, D, E.Each plays 4 matches. Total matches: 10.Key points:1. All teams have different points.2. A has the most points, despite losing to B.3. B and C have no losses (so all their matches are wins or draws).4. D > C.Let's start with the match between A and B: B won, so B gets 3, A gets 0.Since B didn't lose any other matches, all B's other matches (vs C, D, E) are either wins or draws.Similarly, C didn't lose any matches, so all their matches (vs A, B, D, E) are wins or draws.But since A lost to B, and B didn't lose any, but A is supposed to have the highest points, A must have done very well in their other matches.But A's other matches are vs C, D, E. Let's see:Since C didn't lose any matches, the match between A and C must be a draw or a win for C. But if A lost to C, that's another loss for A, which would give A 0 points from that match. But then A's points would be from matches against D and E. If A wins both, that's 6 points. Plus maybe a draw with C? Wait, if A drew with C, then A has 1 point from that match. So total A: 0 (B) +1 (C) +3 (D) +3 (E) =7 points.But then B's points: 3 (A) + at least 1 (draw with C, since B didn't lose to C) and then points from D and E. If B draws with C, then B's points from C:1. Then B's total so far: 3+1=4. Then matches against D and E. If B wins both, total points:4+3+3=10. So B has 10, which is more than A's 7. Contradiction. Therefore, this scenario is invalid.Therefore, the only way for A to have higher points than B is if B doesn't get too many points from D and E. So B must have drawn some of those matches.Wait, let's try another angle. Since all teams have distinct points and A is highest, let's assume the points are in descending order: A > D > B > C > E (since D > C). But not sure yet. Let's think of possible point totals.Total points in the tournament can be calculated by considering all matches. Each match contributes 2 points if drawn, 3 if one team wins. Let's find total points.But we need to find the distribution such that all teams have distinct points, A > D > B > C > E (maybe), but B and C have no losses.Alternatively, maybe the order is A > D > B > C > E, but need to confirm.Alternatively, maybe A > D > B > E > C. But C has to be above E? No, the only condition is D > C. C could be anywhere below D.But let's think about B's matches:B has already beaten A (3 points). The match against C must be a draw (since B can't lose, and C can't lose). So B gets 1 point from C. Then B has matches against D and E. Let's denote those as W or D.Similarly, C's matches: drew with B, and matches against A, D, E. Since C didn't lose any, these must be W or D.But A's match against C: since C didn't lose, it's either a W for C or a D. But if C beat A, then A has another loss, which would lower A's points. Let's see:If C beat A: A has 0 from B, 0 from C, then needs to win against D and E for 6 points. Total A=6.But B would have 3 (A) +1 (C) + results against D and E. If B wins both D and E, that's 3+3=6, total B=3+1+6=10. Then A=6 < B=10. Not possible. So A must have drawn with C, giving A 1 point from that match. Thus, A's total: 0 (B) +1 (C) + results against D and E. If A wins both D and E, that's 6 points, total A=7.Now, B's total:3 (A) +1 (C) + results against D and E. Let's say B draws both D and E: 1+1=2, total B=3+1+2=6. Then A=7 > B=6. That works.Now, C's points: drew with A (1), drew with B (1), and matches against D and E. If C draws both, that's 1+1=2, total C=1+1+2=4. But D needs to have more than C. So D's points must be >4.D's matches: played against A, B, C, E.D lost to A (since A beat D), D's match against B: if B drew with D, then D gets 1 point from B. D's match against C: since C didn't lose, if C drew with D, then D gets 1 point. Then D's match against E: let's say D beat E, so 3 points. Total D:0 (A) +1 (B) +1 (C) +3 (E) =5. So D=5, which is > C=4. That works.E's matches: lost to A, lost to D, and matches against B and C. E's match against B: if B drew with E, E gets 1. Match against C: if C drew with E, E gets 1. So E's total:0 (A) +0 (D) +1 (B) +1 (C) =2. But all teams must have distinct points. So far:A=7, B=6, C=4, D=5, E=2. Wait, but D=5 and B=6. Then the order is A=7, B=6, D=5, C=4, E=2. But the problem states D > C, which is satisfied here (5 >4). However, all points are distinct, which is good. But wait, let's check if all matches are accounted for correctly.Let's verify each team's points:Team A: lost to B (0), drew with C (1), beat D (3), beat E (3). Total:0+1+3+3=7. Correct.Team B: beat A (3), drew with C (1), drew with D (1), drew with E (1). Total:3+1+1+1=6. Correct.Team C: drew with A (1), drew with B (1), drew with D (1), drew with E (1). Total:1+1+1+1=4. But wait, this would mean C drew all four matches, but the problem says teams have different points, and C has 4. However, D has 5, which is okay. But let's check Team D's results:Team D: lost to A (0), drew with B (1), drew with C (1), beat E (3). Total:0+1+1+3=5. Correct.Team E: lost to A (0), lost to D (0), drew with B (1), drew with C (1). Total:0+0+1+1=2. Correct.But wait, in this scenario, Team C drew all their matches. But the problem states that all teams have different points, which they do: 7,6,5,4,2. And C has fewer than D. This seems to satisfy all conditions.But wait, Team B has 6 points, which is more than D's 5. But the problem doesn't say anything about B's position relative to D, only that D > C. So this seems acceptable.However, let's check if there's another possible distribution. For instance, if C had more points by winning some matches, but still less than D.Suppose in C's matches against D and E, instead of drawing both, C won both. Then C's points would be 1 (A) +1 (B) +3 (D) +3 (E) =8. Then D's points would need to be higher than 8. Let's see.But D's matches: lost to A (0), drew with B (1), lost to C (0), and beat E (3). Total:0+1+0+3=4. But that's only 4, which is less than C's 8. That doesn't work. So this scenario is invalid.Alternatively, if C beat E and drew with D. Then C's points:1 (A) +1 (B) +1 (D) +3 (E)=6. Then D's points: lost to A (0), drew with B (1), drew with C (1), beat E (3). Total:0+1+1+3=5. So D=5, C=6. But C > D, which violates the condition. So invalid.Alternatively, C beats D and draws with E. Then C's points:1+1+3+1=6. D's points:0 (A) +1 (B) +0 (C) +3 (E)=4. Again, D=4 < C=6. Invalid.Alternatively, C draws with D and beats E. Then C=1+1+1+3=6. D=0+1+1+3=5. D=5 < C=6. Still invalid.Therefore, the only way for D to have more than C is if C's points are lower, which happens when C draws all matches. Then D can have 5, C=4.But wait, let's check if there's a scenario where C wins one match and draws others, but D still has more.Suppose C beats E and draws with D. Then C's points:1 (A) +1 (B) +1 (D) +3 (E)=6. D's points:0 (A) +1 (B) +1 (C) +3 (E)=5. Still D=5 < C=6. No good.Alternatively, C beats D and draws with E. Then C=1+1+3+1=6. D=0+1+0+3=4. D=4 < C=6. Still no.So it seems the only valid scenario is where C draws all matches, leading to C=4, D=5. Then the points are A=7, B=6, D=5, C=4, E=2. This satisfies all conditions:- All teams have distinct points.- A has the most points (7).- B and C didn't lose any games (B has 3 wins and 3 draws? Wait, in this scenario, B beat A, drew with C, D, and E. So B has 1 win and 3 draws, total 6 points. Correct, and no losses.- C has 4 points (four draws), D has 5 (one win, two draws, one loss).- D > C.Yes, this seems to fit.But wait, let's confirm the total points. Let's calculate total points from all matches.Matches:A vs B: B wins (3 points to B)A vs C: draw (1 each)A vs D: A wins (3 to A)A vs E: A wins (3 to A)B vs C: draw (1 each)B vs D: draw (1 each)B vs E: draw (1 each)C vs D: draw (1 each)C vs E: draw (1 each)D vs E: D wins (3 to D)Now, let's tally the points:A: 0 (B) +1 (C) +3 (D) +3 (E) =7B:3 (A) +1 (C) +1 (D) +1 (E) =6C:1 (A) +1 (B) +1 (D) +1 (E) =4D:0 (A) +1 (B) +1 (C) +3 (E) =5E:0 (A) +0 (D) +1 (B) +1 (C) =2Total points:7+6+4+5+2=24.Now, let's check if this matches the total points calculated from the matches. There are 10 matches. How many draws and wins are there?From the match results:Wins:B beats A (3)A beats D (3)A beats E (3)D beats E (3)Total wins:4, so 4*3=12 points from wins.Draws:A-C, B-C, B-D, B-E, C-D, C-E: total 6 draws, each contributing 2 points:6*2=12 points.Total points:12+12=24. Correct.So this adds up.Therefore, the final points are:A:7B:6C:4D:5E:2But wait, in the order, D has 5 and C has4, which is correct. All points are distinct and satisfy the conditions.But let me just check if there's another possible distribution where A has more than 7. Suppose A can get more points by not drawing with C. But earlier analysis shows that if A drew with C, that's necessary because C can't lose. If A didn't draw with C, then C must have beaten A, but that would give A another loss, making A's points even lower. For example, if A lost to B and C, then A's points would be 0+0+3+3=6, which is worse. So no.Therefore, the only valid solution is:A:7B:6D:5C:4E:2Yes, this seems to satisfy all conditions.

✅:Final Answer:Team ( A ) scored: (boxed{7})Team ( B ) scored: (boxed{6})Team ( C ) scored: (boxed{4})Team ( D ) scored: (boxed{5})Team ( E ) scored: (boxed{2})---Step-by-Step Explanation:1. Match Results and Constraints: - Each team plays 4 matches. - Teams ( B ) and ( C ) did not lose any games (all draws or wins). - Team ( A ) lost to ( B ) but has the highest points. - All teams have distinct points, and ( D > C ).2. Key Matches and Outcomes: - ( A ) vs ( B ): ( B ) wins (3 points to ( B ), 0 to ( A )). - ( A ) vs ( C ): Draw (1 point each) since ( C ) cannot lose. - ( A ) vs ( D ), ( A ) vs ( E ): ( A ) wins both (3 points each, total 6 for ( A )). - ( B ) vs ( C ): Draw (1 point each) since neither can lose. - ( B ) vs ( D ), ( B ) vs ( E ): Both draws (1 point each for ( B )). - ( C ) vs ( D ), ( C ) vs ( E ): Draws (1 point each for ( C )). - ( D ) vs ( E ): ( D ) wins (3 points to ( D ), 0 to ( E )).3. Points Calculation: - Team ( A ): ( 0 ) (loss to ( B )) + ( 1 ) (draw with ( C )) + ( 3 ) (win vs ( D )) + ( 3 ) (win vs ( E )) = 7. - Team ( B ): ( 3 ) (win vs ( A )) + ( 1 ) (draw with ( C )) + ( 1 ) (draw vs ( D )) + ( 1 ) (draw vs ( E )) = 6. - Team ( C ): ( 1 ) (draw with ( A )) + ( 1 ) (draw with ( B )) + ( 1 ) (draw vs ( D )) + ( 1 ) (draw vs ( E )) = 4. - Team ( D ): ( 0 ) (loss to ( A )) + ( 1 ) (draw with ( B )) + ( 1 ) (draw with ( C )) + ( 3 ) (win vs ( E )) = 5. - Team ( E ): ( 0 ) (loss to ( A )) + ( 0 ) (loss to ( D )) + ( 1 ) (draw with ( B )) + ( 1 ) (draw with ( C )) = 2.4. Verification: - All points are distinct: ( 7, 6, 5, 4, 2 ). - ( A ) has the highest points (7). - ( B ) and ( C ) have no losses. - ( D ) (5) > ( C ) (4).This distribution satisfies all given conditions.

🔑:1. Let's denote the number of points for teams A, B, C, D, and E as a, b, c, d, and e respectively.2. Given that each team played with every other team exactly once, and since there are 5 teams, the total number of matches played is: [ frac{5 times (5-1)}{2} = 10 ] This is calculated using the combination formula ( binom{n}{2} ).3. For each match, 3 points are awarded in total (3 points for a win or 1 point each for a draw). Therefore, if all 10 matches had a winner, the total points distributed would be: [ 10 times 3 = 30 ]4. By subtracting the number of drawn matches from 30, we get the total points distributed. Let’s assume x is the number of drawn matches: [ a + b + c + d + e = 30 - x ] where x is the number of drawn matches contributing 2 points each (since each draw gives 1 point to each team).5. From the problem's conditions: - All teams have a different number of points. - Team A scored the highest points even though it lost to team B. - Teams B and C did not lose any games. - Team C scored fewer points than team D.6. Teams B and C did not lose any matches, but given A > D > C in points and all teams have different point totals, this implies: [ c leq c_{text{max}} ] where c_{text{max}} = text{4} (since C plays draws and must finish below D). ] 7. Let’s investigate the points B and C hypothetically: [ b geq c Rightarrow text{let} b geq 6 and c geq 4 ]8. Now, from the given conditions and to ensure they are distinct: [ c < d Rightarrow text{then } a > d > c Rightarrow a,b,c,d are different and unique set ] [ b geq 6 and c may be initially considered out of `speciality next calcs ]9. Compute: [ a = text{7 } winning since b = 6 mean no drawn c = 4 = all drawn, it say d = text {5 }. then held computing e =2. 2 complement points awarded varying 0, full winning hence, (having variations possible ) with minimum fits in (6< points ) ) ]10. Now applying verifies fits seen in records earlier held drawn and playing criteria11. a = 7 b = 6 c = 4 d = 5 e = 2 ensuring non conflict combining drawn must verified points seen...12. Then cancelling substitute drawn counting seen held distinctively. ensuring validility a then affixing fits )13. Finally get yields the scores for: ensuring tallies correct verifying Thus, conclusions: Finally, the points scored by each team are:[boxed{a = 7, , b = 6, , c = 4, , d = 5, , e = 2}]

🧠:Okay, so I need to figure out how many zeros are at the end of the product of all natural numbers from 1 to 60. Hmm, right, the factorial of 60, written as 60!. The question is about trailing zeros. Let me remember how trailing zeros are formed in a number. Each trailing zero is created by multiplying 10, which is 2 times 5. So, essentially, each pair of 2 and 5 in the prime factors of the number contributes to a trailing zero. Therefore, the number of trailing zeros is determined by the number of such pairs in the prime factorization of 60!.But since there are usually more factors of 2 than 5 in a factorial, the number of trailing zeros is typically determined by the number of times 5 is a factor in the numbers from 1 to 60. So, I should count how many times the factor 5 appears in the numbers from 1 to 60. That makes sense.Let me recall the method for counting the number of times a prime number p divides into n!. It's called Legendre's formula, right? The formula is the sum of the floor of n divided by p, plus the floor of n divided by p squared, plus the floor of n divided by p cubed, and so on, until the division yields zero. So, applying that for p = 5 and n = 60.So, first step: floor(60 / 5) = floor(12) = 12. That counts the multiples of 5 between 1 and 60. Then, floor(60 / 25) = floor(2.4) = 2. That counts the multiples of 25 (since 25 is 5 squared) between 1 and 60. Then, floor(60 / 125) = floor(0.48) = 0. Since 125 is greater than 60, this term and all higher ones will be zero. So, adding those together: 12 + 2 = 14. Therefore, there should be 14 trailing zeros in 60!.Wait, let me verify that. Let me go through each step again to make sure I haven't missed anything.Starting with the multiples of 5: from 5 to 60, stepping by 5. There's 60 divided by 5, which is 12 numbers: 5, 10, 15, 20, 25, 30, 35, 40, 45, 50, 55, 60. Yep, 12 numbers. Each contributes at least one 5.Then, multiples of 25: those are numbers divisible by 25. In 1 to 60, those are 25, 50. So, two numbers. Each of these contributes an extra 5, so we need to add 2 more.Multiples of 125: 125 is 5 cubed, but 125 is greater than 60, so there are none. Therefore, no further contributions.So total number of 5s is 12 + 2 = 14. Then, since we need pairs of 2 and 5, but the number of 2s will be more than 14. Let me check that as well, just to confirm that the number of 2s doesn't limit the trailing zeros.Calculating the number of 2s in 60! using Legendre's formula:floor(60 / 2) = 30floor(60 / 4) = 15floor(60 / 8) = 7floor(60 / 16) = 3floor(60 / 32) = 1floor(60 / 64) = 0Adding those up: 30 + 15 = 45, 45 +7=52, 52+3=55, 55+1=56. So, 56 factors of 2. Definitely more than the 14 factors of 5, so the number of trailing zeros is indeed determined by the number of 5s, which is 14. Therefore, there should be 14 trailing zeros in 60!.But let me double-check this with a concrete example. Suppose I take a smaller factorial, say 10!.10! = 3628800, which has two trailing zeros. Using the formula:Number of 5s: floor(10/5) = 2, floor(10/25) = 0. Total 2. Which matches. The number of 2s would be floor(10/2) + floor(10/4) + floor(10/8) + ... = 5 + 2 + 1 = 8. So, 8 factors of 2 vs. 2 factors of 5. Trailing zeros = 2, correct.Another test: 25!.Number of 5s: floor(25/5)=5, floor(25/25)=1, floor(25/125)=0. Total 6. Number of 2s would be floor(25/2)=12, floor(25/4)=6, floor(25/8)=3, floor(25/16)=1, rest 0. Total 12+6+3+1=22. So trailing zeros=6. Indeed, 25! ends with six zeros. So the formula works here.Hence, applying it to 60! gives 14. Therefore, the answer should be 14.But just to make sure, maybe I can compute the exact number of trailing zeros by computing 60! and checking. However, 60! is an extremely large number, so that's impractical. Alternatively, maybe I can use logarithms to estimate the number of digits, but that also might not help. Alternatively, use a calculator or programming, but since I don't have access to that, I need to rely on the formula.Alternatively, think through another angle. Let's list all numbers from 1 to 60 and count the number of times 5 appears as a factor.Every multiple of 5 contributes at least one 5. So starting from 5:5: 110:115:120:125:2 (since 25=5x5)30:135:140:145:150:2 (50=2x5x5)55:160:1So total number of 5s: For numbers 5,10,15,20,30,35,40,45,55,60: each contributes 1, that's 10 numbers. Then 25 and 50 each contribute an extra 1, so total 10 + 2*1 = 12? Wait, wait. Wait, hold on. Wait, 25 and 50 are already included in the first list. Wait, no. Let me recount properly.Wait, let's list them all:Numbers from 1 to 60 that are divisible by 5:5,10,15,20,25,30,35,40,45,50,55,60. That's 12 numbers.Now, for each of these numbers, how many times does 5 divide into them?5: 5^1, so 110: 5^1, so 115:5^1, 120:5^1,125:5^2, so 230:5^1,135:5^1,140:5^1,145:5^1,150:5^2,255:5^1,160:5^1,1Now, adding up all these:Numbers contributing 1: 5,10,15,20,30,35,40,45,55,60. That's 10 numbers, each contributing 1, so 10.Numbers contributing 2: 25 and 50. That's 2 numbers, each contributing 2, so 4.Total factors of 5: 10 + 4 = 14. Yep, same as before. So that confirms it. Therefore, the number of trailing zeros is 14.Therefore, the answer is 14.Final AnswerThe number of trailing zeros in the product of all natural numbers from 1 to 60 is boxed{14}.

✅:To determine how many trailing zeros are in the product of all natural numbers from 1 to 60 (60!), we need to count the number of times 10 is a factor in the product. Each trailing zero is created by a pair of factors 2 and 5. Since there are usually more factors of 2 than 5, the number of trailing zeros is determined by the number of factors of 5.Using Legendre's formula, we count the factors of 5 in the numbers from 1 to 60:1. Count the multiples of 5: [ leftlfloor frac{60}{5} rightrfloor = 12 ]2. Count the multiples of 25 (since 25 contributes an extra factor of 5): [ leftlfloor frac{60}{25} rightrfloor = 2 ]3. Count the multiples of 125 (which is greater than 60, so this term is 0): [ leftlfloor frac{60}{125} rightrfloor = 0 ]Adding these together gives the total number of factors of 5:[12 + 2 = 14]Verification by listing all multiples of 5 and their contributions confirms this result. Each multiple of 5 contributes at least one factor of 5, and multiples of 25 contribute an additional factor. The total number of factors of 5 is indeed 14.Thus, the number of trailing zeros in 60! is (boxed{14}).

🔑:To determine how many zeros are at the end of the product of all natural numbers from 1 to 60, we can observe that zeros in the decimal representation of a number are produced by factors of 10. Each factor of 10 is the product of a prime factor 2 and a prime factor 5.Hence, the number of zeros at the end of the product is determined by the number of pairs of factors 2 and 5. Since factors of 2 are generally more abundant than factors of 5, we just need to count the number of factors of 5 in the decomposition of the factorial (60!).# Steps:Step 1: Identify the general formula for finding the number of occurrences of a prime ( p ) in ( n! ).[e_p(text{n!}) = leftlfloor frac{n}{p_1} rightrfloor + leftlfloor frac{n}{p_1^2} rightrfloor + leftlfloor frac{n}{p_1^3} rightrfloor + cdots]In this problem, ( p = 5 ) and ( n = 60 ).Step 2: Calculate how many times 5 is a factor in numbers 1 through 60.[leftlfloor frac{60}{5} rightrfloor + leftlfloor frac{60}{5^2} rightrfloor + leftlfloor frac{60}{5^3} rightrfloor + cdots]Step 3: Evaluate each term separately:[leftlfloor frac{60}{5} rightrfloor = leftlfloor 12 rightrfloor = 12][leftlfloor frac{60}{25} rightrfloor = leftlfloor 2.4 rightrfloor = 2][leftlfloor frac{60}{125} rightrfloor = leftlfloor 0.48 rightrfloor = 0]Since higher powers of 5 do not contribute (as (5^3 = 125) which is greater than 60), we can stop here.Step 4: Sum the results:[12 + 2 + 0 = 14]Therefore, there are 14 factors of 5 in the product of all natural numbers from 1 to 60. This means there will be 14 pairs of (2, 5) contributing to the number of trailing zeros. Conclusion:[boxed{14}]

🧠:Okay, let's try to tackle these two questions. The problem states that for any integer x, the trinomial ax² + bx + c is an integer. We need to determine two things: first, whether at least one of a, b, or c must be an integer, and second, whether all three coefficients a, b, and c must necessarily be integers.Starting with question (1): Is it necessary that at least one of the coefficients a, b, c is an integer?Hmm. So even if some coefficients are not integers, as long as for every integer x, the whole expression ax² + bx + c is an integer, does that force at least one of them to be an integer? Let's think.Suppose all three coefficients a, b, c are non-integers. But when we plug in any integer x, the combination ax² + bx + c becomes an integer. Is that possible?Let me try some examples. Maybe take specific values for a, b, c that are not integers, but when multiplied by x², x, and added to c, they result in integers for all integer x.Let’s consider fractions. For example, suppose a is 0.5, b is 0.5, and c is 0. Then the trinomial becomes 0.5x² + 0.5x. Let's test some integer x values.If x = 1: 0.5(1) + 0.5(1) = 0.5 + 0.5 = 1, which is integer.x = 2: 0.5(4) + 0.5(2) = 2 + 1 = 3, integer.x = 3: 0.5(9) + 0.5(3) = 4.5 + 1.5 = 6, integer.x = 0: 0 + 0 = 0, integer.x = -1: 0.5(1) + 0.5(-1) = 0.5 - 0.5 = 0, integer.Wait, this seems to work. But here, none of the coefficients are integers (a=0.5, b=0.5, c=0), but c is actually zero, which is an integer. So in this case, c is an integer, so this example doesn't satisfy the case where all coefficients are non-integers. Hmm.Let me adjust. Suppose c is also non-integer. Let's try a different set. Let's take a = 1/2, b = 1/2, c = 1/2. Then the trinomial is (1/2)x² + (1/2)x + 1/2. Let's test x=1: 0.5 + 0.5 + 0.5 = 1.5, which is not an integer. So that doesn't work. So that example fails.Wait, maybe we can find another combination where all a, b, c are non-integers, but ax² + bx + c is always integer. Let's think differently. Maybe a and b are fractions that when combined with integer x, produce integers. For example, suppose a = 1/3, b = 1/3, c = 1/3. Testing x=3: (1/3)(9) + (1/3)(3) + 1/3 = 3 + 1 + 0.333... = 4.333..., not integer. So that doesn't work.Alternatively, think of a and b such that a(x²) + b(x) is an integer for all integer x, and then c is adjusted to be an integer. Wait, but c is just a constant term. If a(x²) + b(x) is always integer, then c must be an integer because when x=0, the expression becomes c. Therefore, if x=0, the trinomial is c, which must be integer. So c must be an integer.Wait a minute! That seems important. If x=0, then ax² + bx + c = a*0 + b*0 + c = c. So c must be an integer because the problem states that for any integer x, the trinomial is an integer. Therefore, when x=0, c is the result, so c must be integer. Therefore, regardless of a and b, c is forced to be integer. Therefore, even if a and b are non-integers, c must be integer. Therefore, the answer to question (1) is yes, because c must be integer. Therefore, at least one coefficient (specifically c) must be integer.Wait, so maybe the answer to (1) is yes. Because when x=0, c has to be integer. So even if a and b are non-integers, c must be integer. Therefore, in all cases, c is integer, so at least one of the coefficients is necessarily integer.But let me check this reasoning again. If x=0, then trinomial is c. So c must be integer. Therefore, regardless of a and b, c must be integer. Therefore, answer to question (1) is yes, because c is always integer. Therefore, at least one coefficient (c) is necessarily integer.But then, moving to question (2): Do all coefficients a, b, c have to be integers?From the above, c must be integer. What about a and b?Suppose a and b are non-integers, but such that for all integer x, ax² + bx is integer. Then since c is integer, the whole trinomial would be integer.So, can we have a and b non-integers, but ax² + bx is integer for all integer x?For example, let me think. Let's take a = 0.5 and b = 0.5. Then ax² + bx = 0.5x² + 0.5x = 0.5x(x + 1). Since x and x + 1 are consecutive integers, one of them is even. Therefore, x(x + 1) is even, so 0.5 times even is integer. Therefore, 0.5x(x + 1) is integer. Therefore, in this case, a and b are both 0.5, which are non-integers, but ax² + bx is integer for all integer x. Then, adding c which is integer (since c must be integer from x=0), the entire trinomial is integer.Therefore, in this case, c is integer, but a and b are non-integers, yet the trinomial is integer for all integer x. Therefore, answer to question (2) is no; not all coefficients have to be integers.So, to recap:For question (1): Since x=0 gives c, which must be integer, so yes, at least one coefficient (c) must be integer.For question (2): The example with a = 0.5, b = 0.5, c = integer shows that a and b can be non-integers as long as their combination with x² and x results in integers, so no, all coefficients don't have to be integers.But wait, let me check another example to be sure. Suppose a = 1/3 and b = 2/3. Then, ax² + bx = (1/3)x² + (2/3)x. Let's test x=1: (1/3) + (2/3) = 1, integer. x=2: (4/3) + (4/3) = 8/3, which is not integer. So that doesn't work. So the previous example with a and b = 0.5 is special. Because x(x + 1) is always even, so 0.5x(x + 1) is integer.Another example: Let a = sqrt(2), b = -sqrt(2), c = integer. Then, ax² + bx + c = sqrt(2)x² - sqrt(2)x + c. Let x=1: sqrt(2) - sqrt(2) + c = c, which is integer. x=2: 4sqrt(2) - 2sqrt(2) + c = 2sqrt(2) + c. But 2sqrt(2) is irrational, so unless c cancels it, but c is integer. So this would not be integer. Therefore, this example doesn't work. So my previous example with a=0.5, b=0.5, c integer seems to be a valid case where a and b are non-integers, but the trinomial is always integer.Therefore, to answer question (2), the coefficients a, b don't have to be integers as long as their combination for any integer x yields an integer. So in the case of a = 0.5, b = 0.5, c integer, this works. Therefore, the answer to (2) is no.But let's think if there are other possibilities. For instance, if a is irrational but such that a(x²) + b(x) is integer for all integer x. Is that possible?Suppose a = 1/2 and b = 1/2, as before. Then 0.5x² + 0.5x = 0.5x(x + 1). Since x and x +1 are consecutive integers, one is even, so x(x +1 ) is even, so 0.5 times even is integer. Therefore, this works. So even though a and b are fractions, the combination works. So maybe the key is that a and b are rational numbers such that when multiplied by x² and x respectively, their sum is integer.But what if a and b are irrational? For example, suppose a = sqrt(2) - sqrt(2), which is 0. Then a is 0, b could be any non-integer, but a=0, which is integer. Wait, but if a=0, then the trinomial becomes bx + c. Since for all integer x, bx + c is integer. Since c is integer (from x=0), then bx must be integer for all integer x. Therefore, b must be rational with denominator 1, i.e., integer. Because if b is a non-integer rational number, say 1/2, then when x=1, bx = 1/2, which is not integer. Therefore, if a=0, then b must be integer. So in that case, a=0 (integer), b integer, c integer. So all coefficients are integers.But in the earlier case where a and b are 0.5, which are non-integers, but when combined, 0.5x(x +1 ) is integer. So that seems acceptable.Therefore, in order for ax² + bx + c to be integer for all integer x, c must be integer (from x=0), and ax² + bx must be integer for all integer x. Therefore, the question reduces to: can ax² + bx be integer for all integer x even if a and b are non-integers?The example shows yes. So in that case, answer to (2) is no. But what constraints must a and b satisfy?Let’s think algebraically. Suppose we have ax² + bx must be integer for all integer x, given that c is integer.Let’s set x = 1: a + b must be integer.x = 2: 4a + 2b must be integer.x = -1: a + (-b) = a - b must be integer.x = 3: 9a + 3b must be integer.We can set up equations.Let’s denote:For x=1: a + b = k (integer)x=2: 4a + 2b = m (integer)x=-1: a - b = n (integer)We can solve these equations.From x=1 and x=-1:a + b = ka - b = nAdding these: 2a = k + n ⇒ a = (k + n)/2Subtracting: 2b = k - n ⇒ b = (k - n)/2Therefore, a and b must be averages of integers, so they must be rational numbers with denominator dividing 2. So a and b must be half-integers (either integers or half-integers).Then, check for x=2: 4a + 2b = 4*( (k + n)/2 ) + 2*( (k - n)/2 ) = 2(k + n) + (k - n) = 3k + n.But 3k + n must be integer. Since k and n are integers, this is integer. Similarly, for x=3:9a + 3b = 9*( (k + n)/2 ) + 3*( (k - n)/2 ) = (9k + 9n + 3k - 3n)/2 = (12k + 6n)/2 = 6k + 3n, which is integer.Similarly, for x=4: 16a + 4b = 16*( (k + n)/2 ) + 4*( (k - n)/2 ) = 8(k + n) + 2(k - n) = 10k + 6n, which is integer.Therefore, if a and b are half-integers (of the form integer/2), then ax² + bx will be integer for all integer x. Because for any integer x, ax² + bx = ( (k + n)/2 )x² + ( (k - n)/2 )x = (k x² + n x² + k x - n x)/2 = (k x(x +1) + n x(x -1))/2. Since x(x +1) and x(x -1) are both products of consecutive integers, hence even, so divided by 2 gives integers. Therefore, this combination works.Therefore, the necessary condition for a and b is that they are rational numbers with denominator dividing 2, i.e., a and b are half-integers. Therefore, a and b don't need to be integers themselves, but they must be half-integers. For example, a=0.5, b=0.5 works, a=1.5, b=0.5, etc.Therefore, the answers are:(1) Yes, because c must be integer.(2) No, because a and b can be half-integers (non-integers) as long as their combination yields integers for all integer x.

✅:Answer:(1) Yes, at least one coefficient must be an integer. *Reasoning:* When ( x = 0 ), the trinomial evaluates to ( c ). Since the problem states the trinomial is an integer for all integer ( x ), ( c ) must be an integer. Thus, ( c ) is always an integer, satisfying the requirement that at least one coefficient is an integer.(2) No, all coefficients ( a, b, c ) do not need to be integers. *Reasoning:* While ( c ) must be an integer, ( a ) and ( b ) can be non-integers if their combination ensures ( ax^2 + bx ) is an integer for all integer ( x ). For example, let ( a = frac{1}{2} ) and ( b = frac{1}{2} ). For any integer ( x ), ( frac{1}{2}x^2 + frac{1}{2}x = frac{1}{2}x(x+1) ), which is always an integer because ( x(x+1) ) is even. This shows ( a ) and ( b ) can be non-integers while the trinomial remains integer-valued for all integer ( x ).Final Answers: (1) (boxed{text{Yes}}) (2) (boxed{text{No}})

🔑:# Statement (1): There is at least one integer among a, b, and c.1. Given Information: For any integer x, the value of the trinomial a x^{2} + b x + c is an integer. That is, [ a x^{2} + b x + c in mathbb{Z} ]2. Taking a Special Case: Consider x = 0. [ a cdot 0^{2} + b cdot 0 + c = c ] 3. Conclusion from the Special Case: From our calculation, we see that: [ c in mathbb{Z} ] Hence, (c) is an integer. Therefore, among the coefficients (a, b, c), there is at least one integer, specifically (c). Thus, we can conclude:[boxed{text{Yes, at least one of } a, b, text{ and } c text{ is an integer.}}]# Statement (2): All coefficients a, b, c are not necessarily integers.1. Understanding the Requirement: We need to check if (a, b, c) must all be integers. Let's consider a specific form of the trinomial for counterexample.2. Counterexample: Consider the polynomial: [ f(x) = frac{1}{2} x^{2} + frac{1}{2} x + 0 ]3. Checking Polynomial for Integer Values: Calculate ( f(x) ) for any integer (x): [ f(x) = frac{1}{2} x^{2} + frac{1}{2} x = frac{x(x+1)}{2} ]4. Analyzing ( f(x) = frac{x(x+1)}{2} ): - For any integer (x), the product (x(x+1)) is always even. - Thus, ( frac{x(x+1)}{2} ) is always an integer.5. Conclusion from the Counterexample: Therefore, for all integer (x): [ frac{1}{2} x^{2} + frac{1}{2} x in mathbb{Z} ] This shows that the polynomial can take integer values for integer (x) without (a) or (b) necessarily being integers.Thus, we can conclude:[boxed{text{No, } a, b, text{ and } c text{ do not all need to be integers.}}]

🧠:Okay, let's try to solve this problem. We need to find the maximum and minimum values of |z₁| given that |z₁·z₂| = 3 and z₁ + z₂ = 2i. Hmm, complex numbers can sometimes be tricky, but maybe if I break it down step by step, it will make more sense.First, let me recall some properties of complex numbers. The modulus of a product is the product of the moduli, so |z₁·z₂| = |z₁|·|z₂|. The problem states that this is equal to 3. So, |z₁|·|z₂| = 3. That might come in handy.We also know that z₁ + z₂ = 2i. If I can express z₂ in terms of z₁, maybe I can substitute it into the modulus equation. Let's try that. Let me denote z₂ = 2i - z₁. Then, substitute this into the modulus product equation.So, |z₁|·|z₂| = |z₁|·|2i - z₁| = 3. Therefore, |2i - z₁| = 3 / |z₁|. Hmm, maybe I can let |z₁| = r, then the equation becomes |2i - z₁| = 3 / r.Now, if I consider z₁ as a complex number, let's write it in terms of real and imaginary parts. Let z₁ = x + yi, where x and y are real numbers. Then z₂ = 2i - z₁ = -x + (2 - y)i.The modulus of z₁ is |z₁| = √(x² + y²) = r. The modulus of z₂ is |z₂| = √(x² + (2 - y)²). Then, according to the given product modulus, we have:√(x² + y²) * √(x² + (2 - y)²) = 3.But this seems a bit complicated with two variables. Maybe there's a geometric interpretation here. Since z₁ and z₂ are complex numbers, they can be represented as points in the plane. The sum z₁ + z₂ = 2i means that the vector from the origin to z₂ is the reflection of z₁ over the midpoint of the segment connecting 0 and 2i. Wait, the midpoint of z₁ and z₂ is (z₁ + z₂)/2 = (2i)/2 = i. So, z₂ is the reflection of z₁ over the point i. That might not be directly helpful, but it's an interesting observation.Alternatively, since z₂ = 2i - z₁, maybe I can consider the modulus |2i - z₁| = 3/r. Let me write this modulus squared to eliminate the square roots. So, |2i - z₁|² = (3/r)².Expanding the left side: |2i - z₁|² = | -z₁ + 2i |² = |z₁ - (-2i)|² = (x - 0)² + (y - 2)² = x² + (y - 2)².So, we have x² + (y - 2)² = 9 / r².But we also know that r² = x² + y². Let's write both equations:1. x² + y² = r²2. x² + (y - 2)² = 9 / r²Subtracting equation 1 from equation 2 gives:x² + (y - 2)² - x² - y² = 9/r² - r²Simplifying the left side:(y - 2)² - y² = 9/r² - r²Expanding (y - 2)²:y² - 4y + 4 - y² = 9/r² - r²So, -4y + 4 = 9/r² - r²Rearranging terms:-4y = 9/r² - r² - 4Divide both sides by -4:y = (r² + 4 - 9/r²)/4Hmm, this seems a bit messy, but maybe we can substitute back into equation 1. Let me express y in terms of r and plug into x² + y² = r².From equation 1: x² = r² - y²But we need to express x² in terms of r. Alternatively, let's use the expression for y:y = (r² + 4 - 9/r²)/4Then, substitute this into x² + y² = r². Let's compute y²:y² = [(r² + 4 - 9/r²)/4]^2So,x² + [(r² + 4 - 9/r²)/4]^2 = r²But x² = r² - y², so:r² - [(r² + 4 - 9/r²)/4]^2 + [(r² + 4 - 9/r²)/4]^2 = r²Wait, that cancels out. Hmm, maybe this approach isn't working. Perhaps I need another strategy.Alternatively, let's consider that both z₁ and z₂ lie on some circles. Since z₂ = 2i - z₁, then if we fix z₁, z₂ is determined. The product of their moduli is 3. Maybe using the triangle inequality or some geometric approach.Wait, the problem is asking for the maximum and minimum of |z₁|. Let me denote |z₁| = r, so |z₂| = 3/r. Then, we have two complex numbers z₁ and z₂ with moduli r and 3/r, respectively, such that their sum is 2i.This seems similar to the problem of finding possible lengths of sides of a triangle with fixed sum and product of lengths? But in this case, the sum is a vector sum, not scalar. Hmm.Alternatively, using the law of cosines. If we think of z₁ and z₂ as vectors adding up to 2i, then we can model this as a triangle with sides of length r, 3/r, and hypotenuse (or rather, the length of the resultant vector) |2i| = 2. Wait, but 2i has modulus 2, so the length of the sum is 2. So, in the complex plane, vectors z₁ and z₂ add up to a vector of length 2. The moduli of z₁ and z₂ are r and 3/r. So, using the triangle inequality.Wait, the triangle inequality tells us that |z₁ + z₂| ≤ |z₁| + |z₂|. Here, |z₁ + z₂| = 2, and |z₁| + |z₂| = r + 3/r. So, 2 ≤ r + 3/r. Similarly, the reverse triangle inequality says that | |z₁| - |z₂| | ≤ |z₁ + z₂|. So, |r - 3/r| ≤ 2.Therefore, we have two inequalities:1. r + 3/r ≥ 22. |r - 3/r| ≤ 2These can help us find the possible values of r. Let's solve these inequalities.First inequality: r + 3/r ≥ 2.Multiply both sides by r (assuming r > 0, since modulus is non-negative):r² + 3 ≥ 2rRearranged:r² - 2r + 3 ≥ 0The quadratic equation r² - 2r + 3 = 0 has discriminant 4 - 12 = -8 < 0, so the quadratic is always positive. Therefore, the inequality r + 3/r ≥ 2 holds for all r > 0. So that doesn't restrict r.Second inequality: |r - 3/r| ≤ 2This splits into two inequalities:-2 ≤ r - 3/r ≤ 2Let's solve the left inequality first: r - 3/r ≥ -2Multiply both sides by r (again, r > 0):r² - 3 ≥ -2rr² + 2r - 3 ≥ 0Solving r² + 2r - 3 = 0: discriminant is 4 + 12 = 16, roots at (-2 ± 4)/2. So (2)/2=1 and (-6)/2=-3. Since r > 0, the critical point is r = 1. The quadratic opens upwards, so the inequality holds when r ≤ -3 or r ≥ 1. But since r > 0, the inequality r² + 2r - 3 ≥ 0 holds when r ≥ 1.Now the right inequality: r - 3/r ≤ 2Multiply by r:r² - 3 ≤ 2rr² - 2r - 3 ≤ 0Solving r² - 2r - 3 = 0: discriminant 4 + 12 = 16, roots (2 ± 4)/2, so 3 and -1. Since r > 0, the critical point is r = 3. The quadratic opens upwards, so the inequality holds when -1 ≤ r ≤ 3. Since r > 0, this translates to 0 < r ≤ 3.Combining both parts of the inequality |r - 3/r| ≤ 2, we have:From the left inequality: r ≥ 1From the right inequality: r ≤ 3Therefore, r ∈ [1, 3]So the possible values of |z₁| = r are between 1 and 3. But we need to check if these bounds are achievable. That is, does there exist z₁ and z₂ satisfying the given conditions when r = 1 or r = 3?Let's check for r = 1. Then |z₁| = 1, |z₂| = 3/1 = 3. The sum z₁ + z₂ = 2i. So we need two complex numbers with moduli 1 and 3 adding up to 2i. Let's see if this is possible.Geometrically, this would mean that the vectors z₁ (length 1) and z₂ (length 3) add up to a vector of length 2 pointing in the imaginary direction. In the complex plane, we can represent this as two vectors forming a triangle with sides 1, 3, and 2. Wait, but in the triangle inequality, the sum of two sides must be greater than the third. 1 + 3 = 4 > 2, so that's okay. The difference of two sides must be less than the third: |3 - 1| = 2, which is equal to the third side. So this would be a degenerate triangle, where the vectors are colinear but pointing in opposite directions. However, in our case, the sum is 2i, which is non-degenerate. Hmm, perhaps this is a straight line case.Wait, if |z₁| = 1, |z₂| = 3, and |z₁ + z₂| = 2. Let me check if such vectors exist.Using the law of cosines:|z₁ + z₂|² = |z₁|² + |z₂|² + 2|z₁||z₂|cosθ, where θ is the angle between z₁ and z₂.So, 2² = 1² + 3² + 2*1*3*cosθ4 = 1 + 9 + 6cosθ4 = 10 + 6cosθThen, 6cosθ = 4 - 10 = -6cosθ = -1So θ = π radians. So z₁ and z₂ are in opposite directions. But their sum is 2i. So if they are colinear but opposite, how can their sum be 2i?Wait, if z₁ and z₂ are colinear and opposite, then their sum would be along the line they are colinear with. So if their sum is 2i, they must be colinear along the imaginary axis. So let's suppose z₁ is along the positive imaginary axis: z₁ = ai, then z₂ = 2i - ai = (2 - a)i. Then |z₁| = |a| = 1, so a = 1 or -1. If a = 1, then z₁ = i, z₂ = i. Then |z₂| = 1, but we need |z₂| = 3. So that doesn't work. If a = -1, z₁ = -i, z₂ = 3i. Then |z₁| = 1, |z₂| = 3, and their sum is 2i. That works! So when r = 1, z₁ = -i, z₂ = 3i. Therefore, r = 1 is achievable.Similarly, check r = 3. Then |z₂| = 3 / 3 = 1. Similarly, we can have z₁ = 3i, z₂ = -i. Then z₁ + z₂ = 2i. So that also works. Thus, r = 3 is achievable.Therefore, the minimum value of |z₁| is 1 and the maximum is 3.Wait, but just to confirm, let's make sure there are no other extrema between 1 and 3. Maybe using calculus to see if there are other maximum or minima.Let me consider r as a variable between 1 and 3, and perhaps find the possible values. Wait, but the problem is to find the maximum and minimum of |z₁|, which we have found to be 1 and 3. But let me verify using another approach.Alternatively, let's parametrize z₁ and z₂. Since z₁ + z₂ = 2i, let me write z₁ = i + t, z₂ = i - t, where t is a complex number. Then, z₁ + z₂ = (i + t) + (i - t) = 2i, which satisfies the sum condition.Now, compute |z₁·z₂| = |(i + t)(i - t)| = |i² - t²| = |-1 - t²| = |1 + t²|. The problem states that this modulus is 3. Therefore, |1 + t²| = 3.Let me write t = a + bi, where a and b are real numbers. Then t² = (a + bi)² = a² - b² + 2abi. So 1 + t² = (1 + a² - b²) + 2abi. Then |1 + t²| = √[(1 + a² - b²)^2 + (2ab)^2] = 3.Simplify the expression inside the square root:(1 + a² - b²)^2 + (2ab)^2Expand (1 + a² - b²)^2:= 1 + 2a² - 2b² + a^4 + b^4 - 2a²b²And (2ab)^2 = 4a²b²Add them together:1 + 2a² - 2b² + a^4 + b^4 - 2a²b² + 4a²b²Simplify:1 + 2a² - 2b² + a^4 + b^4 + 2a²b²Notice that a^4 + 2a²b² + b^4 = (a² + b²)^2So the expression becomes:1 + 2a² - 2b² + (a² + b²)^2Let me denote s = a² + b². Then the expression is 1 + 2a² - 2b² + s²But 2a² - 2b² = 2(a² - b²) = 2( (a² + b²) - 2b² ) = 2s - 4b². Hmm, maybe not helpful. Alternatively, since s = a² + b², then we can express a² = s - b². So 2a² - 2b² = 2(s - b²) - 2b² = 2s - 4b². Still complicated.Alternatively, perhaps we can write in terms of s and another variable. Let me see:We have:1 + 2a² - 2b² + (a² + b²)^2 = 3² = 9So,(a² + b²)^2 + 2a² - 2b² + 1 = 9Let s = a² + b². Then:s² + 2(a² - b²) + 1 = 9But a² - b² = (a² + b²) - 2b² = s - 2b², or alternatively, a² - b² = t where t is some variable. But maybe this approach isn't helping. Let's try another angle.We have |1 + t²| = 3. Let me consider t as a complex number. Then 1 + t² is a complex number with modulus 3. So t² is a complex number with modulus 3 and such that 1 + t² has modulus 3. Wait, maybe this is a circle in the complex plane.Alternatively, set t² = w, so the equation becomes |1 + w| = 3. So w lies on a circle of radius 3 centered at -1 in the complex plane. Also, since t is a complex number, w = t² can be any complex number except with a restriction based on t's modulus. Wait, but t can be any complex number, so w can be any complex number (since squaring covers the entire complex plane except for the negative real axis if we consider t as real, but with t complex, w can indeed be any complex number). Hmm, but maybe not. Wait, t² can be any complex number. For any complex number w, there exists a t such that t² = w. So w can indeed be any complex number. Therefore, the equation |1 + w| = 3 is satisfied by all w on the circle with center at -1 and radius 3. Then, t is any complex number such that t² is on this circle.But perhaps this is overcomplicating. Let's try to parametrize t. Suppose t is a real number. Then t² is real and non-negative. Then 1 + t² is real, so |1 + t²| = |1 + t²| = 1 + t². Setting this equal to 3 gives t² = 2, so t = √2 or -√2. Then z₁ = i + √2, z₂ = i - √2. Then |z₁| = √( (√2)^2 + 1^2 ) = √(2 + 1) = √3 ≈ 1.732. Similarly for z₂.But this gives |z₁| = √3, which is between 1 and 3, so maybe this is a possible value but not the maximum or minimum. Hmm. If t is purely imaginary, let t = bi, where b is real. Then t² = -b². Then 1 + t² = 1 - b², which is real. |1 + t²| = |1 - b²| = 3. So 1 - b² = ±3. If 1 - b² = 3, then b² = -2, which is impossible. If 1 - b² = -3, then b² = 4, so b = ±2. Therefore, t = ±2i. Then z₁ = i + 2i = 3i, |z₁| = 3, or z₁ = i - 2i = -i, |z₁| = 1. So this gives us the maximum and minimum values we found earlier. So when t is purely imaginary, we reach the extremes. When t is real, we get |z₁| = √3. Therefore, the maximum and minimum are indeed 3 and 1.Alternatively, consider the parametrization using angles. Let me write z₁ = r e^{iθ}, then z₂ = 2i - r e^{iθ}. Then |z₂| = |2i - r e^{iθ}|. Then the product |z₁|·|z₂| = r·|2i - r e^{iθ}| = 3. So we have:r·|2i - r e^{iθ}| = 3Let me square both sides:r²·|2i - r e^{iθ}|² = 9Compute |2i - r e^{iθ}|²:Express 2i as 0 + 2i, and r e^{iθ} as r cosθ + i r sinθ. Therefore,|2i - r e^{iθ}|² = | - r cosθ + i(2 - r sinθ) |² = ( - r cosθ )² + (2 - r sinθ )² = r² cos²θ + (2 - r sinθ )²Expanding the second term:4 - 4 r sinθ + r² sin²θTherefore, total:r² cos²θ + 4 - 4 r sinθ + r² sin²θ = r² (cos²θ + sin²θ) + 4 - 4 r sinθ = r² + 4 - 4 r sinθThus, we have:r² (r² + 4 - 4 r sinθ ) = 9So:r² (r² + 4 - 4 r sinθ ) = 9But this equation relates r and θ. To find the extrema of r, we need to consider the possible values of θ.Let me rearrange the equation:r² (r² + 4) - 4 r³ sinθ = 9But sinθ varies between -1 and 1. Therefore, for a fixed r, the left-hand side can take different values depending on θ. To satisfy the equation, the term -4 r³ sinθ must adjust so that the entire left-hand side equals 9.But since we are looking for real solutions, the equation must hold for some θ. Therefore, for a given r, the equation:r² (r² + 4) - 4 r³ sinθ = 9must have a solution for sinθ. Let's solve for sinθ:-4 r³ sinθ = 9 - r² (r² + 4)sinθ = [ r² (r² + 4) - 9 ] / (4 r³ )Since sinθ must lie in [-1, 1], we have:-1 ≤ [ r² (r² + 4) - 9 ] / (4 r³ ) ≤ 1Multiply numerator and denominator by 1/(4 r³):So,-1 ≤ [ r^4 + 4 r² - 9 ] / (4 r³ ) ≤ 1Multiply all parts by 4 r³ (assuming r > 0):-4 r³ ≤ r^4 + 4 r² - 9 ≤ 4 r³Let me split into two inequalities:1. r^4 + 4 r² - 9 ≤ 4 r³2. -4 r³ ≤ r^4 + 4 r² - 9First inequality:r^4 + 4 r² - 9 - 4 r³ ≤ 0r^4 - 4 r³ + 4 r² - 9 ≤ 0Let me write this as r^4 -4 r³ +4 r² -9 ≤0Second inequality:-4 r³ ≤ r^4 +4 r² -9Which rearranged is:0 ≤ r^4 +4 r² -9 +4 r³r^4 +4 r³ +4 r² -9 ≥0Hmm, these are quartic inequalities. This seems complicated, but maybe we can factor them.First, let's tackle the first inequality: r^4 -4 r³ +4 r² -9 ≤0Let me try to factor this polynomial. Let's look for rational roots using the Rational Root Theorem. Possible roots are ±1, ±3, ±9.Test r=1: 1 -4 +4 -9 = -8 ≠0r=3: 81 - 108 + 36 -9 = 0. Oh, r=3 is a root.So we can factor (r -3) out. Let's perform polynomial division.Divide r^4 -4 r³ +4 r² -9 by (r -3):Using synthetic division:3 | 1 -4 4 0 -9 3 -3 3 9 1 -1 1 3 0So the quotient is r³ - r² + r +3. Therefore, the polynomial factors as (r -3)(r³ - r² + r +3)Now let's factor the cubic polynomial r³ - r² + r +3.Again, use Rational Root Theorem. Possible roots: ±1, ±3.Test r=-1: (-1)^3 - (-1)^2 + (-1) +3 = -1 -1 -1 +3 =0. So r=-1 is a root.Factor out (r +1):Using synthetic division:-1 | 1 -1 1 3 -1 2 -3 1 -2 3 0So the cubic factors as (r +1)(r² -2r +3). Therefore, the original polynomial factors as (r -3)(r +1)(r² -2r +3)The quadratic r² -2r +3 has discriminant 4 -12 = -8 <0, so it doesn't factor over reals. Therefore, the first inequality becomes:(r -3)(r +1)(r² -2r +3) ≤0Since r >0 (as modulus), we can ignore the factor (r +1) which is positive. The quadratic factor r² -2r +3 is always positive. Therefore, the sign of the entire expression is determined by (r -3). So (r -3) ≤0, which gives r ≤3.So the first inequality gives r ≤3.Now the second inequality: r^4 +4 r³ +4 r² -9 ≥0Again, try to factor. Let's check possible roots. Try r=1: 1 +4 +4 -9=0. So r=1 is a root.Factor out (r -1). Wait, but r=1 gives 0. Wait, wait: Let's check:r=1: 1 +4 +4 -9=0. Yes, so (r -1) is a factor.Perform polynomial division:Divide r^4 +4 r³ +4 r² -9 by (r -1). Wait, but if r=1 is a root, then synthetic division:1 | 1 4 4 0 -9 1 5 9 9 1 5 9 9 0Therefore, the quotient is r³ +5 r² +9 r +9.So the polynomial factors as (r -1)(r³ +5 r² +9 r +9)Try to factor the cubic. Check r=-3:(-3)^3 +5*(-3)^2 +9*(-3) +9= -27 +45 -27 +9=0. So r=-3 is a root.Factor out (r +3):Using synthetic division:-3 | 1 5 9 9 -3 -6 -9 1 2 3 0So the cubic factors as (r +3)(r² +2r +3). Therefore, the original polynomial factors as (r -1)(r +3)(r² +2r +3)Since r >0, (r +3) is positive, and r² +2r +3 is always positive. So the sign of the expression is determined by (r -1). Therefore, the inequality (r -1) ≥0, since the other factors are positive. Therefore, r ≥1.Thus, the second inequality gives r ≥1.Combining both inequalities, from the first we have r ≤3, from the second r ≥1. Hence, r ∈ [1,3], which matches our previous result.Therefore, the minimum value of |z₁| is 1 and the maximum is 3.To confirm, let's check when r=1 and r=3:- For r=1: z₁ = -i, z₂ = 3i. Then |z₁·z₂| = |(-i)(3i)| = | -3i² | = |3| =3, and z₁ + z₂ = -i +3i=2i. Checks out.- For r=3: z₁=3i, z₂=-i. Similarly, |z₁·z₂|= |3i*(-i)|= | -3i²|= |3|=3, and z₁ + z₂=3i -i=2i. Also checks out.Therefore, the maximum value of |z₁| is 3 and the minimum is 1.Final AnswerThe maximum value of ( left|z_{1}right| ) is (boxed{3}) and the minimum value is (boxed{1}).

✅:Given complex numbers ( z_1 ) and ( z_2 ) such that ( left|z_1 cdot z_2right| = 3 ) and ( z_1 + z_2 = 2i ), we need to find the maximum and minimum values of ( left|z_1right| ).1. Modulus Product Condition: - The product of the moduli is given by ( |z_1| cdot |z_2| = 3 ). - Let ( |z_1| = r ). Then ( |z_2| = frac{3}{r} ).2. Sum Condition: - ( z_1 + z_2 = 2i ) implies ( z_2 = 2i - z_1 ). - The modulus of the sum is ( |z_1 + z_2| = |2i| = 2 ).3. Triangle Inequality: - Using the triangle inequality, we have ( |z_1| + |z_2| geq |z_1 + z_2| ). - Substituting values, ( r + frac{3}{r} geq 2 ). This inequality holds for all ( r > 0 ).4. Reverse Triangle Inequality: - Using the reverse triangle inequality, ( left| |z_1| - |z_2| right| leq |z_1 + z_2| ). - Substituting values, ( left| r - frac{3}{r} right| leq 2 ).5. Solving Inequalities: - The inequality ( left| r - frac{3}{r} right| leq 2 ) splits into two parts: - ( -2 leq r - frac{3}{r} leq 2 ). - Solving these inequalities, we find ( 1 leq r leq 3 ).6. Verification: - For ( r = 1 ), ( z_1 = -i ) and ( z_2 = 3i ) satisfy both conditions. - For ( r = 3 ), ( z_1 = 3i ) and ( z_2 = -i ) satisfy both conditions.Thus, the maximum value of ( left|z_1right| ) is (boxed{3}) and the minimum value is (boxed{1}).

🔑:Given the following details about complex numbers ( z_1 ) and ( z_2 ):1. (left|z_{1} cdot z_{2}right| = 3 ) 2. ( z_{1} + z_{2} = 2 mathrm{i} )We are asked to find the maximum and minimum values of ( left| z_1 right| ).1. Express ( z_1 ) and ( z_2 ) in polar form: Let [ z_1 = r (cos theta + i sin theta) ] and [ z_2 = frac{3}{r} (cos varphi + i sin varphi) ]2. Use the given condition ( z_{1} + z_{2} = 2 mathrm{i} ): Therefore, [ r cos theta + frac{3}{r} cos varphi = 0 ] and [ r sin theta + frac{3}{r} sin varphi = 2 ]3. Square and add both equations: [ left( r cos theta + frac{3}{r} cos varphi right)^2 + left( r sin theta + frac{3}{r} sin varphi right)^2 = 0^2 + 2^2 ] Simplifying, we get: [ left( r cos theta + frac{3}{r} cos varphi right)^2 + left( r sin theta + frac{3}{r} sin varphi right)^2 = 4 ]4. Expand the squared terms: [ left( r cos theta + frac{3}{r} cos varphi right)^2 = r^2 cos^2 theta + frac{9}{r^2} cos^2 varphi + 2 cdot r cos theta cdot frac{3}{r} cos varphi ] and [ left( r sin theta + frac{3}{r} sin varphi right)^2 = r^2 sin^2 theta + frac{9}{r^2} sin^2 varphi + 2 cdot r sin theta cdot frac{3}{r} sin varphi ]Thus: [ r^2 cos^2 theta + frac{9}{r^2} cos^2 varphi + 2 cdot 3 cos theta cos varphi + r^2 sin^2 theta + frac{9}{r^2} sin^2 varphi + 2 cdot 3 sin theta sin varphi = 4 ]5. Combine like terms: Notice that [ cos^2 theta + sin^2 theta = 1 quad text{and} quad cos^2 varphi + sin^2 varphi = 1 ] So, [ r^2 (cos^2 theta + sin^2 theta) + frac{9}{r^2} (cos^2 varphi + sin^2 varphi) + 6 (cos theta cos varphi + sin theta sin varphi) = 4 ]Therefore, [ r^2 + frac{9}{r^2} + 6 (cos theta cos varphi + sin theta sin varphi) = 4 ]6. Use the cosine addition formula: [ cos theta cos varphi + sin theta sin varphi = cos (theta - varphi) ]So we obtain: [ r^2 + frac{9}{r^2} + 6 cos (theta - varphi) = 4 ]7. Bound the ( cos ) term: Since (-1 leq cos (theta - varphi) leq 1), we have: [ r^2 + frac{9}{r^2} + 6 cos (theta - varphi) leq 4 + 6 = 10 ]Thus: [ r^2 + frac{9}{r^2} leq 10 ]8. Minimize the function ( f(r) = r^2 + frac{9}{r^2} ): Define [ f(r) = r^2 + frac{9}{r^2} ]To find the minimum, take the derivative: [ f'(r) = 2r - frac{18}{r^3} ]Set the derivative to zero: [ 2r - frac{18}{r^3} = 0 ]Solving for ( r ): [ 2r^4 = 18 implies r^4 = 9 implies r^2 = 3 implies r = sqrt{3} ]Evaluate (f) at critical points (r = sqrt{3}): [ f(sqrt{3}) = 3 + frac{9}{3} = 6 ]Since (f(r) = 6) is the minimum value at critical points, and the inequality gives: [ 1 leq r^2 leq 9 implies 1 leq r leq 3 ]So, for (1 leq r leq 3), the extreme values are: [ text{Minimum at } r = 1 implies left|z_1right| = 1 ] [ text{Maximum at } r = 3 implies left|z_1right| = 3 ] Conclusion:The maximum value of (left| z_1 right|) is 3 and the minimum value is 1. Thus:[boxed{1 text{ and } 3}]