🧠:Okay, so I have this problem here where I need to figure out the sum of five fractions: П,Я + Т,Ь + Д,Р + О,Б + Е,Й. Each letter represents a different digit, and even though all the addends are not integers, their sum is an integer. The question is asking for the possible sum(s) and examples of such addends, along with why other sums aren't possible. Hmm, let me break this down step by step.First, let me parse the problem correctly. The sum is written as П,Я + Т,Ь + Д,Р + О,Б + Е,Й. I think this notation means that each of these is a decimal number where the integer part is a single digit (represented by the first letter) and the fractional part is another digit (the second letter). For example, П,Я would be the number П.Я, where П and Я are digits from 0 to 9, and different letters represent different digits. So all ten letters (П, Я, Т, Ь, Д, Р, О, Б, Е, Й) must be unique digits.Also, each addend (like П,Я) is not an integer, which means that the fractional part (the second letter) can't be zero. So Я, Ь, Р, Б, Й are all non-zero digits. The sum of these five numbers is an integer. So when we add them up, the total of the fractional parts must add up to an integer. Since each fractional part is a single digit divided by 10 (because it's the first decimal place), the sum of the fractional parts is (Я + Ь + Р + Б + Й)/10. For this sum to be an integer, (Я + Ь + Р + Б + Й) must be a multiple of 10. So the total of the five fractional digits must be 10, 20, 30, 40, or 50. But since each fractional digit is unique and non-zero, their possible maximum sum is 9+8+7+6+5 = 35. So possible sums are 10, 20, 30. 40 and 50 are too high because the maximum sum is 35. Therefore, the sum of the fractional parts must be 10, 20, or 30. Therefore, the total fractional part contributes 1, 2, or 3 to the integer sum. Therefore, the integer part of the sum (the total of П + Т + Д + О + Е) plus this 1, 2, or 3 must be the final integer result.But we need to determine what exactly is the sum. The problem states "what exactly is the sum" and asks to provide examples for each possible answer, but also explain why other sums are not possible. Wait, but maybe the sum is unique? Let me see.Let me consider the total of the integer parts. Each addend's integer part is a single digit, so the integer parts are П, Т, Д, О, Е. Their sum is S = П + Т + Д + О + Е. The fractional parts sum up to F = (Я + Ь + Р + Б + Й)/10. The total sum is S + F. Since F must be an integer (because the total sum is an integer), then F must be 1, 2, or 3 as we established. Therefore, the total sum is S + 1, S + 2, or S + 3.But S itself is the sum of five single digits. The minimum possible S is 0+1+2+3+4=10 (if digits can be zero) but wait, the integer parts: are the digits allowed to be zero? The problem says different letters represent different digits, but it doesn't specify that the digits can't be zero. However, in decimal numbers, the integer part can be zero, but if the integer part is zero, then the number would be written as 0.Я, but the problem lists the addends as П,Я etc., which perhaps implies that the integer part is a single digit, which could be zero. However, if the integer part is zero, then the number is less than 1, but the fractional part is non-zero. However, since all addends are non-integers, their fractional parts are non-zero, but their integer parts can be zero or non-zero. So, the integer parts (П, Т, Д, О, Е) can be 0-9, but they all have to be unique digits, and also distinct from the fractional digits (Я, Ь, Р, Б, Й). Since all ten letters are different digits, the integer parts and fractional parts are completely distinct sets. So we have 5 integer digits and 5 fractional digits, all different.So the integer parts (П, Т, Д, О, Е) are five distinct digits, which can include zero, and the fractional parts (Я, Ь, Р, Б, Й) are five distinct non-zero digits (since they can't be zero because the addends are non-integers). Therefore, the sum S = П + Т + Д + О + Е can range from, if we use the smallest five digits (0,1,2,3,4) sum to 10, and the largest five digits (5,6,7,8,9) sum to 35. Wait, but wait, the integer parts can be any five distinct digits, so possible sums S can be from 0+1+2+3+4=10 up to 5+6+7+8+9=35. But actually, if we include 0 in the integer parts, the minimum sum is 0+1+2+3+4=10, and the maximum is 5+6+7+8+9=35. Wait, but 5 digits from 0-9, all distinct, so the maximum sum is indeed 5+6+7+8+9=35. The minimum is 0+1+2+3+4=10.Similarly, the fractional parts (Я, Ь, Р, Б, Й) are five distinct non-zero digits (1-9). Their sum F_total = Я + Ь + Р + Б + Й must be 10, 20, or 30 as established before. The minimum sum of five distinct non-zero digits is 1+2+3+4+5=15, and the maximum is 5+6+7+8+9=35. Wait, but 1+2+3+4+5=15, so possible sums for the fractional parts can be from 15 to 35. But earlier, we thought F_total must be 10, 20, 30. But 15 is already higher than 10, so perhaps my initial thought was wrong. Wait, hold on.Wait, the fractional parts are each single digits, so each fractional part is, for example, Я/10. So when you add them all up, the total fractional part is (Я + Ь + Р + Б + Й)/10. For this to be an integer, the sum of the numerators must be divisible by 10. So F_total must be a multiple of 10. So the sum of the five fractional digits must be 10, 20, or 30. But as we just saw, the minimal sum of five distinct non-zero digits is 15, so 10 is impossible. Therefore, the possible sums for F_total are 20 or 30. Because 15 is too low for 20? Wait, 15 is less than 20, but 20 is possible? Wait, can we get 20 as the sum of five distinct non-zero digits?Yes, for example, 1+2+3+4+10, but wait digits can't exceed 9. So the maximum digit is 9. Let's see: To get a sum of 20 with five distinct non-zero digits. Let's try 1+2+3+4+10, but 10 is invalid. 1+2+3+5+9=20. Yes, that works. 1+2+3+5+9=20. So that's possible. Similarly, 1+2+4+6+7=20. So 20 is achievable.Similarly, 30 as the sum of five distinct non-zero digits. Let's check: 5+6+7+8+4=30. Wait, 4+5+6+7+8=30. Yes, that's 30. So 30 is achievable. So possible F_total is 20 or 30. Therefore, the fractional parts sum to either 2 or 3 when divided by 10. Therefore, the total sum is S + 2 or S + 3, where S is the sum of the integer parts.But then the total sum must be S + 2 or S + 3. Now, S can range from 10 to 35 (as the sum of five distinct digits from 0-9). So the total sum can be from 10 + 2 =12 up to 35 +3=38. But we need more constraints. Let me think.But actually, the integer parts (П, Т, Д, О, Е) are five distinct digits, and the fractional parts (Я, Ь, Р, Б, Й) are another five distinct non-zero digits, with no overlap between the two sets. So all ten digits are used, each exactly once. Therefore, the integer parts and fractional parts together use all digits from 0 to 9 exactly once, with the fractional parts using non-zero digits. Therefore, the integer parts must include 0 (since the fractional parts cannot). Wait, not necessarily. The integer parts can include 0 or not, but since all ten digits are used, and fractional parts are five non-zero digits, so the integer parts must consist of the remaining five digits, which include 0. Therefore, 0 must be in the integer parts, because the fractional parts are all non-zero. Therefore, the integer parts include 0 and four other digits, or maybe 0 is included or not? Wait, total digits are ten. The fractional parts are five non-zero digits (1-9), so the integer parts must be the remaining five digits, which must include 0. Because if fractional parts are five non-zero digits, then the integer parts must include 0 and four other digits. Therefore, the integer parts are 0 and four digits from 1-9 not used in the fractional parts. So the integer parts must include 0. Therefore, the sum S of the integer parts is 0 + a + b + c + d, where a, b, c, d are four distinct digits from 1-9 not used in the fractional parts. Therefore, S can range from 0+1+2+3+4=10 to 0+6+7+8+9=30. Wait, because the fractional parts take up five digits, so the integer parts take the remaining five digits, which must include 0. Therefore, the integer parts are 0 plus four digits not in the fractional parts. If the fractional parts are, say, 1,2,3,4,5, then the integer parts are 0,6,7,8,9, summing to 30. If fractional parts are 5,6,7,8,9 (sum 35, which is not a multiple of 10, but if fractional parts sum to 30, which is allowed), then the integer parts would be 0,1,2,3,4 summing to 10. Wait, so depending on the fractional parts, the integer parts can vary.But since the fractional parts sum to either 20 or 30, let's consider both cases:Case 1: Fractional parts sum to 20, so F_total =20, contributing 2 to the total sum. Then S (sum of integer parts) is 0 + a + b + c + d, where a,b,c,d are four digits not in the fractional parts. The fractional parts are five digits summing to 20. The remaining digits (for integer parts) are 0 and the other five digits not in the fractional parts. Wait, no: total digits are ten. If fractional parts are five digits, the integer parts are five digits (including 0). So if fractional parts are five digits (non-zero), the integer parts are the remaining five digits, which must include 0. Therefore, S = sum of five digits: 0 + four digits not in the fractional parts.Similarly, if fractional parts sum to 30, then F_total =30, contributing 3 to the total sum. Then the integer parts are 0 and four digits not in the fractional parts. So let's see:If F_total =20:Then fractional digits sum to 20, and the integer parts are 0 and the remaining four digits (since total digits are ten). The remaining four digits must be five digits (since fractional parts are five), but wait, fractional parts are five digits, so integer parts are five digits (including 0). Therefore, the integer parts are 0 plus four digits not in the fractional parts. So S = 0 + four digits. The sum of those four digits would be S_total -0 = S. So S = sum of four digits plus 0. The four digits are from the remaining digits not in the fractional parts.Similarly, if F_total =30, the fractional digits sum to 30, and the integer parts are 0 plus four digits not in the fractional parts. Let's try to find possible S in both cases.First, let's think about the possible S when F_total is 20 or 30.Case 1: F_total =20.Then the fractional digits sum to 20, which are five distinct non-zero digits. Let's find possible combinations. For example, as I mentioned before, 1+2+3+5+9=20. Then the integer parts would be 0 plus the remaining digits: 4,6,7,8. So S =0+4+6+7+8=25. Then total sum would be 25 + 2 =27.Another example for F_total=20: 1+2+4+6+7=20. Then remaining digits for integer parts: 0,3,5,8,9. So S=0+3+5+8+9=25. Total sum 25 +2=27. Wait, same total. Hmm.Wait, maybe different combinations? Let's see another one: 1+3+4+5+7=20. Then remaining digits: 0,2,6,8,9. Sum S=0+2+6+8+9=25. Total sum again 27. Hmm.Wait, so is 27 the only possible total sum when F_total=20? Let's check another example. Suppose fractional parts are 2+3+4+5+6=20. Then remaining digits are 0,1,7,8,9. Sum S=0+1+7+8+9=25. Again total sum 25+2=27. So it seems that regardless of how the fractional digits sum to 20, the remaining integer digits (including 0) sum to 25, leading to a total sum of 27. Wait, why?Wait, the sum of all digits from 0 to 9 is 45. If the fractional digits sum to 20, then the integer digits must sum to 45 -20=25. Therefore, the total sum is S + F_total/10 =25 +20/10=25 +2=27. So regardless of how the fractional digits are arranged, if they sum to 20, the total sum is 27. Similarly, if fractional digits sum to 30, then the integer digits sum to 45 -30=15. Then total sum would be 15 +30/10=15+3=18. So the total sum must be either 27 or 18. Wait, but the problem says "what exactly is the sum" and asks for each possible answer with an example. But according to this, there are two possible sums: 18 and 27. However, the problem says "all five addends are not integers, but the sum itself is an integer". So maybe both 18 and 27 are possible? But wait, let's check.Wait, but if F_total=30, then fractional parts sum to 30, so the remaining digits (integer parts) sum to 15. But the integer parts are 0 plus four digits. So 0+a+b+c+d=15. What's the minimal sum of four distinct digits (from 1-9) not used in fractional parts. Wait, if fractional parts sum to 30, they must be five digits summing to 30. For example, 5+6+7+8+4=30 (but 4+5+6+7+8=30). Then the remaining digits are 0,1,2,3,9. So sum S=0+1+2+3+9=15. Then total sum is 15 +3=18. But another example: fractional parts as 9+8+7+5+1=30. Wait, 9+8+7+5+1=30. Then remaining digits are 0,2,3,4,6. Sum S=0+2+3+4+6=15. Total sum 18. Similarly, any combination where fractional parts sum to 30 would leave the integer parts summing to 15. So total sum is 18.Therefore, the possible sums are 18 and 27. But wait, the problem says "what exactly is the sum" and "provide one example with these five addends". So maybe both 18 and 27 are possible? But the problem might have a unique answer. Wait, but let's check whether fractional parts can sum to 30 or 20.Wait, when the fractional parts sum to 30, the integer parts sum to 15, which includes 0. But 0 is part of the integer parts. So the integer parts are numbers like 0.x, but wait, no: the integer parts are the digits before the decimal. For example, if an addend is 0.1, then the integer part is 0, and the fractional part is 1. But in the problem, the addends are written as П,Я which is П.Я, so the integer part is П and the fractional part is Я. So if the integer part is 0, then the addend is 0.Я. So even with 0 as an integer part, the number is non-integer because Я is non-zero. So that's acceptable.But then, when the fractional parts sum to 30, the integer parts sum to 15, and the total sum is 18. Similarly, when fractional parts sum to 20, integer parts sum to 25, total sum 27. So both 18 and 27 are possible? But the problem says "the sum itself is an integer. What exactly is the sum?" So maybe both are possible. But the answer might be unique. Let me think.Wait, but let's verify with examples.First, let's try to find an example where the total sum is 18.For fractional digits summing to 30: Let's pick fractional digits as 4,5,6,7,8 (sum=30). Then integer digits are 0,1,2,3,9. So the integer parts sum to 0+1+2+3+9=15. So total sum is 15 +3=18. Now, we need to assign these digits to the addends. The addends are П.Я, Т.Ь, Д.Р, О.Б, Е.Й. Each of П,Т,Д,О,Е are from the integer digits: 0,1,2,3,9. Each of Я,Ь,Р,Б,Й are from fractional digits:4,5,6,7,8.We need to assign each integer digit to an addend, and each fractional digit to an addend, making sure all letters are unique. Let's construct such an example.Let's assign:П=0, Я=4 → 0.4Т=1, Ь=5 →1.5Д=2, Р=6 →2.6О=3, Б=7 →3.7Е=9, Й=8 →9.8Now, sum these up:0.4 +1.5 +2.6 +3.7 +9.8Calculating:0.4 +1.5 =1.91.9 +2.6=4.54.5 +3.7=8.28.2 +9.8=18.0Yes, the sum is 18. So this works. All digits are unique: integer digits are 0,1,2,3,9; fractional digits are4,5,6,7,8. All different, and the sum is integer 18.Now, let's construct an example for sum 27.Fractional digits sum to 20. Let's choose fractional digits 1,2,3,5,9 (sum=20). Then integer digits are 0,4,6,7,8. Sum of integer digits:0+4+6+7+8=25. Total sum 25 +2=27.Assigning:П=0, Я=1 →0.1Т=4, Ь=2 →4.2Д=6, Р=3 →6.3О=7, Б=5 →7.5Е=8, Й=9 →8.9Summing:0.1 +4.2 +6.3 +7.5 +8.9Calculating:0.1 +4.2=4.34.3 +6.3=10.610.6 +7.5=18.118.1 +8.9=27.0Perfect. All digits are unique: integer digits 0,4,6,7,8; fractional digits1,2,3,5,9. Sum is 27.Therefore, both 18 and 27 are possible sums. However, the problem asks "what exactly is the sum" and to provide examples for each possible answer. But the problem might have only one possible answer. Wait, maybe I made a mistake here. Let me check again.Wait, the problem states "all five addends are not integers, but the sum itself is an integer". It doesn't specify that there's only one possible sum. So both 18 and 27 are possible. But let's confirm if there are other possibilities. Wait, the fractional parts can only sum to 20 or 30, leading to total sums of 27 or 18. There's no other multiple of 10 possible between 15 and 35 for fractional parts. Because the sum of five distinct non-zero digits must be between 15 and 35, and the only multiples of 10 in that range are 20 and 30. Hence, only two possible total sums: 18 and 27.But let me check if fractional parts can sum to 10. Wait, fractional parts are five distinct non-zero digits. The minimum sum is 1+2+3+4+5=15, so 10 is impossible. So indeed, only 20 and 30 are possible. Therefore, total sums are 18 and 27.But the problem says "what exactly is the sum". If it's expecting a unique answer, maybe I missed something. Let me check the problem statement again.Wait, the problem is in the sum:П,Я + Т,Ь + Д,Р + О,Б + Е,ЙAll digits are encoded by letters (different letters represent different digits). All five addends are not integers, but the sum is an integer. What exactly is the sum? For each possible answer, provide one example with these five addends. Explain why other sums are not possible.So it's possible that both 18 and 27 are valid, and we need to provide examples for each and explain why others are impossible. But the explanation would be that the fractional parts must sum to 20 or 30, leading to total sums of 27 or 18, and since fractional parts can't sum to 10 or 40, there are no other possibilities.However, in my previous analysis, I considered that the sum of all digits (0-9) is 45. If fractional parts sum to F_total, then integer parts sum to 45 - F_total. Then total sum is (45 - F_total) + F_total/10 =45 - F_total + F_total/10=45 - (9/10)F_total.Therefore, the total sum must be 45 - (9/10)F_total. But since F_total is 20 or 30:If F_total=20: total sum=45 - (9/10)*20=45 -18=27.If F_total=30: total sum=45 - (9/10)*30=45 -27=18.Hence, only two possible sums:18 and27. Therefore, these are the only possible sums, and all other sums are not possible because F_total must be a multiple of 10, and the only feasible multiples of 10 for the sum of five distinct non-zero digits are 20 and30.Therefore, the answer is that the possible sums are 18 and27, each achievable with different assignments. The examples are as constructed above.But wait, the problem says "what exactly is the sum". Maybe there's a unique answer. Wait, maybe I miscalculated the possible sums. Let me verify again.Suppose F_total=20: sum of fractional parts is20 → total sum is (45 -20) +20/10=25 +2=27.If F_total=30: sum is (45 -30)+30/10=15 +3=18.So yes, these are the only two possibilities. Therefore, the sums are exactly 18 and27.But maybe the problem is translated from Russian, and the letters used might have some constraints. Wait, looking at the letters: П, Я, Т, Ь, Д, Р, О, Б, Е, Й. These are Cyrillic letters. But the problem states that different letters represent different digits. Since all letters are distinct, but in the Cyrillic alphabet, some letters might correspond to the same digit in another script, but here it's specified that different letters are different digits. So each Cyrillic letter is a unique digit from0-9. But that doesn't add any constraints beyond what's already considered.Therefore, the conclusion is that the sum can be either18 or27. For each, examples can be constructed as shown. Other sums are impossible because the fractional parts must sum to a multiple of10, and the only possible such sums are20 and30, leading to total sums of27 and18 respectively.But wait, let me check if fractional parts can actually sum to30. Let's take the digits9,8,7,6,5 which sum to35. Wait, 9+8+7+6+5=35. That's too much. Wait, no, 4+5+6+7+8=30. Yes, that works. So for example, fractional digits4,5,6,7,8 sum to30, integer digits0,1,2,3,9 sum to15, total sum18.Similarly, fractional digits1,2,3,5,9 sum to20, integer digits0,4,6,7,8 sum to25, total sum27.Therefore, both sums are possible. Hence, the answer is that the possible sums are18 and27, each achievable with different assignments, and no other sums are possible because the fractional parts can only sum to20 or30.

✅:The possible sums are 18 and 27. Here's why: Key Analysis:1. Fractional Parts Sum Constraint: - Each addend is a decimal number with one fractional digit (non-zero, distinct digits). - The sum of fractional digits ((Я + Ь + Р + Б + Й)) must be a multiple of 10 (to make the total sum an integer). - The minimum sum of five distinct non-zero digits is (1+2+3+4+5=15), and the maximum is (5+6+7+8+9=35). Thus, only 20 or 30 are valid (multiples of 10 in this range).2. Total Sum Calculation: - The sum of all digits (0)–(9) is (45). If fractional digits sum to (F), the integer digits (including 0) sum to (45 - F). - Total sum (= text{(integer sum)} + frac{text{fractional sum}}{10} = (45 - F) + frac{F}{10}). - For (F = 20): Total sum (= 25 + 2 = 27). - For (F = 30): Total sum (= 15 + 3 = 18). Examples:# Sum = 18:- Fractional digits: (4, 5, 6, 7, 8) (sum = 30).- Integer digits: (0, 1, 2, 3, 9) (sum = 15).- Addends: - (0.4 + 1.5 + 2.6 + 3.7 + 9.8 = 18.0).# Sum = 27:- Fractional digits: (1, 2, 3, 5, 9) (sum = 20).- Integer digits: (0, 4, 6, 7, 8) (sum = 25).- Addends: - (0.1 + 4.2 + 6.3 + 7.5 + 8.9 = 27.0). Why Other Sums Are Impossible:- Fractional digits cannot sum to 10 (minimum sum is 15) or 40/50 (exceeds maximum possible sum of 35). Only 20 and 30 are valid, leading uniquely to 18 and 27.Final Answer: The possible sums are (boxed{18}) and (boxed{27}).

🔑:To solve this problem, we need to determine the possible sums of five fractions, each represented by different digits, that would result in an integer after adding them together. Let's break this down step-by-step.1. Identify the requirement: - Each digit after the comma (i.e., fraction part) must add up to a number ending in 0 when combined (since the sum is a whole number).2. Sum of decimal places: - The sum of the digits after the comma must end in 0. This implies that sum (text{fractional parts}) must be either 10, 20, or 30, where each digit after the decimal place is unique.3. Determine the possible sums: - The total sum of digits from 0 to 9 is ( sum_{i=0}^{9} i = 45 ).4. Case: Sub-sum equals 20: - Sum of digits after the comma equals 20 implies the digits before the comma must sum to ( 45 - 20 = 25 ). - There will be a contribution of 2 to the integer part due to the carryover from the sum of the fractional parts equating to 20. - Hence the final sum of the integer parts would be ( 25 + 2 = 27 ).5. Case: Sub-sum equals 30: - Sum of digits after the comma equals 30 implies the digits before the comma must sum to ( 45 - 30 = 15 ). - There will be a contribution of 3 to the integer part due to the carryover from the sum of the fractional parts equating to 30. - Hence the final sum of the integer parts would be ( 15 + 3 = 18 ). Conclusion:We need to provide examples for each possible sum and explain why no other sum is possible. Examples:- For the sum 27: Use the fractional parts adding up to 20: [ 0.5 + 1.6 + 7.4 + 8.3 + 9.2 = 27 ] Here, the fractional parts sum up to 20, and the integer parts to 25, carrying over 2, resulting in a total sum of 27.- For the sum 18: Use the fractional parts adding up to 30: [ 0.9 + 1.8 + 3.7 + 5.4 + 6.2 = 18 ] Here, the fractional parts sum up to 30, and the integer parts to 15, carrying over 3, resulting in a total sum of 18. Explanation for exclusion of other sums:The only other sums that could end in 0 are 10, 20, 30, etc. The sum 10 is excluded because it would require the digit 0 to be in the fractional part which contradicts the problem’s requirement of having non-integral summands. Thus, the only viable sums are 27 and 18.[boxed{27} quad text{and} quad boxed{18}]

🧠:Okay, let's tackle this problem step by step. The question is about finding the next common term after 71 in two given arithmetic sequences. The first sequence is 3, 20, 37, 54, 71, ..., and the second is 16, 27, 38, 49, 60, 71, .... The options provided are A) 115, B) 187, C) 258, D) 445, E) 1006. I need to figure out which one is the correct answer. Let me start by understanding each sequence's pattern.First, let's look at the first sequence: 3, 20, 37, 54, 71, ... To find the common difference, I can subtract consecutive terms. 20 - 3 = 17, 37 - 20 = 17, 54 - 37 = 17, 71 - 54 = 17. So, this is an arithmetic sequence with first term a₁ = 3 and common difference d₁ = 17. Therefore, the nth term of this sequence can be written as:a(n) = a₁ + (n - 1)d₁a(n) = 3 + (n - 1) * 17a(n) = 17n - 14Wait, let me check that formula. When n = 1, 17(1) - 14 = 3, which matches. n=2: 34 -14=20, correct. Yes, that formula works.Now the second sequence: 16, 27, 38, 49, 60, 71, ... Let's find its common difference as well. 27 - 16 = 11, 38 - 27 = 11, 49 - 38 = 11, etc. So, this is also an arithmetic sequence with first term b₁ = 16 and common difference d₂ = 11. The nth term would be:b(n) = b₁ + (n - 1)d₂b(n) = 16 + (n - 1)*11b(n) = 11n + 5Wait, let me verify. For n=1: 11(1) +5 = 16, correct. n=2: 22 +5=27, right. Yes, that works.So, we need to find the next common term after 71 in these two sequences. The problem is essentially asking for the next term that appears in both sequences. Since both are arithmetic sequences, their common terms form another arithmetic sequence. The first common term is 71, and we need to find the next one.To find the common terms of two arithmetic sequences, we can model this as solving the equation a(n) = b(m) for integers n and m. So:17n - 14 = 11m + 5Let me write that equation again:17n - 14 = 11m + 5Rearranging terms:17n - 11m = 19So, we need to solve the Diophantine equation 17n - 11m = 19 for integers n and m. Then, the solutions will give us the terms where the two sequences intersect. Since we already know that 71 is a common term, that corresponds to one solution (n1, m1), and we need to find the next solution (n2, m2) which will give the next common term.First, let's find the particular solution for n and m that gives 71. Let's see, in the first sequence, 71 is the 5th term. Let's check n:a(n) = 17n -14 = 7117n = 71 +14 = 85n = 85 /17 =5. Correct, n=5.In the second sequence:b(m) =11m +5=7111m=71-5=66m=66/11=6. So, m=6.So, the solution (n=5, m=6) gives the term 71. Now, we need the next solution. To find all solutions of the Diophantine equation 17n -11m =19, we can use the method of finding general solutions once a particular solution is found.First, let's write the equation as:17n -11m =19We know that (n, m) = (5,6) is a solution. Let's find the general solution.The coefficients are 17 and -11. The greatest common divisor (gcd) of 17 and 11 is 1, since they are both primes. Since 1 divides 19, there are solutions.The general solution can be written as:n = n0 + (11)km = m0 + (17)kWhere k is an integer, and n0=5, m0=6.Wait, let me confirm this. For linear Diophantine equations of the form ax + by = c, once a particular solution (x0, y0) is found, the general solution is given by:x = x0 + (b/gcd(a,b)) * ty = y0 - (a/gcd(a,b)) * tBut in our case, the equation is 17n -11m =19. Let me write it as 17n + (-11)m =19. So a=17, b=-11.The gcd of 17 and 11 is 1. Therefore, the general solution is:n = n0 + (11/1)*t =5 +11tm = m0 + (17/1)*t=6 +17tWait, but the signs can be tricky. Let me double-check.The standard form is ax + by = c. Here, it's 17n -11m =19. So, a=17, b=-11, c=19.The standard solution is:x = x0 + (b/gcd(a,b)) * ty = y0 - (a/gcd(a,b)) * tBut here, variables are n and m. So, let's write:n = n0 + (-11)/1 * t =5 -11tm = m0 - (17)/1 * t =6 -17tWait, this is confusing. Maybe I should recall the method properly.Alternatively, since the equation is 17n -11m =19, we can think of it as 17(n) =11m +19.Given that 17 and 11 are coprime, the solutions will be of the form n = n0 +11k and m = m0 +17k for integer k. Let's check.If n =5 +11k and m=6 +17k, then substituting into the equation:17*(5 +11k) -11*(6 +17k) =17*5 +17*11k -11*6 -11*17k=85 +187k -66 -187k=19. So yes, that works.Therefore, the general solution is n=5+11k and m=6+17k for integer k.Therefore, the common terms occur when k is an integer. The term corresponding to k=0 is 71. The next term will be when k=1, then k=2, etc.So, substituting k=1:n=5+11(1)=16m=6+17(1)=23Then, the term would be a(16)=17*16 -14=272 -14=258Alternatively, check b(23)=11*23 +5=253 +5=258. So, 258 is the next common term.Therefore, the answer should be C) 258.Wait, but let's confirm this step by step.First, for k=1: n=16, so a(16)=17*16 -14. 17*16: 10*16=160, 7*16=112, so 160+112=272. 272-14=258.Similarly, m=23, so b(23)=11*23 +5. 10*23=230, 1*23=23, total 253. 253+5=258. Correct.So, 258 is indeed the next common term. Let me check if any of the options is 258. The options are A)115, B)187, C)258, D)445, E)1006. So, option C is 258. Therefore, the answer is C.Wait, but just to be thorough, let's check if there's another term between 71 and 258. For k=-1, which would give n=5-11= -6, which is invalid as term numbers can't be negative. So, k=0 gives 71, k=1 gives 258. Then k=2 would give n=5+22=27, m=6+34=40. Then a(27)=17*27 -14. 17*20=340, 17*7=119, so 340+119=459. 459-14=445. Similarly, b(40)=11*40 +5=440+5=445. So, 445 is the next term after 258. But 445 is one of the options (D). But the question asks for the next term after 71, so 258 is the immediate next. Therefore, the answer is C)258.Alternatively, maybe I made a mistake here. Let's check if 187 is a possible term. Let's see.Suppose someone might think the common difference is LCM(17,11)=187, so adding 187 to 71 gives 258. Wait, 71 +187=258. So, the common terms would be 71, 258, 445, etc., each 187 apart. Therefore, the next term is 258. Hence, option C. That's another way to see it. The common difference of the combined sequence is LCM(17,11)=187. Therefore, starting at 71, the next term is 71+187=258.Yes, that's a quicker way. Since both sequences are arithmetic with differences 17 and 11, which are coprime, the common terms form an arithmetic sequence with difference LCM(17,11)=187. Therefore, the next term after 71 is 71 +187=258. So, answer C. That's a more straightforward approach.But let me confirm with the previous method. The first method using Diophantine equations gives the same result, so both methods confirm 258 is the next term.Therefore, the correct answer is C) 258.

✅:The sequences given are arithmetic. The first sequence has a common difference of 17, starting at 3, so its nth term is ( a(n) = 17n - 14 ). The second sequence has a common difference of 11, starting at 16, so its nth term is ( b(m) = 11m + 5 ).To find common terms, solve ( 17n - 14 = 11m + 5 ), which simplifies to the Diophantine equation ( 17n - 11m = 19 ). A particular solution is ( n = 5, m = 6 ) (giving 71). The general solution is:[n = 5 + 11k, quad m = 6 + 17k quad text{for integer } k.]For the next term after 71, set ( k = 1 ):- ( n = 16 Rightarrow a(16) = 17(16) - 14 = 258 )- ( m = 23 Rightarrow b(23) = 11(23) + 5 = 258 )Alternatively, the common terms form an arithmetic sequence with a common difference of ( text{LCM}(17, 11) = 187 ). Starting at 71, the next term is ( 71 + 187 = 258 ).Answer: (boxed{C})

🔑:1. Identify the two sequences: - The first sequence is (3, 20, 37, 54, 71, ldots). - The second sequence is (16, 27, 38, 49, 60, 71, ldots).2. Determine the pattern: - The first sequence increases by a constant amount. Let's verify: [ 20 - 3 = 17, quad 37 - 20 = 17, quad 54 - 37 = 17, quad 71 - 54 = 17 ] Hence, the common difference is (d_1 = 17). - The second sequence also increases by a constant amount. Let's verify: [ 27 - 16 = 11, quad 38 - 27 = 11, quad 49 - 38 = 11, quad 60 - 49 = 11, quad 71 - 60 = 11 ] Hence, the common difference is (d_2 = 11).3. Formulate the general term of each sequence: - The general term of the first sequence can be written as: [ a_n = 3 + (n-1) cdot 17 ] - Simplify it: [ a_n = 3 + 17n - 17 = 17n - 14 ] - The general term of the second sequence can be written as: [ b_m = 16 + (m-1) cdot 11 ] - Simplify it: [ b_m = 16 + 11m - 11 = 11m + 5 ]4. Identify when the terms are equal: - Set the general terms equal to find the common terms: [ 17n - 14 = 11m + 5 ]5. Solve for the next common term: - Rearrange the equation to isolate (m): [ 17n - 11m = 19 ] - We need to find (n) and (m) such that the above equation holds. Previously, we know that: [ 71 = 17n - 14 = 11m + 5 ] which means: [ n = 5 quad text{and} quad m = 6 ]6. Find the next common term after 71: - Add the least common multiple (LCM) of the common differences of the sequences: - (d_1 = 17) and (d_2 = 11) - Since 17 and 11 are co-prime numbers, their LCM is their product: [ text{LCM}(17, 11) = 17 times 11 = 187 ] - Therefore, the next common term after 71: [ 71 + 187 = 258 ] Conclusion:The next common term in both sequences after 71 is:[boxed{258}]

🧠:Okay, let's see. The problem is about a class of 20 students where each student is friends with at least 14 others. We need to determine if there must be four students who are all friends with each other (a clique of size 4). Hmm, this seems like a graph theory problem. Let me recall some concepts.First, translating the problem into graph terms: each student is a vertex, and each friendship is an edge. So, we have a graph with 20 vertices where each vertex has a degree of at least 14. The question is whether this graph necessarily contains a complete subgraph (clique) of size 4.I remember that Turán's theorem might be relevant here. Turán's theorem gives the maximum number of edges a graph can have without containing a complete subgraph of a given size. Let me check the formula. The Turán number T(n, r-1) is the maximum number of edges in an n-vertex graph that does not contain a complete r-graph. The formula is (1 - 1/(r-1)) * n² / 2. So, if a graph has more edges than the Turán number, it must contain a K_r.In our case, r = 4, so we need to compute T(20, 3). The Turán number T(20, 3) would be (1 - 1/3) * 20² / 2 = (2/3) * 400 / 2 = (2/3)*200 ≈ 133.33. Since the number of edges must be an integer, the maximum number of edges without a K₄ is 133. If our graph has more than 133 edges, then it must contain a K₄.Now, let's calculate the number of edges in our graph. Each student has at least 14 friends, so the total degree is at least 20*14 = 280. Since each edge is counted twice, the number of edges is at least 280/2 = 140. Comparing this to Turán's number 133, since 140 > 133, Turán's theorem tells us that the graph must contain a K₄. Therefore, the answer should be yes, there must be four students who are all friends with each other.Wait, let me make sure I applied Turán's theorem correctly. Turán's theorem states that any graph with more edges than T(n, r-1) contains a K_r. Here, n=20, r=4, so T(20,3)=133.33. Our graph has at least 140 edges, which exceeds 133, so by Turán's theorem, it must contain a K₄. That seems right.Alternatively, maybe there's another approach using the pigeonhole principle or Ramsey numbers? Let's see. The Ramsey number R(s,t) is the smallest number n such that any graph of n vertices contains a clique of size s or an independent set of size t. However, Ramsey numbers are generally difficult to compute, and I don't think R(4, something) is known exactly here. But Turán's theorem seems more directly applicable here because we have a density condition (minimum degree) leading to a guaranteed clique.Another way to think about it: in a graph with minimum degree δ, what's the largest clique we can guarantee? There's a theorem called Dirac's theorem, but that's about Hamiltonian cycles. Maybe another theorem? Wait, maybe using induction or constructive reasoning.Suppose we pick any vertex, say vertex A. Since A has at least 14 friends, let's consider the subgraph induced by these 14 friends. If within this subgraph, each vertex has a high enough degree, we might find a clique of size 3, which together with A would form a clique of size 4.But wait, the problem is that the original graph has minimum degree 14, but when we look at the neighbors of A, their degrees within the neighbor set might be lower. Each neighbor of A has degree at least 14 in the entire graph, but some of their friends could be outside the neighbor set of A. Since the total number of vertices is 20, each neighbor of A has at least 14 - 1 = 13 edges (excluding A). However, there are 14 neighbors of A and 5 non-neighbors. So each neighbor of A has at most 5 friends outside the neighbor set. Therefore, each neighbor of A has at least 14 - 1 - 5 = 8 friends within the neighbor set of A. So in the subgraph induced by A's friends, each vertex has degree at least 8.Now, we have a 14-vertex graph with minimum degree 8. Can we guarantee a clique of size 3 in this subgraph? If so, then together with A, we have a clique of size 4. To check if a graph with 14 vertices and minimum degree 8 contains a triangle (K₃), we can use Turán's theorem again. The Turán number for triangle-free graphs (r=3) on n=14 vertices is T(14,2) = floor(14² /4) = floor(196/4)=49. So a triangle-free graph on 14 vertices can have at most 49 edges. If our subgraph has more than 49 edges, it must contain a triangle.What's the number of edges in this subgraph? With 14 vertices and minimum degree 8, the total degree is at least 14*8=112, so edges are at least 56. Since 56 > 49, Turán's theorem says it contains a triangle. Therefore, there is a triangle among A's friends, and adding A gives a clique of size 4. So this also confirms that such a clique exists.So both approaches via Turán's theorem on the whole graph and by looking at the neighborhood of a vertex lead to the conclusion that there must be a K₄. Therefore, the answer is yes, it can be asserted that there will be four students who are all friends with each other.Wait, but let me check if Turán's theorem is tight. For example, the Turán graph T(n, r-1) is the complete (r-1)-partite graph with partitions as equal as possible. For r=4, n=20, the Turán graph T(20,3) would be a complete 3-partite graph with partitions of size 7,7,6. The number of edges is calculated as follows: each vertex in the first two partitions (size 7) connects to all vertices not in their own partition, so each has 13 + 6 = 19 edges? Wait, no. Wait, in a complete 3-partite graph, edges exist between different partitions only.Wait, the Turán graph T(n,3) is a complete 3-partite graph with partitions as equal as possible. For n=20, the partitions would be 7,7,6. The number of edges is then 7*7 + 7*6 + 7*6 = 49 + 42 + 42 = 133. Yes, that's the same as before. So the Turán graph has 133 edges and no K₄. However, in our problem, the graph has at least 140 edges, which is more than 133. Therefore, it cannot be the Turán graph, so it must contain a K₄.Therefore, the answer is yes. The minimum degree condition ensures that the graph is dense enough to surpass the Turán threshold, forcing a K₄.

✅:Step-by-Step Explanation and Answer:1. Graph Representation: Model the class as a graph ( G ) with 20 vertices (students), where edges represent friendships. Each vertex has a degree of at least 14.2. Turán's Theorem Application: Turán's theorem states that a graph with ( n ) vertices and more than ( T(n, r-1) ) edges must contain a ( K_r ) (complete subgraph of size ( r )). - For ( r = 4 ), compute ( T(20, 3) ): [ T(20, 3) = left(1 - frac{1}{3}right) frac{20^2}{2} = frac{2}{3} cdot 200 approx 133.33 ] - The maximum edges without a ( K_4 ) is 133. 3. Edge Count in ( G ): - Each vertex has degree ≥ 14. Total degrees sum to ≥ ( 20 times 14 = 280 ). - Total edges ≥ ( 280 / 2 = 140 ). - Since ( 140 > 133 ), ( G ) exceeds the Turán threshold and must contain a ( K_4 ). 4. Alternative Neighborhood Analysis: - Pick a vertex ( A ); it has ≥14 neighbors. - The subgraph of ( A )'s neighbors has 14 vertices. Each neighbor has ≥8 edges within this subgraph (since they can have ≤5 edges outside). - In this 14-vertex subgraph, the minimum degree (8) ensures enough edges to force a triangle (via Turán's theorem for ( K_3 )): - ( T(14, 2) = lfloor frac{14^2}{4} rfloor = 49 ). - Subgraph edges ≥ ( frac{14 times 8}{2} = 56 ), exceeding 49, thus containing a triangle. - Combining ( A ) with this triangle forms a ( K_4 ). Conclusion: Yes, in a class of 20 students where each is friends with at least 14 others, there must exist four students who are all friends with each other. This is guaranteed by exceeding the Turán threshold for ( K_4 )-free graphs. Answer: boxed{Yes}

🔑:1. Define the Problem in Graph Theory Terms: - Consider a graph ( G ) where each vertex represents a student and an edge between two vertices indicates that the corresponding students are friends. - The problem states that every student in the class of 20 students has at least 14 friends.2. Analyze Vertex Degrees: - Let ( delta(G) ) denote the minimum degree of ( G ). - Given that each student has at least 14 friends, ( delta(G) geq 14 ).3. Use the Friendship Constraints: - We need to show that there exists a clique of size 4, i.e., a complete subgraph ( K_4 ).4. Consider any Student ( A ): - Let ( A ) be any student. According to the problem, ( A ) has at least 14 friends. - Remove the students who are not friends with ( A ). The number of such students cannot exceed ( 20 - 14 - 1 = 5 ).5. Count Students Still in the Room: - After removing those 5 students, at least ( 20 - 5 = 15 ) students, including ( A ), remain in the room (since ( A ) has at least 14 friends).6. Pick Another Student ( B ) from the Remaining: - Choose another student ( B ) from these 15 students who is different from ( A ). - Similar to ( A ), ( B ) also has at least 14 friends. - Remove students who are not friends with ( B ), which can be at most 5 students.7. Count Students Still in the Room: - At least ( 15 - 5 = 10 ) students, including ( B ), remain in the room.8. Pick Another Student ( C ): - Choose another student ( C ) among these 10 students different from ( A ) and ( B ). - Similar to the previous steps, ( C ) also has at least 14 friends. - Remove students who are not friends with ( C ), making at most 5 students leave.9. Count Students Still in the Room: - At least ( 10 - 5 = 5 ) students remain in the room.10. Identify the Clique: - These 5 students include ( A ), ( B ), ( C ), and at least 2 more students ( D ) and ( E ) who are friends with ( A ), ( B ), and ( C ). - Therefore, ( A ), ( B ), ( C ), and ( D ) form the clique of 4 students who all know each other. ConclusionTherefore, it can be concluded that there exists a subset of 4 students in the class who are all friends with each other.[boxed{text{Yes}}]

🧠:Okay, so the problem is asking: How many minutes after 17:00 will the angle between the hour and minute hands be exactly the same again? Hmm, let's break this down. I need to find a time after 5:00 PM when the angle between the hour and minute hands is the same as it was at 5:00 PM. Wait, but actually, maybe not exactly the same angle but when the angle between them repeats again? Wait, the question says "exactly the same again," so perhaps after 17:00, when is the next time the angle between the hands is equal to the angle that existed at 17:00. Let me confirm that.At 17:00, which is 5:00 PM, the hour hand is exactly on the 5, and the minute hand is exactly on the 12. The angle between them at that time is... Well, each hour mark is 30 degrees because 360 degrees divided by 12 hours is 30 degrees per hour. So from 12 to 5 is 5 hours, so 5*30 = 150 degrees. So the initial angle at 17:00 is 150 degrees. So we need to find the next time after 17:00 when the angle between the hour and minute hands is also 150 degrees again.Alright, so how do the angles of the hands change over time? The minute hand moves 360 degrees in 60 minutes, so that's 6 degrees per minute. The hour hand moves 30 degrees per hour, which is 0.5 degrees per minute. So the minute hand moves at 6 degrees per minute, and the hour hand moves at 0.5 degrees per minute. The angle between them changes over time as the minute hand catches up to and passes the hour hand.The formula for the angle between the two hands at any given time can be expressed as |30H - 5.5M| degrees, where H is the hour and M is the minutes past the hour. Wait, let me verify that. Since the hour hand moves 0.5 degrees per minute, after M minutes past the hour, it's at 30H + 0.5M degrees. The minute hand is at 6M degrees. So the angle between them is |30H + 0.5M - 6M| = |30H - 5.5M|. That's correct. And this angle should be equal to 150 degrees again after 17:00.But since we're starting at 17:00, which is 5:00, H is 5. So we can set up the equation |30*5 - 5.5M| = 150. Let's compute 30*5: that's 150. So the equation becomes |150 - 5.5M| = 150. Now, solving for M, where M is the number of minutes after 17:00.The absolute value equation splits into two cases:1. 150 - 5.5M = 1502. 150 - 5.5M = -150Let's solve each case.Case 1: 150 - 5.5M = 150Subtract 150 from both sides: -5.5M = 0 => M = 0. Which makes sense because that's the initial time 17:00. So we need the next occurrence, so we can ignore this solution.Case 2: 150 - 5.5M = -150Subtract 150 from both sides: -5.5M = -300Divide both sides by -5.5: M = (-300)/(-5.5) = 300/5.5Compute 300 divided by 5.5. Let's see. 5.5 is equal to 11/2, so 300 divided by (11/2) is 300*(2/11) = 600/11 ≈ 54.545454... minutes. So approximately 54 and 6/11 minutes.So the angle between the hour and minute hands will be 150 degrees again approximately 54 and 6/11 minutes after 17:00. Therefore, the answer should be 600/11 minutes, which is 54 and 6/11 minutes.But wait, let me verify this because sometimes when dealing with angles on the clock, there can be another scenario where the angle is the same but in the other direction. Wait, but the absolute value takes care of that. So the angle between the hands is always the smallest angle, but since the problem says "the angle," maybe it refers to the absolute difference, not considering the direction.Wait, at 17:00, the angle is 150 degrees. Then, as time passes, the minute hand moves faster than the hour hand. The angle between them decreases until they overlap, then increases again. But when does it reach 150 degrees again?Wait, another way to think about it is that the hands form a certain angle, and the next time they form the same angle is when the minute hand has lapped the hour hand in terms of the angle difference. Hmm, but maybe the general formula is that the time between consecutive occurrences of a particular angle (other than 0 and 180 degrees) is 12/11 hours or 720/11 minutes. But wait, that might be for the time between overlaps or something else.Wait, perhaps the time between two occurrences of the same angle is 12/11 hours, but since 12/11 hours is approximately 65.4545 minutes. But in this case, we're getting 54.5454 minutes. So maybe my previous answer is wrong?Wait, maybe I need to check the formula again.Alternatively, maybe there's a different approach. Let's model the movement of the hands.Let t be the number of minutes after 17:00.The position of the hour hand in degrees: starting at 150 degrees (since at 5:00, it's at 150 degrees), and moving at 0.5 degrees per minute. So hour angle = 150 + 0.5t.The position of the minute hand in degrees: starting at 0 degrees, moving at 6 degrees per minute. So minute angle = 6t.The angle between them is the absolute difference between these two angles, modulo 360. But since angles on the clock are typically taken as the smallest angle, we take the minimum between the absolute difference and 360 - absolute difference. But in this problem, since the original angle is 150 degrees, which is already the smaller angle (as 360 - 150 = 210, which is larger), so the angle we want is 150 degrees again, either in the same direction or the other direction.But the problem says "the angle between the hour and minute hands be exactly the same again," so it could be 150 degrees or 210 degrees, but since 150 is the smaller angle, maybe the question is referring to 150 degrees. But to be precise, let's check.The formula for the angle between the two hands is |30H - 5.5M|, as before, and we take the minimum between that and 360 - that. So when the angle calculated is over 180, we subtract it from 360. So at 17:00, it's 150 degrees, which is less than 180, so that's the angle we want.So we need to find t minutes after 17:00 when |30*5 - 5.5t| = 150 or 210 (since 360 - 150 = 210), but the problem says "exactly the same again," so likely 150 degrees. Wait, but the equation |30H - 5.5t| can result in 150 or 210, but since H is 5, but t is the minutes after 5:00. Wait, actually, H is fixed at 5 because we are in the 5 o'clock hour. Wait, no, actually, once the time passes 5:00, the hour hand moves as well.Wait, perhaps the formula should be adjusted. Wait, the formula is |30H - 5.5M| where H is the hour, and M is the minutes past H o'clock. But in our case, after 17:00, as time passes, H remains 5 until 18:00. But once t minutes pass 17:00, the hour is still 5, but M is t. Wait, but the formula is for any time H:M. So if we're in the 5 o'clock hour, H is 5, M is t. So the angle is |30*5 - 5.5t| = |150 - 5.5t|. So we set this equal to 150 degrees (since that's the original angle) and solve for t. But as before, the absolute value gives two equations.But in that case, solving |150 - 5.5t| = 150 gives t = 0 (which is 17:00) and t = 300/5.5 ≈ 54.5454 minutes. So that's the first time after 17:00 when the angle is again 150 degrees. However, another thought: the angle between the hands can also reach 150 degrees when the minute hand is behind the hour hand. Wait, but in that case, maybe the formula would give 150 degrees again.Wait, perhaps. Let's see. Let's think in terms of the relative speed. The minute hand moves at 6 degrees per minute, the hour hand at 0.5 degrees per minute, so the relative speed of the minute hand with respect to the hour hand is 6 - 0.5 = 5.5 degrees per minute.At 17:00, the angle between them is 150 degrees. To reach the same angle again, the minute hand needs to gain 360 - 150*2 = 60 degrees? Wait, maybe not. Alternatively, when the minute hand has gained 360 degrees relative to the hour hand, the angles repeat. But since we want the angle to be the same again, the time between two consecutive same angles is 360 / (5.5) ≈ 65.4545 minutes. But that's the general case. However, in this case, since we start at 150 degrees, which is a specific angle, maybe the first occurrence is after 54.5454 minutes, and then again after another 65.4545 minutes. Hmm, maybe.Wait, let's try to model the angle as a function of time.Let t be minutes after 17:00.Hour hand position: 150 + 0.5t degrees.Minute hand position: 0 + 6t degrees.The angle between them is |(150 + 0.5t) - 6t| = |150 - 5.5t|. Then, since angles are modulo 360, we take the minimum of this value and 360 - this value.But we need this angle to be 150 degrees again. So either:1. |150 - 5.5t| = 150, which gives two solutions:150 - 5.5t = 150 ⇒ t = 0150 - 5.5t = -150 ⇒ -5.5t = -300 ⇒ t = 300/5.5 = 54.5454...Or2. 360 - |150 - 5.5t| = 150 ⇒ |150 - 5.5t| = 210Which would mean:150 - 5.5t = 210 ⇒ -5.5t = 60 ⇒ t = -60/5.5 ≈ -10.909 minutes (invalid as t must be positive)Or 150 - 5.5t = -210 ⇒ -5.5t = -360 ⇒ t = 360/5.5 ≈ 65.4545 minutesSo here, t ≈ 65.4545 minutes is another solution. Wait, so does that mean there are two times after 17:00 when the angle is 150 degrees? At approximately 54.5454 minutes and 65.4545 minutes? Wait, but how does that work?Wait, the angle between the hands is decreasing initially as the minute hand catches up to the hour hand. The angle reaches 0 degrees when they overlap, then starts increasing again. So after 17:00, the angle decreases from 150 degrees to 0 degrees (when they overlap), then increases back to 180 degrees (when they are opposite), and then starts decreasing again. So the angle of 150 degrees will occur once when the minute hand is approaching the hour hand (before overlapping) and once when it's moving away after overlapping. Wait, but that contradicts the previous equations.Wait, let's think step by step.At 17:00, angle is 150 degrees.As the minute hand moves faster, the angle decreases until the hands overlap. The time it takes to overlap can be calculated. The initial angle is 150 degrees, and the relative speed is 5.5 degrees per minute. So time to overlap is 150 / 5.5 ≈ 27.2727 minutes. So at around 17:27:16, the hands overlap, angle 0 degrees.After that, the minute hand continues moving, and the angle between them starts increasing. It will reach 180 degrees at some point, then start decreasing again. Wait, no. Actually, after overlapping, the minute hand is ahead of the hour hand. The angle between them will increase until the minute hand is 180 degrees ahead, then starts decreasing as the minute hand approaches the hour hand from the other side.But in order for the angle to be 150 degrees again, it can happen when the minute hand is ahead of the hour hand by 150 degrees or behind by 150 degrees. But since the minute hand is moving faster, it can only be behind initially, then after overlapping, it's ahead.Wait, perhaps another approach. Let's consider two scenarios:1. The minute hand is behind the hour hand by 150 degrees.2. The minute hand is ahead of the hour hand by 150 degrees.At 17:00, scenario 1 is true. Then, as time passes, the minute hand catches up, reducing the angle. Then, after overlapping, the minute hand is ahead, and the angle increases again. So the angle of 150 degrees can occur again when the minute hand is ahead by 150 degrees.So in total, there are two times when the angle is 150 degrees: once when the minute hand is behind (at 17:00) and once when it's ahead (later).But since the problem says "after 17:00", we need the next occurrence, which would be when the minute hand is ahead by 150 degrees. So perhaps my earlier calculation of 54.5454 minutes is incorrect, and the correct answer is 65.4545 minutes?Wait, let's re-examine the equations.We have the angle between the hands given by |150 - 5.5t|. We want this angle to be 150 degrees. So:Case 1: 150 - 5.5t = 150 ⇒ t = 0 (the starting time)Case 2: 150 - 5.5t = -150 ⇒ -5.5t = -300 ⇒ t = 300/5.5 ≈ 54.5454 minutes.But wait, in this case, the absolute value gives 150 degrees regardless of direction. So if 150 - 5.5t is negative, then the absolute value is 5.5t - 150, which would be the angle when the minute hand is ahead.Wait, hold on. Let's clarify:The formula |150 - 5.5t| gives the absolute difference between the hour and minute hands. However, the actual angle on the clock is the minimum between this value and 360 minus this value. For example, if the difference is 250 degrees, the actual angle considered is 110 degrees. So in our problem, when we set |150 - 5.5t| = 150, we are not considering the minimum angle. Wait, but in reality, the angle between the hands is defined as the smallest angle, so it's always between 0 and 180 degrees. Therefore, if the absolute difference is greater than 180, we subtract it from 360 to get the smaller angle.But in our case, at t = 0, the difference is 150 degrees. Then, as t increases, the difference decreases until t ≈ 27.27 minutes (overlap), then the difference increases again but now the minute hand is ahead. So when does the difference (without considering the minimum) reach 150 again? Let's think.At t = 54.5454 minutes, the absolute difference |150 - 5.5*54.5454| = |150 - 300| = 150. But since the minute hand is ahead by 150 degrees, the actual angle is 150 degrees, which is acceptable because it's less than 180. So this is a valid time.Alternatively, when t = 65.4545 minutes, the absolute difference would be |150 - 5.5*65.4545| = |150 - 360| = 210. Then, the actual angle is 360 - 210 = 150 degrees. So that's also a valid time. Wait, so here's the confusion.If we solve |150 - 5.5t| = 150, we get t = 0 and t = 54.5454. But solving for when the angle (the minimal angle) is 150 degrees, we need to consider |150 - 5.5t| and 360 - |150 - 5.5t|, whichever is smaller. But since 150 is already less than 180, the equation we need is either |150 - 5.5t| = 150 or 360 - |150 - 5.5t| = 150. Wait, 360 - |150 - 5.5t| = 150 ⇒ |150 - 5.5t| = 210. Which would lead to the solutions t = (150 - 210)/-5.5 = (-60)/-5.5 ≈ 10.909 minutes (invalid as t must be positive after 17:00) and t = (150 + 210)/5.5 = 360/5.5 ≈ 65.4545 minutes. So that's another solution.Therefore, the two times after 17:00 when the angle between the hands is 150 degrees are approximately 54.5454 minutes and 65.4545 minutes. But since we are looking for the first time after 17:00 when this happens, it's the smaller t, which is 54.5454 minutes. However, wait a second. At t = 54.5454 minutes, which is 54 minutes and ~32.7 seconds after 17:00, the absolute difference is 150 degrees. But since the minute hand is moving ahead, at that time, is the angle between the hands 150 degrees in the other direction?Wait, let's compute the actual positions.At t = 54.5454 minutes:Hour hand position: 150 + 0.5*54.5454 ≈ 150 + 27.2727 ≈ 177.2727 degrees.Minute hand position: 6*54.5454 ≈ 327.2727 degrees.The difference between them is |327.2727 - 177.2727| = 150 degrees. Since this is less than 180, it is the angle between them. So this is valid.At t = 65.4545 minutes:Hour hand position: 150 + 0.5*65.4545 ≈ 150 + 32.7273 ≈ 182.7273 degrees.Minute hand position: 6*65.4545 ≈ 392.7273 degrees. But since 392.7273 is more than 360, subtract 360 to get 32.7273 degrees.Difference is |32.7273 - 182.7273| = 150 degrees. Which is again the minimal angle. So both times are valid. So after 17:00, the angle between the hands is 150 degrees at approximately 54.5454 minutes and again at 65.4545 minutes. Therefore, the first occurrence is at 54.5454 minutes, which is 54 and 6/11 minutes.But wait, the problem says "exactly the same again". Does that mean the same configuration, i.e., the same angle in the same direction? At 17:00, the hour hand is at 150 degrees, minute at 0. So the angle is 150 degrees with the hour hand leading. At 54.5454 minutes later, the hour hand is at ~177.27 degrees, minute at ~327.27 degrees. The angle between them is 150 degrees with the minute hand now leading by 150 degrees. So the angle is the same in magnitude but opposite in direction. If the question counts that as the same angle, then 54.5454 minutes is the answer. However, if it requires the same orientation (hour hand leading by 150 degrees), then maybe the 65.4545 minutes is the answer. Wait, but let's check that.Wait, in the case of t = 65.4545 minutes:Hour hand is at ~182.73 degrees, minute at ~32.73 degrees. The angle between them is |32.73 - 182.73| = 150 degrees. So here, the hour hand is at 182.73, which is past 6 (180 degrees), and the minute hand is at 32.73, which is just past 12. So the angle is 150 degrees with the hour hand still leading but now past 6. But since 182.73 - 32.73 = 150, but modulo 360, the minimal angle is 150. So actually, regardless of direction, the angle is 150. So both times are valid. Therefore, the first time after 17:00 is 54 and 6/11 minutes.But let me confirm with another method. The general formula for the times when the angle θ occurs is given by:t = (θ + 30H)/5.5 and t = (θ - 30H)/5.5, but adjusted for the modulus.Wait, perhaps not. Let me recall that the formula to find the times when the angle θ is formed is:θ = |30H - 5.5M|So solving for M:30H - 5.5M = ±θSo M = (30H ± θ)/5.5At H = 5, θ = 150 degrees.So M = (30*5 ± 150)/5.5 = (150 ± 150)/5.5So:M = (150 + 150)/5.5 = 300/5.5 ≈ 54.5454 minutesM = (150 - 150)/5.5 = 0/5.5 = 0 minutesSo that's the two solutions. Therefore, the next time is 300/5.5 minutes after 17:00, which is 600/11 minutes ≈54.5454 minutes.Therefore, the answer should be 600/11 minutes, which is 54 and 6/11 minutes.Therefore, the correct answer is boxed{dfrac{600}{11}} minutes.But to confirm, let's check another resource or method.Another way: the hands overlap every 12/11 hours, which is approximately 65.4545 minutes. But overlaps are when the angle is 0. Similarly, for other angles, the time between occurrences is 12/11 hours, but the first occurrence after a given time might be different.Wait, but in our case, the angle is 150 degrees. The formula for the time between two consecutive angles of θ is 12/11 hours if θ is 0, but for other angles, it's different. Wait, no, actually, for any angle θ (except 0 and 180 which have different frequencies), the time between two consecutive occurrences is 12/11 hours or 24/11 hours depending on the angle. Wait, no, that might not be accurate.Alternatively, since the relative speed is 5.5 degrees per minute, to cover 360 degrees, it takes 360/5.5 = 65.4545 minutes. So every 65.4545 minutes, the minute hand gains 360 degrees on the hour hand. Therefore, any particular angle (other than 0) will occur twice every 65.4545 minutes. Wait, but that might not be the case.Wait, for example, between two overlaps (0 degrees), there's a period of 65.4545 minutes. But in between, the angle reaches 180 degrees once. So for angles like 90 degrees, they occur twice every hour. Wait, perhaps for each angle θ (0 < θ < 180), there are two occurrences in every 12 hour period. But how does that translate to time between them?Alternatively, the time between occurrences of the same angle is 12/11 hours ≈65.4545 minutes. But if we start at a specific angle, the first occurrence after that could be less.Wait, but in our specific problem, we started at 150 degrees. The minute hand is behind the hour hand by 150 degrees. Then, after some time, the minute hand catches up, overtakes the hour hand, and then later is ahead by 150 degrees. Then, as time goes on, the minute hand laps the hour hand, and the angle cycles. So the time between being behind by θ and ahead by θ is 360/(5.5) ≈65.4545 minutes. But in this case, since we start at behind by 150 degrees, the next time it's ahead by 150 degrees would be after 300/5.5 minutes (the time to cover 150 degrees to catch up and then 150 degrees ahead). Wait, 150 + 150 = 300 degrees. So 300/5.5 ≈54.5454 minutes. Wait, that matches our earlier answer. So the first time when the angle is again 150 degrees, but in the opposite direction, is after 54.5454 minutes. Then, the next time it's 150 degrees again (behind) would be after another 65.4545 minutes. So the cycle is 54.5454 + 65.4545 ≈120 minutes, which is 2 hours, which makes sense because in 12 hours, it happens 22 times (11 times for each direction). But perhaps I'm overcomplicating.In any case, based on the equations, the first time after 17:00 when the angle between the hour and minute hands is again 150 degrees is 600/11 minutes, which is approximately 54.5454 minutes. Therefore, the answer is 600/11 minutes, which is the exact value.I need to ensure that there isn't an error in assuming that the absolute angle is the same regardless of direction. Since the problem states "the angle between the hour and minute hands," which is defined as the smallest angle between them, and 150 degrees is less than 180, both instances when the hands are 150 degrees apart in either direction count as the same angle. Therefore, the first occurrence is indeed at 600/11 minutes.Therefore, the answer is boxed{dfrac{600}{11}}.

✅:To determine how many minutes after 17:00 the angle between the hour and minute hands will be exactly the same again, we start by noting that at 17:00, the angle is 150 degrees. The hour hand moves at 0.5 degrees per minute, and the minute hand moves at 6 degrees per minute. The angle between them can be modeled by the formula:[|30H - 5.5M|]where ( H ) is the hour and ( M ) is the minutes past the hour. At 17:00, ( H = 5 ) and we need to find ( M ) such that the angle is 150 degrees again. This gives us the equation:[|150 - 5.5M| = 150]Solving this absolute value equation, we consider two cases:1. ( 150 - 5.5M = 150 )2. ( 150 - 5.5M = -150 )For the first case, solving ( 150 - 5.5M = 150 ) gives ( M = 0 ), which is the initial time 17:00. For the second case, solving ( 150 - 5.5M = -150 ):[-5.5M = -300 implies M = frac{300}{5.5} = frac{600}{11} approx 54.5454]Thus, the angle between the hour and minute hands will be 150 degrees again at ( frac{600}{11} ) minutes after 17:00. This is the first occurrence after 17:00 when the angle is the same, considering the smallest angle between the hands.[boxed{dfrac{600}{11}}]

🔑:1. Determine the initial angle: At 17:00, the hour hand is at the 5-hour mark, and the minute hand is at 0 minutes. The angle between the hour and minute hands is calculated as follows: [ text{Angle} = left(frac{5}{12}right) times 360^circ = 150^circ ]2. Set up the time when the angle is the same again: Let ( x ) be the number of minutes after 17:00 when the angle between the minute and hour hands returns to exactly 150 degrees.3. Express the positions of the hands after ( x ) minutes: - The minute hand moves 6 degrees per minute, so in ( x ) minutes, it moves ( 6x ) degrees. - The hour hand moves 0.5 degrees per minute, so in ( x ) minutes, it moves ( 5 times 30 + 0.5x = 150 + 0.5x ) degrees from the 12 o'clock position. Hence, the difference in positions of the hour and minute hands after ( x ) minutes should be the same 150 degrees: [ left(6x - (150 + 0.5x)right) mod 360 = 150 ]4. Set up and solve the equation: Simplify the above equation: [ 6x - 150 - 0.5x = 150 ] [ 5.5x - 150 = 150 ] [ 5.5x = 300 ] [ x = frac{300}{5.5} = frac{3000}{55} = 54 frac{6}{11} ]5. Conclusion: Therefore, the time in minutes after 17:00 when the angle between the hour and minute hands is again exactly 150 degrees is:[boxed{54 frac{6}{11}}]

🧠:Okay, let me try to figure out this problem. The equation given is ( 3^{2x + 1} + (m-1)left(3^{x+1} - 1right) - (m-3) 3^{x} = 0 ), and we need to find the range of real number ( m ) such that this equation has two distinct real roots. Hmm, exponential equations can be tricky, but maybe I can simplify it by substitution. Let me think.First, exponential equations with the same base can often be simplified by substituting ( y = 3^x ), since ( 3^{2x} ) would then be ( y^2 ), and so on. Let me try that substitution here.Given ( y = 3^x ), then ( 3^{2x} = y^2 ), and ( 3^{2x + 1} = 3 cdot 3^{2x} = 3y^2 ). Similarly, ( 3^{x + 1} = 3 cdot 3^x = 3y ). Let's substitute these into the original equation.Original equation:( 3^{2x + 1} + (m - 1)(3^{x + 1} - 1) - (m - 3)3^x = 0 )Substituting:( 3y^2 + (m - 1)(3y - 1) - (m - 3)y = 0 )Now let's expand the terms step by step.First term: ( 3y^2 )Second term: ( (m - 1)(3y - 1) = 3(m - 1)y - (m - 1) )Third term: ( - (m - 3)y )So combining all these:( 3y^2 + 3(m - 1)y - (m - 1) - (m - 3)y = 0 )Now, let's combine like terms. Let's look at the terms with ( y ):First, ( 3(m - 1)y ) and ( - (m - 3)y ). Combining these:( [3(m - 1) - (m - 3)]y )Let me compute the coefficient:( 3(m - 1) - (m - 3) = 3m - 3 - m + 3 = 2m )So the linear term becomes ( 2m y ).The constant term is just ( - (m - 1) ).So now the equation is:( 3y^2 + 2m y - (m - 1) = 0 )So, the equation simplifies to ( 3y^2 + 2m y - (m - 1) = 0 ). Now, since ( y = 3^x ), and ( 3^x ) is always positive for any real x, we need to ensure that any solutions for y must be positive. Therefore, the quadratic equation in y must have two distinct positive real roots. Then, the original equation will have two distinct real roots if and only if the quadratic in y has two distinct positive roots. Therefore, the problem reduces to finding the range of m such that the quadratic equation ( 3y^2 + 2m y - (m - 1) = 0 ) has two distinct positive roots.So, let me recap: original equation in x leads to quadratic in y, which must have two distinct positive roots. Therefore, I need to apply the conditions for a quadratic equation to have two distinct positive real roots. These conditions are:1. The discriminant of the quadratic is positive (to ensure two distinct real roots).2. The sum of the roots is positive.3. The product of the roots is positive.Let me recall that for a quadratic equation ( ax^2 + bx + c = 0 ), the discriminant is ( D = b^2 - 4ac ). The sum of the roots is ( -b/a ), and the product is ( c/a ).So, applying this to our quadratic equation:Quadratic equation: ( 3y^2 + 2m y - (m - 1) = 0 )Here, ( a = 3 ), ( b = 2m ), ( c = - (m - 1) ).First, discriminant D must be positive:( D = (2m)^2 - 4 times 3 times (- (m - 1)) )Compute this:( D = 4m^2 - 4 times 3 times (-m + 1) = 4m^2 + 12(m - 1) )Simplify:( 4m^2 + 12m - 12 )So, discriminant D = ( 4m^2 + 12m - 12 ). For two distinct real roots, D > 0:( 4m^2 + 12m - 12 > 0 )Let's divide both sides by 4 to simplify:( m^2 + 3m - 3 > 0 )Now, let's solve the inequality ( m^2 + 3m - 3 > 0 ). To do this, first find the roots of the quadratic equation ( m^2 + 3m - 3 = 0 ).Using quadratic formula:( m = frac{ -3 pm sqrt{9 + 12} }{ 2 } = frac{ -3 pm sqrt{21} }{ 2 } )So the roots are ( m = frac{ -3 + sqrt{21} }{ 2 } ) and ( m = frac{ -3 - sqrt{21} }{ 2 } ).Since the coefficient of ( m^2 ) is positive, the parabola opens upwards. Therefore, the inequality ( m^2 + 3m - 3 > 0 ) holds when ( m < frac{ -3 - sqrt{21} }{ 2 } ) or ( m > frac{ -3 + sqrt{21} }{ 2 } ).So that's the first condition. Now, moving on to the second and third conditions for the quadratic equation to have two positive roots.Sum of roots: ( -b/a = - (2m)/3 ). For the sum to be positive:( -2m/3 > 0 )Multiply both sides by 3:( -2m > 0 )Divide by -2 (inequality flips):( m < 0 )Product of roots: ( c/a = [ - (m - 1) ] / 3 = (1 - m)/3 ). For the product to be positive:( (1 - m)/3 > 0 )Multiply both sides by 3:( 1 - m > 0 )Which gives:( m < 1 )So summarizing the three conditions:1. ( m < frac{ -3 - sqrt{21} }{ 2 } ) or ( m > frac{ -3 + sqrt{21} }{ 2 } )2. ( m < 0 )3. ( m < 1 )But since conditions 2 and 3 are both inequalities on m, and since condition 2 is stricter (since m < 0 is more restrictive than m < 1), the combined conditions from sum and product are m < 0.But we also need to combine this with the discriminant condition. So the discriminant requires either m < [ -3 - sqrt(21) ] / 2 or m > [ -3 + sqrt(21) ] / 2. However, the sum and product conditions require m < 0.Therefore, the valid range for m is the intersection of the discriminant condition with m < 0. So:Either m < [ -3 - sqrt(21) ] / 2 (which is less than 0) or m > [ -3 + sqrt(21) ] / 2, but m must be less than 0. However, [ -3 + sqrt(21) ] / 2 is approximately ( -3 + 4.583 ) / 2 ≈ 1.583 / 2 ≈ 0.7915, which is positive. Since m has to be less than 0, the second interval m > [ -3 + sqrt(21) ] / 2 is incompatible with m < 0. Therefore, only the interval m < [ -3 - sqrt(21) ] / 2 satisfies both the discriminant condition and the sum and product conditions.Wait, hold on, let me check that. Let me compute the numerical values for clarity.First, compute [ -3 - sqrt(21) ] / 2:sqrt(21) is approximately 4.58366, so:-3 - 4.58366 ≈ -7.58366, divided by 2 is approximately -3.79183.Similarly, [ -3 + sqrt(21) ] / 2 is ( -3 + 4.58366 ) / 2 ≈ 1.58366 / 2 ≈ 0.79183.Therefore, the discriminant condition is m < -3.79183 or m > 0.79183. However, the sum and product conditions require m < 0. Therefore, the intersection is m < -3.79183. So the only valid interval is m < (-3 - sqrt(21))/2.But wait, let me double-check the sum and product conditions.Sum of roots: -2m/3 > 0 => m < 0.Product of roots: (1 - m)/3 > 0 => 1 - m > 0 => m < 1.So the sum is more restrictive here. So combining with discriminant: we need m < (-3 - sqrt(21))/2 (approx -3.7918) to satisfy discriminant and sum and product conditions. Alternatively, if m > (-3 + sqrt(21))/2 (approx 0.7918), but that's conflicting with m < 0, so that interval is discarded.Hence, the range of m is m < (-3 - sqrt(21))/2.But wait, let me check if that's correct. Let me verify with an example. Suppose m is very negative, say m = -10. Then, substitute into the quadratic equation: 3y^2 + 2*(-10)y - (-10 -1) = 3y^2 -20y +11 =0.Compute discriminant: (-20)^2 -4*3*11=400 -132=268>0, so two real roots.Sum of roots: 20/3 ≈ 6.666>0.Product of roots: 11/3 ≈3.666>0. So, both roots positive. So, m=-10 would be acceptable. So, the quadratic equation would have two positive roots, hence original equation has two real roots.Wait, but according to our previous conclusion, m < (-3 - sqrt(21))/2 ≈-3.7918. So m=-10 is indeed less than -3.7918, so included in the range. Then, another example, m=-4, which is less than -3.7918. Let's check.Quadratic equation: 3y^2 + 2*(-4)y - (-4 -1)= 3y^2 -8y +5=0.Discriminant: 64 -60=4>0. Two real roots.Sum: 8/3 ≈2.666>0, product:5/3≈1.666>0. So two positive roots. Thus acceptable.Now, let's check m=-3.79183 approximately. Let's compute exact value of (-3 - sqrt(21))/2≈(-3 -4.58366)/2≈-7.58366/2≈-3.79183. At this point, discriminant is zero. So the quadratic equation would have a repeated root. But we need two distinct roots, so m must be strictly less than (-3 - sqrt(21))/2.Wait, but in our discriminant condition, we have D>0, so m must be in the intervals where D>0, which is m < (-3 - sqrt(21))/2 or m > (-3 + sqrt(21))/2. But since m must also be less than 0, the only valid interval is m < (-3 - sqrt(21))/2. Therefore, the range of m is all real numbers less than (-3 - sqrt(21))/2.But wait, let me confirm once again. Since when m approaches (-3 - sqrt(21))/2 from below, the discriminant approaches zero from above, so we have two distinct roots near that point. But when m equals (-3 - sqrt(21))/2, discriminant is zero, so a repeated root. However, we need two distinct real roots, so m must be strictly less than (-3 - sqrt(21))/2. But also, we need the quadratic equation to have two positive roots. So even if m is slightly less than (-3 - sqrt(21))/2, we need to ensure that the roots are positive. However, since for m < (-3 - sqrt(21))/2, the sum of roots is -2m/3. If m is negative, then -2m/3 is positive. And product of roots is (1 - m)/3. Since m is negative, 1 - m is greater than 1, so positive. Therefore, both sum and product are positive, so both roots are positive. Therefore, as long as discriminant is positive (i.e., m < (-3 - sqrt(21))/2), the quadratic equation has two distinct positive roots. Therefore, the original equation will have two distinct real roots. Therefore, the range of m is m < [ -3 - sqrt(21) ] / 2.But let me check another example. Let m = -5, which is less than (-3 - sqrt(21))/2≈-3.79183. Quadratic equation: 3y^2 + 2*(-5)y - (-5 -1)=3y^2 -10y +6=0.Discriminant: 100 -72=28>0. Two real roots. Sum:10/3≈3.333>0, product:6/3=2>0. Both roots positive. So yes, valid.Another example: m = (-3 - sqrt(21))/2 - 1 (which is less than the critical value). Let's compute:sqrt(21)≈4.58366, so (-3 -4.58366)/2≈-7.58366/2≈-3.79183. Subtract 1: m≈-4.79183.Quadratic equation:3y^2 +2*(-4.79183)y - (-4.79183 -1)=3y^2 -9.58366y +5.79183=0.Discriminant: (9.58366)^2 -4*3*5.79183≈91.84 -69.50≈22.34>0. Sum:9.58366/3≈3.194>0, product:5.79183/3≈1.930>0. So two positive roots. Hence valid.Therefore, conclusion: m must be less than (-3 - sqrt(21))/2. So the range is (-infty, (-3 - sqrt(21))/2 ).But wait, let me check if there's a possibility that when m is between (-3 - sqrt(21))/2 and 0, even though discriminant is negative, but maybe other roots? Wait, no. If m is between (-3 - sqrt(21))/2 and 0, discriminant would be negative (since discriminant is positive only when m < (-3 - sqrt(21))/2 or m > (-3 + sqrt(21))/2). So between those two values, discriminant is negative, so no real roots. So if m is between (-3 - sqrt(21))/2 and (-3 + sqrt(21))/2, the quadratic equation has no real roots. But since m needs to be less than 0, in the interval [ (-3 - sqrt(21))/2 , 0 ), the quadratic equation has no real roots. Therefore, only for m < (-3 - sqrt(21))/2, quadratic has two distinct real roots, which are positive, leading to two real roots in x.Therefore, the correct range is m < (-3 - sqrt(21))/2. So boxed answer is boxed{(-infty, dfrac{ -3 - sqrt{21} }{ 2 })}.But let me just re-examine the problem statement to ensure I didn't miss anything. The original equation is supposed to have two distinct real roots. Since we substituted y = 3^x, each positive root y corresponds to a real root x = log_3 y. Therefore, if the quadratic in y has two distinct positive roots, then the original equation has two distinct real roots. Therefore, our reasoning holds. Therefore, the answer is as above.Final AnswerThe range of the real number ( m ) is boxed{left(-infty, dfrac{-3 - sqrt{21}}{2}right)}.

✅:Given the equation ( 3^{2x + 1} + (m-1)left(3^{x+1} - 1right) - (m-3) 3^{x} = 0 ), we need to find the range of the real number ( m ) such that the equation has two distinct real roots.First, we substitute ( y = 3^x ). This transforms the equation into a quadratic equation in terms of ( y ):[3y^2 + (m - 1)(3y - 1) - (m - 3)y = 0]Simplifying this, we get:[3y^2 + 2my - (m - 1) = 0]For the original equation to have two distinct real roots, the quadratic equation in ( y ) must have two distinct positive roots. The conditions for a quadratic equation ( ax^2 + bx + c = 0 ) to have two distinct positive roots are:1. The discriminant ( D ) is positive.2. The sum of the roots is positive.3. The product of the roots is positive.For the quadratic equation ( 3y^2 + 2my - (m - 1) = 0 ):- The discriminant ( D ) is given by: [ D = (2m)^2 - 4 cdot 3 cdot (-(m - 1)) = 4m^2 + 12(m - 1) = 4m^2 + 12m - 12 ] For distinct real roots, ( D > 0 ): [ 4m^2 + 12m - 12 > 0 implies m^2 + 3m - 3 > 0 ] Solving ( m^2 + 3m - 3 = 0 ) using the quadratic formula: [ m = frac{-3 pm sqrt{21}}{2} ] The inequality ( m^2 + 3m - 3 > 0 ) holds for: [ m < frac{-3 - sqrt{21}}{2} quad text{or} quad m > frac{-3 + sqrt{21}}{2} ]- The sum of the roots ( -frac{2m}{3} ) must be positive: [ -frac{2m}{3} > 0 implies m < 0 ]- The product of the roots ( frac{1 - m}{3} ) must be positive: [ frac{1 - m}{3} > 0 implies m < 1 ]Combining these conditions, the most restrictive is ( m < 0 ). Intersecting this with the discriminant condition, we find that ( m ) must be less than ( frac{-3 - sqrt{21}}{2} ).Thus, the range of the real number ( m ) is:[boxed{left(-infty, dfrac{-3 - sqrt{21}}{2}right)}]

🔑:1. Define (3^x = t) where (t > 0). 2. Substitute (3^x = t) into the given equation: [ 3^{2x+1} + (m-1)(3^{x+1}-1) - (m-3)3^x = 0 ] Rewriting each term using (t): [ 3^{2x+1} = 3 cdot (3^x)^2 = 3t^2, ] [ 3^{x+1} = 3 cdot 3^x = 3t, ] [ (m-1)(3^{x+1}-1) = (m-1)(3t - 1), ] [ (m-3)3^x = (m-3)t. ] Hence, the equation becomes: [ 3t^2 + (m-1)(3t - 1) - (m-3)t = 0. ]3. Simplify the equation: [ 3t^2 + 3(m-1)t - (m-1) - (m-3)t = 0, ] [ 3t^2 + [3(m-1) - (m-3)]t - (m-1) = 0, ] [ 3t^2 + [3m - 3 - m + 3]t - (m-1) = 0, ] [ 3t^2 + 2mt - m + 1 = 0. ]4. For the quadratic equation (3t^2 + 2mt - m + 1 = 0) to have two distinct positive real roots, we need: [ left{ begin{array}{l} Delta > 0, t_1 + t_2 > 0, t_1 cdot t_2 > 0, end{array} right. ] where (Delta) is the discriminant of the quadratic equation.5. Calculate the discriminant: [ Delta = (2m)^2 - 4 cdot 3 cdot (-m + 1), ] [ Delta = 4m^2 + 12m - 12, ] [ Delta = 4(m^2 + 3m - 3). ] For two distinct real roots, (Delta > 0): [ 4(m^2 + 3m - 3) > 0, ] [ m^2 + 3m - 3 > 0. ]6. Solve the quadratic inequality (m^2 + 3m - 3 > 0) by finding the roots of the corresponding quadratic equation (m^2 + 3m - 3 = 0): [ m = frac{-3 pm sqrt{3^2 + 4 cdot 3}}{2 cdot 1}, ] [ m = frac{-3 pm sqrt{21}}{2}. ] The roots are: [ m = frac{-3 + sqrt{21}}{2} quad text{and} quad m = frac{-3 - sqrt{21}}{2}. ]7. Determine the intervals where (m^2 + 3m - 3 > 0) using a sign chart or test points: [ m < frac{-3 - sqrt{21}}{2} quad text{or} quad m > frac{-3 + sqrt{21}}{2}. ]8. To ensure the roots are positive: [ t_1 + t_2 = -frac{(2m)}{3} > 0, ] [ -frac{2m}{3} > 0, ] [ m < 0. ] Also: [ t_1 cdot t_2 = frac{1 - m}{3} > 0, ] [ 1 - m > 0, ] [ m < 1. ] 9. Combine the conditions: [ m < min(frac{-3 - sqrt{21}}{2}, 1). ] Since (frac{-3 - sqrt{21}}{2} < 1): [ m < frac{-3 - sqrt{21}}{2}. ] Conclusion:[boxed{m < frac{-3 - sqrt{21}}{2}}]

🧠:Okay, so I have this problem here about a sequence of non-negative integers ( c_1, c_2, ldots, c_{1997} ) that satisfies two conditions: first, ( c_1 geq 0 ), and second, for all positive integers ( m, n ) with ( m + n < 1998 ), the inequality ( c_m + c_n leq c_{m+n} leq c_m + c_n + 1 ) holds. The task is to show that there exists a real number ( k ) such that ( c_n = lfloor nk rfloor ) for all ( 1 leq n leq 1997 ).Hmm, let's start by understanding what this inequality is saying. For any two indices ( m ) and ( n ), the value at ( m + n ) is at least the sum of the values at ( m ) and ( n ), but at most that sum plus one. This seems like a subadditive condition but with a slight relaxation—allowing the value at ( m + n ) to exceed the sum by at most one. The problem is asking to show that such a sequence can be expressed as the floor of ( nk ) for some real number ( k ).I remember that sequences which are subadditive (i.e., ( c_{m+n} leq c_m + c_n )) often relate to linear functions or functions with a linear bound. Here, the condition is a bit different because there's also a lower bound. So it's almost additive, but with a possible error of 1. This reminds me of properties of floor functions. For example, if ( c_n = lfloor nk rfloor ), then ( c_m + c_n leq lfloor mk rfloor + lfloor nk rfloor leq lfloor (m + n)k rfloor + 1 ), which is similar to the given inequality. Wait, maybe that's the key?Let me check. Suppose ( c_n = lfloor nk rfloor ). Then, ( c_{m + n} = lfloor (m + n)k rfloor ). Also, ( c_m + c_n = lfloor mk rfloor + lfloor nk rfloor ). We know that ( lfloor mk rfloor + lfloor nk rfloor leq lfloor (m + n)k rfloor leq lfloor mk rfloor + lfloor nk rfloor + 1 ). Wait, is that always true? Let's see.Actually, ( lfloor a rfloor + lfloor b rfloor leq lfloor a + b rfloor leq lfloor a rfloor + lfloor b rfloor + 1 ). This is a standard inequality for floor functions. So yes, if ( c_n = lfloor nk rfloor ), then the given inequalities hold. Therefore, the problem is essentially asking to prove the converse: that any sequence satisfying these inequalities must be of the form ( lfloor nk rfloor ).So how do we approach this? Maybe by constructing such a ( k ). One common approach with floor functions is to consider the limit of ( c_n / n ) as ( n ) approaches infinity. But here, the sequence is finite (up to 1997), so we can't take a limit. However, maybe we can define ( k ) as the supremum of ( c_n / n ) or something like that.Alternatively, we can use induction or recursive relations. Let's think step by step.First, let's note that ( c_1 geq 0 ). Let's denote ( c_1 = lfloor k rfloor ). Wait, but if ( k ) is between, say, ( c_1 ) and ( c_1 + 1 ), then ( lfloor k rfloor = c_1 ). So maybe ( k ) is chosen such that ( c_1 = lfloor k rfloor ), ( c_2 = lfloor 2k rfloor ), etc. But how do we ensure that such a ( k ) exists for all terms up to 1997?Perhaps we need to define ( k ) as the infimum of ( (c_n + 1)/n ) over all ( n ), or something like that. Let me recall that for subadditive functions, Fekete's lemma says that the limit ( lim_{n to infty} c_n / n ) exists, but again, here it's finite. But maybe a similar idea applies.Suppose we define ( k = sup { c_n / n | 1 leq n leq 1997 } ). Then, for each ( n ), ( c_n leq nk ). But we need ( c_n = lfloor nk rfloor ). Alternatively, maybe ( k ) should be the infimum of ( (c_n + 1)/n ), so that ( nk leq c_n + 1 ).Wait, here's a thought. If we can find a ( k ) such that for all ( n ), ( c_n leq nk < c_n + 1 ), then ( lfloor nk rfloor = c_n ). So such a ( k ) must satisfy ( c_n / n leq k < (c_n + 1)/n ) for all ( n ).Therefore, the problem reduces to showing that the intersection of all intervals ( [c_n / n, (c_n + 1)/n) ) for ( 1 leq n leq 1997 ) is non-empty. In other words, there exists a real number ( k ) that lies in all these intervals simultaneously.So, to show that ( bigcap_{n=1}^{1997} [c_n / n, (c_n + 1)/n) ) is non-empty. For this, it's sufficient to show that for any ( m, n ), ( c_m / m leq (c_n + 1)/n ). Because then the maximum of all lower bounds ( c_n / n ) would be less than or equal to the minimum of all upper bounds ( (c_n + 1)/n ), so the interval between them would be non-empty.Therefore, if we can show that for all ( m, n ), ( c_m / m leq (c_n + 1)/n ), then the intersection is non-empty.Alternatively, the necessary and sufficient condition for the intersection to be non-empty is that for any ( m, n ), ( c_m / m leq (c_n + 1)/n ). Let's check if this holds under the given conditions.Given the inequalities ( c_{m + n} leq c_m + c_n + 1 ), perhaps we can derive bounds on ( c_n / n ).Let me try to analyze the behavior of ( c_n ). Let's start with some small ( n ).For ( n = 1 ), ( c_1 geq 0 ). Let's denote ( k_1 = c_1 ). If we suppose ( c_1 = lfloor k rfloor ), then ( k ) must be in [c1, c1 +1).For ( n = 2 ), ( c_2 leq c_1 + c_1 +1 = 2c1 +1 ), and ( c_2 geq c1 + c1 = 2c1 ). So ( c2 in [2c1, 2c1 +1] ). Since ( c2 ) is an integer, it's either ( 2c1 ) or ( 2c1 +1 ).Similarly, if we take ( k ) such that ( c2 = lfloor 2k rfloor ), then ( 2k ) must be in [c2, c2 +1). So ( k in [c2 / 2, (c2 +1)/2) ).But since ( c1 leq k < c1 +1 ), and ( c2 /2 leq k < (c2 +1)/2 ), we need these intervals to overlap. Since ( c2 in [2c1, 2c1 +1] ), then c2 /2 is in [c1, c1 + 0.5], and (c2 +1)/2 is in [c1 + 0.5, c1 +1]. So the interval [c2 /2, (c2 +1)/2) is within [c1, c1 +1). Therefore, the intersection of [c1, c1 +1) and [c2 /2, (c2 +1)/2) is [max(c1, c2 /2), min(c1 +1, (c2 +1)/2)). But given that c2 is either 2c1 or 2c1 +1, let's check:If c2 = 2c1: then [c2 /2, (c2 +1)/2) = [c1, c1 + 0.5). So overlapping with [c1, c1 +1) gives [c1, c1 +0.5).If c2 = 2c1 +1: then [c2 /2, (c2 +1)/2) = [c1 +0.5, c1 +1). So overlapping with [c1, c1 +1) gives [c1 +0.5, c1 +1).Therefore, depending on c2, the overlapping interval is either [c1, c1 +0.5) or [c1 +0.5, c1 +1). So k must lie in this narrower interval.Similarly, moving to n=3:c3 satisfies c1 + c2 <= c3 <= c1 + c2 +1.But c1 is known, c2 is either 2c1 or 2c1 +1. Let's suppose c2 = 2c1. Then c3 >= c1 + 2c1 = 3c1, and c3 <= 3c1 +1. Similarly, if c2 = 2c1 +1, then c3 >= c1 + (2c1 +1) = 3c1 +1, and c3 <= 3c1 + 2.But since c3 must be an integer, if c2 = 2c1, then c3 can be 3c1 or 3c1 +1. If c2 = 2c1 +1, then c3 can be 3c1 +1 or 3c1 +2.So, in terms of k:If c3 = 3c1, then k must satisfy 3c1 <= 3k < 3c1 +1, so k in [c1, c1 + 1/3).If c3 = 3c1 +1, then k in [c1 +1/3, c1 + 2/3).But previously, from n=2, when c2=2c1, k is in [c1, c1 +0.5). So the overlap with n=3 would be either [c1, c1 +1/3) or [c1 +1/3, c1 +0.5), depending on c3.Similarly, if c2=2c1 +1, then k is in [c1 +0.5, c1 +1), and c3 can be 3c1 +1 or 3c1 +2, leading to k in [c1 +1/3, c1 +2/3) or [c1 +2/3, c1 +1). But overlapping with [c1 +0.5, c1 +1) would give [c1 +0.5, c1 +2/3) if c3=3c1 +1, or [c1 +2/3, c1 +1) if c3=3c1 +2.So as n increases, each term c_n restricts k to a certain interval, which must be compatible with previous intervals. The key is that these intervals are overlapping in a way that they have a non-empty intersection.But how do we ensure that for all n up to 1997, the intervals [c_n /n, (c_n +1)/n) have a common intersection? The problem is equivalent to showing that the system of inequalities c_n /n ≤ k < (c_n +1)/n for all n has a solution k.This is equivalent to showing that for any m and n, c_m /m < (c_n +1)/n. Because if that's true, then the maximum of all lower bounds c_n /n is less than the minimum of all upper bounds (c_n +1)/n, so the interval between them is non-empty.Therefore, we need to prove that for any m, n ≤ 1997, c_m /m < (c_n +1)/n. Let's see.Given the original condition: for any m, n with m +n <1998, c_{m +n} <= c_m + c_n +1 and c_{m +n} >= c_m + c_n.Let’s try to relate c_n to c_m. For example, take m = n. Then c_{2n} <= 2c_n +1 and c_{2n} >= 2c_n. Similarly, for m = 2n, n, we can get c_{3n} <= c_{2n} + c_n +1 <= 3c_n +2, etc. It seems that c_n grows roughly linearly, which is consistent with the floor function of nk.Alternatively, let's consider that for any n, c_n <= nk < c_n +1. Then, if we can express k in terms of the given c_n's, maybe through induction.Let me try to construct k step by step. Suppose we start with n=1. Then k must be in [c1, c1 +1). For n=2, k must be in [c2 /2, (c2 +1)/2). The intersection of these intervals must be non-empty. As we saw earlier, since c2 is either 2c1 or 2c1 +1, the interval [c2 /2, (c2 +1)/2) is either [c1, c1 +0.5) or [c1 +0.5, c1 +1), which overlaps with [c1, c1 +1). So the intersection is either [c1, c1 +0.5) or [c1 +0.5, c1 +1). Let's call this interval I2.Next, for n=3, we need k to be in [c3 /3, (c3 +1)/3). The interval I3 must intersect with I2. Let's see.Suppose c2 = 2c1. Then I2 = [c1, c1 +0.5). Now c3 must satisfy c3 >= c1 + c2 = 3c1 and c3 <= c1 + c2 +1 = 3c1 +1. So c3 is either 3c1 or 3c1 +1.If c3 = 3c1, then I3 = [c1, (3c1 +1)/3) = [c1, c1 + 1/3). The intersection with I2 [c1, c1 +0.5) is [c1, c1 +1/3).If c3 = 3c1 +1, then I3 = [(3c1 +1)/3, (3c1 +2)/3) = [c1 +1/3, c1 +2/3). The intersection with I2 [c1, c1 +0.5) is [c1 +1/3, c1 +0.5).So in either case, the intersection is non-empty.Similarly, if c2 = 2c1 +1, then I2 = [c1 +0.5, c1 +1). c3 must satisfy c3 >= c1 + c2 = 3c1 +1 and c3 <= c1 + c2 +1 = 3c1 +2. So c3 is either 3c1 +1 or 3c1 +2.If c3 = 3c1 +1, then I3 = [(3c1 +1)/3, (3c1 +2)/3) = [c1 +1/3, c1 +2/3). Intersection with I2 [c1 +0.5, c1 +1) is [c1 +0.5, c1 +2/3).If c3 = 3c1 +2, then I3 = [(3c1 +2)/3, (3c1 +3)/3) = [c1 +2/3, c1 +1). Intersection with I2 [c1 +0.5, c1 +1) is [c1 +2/3, c1 +1).Again, the intersection is non-empty.So, inductively, each new n restricts the interval for k, but the intersection remains non-empty. So perhaps, continuing this way, even up to n=1997, the intervals will keep overlapping, ensuring a non-empty intersection. Therefore, there exists such a k.But how to formalize this? Maybe using induction on n.Assume that for all m <= n, the intervals [c_m /m, (c_m +1)/m) have a common intersection. Then for m = n +1, we need to show that the interval [c_{n+1}/(n+1), (c_{n+1} +1)/(n+1)) intersects with the previous intervals.But how is c_{n+1} related to previous c_m?Wait, we can write c_{n+1} using the given inequality. For example, take m = n and m' =1. Then c_{n+1} >= c_n + c1, and c_{n+1} <= c_n + c1 +1.But c1 is fixed. So if we know the interval for k from previous terms, can we relate c_{n+1}?Alternatively, since k must lie in [c_m /m, (c_m +1)/m) for all m <=n, then we can express bounds on k. Then, for m =n, k >= c_n /n and k < (c_n +1)/n.Similarly, for m=1, k >= c1 and k < c1 +1.Then, c_{n+1} is bounded by c_n + c1 <= c_{n+1} <= c_n + c1 +1.But c_n <= nk < c_n +1, and c1 <=k <c1 +1.Therefore, c_n + c1 <= nk +k = (n +1)k < c_n +1 + c1 +1.So, (n +1)k is in [c_n + c1, c_n + c1 + 2).But c_{n+1} is in [c_n + c1, c_n + c1 +1], so (n +1)k must be in [c_{n+1}, c_{n+1} +1) to have c_{n+1} = floor((n +1)k).But we have (n +1)k >= c_n + c1 >= c_{n+1} (from the lower bound of the given inequality). Wait, no. Wait, the given inequality says c_{n+1} >= c_n + c1. Therefore, (n +1)k >= c_{n+1}? Not necessarily. Because (n +1)k >= c_n + c1, but c_{n+1} >= c_n + c1. So (n +1)k >= c_{n+1} - something?Wait, maybe we need to use the previous intervals. Since k is in [c_m /m, (c_m +1)/m) for all m <=n.Therefore, k >= c_n /n, so (n +1)k >= (n +1)(c_n /n) = c_n + c_n /n.Similarly, k < (c_n +1)/n, so (n +1)k < (n +1)(c_n +1)/n = c_n +1 + (c_n +1)/n.But I'm not sure if this is helpful.Alternatively, let's use the given inequality for m =n and m' =1. Then c_{n+1} >= c_n + c1 and c_{n+1} <= c_n + c1 +1. Since c1 = floor(k), we have c1 <=k <c1 +1. Also, c_n <=nk <c_n +1. Therefore, c_n + c1 <= nk +k = (n +1)k <c_n +1 +k +1.But wait, c_{n+1} <= c_n +c1 +1, so (n +1)k >= c_n +c1 <= c_{n+1}, but (n +1)k <c_n +1 +k +1.This seems a bit messy. Maybe another approach.Suppose we define k as the limit superior of c_n /n. Since the sequence is finite, we can take k = max_{1 <=n <=1997} c_n /n. Wait, but if we take k to be the maximum of c_n /n, then for each n, c_n /n <=k. But we need k < (c_n +1)/n for all n, meaning that k must be less than all (c_n +1)/n. So if k is the maximum of c_n /n, then we need that for all n, c_n /n <=k < (c_n +1)/n. Which would mean that k is in [c_n /n, (c_n +1)/n) for all n. Therefore, k must be at least as large as every c_n /n, but less than every (c_n +1)/n. So this is equivalent to k >= sup c_n /n and k < inf (c_n +1)/n. Therefore, if sup c_n /n < inf (c_n +1)/n, then such a k exists.Hence, the problem reduces to showing that sup c_n /n < inf (c_n +1)/n. Which would follow if for all m, n, c_m /m < (c_n +1)/n.So, if for all m, n, c_m /m < (c_n +1)/n, then the supremum of c_m /m is less than the infimum of (c_n +1)/n, so there is a k in between.Therefore, we need to show that for any m, n, c_m /m < (c_n +1)/n.Let’s assume m and n are arbitrary indices between 1 and 1997. Let's try to relate c_m and c_n using the given inequalities.Let’s consider the case where m < n. Let’s write n = qm + r, where q is the quotient and r is the remainder, 0 <= r <m. Then, using the given inequality repeatedly, we can express c_n in terms of c_m and c_r.For example, let's use induction on q. For n = qm + r:c_{qm + r} <= c_{(q-1)m + r} + c_m +1<= c_{(q-2)m + r} + 2c_m +2...<= c_r + q c_m + qSimilarly, the lower bound:c_{qm + r} >= c_{(q-1)m + r} + c_m>= c_{(q-2)m + r} + 2c_m...>= c_r + q c_mTherefore, c_r + q c_m <= c_n <= c_r + q c_m + qBut since n = qm + r, we can write q = floor(n/m). Let's make this precise.Let’s divide n by m: n = m * q + r, where 0 <= r <m. Then, using the inequalities:c_n <= q c_m + c_r + qandc_n >= q c_m + c_rTherefore, c_n /n >= (q c_m + c_r)/nBut n = m q + r, so:(q c_m + c_r)/n = (q c_m + c_r)/(m q + r)Similarly, c_n /n <= (q c_m + c_r + q)/n = (q(c_m +1) + c_r)/nBut we need to relate this to c_m /m and c_n /n.Alternatively, let's express the inequalities in terms of k.Suppose we have k such that c_n = floor(nk). Then, nk -1 < c_n <= nk. So, c_n <= nk < c_n +1.Similarly, c_m <= mk <c_m +1.Therefore, c_n /n <=k < (c_n +1)/n and similarly for m.To show that c_m /m < (c_n +1)/n, we need to show that c_m /m < (c_n +1)/n for all m, n.Assume for contradiction that there exists m, n such that c_m /m >= (c_n +1)/n. Then, cross-multiplying, n c_m >= m (c_n +1).But n c_m >= m c_n + m.But using the given inequality, can we derive a contradiction?Let's take m and n such that n c_m >= m c_n + m.Consider the number of times m fits into n. Let q = floor(n/m), so n = q m + r with 0 <= r <m.From earlier, we have c_n <= q c_m + c_r + q.But we have n c_m >= m c_n + m.Substitute the upper bound of c_n:n c_m >= m (q c_m + c_r + q) + m=> n c_m >= m q c_m + m c_r + m q + mBut n = q m + r, so substitute:(q m + r) c_m >= m q c_m + m c_r + m q + mSimplify left side: q m c_m + r c_mRight side: q m c_m + m c_r + m q + mTherefore, subtract q m c_m from both sides:r c_m >= m c_r + m q + m=> r c_m >= m(c_r + q +1)But since r <m, and c_r is non-negative, the right side is at least m(q +1). The left side is r c_m. Since r <m and c_m >=0, the left side is at most (m -1)c_m.So:(m -1)c_m >= m(q +1)But q = floor(n/m). Let's think about specific values. For example, if n =m, then q =1, r=0. Then the inequality becomes 0 >= m(1 +1) => 0 >= 2m, which is impossible. So contradiction.If n =2m, then q=2, r=0. Then:0 >= m(2 +1) => 0 >= 3m, which is impossible.If n =m +1, then q=1, r=1. Then:1 *c_m >= m(c_1 +1 +1) => c_m >= m(c_1 +2). But since c_{m +1} >= c_m +c_1, so c_{m +1} >= c_m +c_1.But if c_m >= m(c_1 +2), then c_{m +1} >= m(c_1 +2) +c_1. However, the upper bound for c_{m +1} is c_m +c_1 +1. But if c_m is so large, then c_{m +1} <= c_m +c_1 +1 < m(c_1 +2) +c_1 +1. But this is possible only if m(c_1 +2) <= c_m +c_1 +1. But if c_m >=m(c_1 +2), then m(c_1 +2) <= c_m <= c_{m +1} -c_1 <= (c_m +c_1 +1) -c_1 =c_m +1. Therefore, m(c_1 +2) <=c_m +1. But since c_m >=m(c_1 +2), this implies m(c_1 +2) <=c_m +1 <= m(c_1 +2) +1, but c_m is an integer, so this would require c_m = m(c_1 +2). Then, c_{m +1} >=c_m +c_1 =m(c_1 +2) +c_1 =c_1(m +1) +2m. However, c_{m +1} <=c_m +c_1 +1 =m(c_1 +2) +c_1 +1 =c_1(m +1) +2m +1. But since c_{m +1} must also satisfy c_{m +1} <= (m +1)k <c_{m +1} +1. But this is getting too convoluted.Perhaps there's a better way. Let's suppose that for some m, n, c_m /m >= (c_n +1)/n. Then:n c_m >= m(c_n +1)But from the given inequality, we can relate c_n and c_m via multiple applications. For example, we can use the inequality multiple times to express c_n in terms of c_m and other terms.Alternatively, use induction. Suppose that for all m < N, c_m = floor(mk). Then for m =N, need to show c_N = floor(Nk). But this requires knowing k, which is what we're trying to find.Alternatively, use the concept of Beatty sequences or something related.Wait, another approach: Define k as the lim sup of c_n /n. Since the sequence is finite, k is just the maximum of c_n /n over n=1,...,1997. Let k = max_{1<=n<=1997} c_n /n. We need to show that for all n, c_n <= nk <c_n +1.First, by definition of k, c_n /n <=k for all n. So c_n <=nk.Now, need to show that nk <c_n +1 for all n. Assume for contradiction that for some n, nk >=c_n +1. Then, k >=(c_n +1)/n. But k is the maximum of c_m /m, so there exists some m where k =c_m /m. Therefore, c_m /m >= (c_n +1)/n => n c_m >=m(c_n +1). But we can use the given inequalities to show this leads to a contradiction.Indeed, from the given inequality, c_{m +n} <=c_m +c_n +1. But if n c_m >=m(c_n +1), then we can consider adding m n times.Wait, let's consider adding m n times. Let's apply the inequality inductively. For example, c_{2m} <=2c_m +1, c_{3m} <=3c_m +2, ..., c_{nm} <=n c_m + (n -1). Similarly, since n is a positive integer, if we set m such that k =c_m /m, then c_{nm} <=n c_m + (n -1) =m n k +n -1.But on the other hand, since k is the maximum, c_{nm} >= nm k. Therefore:nm k <=c_{nm} <=n c_m +n -1 =n m k +n -1Thus:nm k <=n m k +n -1Subtract n m k:0 <=n -1Which is true. But this doesn't lead to a contradiction. Hmm.Alternatively, since we assumed that nk >=c_n +1, so k >=(c_n +1)/n. But k is the maximum of c_m /m, so there exists m such that k =c_m /m >=(c_n +1)/n. Therefore, n c_m >=m(c_n +1). Let's use this in the inequality.Consider the term c_{n +m}. From the given condition:c_{n +m} <=c_n +c_m +1But also, from the lower bound:c_{n +m} >=c_n +c_mSo, c_n +c_m <=c_{n +m} <=c_n +c_m +1But since k =c_m /m, and we have nk >=c_n +1, then c_m =m k. So:c_{n +m} <=c_n +m k +1But also, by the definition of k, c_{n +m} >=(n +m)k. Therefore:(n +m)k <=c_{n +m} <=c_n +m k +1=> (n +m)k <=c_n +m k +1Subtract m k:n k <=c_n +1But we assumed that nk >=c_n +1. Therefore, nk =c_n +1. But nk must be an integer since c_n is an integer and nk =c_n +1. Therefore, k must be rational with denominator dividing n. But since k =c_m /m, which is rational with denominator m. Unless m divides n, this could be a contradiction. However, m and n are arbitrary, so this might not hold.Wait, but if nk =c_n +1, then k=(c_n +1)/n. But k is also equal to c_m /m. Therefore, c_m /m = (c_n +1)/n. So n c_m =m(c_n +1). Let's see if this is possible.Let's take m and n such that n c_m =m(c_n +1). Let's suppose m=1. Then n c_1 =1*(c_n +1). Since c_1 is an integer, c_n =n c_1 -1. But from the lower bound, c_n >=n c_1. So c_n >=n c_1, but here c_n =n c_1 -1, which is less than n c_1. Contradiction. Therefore, m cannot be 1.Similarly, take m=2. Then n c_2 =2(c_n +1). From the lower bound, c_n >=c_{n-1} +c_1 >=c_{n-2} +2c_1 >=...>=n c_1. So c_n >=n c_1. Then 2(c_n +1) >=2(n c_1 +1). So n c_2 >=2(n c_1 +1). But c_2 <=2c_1 +1. Therefore, n(2c_1 +1) >=2(n c_1 +1). Simplifying: 2n c_1 +n >=2n c_1 +2 =>n >=2. Which is possible if n>=2. But c_n >=n c_1, so c_2 >=2c_1. But c_2 <=2c_1 +1. Therefore, c_2 is either 2c_1 or 2c_1 +1.If c_2 =2c_1, then n*2c_1 =2(c_n +1) =>n c_1 =c_n +1. But c_n >=n c_1, so n c_1 <=c_n <=n c_1, which implies c_n =n c_1. But then n c_1 =c_n =n c_1, so 0=1. Contradiction.If c_2 =2c_1 +1, then n*(2c_1 +1) =2(c_n +1). So c_n +1= (n/2)(2c_1 +1). Therefore, c_n =n c_1 +n/2 -1. But c_n must be an integer. Therefore, n must be even. Let n=2k. Then c_n=2k c_1 +k -1. But from the lower bound, c_n >=2k c_1. Therefore, 2k c_1 +k -1 >=2k c_1 =>k -1 >=0 =>k >=1. Which is true. But c_n=2k c_1 +k -1 is not necessarily greater than or equal to 2k c_1, since k -1 could be zero or positive. Wait, if k >=1, then k -1 >=0. But k=1: c_n=2c1 +1 -1=2c1. Which is okay. For k=2: c_n=4c1 +2 -1=4c1 +1. Which is >=4c1. So this works. But in this case, c_n =2k c_1 +k -1. But is this compatible with the upper bound?For example, take n=2 (k=1). c_2=2c1. Which is allowed. For n=4, c4=4c1 +1. From the upper bound, c4 <=c2 +c2 +1=2c1 +2c1 +1=4c1 +1. So equality holds. So c4=4c1 +1 is allowed. Similarly, c3 >=c2 +c1=2c1 +c1=3c1, and c3 <=2c1 +c1 +1=3c1 +1. So c3 is either 3c1 or 3c1 +1. If we take n=3 and m=2, we need to check if 3c_2=2(c3 +1). If c3=3c1 +1, then 2(c3 +1)=2(3c1 +2)=6c1 +4. But 3c2=3*(2c1 +1)=6c1 +3. Not equal. If c3=3c1, then 2(c3 +1)=2(3c1 +1)=6c1 +2. 3c2=6c1 +3. Not equal. So this case doesn't lead to a contradiction. Therefore, even though for certain m and n, the equation n c_m =m(c_n +1) could hold, it doesn't necessarily lead to a contradiction because of the integer constraints. Therefore, this approach might not work.Alternative approach: Let's use the concept of the interval overlap. We need to show that for all m, n, [c_m /m, (c_m +1)/m) and [c_n /n, (c_n +1)/n) overlap. If all these intervals overlap pairwise, then by the Helly's theorem (which states that for intervals on the real line, if every pair has a non-empty intersection, then the whole collection has a non-empty intersection), the entire collection has a common intersection.But Helly's theorem applies to convex sets in higher dimensions, but for intervals on the real line, if every pair overlaps, then all of them overlap. So we need to show that for any m, n, [c_m /m, (c_m +1)/m) and [c_n /n, (c_n +1)/n) have a non-empty intersection.To prove that, we need to show that the upper bound of one interval is greater than the lower bound of the other, and vice versa. That is:c_m /m < (c_n +1)/n and c_n /n < (c_m +1)/m.So we need both inequalities.Let's first show that c_m /m < (c_n +1)/n.Assume for contradiction that c_m /m >= (c_n +1)/n. Then, as before, n c_m >=m(c_n +1).But from the given conditions, we can relate c_m and c_n.Let's use the upper bound inequality. Let's consider the term c_{m +n}. From the given condition, c_{m +n} <=c_m +c_n +1. But also, from the lower bound, c_{m +n} >=c_{m} +c_n.But since m +n <1998 (assuming m +n <1998), the condition applies.But if n c_m >=m(c_n +1), let's see if we can derive a contradiction.Let’s apply the inequality inductively. Suppose we take m copies of n. Wait, but m and n are different. Alternatively, use Euclidean algorithm to express gcd(m,n).Alternatively, note that for any integers m and n, there exist integers a and b such that am + bn = gcd(m,n). But I'm not sure if this helps.Alternatively, consider the sequence c_n and relate it to a linear function.Let’s define d_n = c_n /n. We need to show that d_n -d_m <1/m -1/n for all m, n. Wait, not sure.Alternatively, consider that the maximum difference between d_n and d_m is less than the difference between 1/m and 1/n. This seems vague.Alternatively, think of the problem in terms of the interval graph. Each interval [c_n /n, (c_n +1)/n) corresponds to a node, and edges represent overlapping intervals. If the graph is a complete graph (every pair overlaps), then the intersection is non-empty. So we need to show every pair overlaps.To show that [c_m /m, (c_m +1)/m) and [c_n /n, (c_n +1)/n) overlap, it's equivalent to show that c_m /m < (c_n +1)/n and c_n /n < (c_m +1)/m.We need both inequalities.Let’s first show c_m /m < (c_n +1)/n.Assume for contradiction that c_m /m >= (c_n +1)/n. Then, n c_m >=m(c_n +1).Let’s consider the term c_{m +n}. From the upper bound inequality:c_{m +n} <=c_m +c_n +1.But from the lower bound:c_{m +n} >=c_m +c_n.But if n c_m >=m(c_n +1), then:c_{m +n} >=c_m +c_n =c_m +c_n.But also, using the upper bound:c_{m +n} <=c_m +c_n +1 <= (n c_m)/m +1, since c_n <= (n c_m)/m -1 from n c_m >=m(c_n +1).Wait, if n c_m >=m(c_n +1), then c_n <= (n c_m)/m -1. Therefore:c_{m +n} <=c_m +c_n +1 <=c_m + (n c_m)/m -1 +1 =c_m + (n c_m)/m.But c_m + (n c_m)/m =c_m (1 +n/m) =c_m (m +n)/m.Therefore, c_{m +n} <=c_m (m +n)/m.But from the lower bound, c_{m +n} >= (m +n)k, where k is the sup of d_i. Wait, not necessarily. Alternatively, since c_{m +n} >=c_m +c_n, and c_n <= (n c_m)/m -1, then c_{m +n} >=c_m + (n c_m)/m -1.But c_m + (n c_m)/m -1 =c_m (1 +n/m) -1 =c_m (m +n)/m -1.Therefore:c_{m +n} >=c_m (m +n)/m -1.But we also have:c_{m +n} <=c_m (m +n)/m.Therefore, combining these:c_m (m +n)/m -1 <=c_{m +n} <=c_m (m +n)/m.But c_{m +n} must be an integer. Therefore, c_{m +n} must be equal to floor(c_m (m +n)/m). But c_m (m +n)/m =c_m + (n/m)c_m. Since c_m is an integer and n/m may not be an integer, floor(c_m (m +n)/m) =c_m + floor((n/m)c_m).But (n/m)c_m is an integer only if m divides n c_m. But n c_m >=m(c_n +1), which is m times an integer. So n c_m is divisible by m? Not necessarily. For example, if m=2, n=3, c_m=2, then n c_m=6, which is divisible by m=2. But if m=2, n=3, c_m=3, then n c_m=9, which is not divisible by m=2. Wait, but in our case, n c_m >=m(c_n +1). If m divides n c_m, then (n/m)c_m is an integer. Otherwise, it's a fraction.But this line of thought is leading me nowhere. Let’s instead consider that c_{m +n} is sandwiched between two values that differ by 1. Therefore, c_{m +n} must be either floor(c_m (m +n)/m) or floor(c_m (m +n)/m) -1. But this may not necessarily be an integer.Alternatively, since c_{m +n} is an integer, the inequalities imply that c_{m +n} is either floor(c_m (m +n)/m) or floor(c_m (m +n)/m) -1. But I don't see how this leads to a contradiction.Perhaps a better approach is to consider that the assumption n c_m >=m(c_n +1) implies c_{m +n} <=c_m +c_n +1 <= (n c_m)/m -1 +c_m +1 =c_m (1 +n/m). But also, from the lower bound, c_{m +n} >=c_m +c_n >=c_m + (n c_m)/m -1. So c_{m +n} is in [c_m (1 +n/m) -1, c_m (1 +n/m)].But c_{m +n} must be an integer, so if c_m (1 +n/m) is not an integer, then c_{m +n} could be floor(c_m (1 +n/m)) or floor(c_m (1 +n/m)) -1. But this doesn't necessarily contradict anything unless we can show that it violates the existing bounds.Alternatively, notice that c_{m +n} / (m +n) <= [c_m (1 +n/m)] / (m +n) =c_m /m.But since k is the maximum of c_i /i, this would imply that c_{m +n} / (m +n) <=k, which is consistent. However, if c_m /m =k, then c_{m +n} / (m +n) <=k, so this doesn't necessarily contradict.But from the lower bound:c_{m +n} >=c_m +c_n >=c_m + (m(c_n +1))/n -1 (from the assumption n c_m >=m(c_n +1)).Wait, substituting c_n >= (n c_m)/m -1.So c_{m +n} >=c_m + (n c_m)/m -1 =c_m (1 +n/m) -1.Therefore, c_{m +n} / (m +n) >= [c_m (1 +n/m) -1] / (m +n) =c_m /m -1/(m +n).But since k >=c_m /m, this gives c_{m +n} / (m +n) >=k -1/(m +n).But this doesn't contradict anything either.Hmm, this seems stuck. Maybe another angle. Let’s consider the function f(n) =c_n /n. The problem wants to show that this function is always less than k +1/n, where k is the same for all n. But I'm not sure.Alternatively, consider that the given condition resembles the property of additive functions with a bounded error. In additive number theory, functions that are approximately additive can often be approximated by linear functions. Maybe we can use some form of the Grüss inequality or approximate linearity.Alternatively, let's consider the sequence d_n =c_n - floor(k n) for some k. We need to show that d_n =0 for all n. But since we don't know k yet, this is not helpful.Wait, going back to the original problem: the sequence satisfies c_{m +n} is approximately c_m +c_n, within an error of 1. This is similar to the property of the floor function, as mentioned earlier. So perhaps the key is to construct k such that c_n =floor(nk), and use induction.Let’s try induction on n. For n=1, set k to be in [c1, c1 +1). Assume that for all m <n, c_m =floor(mk). Now, for m +n <1998, we have c_{m +n} <=c_m +c_n +1 and c_{m +n} >=c_m +c_n. If we can show that c_n =floor(nk), then we are done.But how to choose k? Since we need k to satisfy floor(mk) =c_m for all m <n. By the induction hypothesis, k is in the intersection of [c_m /m, (c_m +1)/m) for all m <n.We need to choose k such that it also satisfies [c_n /n, (c_n +1)/n) overlapping with the previous intervals.But this requires that the new interval [c_n /n, (c_n +1)/n) overlaps with the intersection of all previous intervals. For this, it's sufficient that c_n /n < upper bound of previous intervals and (c_n +1)/n > lower bound of previous intervals.But the previous intervals' intersection is a single interval [max_{m <n} c_m /m, min_{m <n} (c_m +1)/m). So, to have overlap with [c_n /n, (c_n +1)/n), we need:c_n /n < min_{m <n} (c_m +1)/mand(c_n +1)/n > max_{m <n} c_m /m.But this is equivalent to:For all m <n, c_n /n < (c_m +1)/mand(c_n +1)/n > c_m /m.Which is the same as:For all m <n, m c_n <n(c_m +1)andm(c_n +1) >n c_m.Which are exactly the conditions we need to verify.So, using the given conditions, we need to show that for all m <n, m c_n <n(c_m +1) and m(c_n +1) >n c_m.Let’s start with the first inequality: m c_n <n(c_m +1). Assume for contradiction that m c_n >=n(c_m +1). Then:m c_n >=n c_m +n.But from the given inequality, since m <n, we can write n =m +k, where k >=1.Then, c_n <=c_{m +k} <=c_m +c_k +1.But k =n -m, so recursively, this might not help.Alternatively, apply the given inequality multiple times. For example, for m and n with m <n, write n =q m +r, as before. Then, using the upper bound:c_n <=q c_m +c_r +q.But m c_n <=q m c_m +m c_r +q m.But from the assumption m c_n >=n c_m +n = (q m +r)c_m +n.Therefore:q m c_m +m c_r +q m >=(q m +r)c_m +n.Simplify:q m c_m +m c_r +q m >=q m c_m +r c_m +n.Subtract q m c_m:m c_r +q m >=r c_m +n.But n =q m +r, so substitute:m c_r +q m >=r c_m +q m +r.Subtract q m:m c_r >=r c_m +r.Divide both sides by r:(m/r)c_r >=c_m +1.But since r <m (from n =q m +r and m <n), then m/r >1. Also, c_r is at least floor(r k), but since k is in the previous intervals, c_r >=r * (lower bound of k).But I'm not sure how this helps. Let's take specific values. Suppose m=1. Then the inequality becomes 1 *c_n >=n(c_1 +1). But from the lower bound, c_n >=n c_1. So c_n >=n c_1 >=n(c_1 +1) implies 0 >=n, which is impossible. Therefore, for m=1, the inequality cannot hold. So for m=1, the first inequality holds: c_n /n < (c_1 +1)/1.Similarly, for m=2 and n=3, if we assume m c_n >=n(c_m +1), then 2c_3 >=3(c_2 +1). But from the upper bound, c_3 <=c_2 +c_1 +1. If c_1 >=0, then c_3 <=c_2 +c_1 +1 <=c_2 +c_2 +1, since c_1 <=c_2 /2 (if k >=c_1). Wait, this is getting too vague.Another approach: Use mathematical induction on n. Assume that for all m <n, there exists a k such that c_m =floor(mk). Now, we need to show that there exists a k such that c_n =floor(nk) and k is consistent with all previous intervals.The key is that the new interval [c_n /n, (c_n +1)/n) must intersect with the intersection of all previous intervals. To ensure this, we need to show that c_n /n < min_{m <n} (c_m +1)/m and (c_n +1)/n > max_{m <n} c_m /m.But how to prove this using the given conditions?From the given conditions, for any m <n, we have:c_n >=c_{n -m} +c_m.Andc_n <=c_{n -m} +c_m +1.Similarly, for m=1:c_n >=c_{n -1} +c_1c_n <=c_{n -1} +c_1 +1Assuming by induction that c_{n -1} =floor((n -1)k), then c_n is either floor((n -1)k) +c1 or floor((n -1)k) +c1 +1. But floor((n -1)k) +c1 =floor((n -1)k) +floor(k) <=floor((n -1)k +k) =floor(nk). Similarly, floor((n -1)k) +c1 +1 <=floor(nk) +1. But this requires more precise analysis.Alternatively, since k is in the intersection of all previous intervals, for each m <n, k >=c_m /m and k <(c_m +1)/m. Therefore, for each m <n, c_m <=mk <c_m +1.Now, consider the lower bound for c_n: c_n >=c_{n -m} +c_m >= (n -m)k +m k -1 =n k -1.Similarly, the upper bound: c_n <=c_{n -m} +c_m +1 <= (n -m)k +m k +1 +1 =n k +2.Wait, this seems off. Let's precise:From the lower bound, c_{n} >=c_{n -m} +c_m >=floor((n -m)k) +floor(mk). But floor((n -m)k) +floor(mk) <=floor((n -m)k +mk) =floor(nk). Therefore, c_n >=floor(nk).From the upper bound, c_n <=c_{n -m} +c_m +1 <=floor((n -m)k) +floor(mk) +1. But floor(a) +floor(b) +1 >=floor(a +b) +1 >=a +b -1 +1 =a +b. So floor((n -m)k) +floor(mk) +1 >= (n -m)k +mk -1 +1 =nk. Therefore, c_n <=nk.But c_n <=nk and c_n >=floor(nk). Since c_n is an integer, this implies that c_n =floor(nk) or floor(nk) +1. Wait, but from the upper bound, c_n <=nk. Since nk >=floor(nk), and c_n is an integer, c_n <=floor(nk) +0. But floor(nk) <=c_n <=nk <floor(nk) +1. Therefore, c_n must be floor(nk).Yes! Let me elaborate.From the lower bound: c_n >=c_{n -m} +c_m >=floor((n -m)k) +floor(mk). But floor(a) +floor(b) <=floor(a +b). So floor((n -m)k) +floor(mk) <=floor((n -m)k +mk) =floor(nk). Therefore, c_n >=floor(nk).From the upper bound: c_n <=c_{n -m} +c_m +1 <=floor((n -m)k) +floor(mk) +1. But floor(a) +floor(b) +1 <=a +b +1. Since (n -m)k +mk =nk, we have c_n <=nk +1.But since k <(c_m +1)/m for all m <n, we can choose m=1. Then k <c_1 +1. But c_1 <=k <c_1 +1, so k <c_1 +1 implies k is in [c_1, c_1 +1). Therefore, nk <n(c_1 +1). But c_n >=n c_1. So c_n is in [n c_1, n(c_1 +1)).But since c_n <=nk +1, and nk <n(c_1 +1), then c_n <n(c_1 +1) +1. But c_n is an integer. However, this line of reasoning might not directly help.Wait, back to the bounds:We have c_n >=floor(nk) and c_n <=nk.But floor(nk) <=c_n <=nk <floor(nk) +1.Since c_n is an integer, the only possibility is c_n =floor(nk).Yes! Because if floor(nk) <=c_n <floor(nk) +1, and c_n is an integer, then c_n must be floor(nk).Therefore, by induction, we can show that for all n, c_n =floor(nk), provided that k is chosen such that for all m <n, k is in [c_m /m, (c_m +1)/m).Since we have shown inductively that each new interval [c_n /n, (c_n +1)/n) must overlap with the previous intervals, and thus there exists a k that lies in all intervals simultaneously.Therefore, such a k exists, and c_n =floor(nk) for all n up to 1997.This completes the proof.

✅:To show that there exists a real number ( k ) such that ( c_n = lfloor nk rfloor ) for ( 1 leq n leq 1997 ), we analyze the given conditions and construct ( k ) by ensuring it lies within specific intervals derived from the sequence ( {c_n} ).Step-by-Step Explanation:1. Understanding the Inequalities: The sequence ( {c_n} ) satisfies ( c_{m+n} leq c_m + c_n + 1 ) and ( c_{m+n} geq c_m + c_n ) for all ( m, n > 0 ) with ( m + n < 1998 ). These inequalities suggest that ( c_{m+n} ) is approximately linear with a bounded error of 1.2. Defining Intervals for ( k ): For each ( n ), define the interval ( I_n = left[ frac{c_n}{n}, frac{c_n + 1}{n} right) ). If there exists a real number ( k ) in the intersection of all ( I_n ), then ( c_n = lfloor nk rfloor ).3. Key Requirement: To ensure the intersection ( bigcap_{n=1}^{1997} I_n ) is non-empty, it suffices that for every pair ( m, n ), ( frac{c_m}{m} < frac{c_n + 1}{n} ). This guarantees overlapping intervals, allowing a common ( k ).4. Inductive Proof of Interval Overlap: - Base Case: For ( n = 1 ), ( k ) must lie in ( [c_1, c_1 + 1) ). - Inductive Step: Assume for all ( m < n ), intervals ( I_m ) overlap. For ( I_n ), using the given inequalities: - ( c_n geq c_{n-m} + c_m ) implies ( frac{c_n}{n} geq frac{c_{n-m} + c_m}{n} ). - ( c_n leq c_{n-m} + c_m + 1 ) implies ( frac{c_n + 1}{n} geq frac{c_{n-m} + c_m + 2}{n} ). By induction, these maintain the overlap with previous intervals.5. Conclusion from Overlap: By Helly's theorem for intervals on the real line, if every pair ( I_m, I_n ) overlaps, the entire collection has a common intersection. Thus, there exists ( k ) such that ( k in I_n ) for all ( n ), yielding ( c_n = lfloor nk rfloor ).Final Answer:By systematically ensuring each interval ( left[ frac{c_n}{n}, frac{c_n + 1}{n} right) ) overlaps with all others through the given inequalities, we conclude there exists a real number ( k ) such that ( boxed{c_n = lfloor nk rfloor} ) for ( 1 leq n leq 1997 ).

🔑: ProblemThe sequence of non-negative integers (c_1, c_2, ldots, c_{1997}) satisfies (c_1 geq 0) and (c_m + c_n leq c_{m+n} leq c_m + c_n + 1) for all (m, n > 0) with (m + n < 1998). Show that there is a real (k) such that (c_n = lfloor nk rfloor) for (1 leq n leq 1997).1. We start by examining the inequalities (c_m + c_n leq c_{m+n} leq c_m + c_n + 1).2. For (c_n) to be of the form (c_n = lfloor nk rfloor) for some (k), it must hold that ( frac{c_n}{n} leq k < frac{c_n}{n} + frac{1}{n}).3. Define [ frac{c_m}{m} < frac{c_n}{n} + frac{1}{n} quad text{or} quad n c_m < m c_n + m ] for all (m, n). Inductive Step: Base Case4. The base case is trivial for (m = n = 1): [ c_1 leq c_1 + 1 quad text{and}quad c_1 geq c_1 ] Thus, (k) must satisfy ( frac{c_1}{1} leq k < frac{c_1}{1} + 1). Inductive Hypothesis5. Suppose for any (i leq m + n - 1), the inequalities hold: [ n c_m < m c_n + m ] Inductive Step6. Consider (m < n). By the inductive hypothesis: [ (n - m) c_m < m c_{n-m} + m ] Since (c_m leq c_n - c_{n-m}), we have: [ m c_m leq m c_n - m c_{n-m} ] Adding these two inequalities yields: [ n c_m < m c_n + m ]7. Now consider (m > n). By the inductive hypothesis: [ n c_{m-n} < (m - n) c_n + (m - n) ] Since ( c_m leq c_m - n c_{m-n}), we have: [ n c_m leq n c_n + n c_{m-n} + n ] Adding these gives us: [ n c_m < m c_n + m ]8. Therefore, we can conclude that for all ( m, n ), the inequalities ( m c_n leq n c_m + m) and ( n c_m leq m c_n + m ) hold. ConclusionThis verifies that the sequences can be written in the form (c_n = lfloor nk rfloor), and we conclude the solution with:[ boxed{c_n = lfloor nk rfloor text{ for some real number } k} ]