🧠:Okay, so I need to figure out how many four-digit numbers, where all digits are non-zero (so digits a, b, c, d are from 1 to 9), satisfy the following conditions:1. The first two digits form a number less than 20. So, the number formed by digits a and b, which is 10a + b, must be less than 20.2. The middle two digits form a number less than 21. So, the number formed by digits b and c, which is 10b + c, must be less than 21.3. The last two digits form a number less than 22. So, the number formed by digits c and d, which is 10c + d, must be less than 22.All digits are non-zero, so a, b, c, d ∈ {1, 2, ..., 9}.First, let's break down each condition step by step.Starting with the first condition: the first two digits, ab (which is 10a + b), must be less than 20. Since a is the first digit of a four-digit number, a cannot be zero. However, the problem already states all digits are non-zero. So a is from 1 to 9, but with the additional constraint that 10a + b < 20.Let me analyze this. 10a + b < 20. Since a is at least 1, if a = 1, then 10*1 + b < 20 => 10 + b < 20 => b < 10. Since b is a non-zero digit, b can be from 1 to 9. So if a=1, then b can be 1-9.If a=2, then 10*2 + b < 20 => 20 + b < 20 => b < 0. But b is a non-zero digit, so this is impossible. Similarly, for a >=2, the condition 10a + b <20 would require b to be negative, which is impossible. Hence, the only possible value for a is 1, and when a=1, b can be 1 through 9. So the first two digits must start with 1, followed by any digit from 1 to 9. Therefore, possible values for a are fixed as 1, and b can be from 1 to 9.So that gives us a=1, and b ∈ {1,2,...,9}.So the first condition restricts a to 1 and allows b to be 1-9.Now moving to the second condition: bc <21. Since b is already from 1-9 (from the first condition), and c is a digit from 1-9, we need to ensure that 10b + c <21.So for each possible b (1-9), c has to be such that c < 21 -10b.Wait, let's write it as 10b + c <21 => c < 21 -10b.But c is a positive integer from 1 to 9. So for each b, c must satisfy 1 ≤ c ≤ floor(20 -10b). Wait, no, 10b + c <21 => c < 21 -10b. Since c must be at least 1, this inequality 21 -10b must be greater than 1 for there to be possible c's.Wait, let's take each possible b (from 1 to 9) and find the possible c.Case 1: b=1. Then 10*1 + c <21 => 10 + c <21 => c <11. But c is a digit from 1-9, so c can be 1-9. So for b=1, c can be 1-9.Case 2: b=2. 10*2 + c <21 => 20 + c <21 => c <1. But c must be at least 1, so no solution. So if b=2, there are no possible c's.Similarly, for b=3: 10*3 +c <21 => 30 +c <21 => c < -9. Impossible. Similarly for b=4 to b=9, 10b is already >=10*4=40, which is way larger than 21. So for b >=2, there are no possible c's. Wait, but this contradicts the earlier thought. Wait, if b=1, c can be 1-9. For b=2, c must be less than 1, which is impossible. So only when b=1, c can be from 1-9. So that's a critical point.Wait, so that means the second condition bc <21 is only possible if b=1, and c=1-9. Therefore, even though the first condition allowed b=1-9, the second condition restricts b to only 1. Because if b is 2-9, then 10b +c >= 10*2 +1=21, which is already equal to 21, which doesn't satisfy bc <21. Therefore, the only possible value for b is 1.Therefore, combining the first two conditions:From the first condition: a=1 and b=1-9.From the second condition: b must be 1 (since otherwise, if b >=2, no c satisfies bc <21). So combining these two, the only possible value for b is 1.Therefore, a=1, b=1.Then, moving to the second condition, since b=1, then c can be 1-9 (since 10*1 + c <21 => c <11, which is automatically satisfied as c is <=9). So c can be 1-9.Now moving to the third condition: cd <22. Which is 10c + d <22. Here, c is from 1-9 (as above), and d is from 1-9. So we need to find for each c (1-9), the possible d's such that 10c + d <22.Let's analyze for each c:c=1: 10*1 + d <22 => d <12. Since d is 1-9, d can be 1-9. So 9 possibilities.c=2: 10*2 + d <22 => 20 + d <22 => d <2. So d=1. Only 1 possibility.c=3: 10*3 + d <22 => 30 + d <22 => d < -8. Impossible. Similarly, for c=3 to 9, 10c is already 30 to 90, which added to any d (>=1) will be >=31, which is way larger than 22. So for c >=3, no possible d's.Therefore, the third condition restricts c to be 1 or 2.But from the second condition, we had c can be 1-9. Wait, but now combining with the third condition, even though in the second condition c could be 1-9, the third condition restricts c to 1 or 2. Because if c=3-9, then d would have to be negative, which is impossible.Therefore, putting it all together:From first condition: a=1, b=1-9.But second condition restricts b=1.Then, c can be 1-9 (from second condition), but third condition restricts c=1 or 2.Thus, c can be 1 or 2.For c=1: d can be 1-9 (since 10*1 + d <22 => d <12).For c=2: d can be 1 (since d <2).Therefore, total possibilities:a=1, b=1, c=1, d=1-9: 9 numbers.a=1, b=1, c=2, d=1: 1 number.So total of 9 + 1 = 10 numbers.Wait, but hold on. Let me verify again step by step.First, the first two digits ab must be less than 20. Since a is non-zero, and 10a + b <20. So a must be 1, and b can be 1-9. Then, the second condition is bc <21. So 10b + c <21. Since b is 1-9 (from the first condition), but if b is 2-9, 10b is 20-90, so even with c=1, 10b +c would be 21-91, which is >=21 for b>=2. Hence, only when b=1, 10*1 +c <21 => c <11. Since c is 1-9, c can be 1-9. Then, the third condition is cd <22, which is 10c +d <22. For c=1-9, so:If c=1: 10*1 +d <22 => d <12. Since d is 1-9, d can be 1-9. So 9 possibilities.If c=2: 10*2 +d <22 => d <2. So d=1. 1 possibility.If c=3: 10*3 +d =30 +d <22 => impossible, since 30 >=22. Similarly for c=4-9: impossible. Therefore, c can only be 1 or 2.Therefore, with b=1, c can be 1 or 2, but from the second condition (bc <21), when b=1, c can be 1-9. Wait, that seems conflicting.Wait, hold on. The second condition is bc <21. When b=1, the bc is 10*1 +c =10 +c. So 10 +c <21 => c <11, so c can be 1-9. But then the third condition (cd <22) further restricts c to 1 or 2. So even though the second condition allows c=3-9, the third condition overrules that. Therefore, even though c could be 1-9 based on the second condition, the third condition restricts c to 1 or 2.Therefore, combining all conditions:a=1 (from first condition),b=1 (from second condition, since if b>=2, no c satisfies bc <21),c=1 or 2 (from third condition),and for each c:If c=1, d=1-9,If c=2, d=1.Therefore, total numbers:For c=1: 1 (a) *1 (b) *1 (c) *9 (d) =9.For c=2:1*1*1*1=1.Total:10.Wait, but this seems very few. Let me check if there is a mistake here.Wait, is there a scenario where b can be different?Wait, the first condition says ab <20. So a=1 and b=1-9. Then, second condition is bc <21. If a=1 and b=2, then bc would be 10*2 +c =20 +c. To have 20 +c <21 => c <1. But c is at least 1. So no possible c. Similarly, if a=1 and b=3, bc=10*3 +c=30 +c >=30 +1=31>21. Hence, for b>=2, no c possible. Therefore, only b=1 is allowed.Hence, indeed, b must be 1. Therefore, the four-digit number is 1 1 c d, where c and d must satisfy cd <22.As established, c=1 or 2. For c=1, d=1-9 (since 11d? Wait, no. Wait, the third condition is cd <22, which is 10c +d <22.So when c=1, 10*1 +d <22 => d <12. Since d is 1-9, so 9 options.When c=2, 10*2 +d <22 => d <2. So d=1.Hence, total numbers: 9 +1=10.But this seems surprisingly low. Is this correct?Wait, let's test with actual numbers.Take a=1, b=1.Then, possible c and d:c=1: d can be 1-9. So numbers 1111, 1112, ..., 1119. That's 9 numbers.c=2: d must be 1. So number 1121.Wait, but 1121: check the conditions.ab=11 <20: yes.bc=12 (digits b=1, c=2, so bc=12) <21: yes.cd=21 (digits c=2, d=1, so cd=21). Wait, 21 is not less than 22. Wait, hold on. cd=21. But the third condition is cd <22. So 21 <22 is true. Therefore, 21 is allowed. Wait, 21 is less than 22. So 21 is acceptable.Wait, but in the calculation above, when c=2, d=1: 10*2 +1=21 <22. So 21 is allowed. So that's okay. So 1121 is acceptable.But wait, the number is 1121. Then, let's check all three conditions:1. ab=11 <20: okay.2. bc=12 <21: okay.3. cd=21 <22: okay.Yes, all conditions are satisfied.Similarly, for c=1, d=9: number 1119.Check:ab=11 <20: okay.bc=11 (digits b=1, c=1) <21: okay.cd=19 (digits c=1, d=9) <22: okay.Yes, all good.But wait, if c=1, d=9: 10*1 +9=19 <22: okay.If c=2, d=1:21 <22: okay.So that is correct.But then, are there only 10 numbers?Wait, let's check for c=2, d=1: only one number.But wait, is there another way? For example, if c=1, d=10, but d cannot be 10. So no.Hence, total numbers are 9 (c=1, d=1-9) +1 (c=2, d=1)=10.Therefore, the answer is 10.But let me double-check if there is any other possible value for b.Wait, in the first condition, ab <20. So a=1, and b can be 1-9. Then, the second condition bc <21. If a=1 and b=1, then bc can be 1c, which is 10 +c <21, so c <11. But c is 1-9, so okay. If a=1 and b=2, then bc=20 +c <21 =>c <1. Impossible. Similarly, for b=3, bc=30 +c <21, which is impossible. Hence, only b=1.Therefore, all four-digit numbers must start with 11, followed by c and d where c=1 or 2, and d as above. So total of 10 numbers. Therefore, the answer is 10. Hmm. That seems correct, but I need to verify once more.Alternatively, maybe I made an error in combining the conditions.Wait, perhaps there's a different approach. Let's model the problem as a chain of digits with dependencies.First digit a: must be 1, since 10a +b <20 and a >=1. Therefore, a=1.Second digit b: when a=1, 10 +b <20 =>b <10. Since b is non-zero, b=1-9. So possible b=1-9.Third digit c: depends on b. For each b, 10b +c <21. If b=1: 10 +c <21 =>c <11 =>c=1-9.If b=2: 20 +c <21 =>c <1. Impossible.Similarly, for b >=2, c would have to be less than (21 -10b), which is negative. So only possible when b=1, c=1-9.Fourth digit d: depends on c. For each c, 10c +d <22.If c=1: 10 +d <22 =>d <12 =>d=1-9 (since d is non-zero digit).If c=2:20 +d <22 =>d <2 =>d=1.If c >=3:10c +d >=30 +1=31 >22. So no possible d.Therefore, for c=1: d=1-9 (9 options), for c=2: d=1 (1 option), and c>=3: no options.Thus, the total number of numbers is:Number of possibilities for a:1Number of possibilities for b:1 (since only b=1 allows c)Number of possibilities for c: when b=1, c=1-9, but c is further restricted by d's possibilities. Wait, no. Wait, when b=1, c can be 1-9 (from bc <21). But then, for each c, d is restricted. So even though c can be 1-9 from the second condition, the third condition may limit it. Therefore, the total number of possible c and d pairs is for c=1-9:If c=1:9 d'sIf c=2:1 dc=3-9:0Total:9 +1=10.Hence, total numbers:1 (a) *1 (b) *10 (c and d combinations)=10.Therefore, the answer is 10. So I think that's correct.But let me try to list all 10 numbers to confirm.Numbers starting with 11, then c and d:When c=1, d=1-9:1111, 1112, 1113, 1114, 1115, 1116, 1117, 1118, 1119. That's 9 numbers.When c=2, d=1:1121. That's 1 number.Total 10 numbers.Check each number:Take 1111:ab=11 <20: yes.bc=11 <21: yes.cd=11 <22: yes.Take 1119:ab=11 <20: yes.bc=11 <21: yes.cd=19 <22: yes.Take 1121:ab=11 <20: yes.bc=12 <21: yes.cd=21 <22: yes.All conditions are satisfied. So yes, 10 numbers. Therefore, the answer is 10.Final AnswerThe number of "good numbers" is boxed{10}.

✅:To determine the number of four-digit numbers overline{a b c d} (where all digits are non-zero) that satisfy the conditions overline{a b} < 20, overline{b c} < 21, and overline{c d} < 22, we analyze each condition step by step:1. Condition overline{a b} < 20: - The first two digits a and b must form a number less than 20. Since a is non-zero, the only possible value for a is 1, and b can be any digit from 1 to 9. Thus, a = 1 and b in {1, 2, ldots, 9}.2. Condition overline{b c} < 21: - The middle two digits b and c must form a number less than 21. Given a = 1 and b in {1, 2, ldots, 9}, we check each possible b: - If b geq 2, then overline{b c} geq 20 + 1 = 21, which does not satisfy the condition. Therefore, b must be 1. For b = 1, overline{b c} = 10 + c < 21 implies c can be any digit from 1 to 9.3. Condition overline{c d} < 22: - The last two digits c and d must form a number less than 22. Given c in {1, 2, ldots, 9}: - If c = 1, then overline{c d} = 10 + d < 22 implies d can be any digit from 1 to 9 (9 possibilities). - If c = 2, then overline{c d} = 20 + d < 22 implies d must be 1 (1 possibility). - For c geq 3, overline{c d} geq 30 + 1 = 31, which does not satisfy the condition.Combining all conditions, we find:- a = 1, b = 1- c = 1 with d in {1, 2, ldots, 9} (9 numbers)- c = 2 with d = 1 (1 number)Thus, the total number of "good numbers" is 9 + 1 = 10.[boxed{10}]

🔑:1. Interpret the constraints:- overline{ab} < 20 implies that the two-digit number formed by a and b must be less than 20.- overline{bc} < 21 implies that the two-digit number formed by b and c must be less than 21.- overline{cd} < 22 implies that the two-digit number formed by c and d must be less than 22.2. List the possible values:- Since overline{ab} < 20, the possible values for overline{ab} are from 10 to 19 (inclusive), because a and b cannot be zero.- Since overline{bc} < 21, the possible values for overline{bc} are from 10 to 20 (inclusive), because b and c cannot be zero.- Since overline{cd} < 22, the possible values for overline{cd} are from 10 to 21 (inclusive), because c and d cannot be zero.3. Find valid combination of digits:- Let’s examine combinations digit by digit.- If overline{ab} < 20, the possible values are: - a = 1 and b = 0 implies overline{ab} = 10 - a = 1 and b = 1 implies overline{ab} = 11 - ldots - a = 1 and b = 9 implies overline{ab} = 19 - If overline{bc} < 21, the possible values are: - b = 1 and c = 0 implies overline{bc} = 10 - b = 1 and c = 1 implies overline{bc} = 11 - b = 1 and c = 2 implies overline{bc} = 12 - ldots - b = 2 and c = 0 implies overline{bc} = 20 - If overline{cd} < 22, the possible values are: - c = 1 and d = 0 implies overline{cd} = 10 - c = 1 and d = 1 implies overline{cd} = 11 - c = 1 and d = 2 implies overline{cd} = 12 - ldots - c = 2 and d = 1 implies overline{cd} = 214. Count valid good numbers:- Let’s count the combinations. Since b and c must ensure overline{bc} < 21, several values should be checked: - Possible values of b and c from step-by-step checking: (b=1, c=0) to (b=2, c=0). - Each valid (b, c) combination should ensure the preceding is non-zero and the next step fits.5. Count total combinations:- Given possibilities for a, b, c, and d: - 10 (a,b values) times 2 ( b,c values fitting both previous b and next c) times 11 (c,d fit into range) = 2 values which form fit into ranges. Conclusion:There are 10 such good numbers. Therefore, boxed{10}

🧠:Okay, so I need to calculate the value of a₁ plus the sum of the squares of a₁ through a₁₀₀. Let me start by understanding what each aₖ is. Given that aₖ is the sum from 1/k up to 1/100, so for each k, aₖ = 1/k + 1/(k+1) + ... + 1/100. That makes sense. So a₁ is the harmonic series from 1 to 100, a₂ is from 1/2 to 1/100, and so on until a₁₀₀ which is just 1/100.The expression we need to compute is a₁ + a₁² + a₂² + ... + a₁₀₀². Wait, so the first term is a₁, and then we add the squares of all aₖ from k=1 to k=100. So the total expression is a₁ + Σₖ=1^100 (aₖ)². Hmm. Let me verify that. The problem says: "a₁ + a₁² + a₂² + ... + a₁₀₀²". So yes, that's a₁ plus the sum of the squares of all aₖ. So the total is a₁ + Σₖ=1^100 aₖ². Got it.First, let's note that aₖ is the sum from n=k to 100 of 1/n. So perhaps we can express aₖ as H_{100} - H_{k-1}, where H_n is the nth harmonic number. That's right, since H_n = 1 + 1/2 + ... + 1/n. Therefore, aₖ = H_{100} - H_{k-1}.But maybe I don't need to use harmonic numbers here. Let me see if there's another way to approach the problem. The challenge is to compute a₁ + sum_{k=1}^{100} aₖ². Wait, but a₁ is already part of the sum. Wait, no. The expression is a₁ + (a₁² + a₂² + ... + a₁₀₀²). So it's a₁ plus the sum of the squares of all aₖ from k=1 to 100. So actually, the expression is a₁ + sum_{k=1}^{100} aₖ². But a₁ is equal to a₁, so it's included as the linear term and also squared in the sum. Wait, no. Wait, the problem is written as:a₁ + a₁² + a₂² + ... + a₁₀₀². So that is a₁ plus the sum of a₁ squared plus a₂ squared plus ... plus a₁₀₀ squared. Therefore, it's a₁ + sum_{k=1}^{100} aₖ². So the total expression is a₁ + sum_{k=1}^{100} aₖ². Therefore, the first term is a₁ (linear), and then we add all the squares from a₁² up to a₁₀₀². So we need to compute both a₁ and the sum of the squares.Let me first compute a₁. Since a₁ is the sum from n=1 to 100 of 1/n, which is H_{100}, the 100th harmonic number. The approximate value is around ln(100) + γ + 0.5/100, where γ is the Euler-Mascheroni constant, approximately 0.5772. So ln(100) is about 4.605, so H_{100} ≈ 4.605 + 0.5772 + 0.005 ≈ 5.1872. But we might need the exact value as a fraction. However, calculating H_{100} exactly is tedious, but maybe we don't need to compute it explicitly if terms cancel out. Wait, but the problem requires the exact value? The problem says "Calculate the value of...", so it's likely expecting an exact fraction. But harmonic numbers up to H_{100} are known, although they are large fractions. But perhaps the expression can be simplified in a way that avoids computing H_{100} directly.Similarly, the sum of squares of aₖ from k=1 to 100. Let's think about how to approach this. Let's write down the expressions for aₖ. For each k, aₖ = sum_{n=k}^{100} 1/n. Then aₖ² is the square of that sum, which can be expanded as sum_{n=k}^{100} 1/n² + 2 sum_{k ≤ i < j ≤ 100} 1/(i j). Therefore, when we sum aₖ² over k from 1 to 100, we have to sum over k of [sum_{n=k}^{100} 1/n² + 2 sum_{k ≤ i < j ≤ 100} 1/(i j)].But perhaps there's a smarter way to compute the sum of aₖ². Let me consider interchanging the order of summation. Let me see. The sum over k=1 to 100 of aₖ² is sum_{k=1}^{100} [sum_{n=k}^{100} 1/n]^2. Expanding this, it's sum_{k=1}^{100} [sum_{n=k}^{100} 1/n sum_{m=k}^{100} 1/m] = sum_{k=1}^{100} sum_{n=k}^{100} sum_{m=k}^{100} 1/(n m). So that's equal to sum_{k=1}^{100} sum_{n=k}^{100} sum_{m=k}^{100} 1/(n m). Let's see if we can change the order of summation here.Instead of summing over k first, maybe we can fix n and m and see for how many k's they are included. For each pair (n, m), the term 1/(n m) appears in the sum for all k such that k ≤ n and k ≤ m. Since k ranges from 1 to 100, the number of k's for which k ≤ min(n, m) is min(n, m). Therefore, the total sum becomes sum_{n=1}^{100} sum_{m=1}^{100} [sum_{k=1}^{min(n,m)} 1] * 1/(n m) = sum_{n=1}^{100} sum_{m=1}^{100} min(n, m)/(n m). Wait, let's verify this step. For each n and m, the inner sum over k from 1 to min(n, m) gives min(n, m) terms. Each term is 1/(n m), and since k runs from 1 to min(n, m), the total contribution for each (n, m) is min(n, m)/(n m). Therefore, the entire sum is sum_{n=1}^{100} sum_{m=1}^{100} min(n, m)/(n m).Hmm, that's an interesting transformation. So now, the sum of aₖ² from k=1 to 100 is equal to the double sum over n and m of min(n, m)/(n m). Therefore, the original expression we need to compute is a₁ + sum_{n=1}^{100} sum_{m=1}^{100} min(n, m)/(n m). But let's also note that a₁ is the sum from n=1 to 100 of 1/n, which is H_{100}. So the entire expression is H_{100} + sum_{n=1}^{100} sum_{m=1}^{100} min(n, m)/(n m). Now, how do we compute this double sum? Let's consider splitting the double sum into two parts: where n ≤ m and where m < n. Since min(n, m) is n when n ≤ m and m when m < n. So we can write the double sum as sum_{n=1}^{100} sum_{m=n}^{100} n/(n m) + sum_{m=1}^{100} sum_{n=m+1}^{100} m/(n m). Simplifying the first term: sum_{n=1}^{100} sum_{m=n}^{100} n/(n m) = sum_{n=1}^{100} sum_{m=n}^{100} 1/m. Similarly, the second term is sum_{m=1}^{100} sum_{n=m+1}^{100} m/(n m) = sum_{m=1}^{100} sum_{n=m+1}^{100} 1/n. Therefore, combining both terms, the double sum becomes sum_{n=1}^{100} sum_{m=n}^{100} 1/m + sum_{m=1}^{100} sum_{n=m+1}^{100} 1/n. Notice that in the first term, n ranges from 1 to 100 and m ranges from n to 100. In the second term, m ranges from 1 to 100 and n ranges from m+1 to 100. However, if we switch the indices in the second term by swapping m and n, since they are dummy variables, the second term becomes sum_{n=1}^{100} sum_{m=1}^{n-1} 1/n. Wait, let's check that. Wait, the second term is sum_{m=1}^{100} sum_{n=m+1}^{100} 1/n. Let's change variables: let's let n be the outer index. For each n from 2 to 100, m can range from 1 to n-1. Therefore, the second term can be rewritten as sum_{n=2}^{100} sum_{m=1}^{n-1} 1/n. So the double sum is sum_{n=1}^{100} sum_{m=n}^{100} 1/m + sum_{n=2}^{100} sum_{m=1}^{n-1} 1/n. Now, combining these two sums, let's consider for each n, the terms where m ≥ n and m < n. For n=1, the first term has m from 1 to 100, contributing sum_{m=1}^{100} 1/m. The second term for n=1 doesn't exist because n starts at 2. For n ≥ 2, the first term contributes sum_{m=n}^{100} 1/m, and the second term contributes sum_{m=1}^{n-1} 1/n. Therefore, the total double sum becomes sum_{n=1}^{100} sum_{m=n}^{100} 1/m + sum_{n=2}^{100} (n-1)/n. Because sum_{m=1}^{n-1} 1/n is just (n-1)/n. So breaking this down, the first part is sum_{n=1}^{100} sum_{m=n}^{100} 1/m. Let's see what this is. For each m from 1 to 100, how many times does 1/m appear in the sum? For each m, the term 1/m is included for all n from 1 to m. Therefore, the sum becomes sum_{m=1}^{100} sum_{n=1}^m 1/m = sum_{m=1}^{100} (m / m) = sum_{m=1}^{100} 1 = 100. Wait, that's interesting. So the first part simplifies to 100. Then the second part is sum_{n=2}^{100} (n - 1)/n. Let's compute that. sum_{n=2}^{100} (n - 1)/n = sum_{n=2}^{100} (1 - 1/n) = sum_{n=2}^{100} 1 - sum_{n=2}^{100} 1/n. Calculating this, sum_{n=2}^{100} 1 is 99 terms of 1, so 99. Then sum_{n=2}^{100} 1/n is H_{100} - 1. Therefore, the second part becomes 99 - (H_{100} - 1) = 99 - H_{100} + 1 = 100 - H_{100}. Therefore, combining both parts, the double sum is 100 + (100 - H_{100}) = 200 - H_{100}. Therefore, the original expression is a₁ + sum_{k=1}^{100} aₖ² = H_{100} + (200 - H_{100}) = 200. Wait, that can't be. Let me verify each step again because this seems too straightforward. Starting from the double sum: sum_{n,m} min(n, m)/(n m) = 200 - H_{100}. Then adding a₁ which is H_{100}, so total is 200. But let me check the steps again. First, we transformed sum_{k=1}^{100} aₖ² into sum_{n=1}^{100} sum_{m=1}^{100} min(n, m)/(n m). Then we split this into two sums: one where n ≤ m and one where m < n. Then we found that each term simplifies, leading to the double sum being 200 - H_{100}. Then adding a₁ (which is H_{100}) gives 200. But that seems correct? Let's do a small case to verify. Let's take the case where instead of 100, we have N=1. Then a₁ = 1/1 = 1. The expression would be a₁ + a₁² = 1 + 1 = 2. According to the formula, it should be 200 in the case N=100, but for N=1, following the same logic: sum_{n,m} min(n,m)/(n m) where n and m go from 1 to 1. Then min(1,1)/1*1 = 1. So sum is 1. Then a₁ = 1. So total is 1 + 1 = 2. Which matches. Another test: N=2. Then a₁ = 1 + 1/2 = 3/2, a₂ = 1/2. The expression is a₁ + a₁² + a₂² = 3/2 + (9/4) + (1/4) = 3/2 + 10/4 = 3/2 + 5/2 = 4. According to the formula, sum_{n,m=1}^2 min(n,m)/(n m) + a₁. Let's compute the double sum. For n=1, m=1: min(1,1)/1*1 = 1n=1, m=2: min(1,2)/1*2 = 1/2n=2, m=1: min(2,1)/2*1 = 1/2n=2, m=2: min(2,2)/2*2 = 1/4Total double sum: 1 + 1/2 + 1/2 + 1/4 = 2.25. Then a₁ = 3/2 = 1.5. So total expression is 1.5 + 2.25 = 3.75. Wait, but according to the formula, it should be 200 - H_N + H_N = 200. But with N=2, according to my previous logic, sum_{k=1}^2 a_k² = 200 - H_2. But H_2 = 1 + 1/2 = 3/2. So sum would be 200 - 3/2, which is not the case here. Wait, clearly, my mistake is that the transformation in the general case was for N=100, but when I derived the formula, I might have made a miscalculation. Wait, let me step back. For N=2, according to my previous steps:The sum of aₖ² from k=1 to 2 is sum_{k=1}^2 [sum_{n=k}^2 1/n]^2. Compute a₁ = 1 + 1/2 = 3/2, a₂ = 1/2. So a₁² + a₂² = (9/4) + (1/4) = 10/4 = 5/2. Then the total expression is a₁ + sum aₖ² = 3/2 + 5/2 = 4. But according to the previous formula, sum_{n,m=1}^2 min(n,m)/(n m). Let's compute this. The matrix for n and m from 1 to 2:For n=1, m=1: min(1,1)/(1*1) = 1/1 = 1n=1, m=2: min(1,2)/(1*2) = 1/2n=2, m=1: min(2,1)/(2*1) = 1/2n=2, m=2: min(2,2)/(2*2) = 2/4 = 1/2Summing these up: 1 + 1/2 + 1/2 + 1/2 = 1 + 1.5 = 2.5. Then adding a₁ which is 3/2 gives 2.5 + 1.5 = 4, which matches. But according to the general formula I derived earlier, the double sum should be 200 - H_{100} when N=100. But for N=2, according to the same steps:First, split into two parts: sum_{n=1}^2 sum_{m=n}^2 1/m + sum_{n=2}^2 sum_{m=1}^{n-1} 1/n. First part: For n=1, sum_{m=1}^2 1/m = 1 + 1/2 = 3/2. For n=2, sum_{m=2}^2 1/m = 1/2. So total first part: 3/2 + 1/2 = 2. Second part: sum_{n=2}^2 (n-1)/n = (2-1)/2 = 1/2. So total double sum: 2 + 1/2 = 5/2 = 2.5. Then adding a₁ = 3/2 gives 4, which is correct. But according to my general formula, the double sum is supposed to be 200 - H_{100} for N=100. Wait, but in the N=2 case, the formula gives 2 + (sum_{n=2}^2 (n-1)/n) = 2 + (1/2) = 2.5. Then adding a₁ (H_2) gives 2.5 + 1.5 = 4. So in general, for N, the double sum is [sum_{n=1}^N sum_{m=n}^N 1/m] + [sum_{n=2}^N (n-1)/n]. The first sum is sum_{m=1}^N sum_{n=1}^m 1/m = sum_{m=1}^N (m * 1/m) ) = N. The second sum is sum_{n=2}^N (1 - 1/n) = (N - 1) - (H_N - 1) = N - H_N. Therefore, total double sum is N + (N - H_N) = 2N - H_N. Then adding a₁ (which is H_N) gives 2N - H_N + H_N = 2N. Wait, that's the key! For general N, the expression a₁ + sum_{k=1}^N aₖ² equals 2N. So in the case N=2, we get 4, which matches. For N=1, 2*1=2, which matches. For N=100, the answer should be 200. So regardless of the value of N, the expression is 2N. Therefore, for N=100, it's 200. That's fascinating! Therefore, the answer is 200. But let me confirm once more with another small N. Let's take N=3. Compute manually:a₁ = 1 + 1/2 + 1/3 ≈ 1.8333a₂ = 1/2 + 1/3 ≈ 0.8333a₃ = 1/3 ≈ 0.3333Compute the expression: a₁ + a₁² + a₂² + a₃²Compute a₁ ≈ 1.8333, a₁² ≈ 3.3611, a₂² ≈ 0.6944, a₃² ≈ 0.1111. Summing these: 1.8333 + 3.3611 + 0.6944 + 0.1111 ≈ 6.0. According to the formula, 2*3=6. Correct. So the formula holds. Therefore, regardless of N, the expression is 2N. Hence, for N=100, the answer is 200. Therefore, the final answer is boxed{200}.Final Answerboxed{200}

✅:Given each positive integer (1 leq k leq 100), let (a_k) denote the sum (frac{1}{k} + frac{1}{k+1} + ldots + frac{1}{100}). We need to calculate the value of (a_1 + a_1^2 + a_2^2 + ldots + a_{100}^2).First, we note that (a_k) is the harmonic series from (k) to 100. We can express (a_k) as (H_{100} - H_{k-1}), where (H_n) is the (n)-th harmonic number. However, instead of directly computing harmonic numbers, we use a different approach by transforming the sum of squares.The expression we need to compute is (a_1 + sum_{k=1}^{100} a_k^2). We start by expanding the sum of squares:[sum_{k=1}^{100} a_k^2 = sum_{k=1}^{100} left( sum_{n=k}^{100} frac{1}{n} right)^2]Expanding the square, we get:[sum_{k=1}^{100} left( sum_{n=k}^{100} frac{1}{n} sum_{m=k}^{100} frac{1}{m} right) = sum_{k=1}^{100} sum_{n=k}^{100} sum_{m=k}^{100} frac{1}{nm}]By interchanging the order of summation, we consider pairs ((n, m)) and count how many times each term (frac{1}{nm}) appears. For each pair ((n, m)), the term (frac{1}{nm}) appears for all (k leq min(n, m)). Thus, the sum becomes:[sum_{n=1}^{100} sum_{m=1}^{100} frac{min(n, m)}{nm}]Splitting this into two parts where (n leq m) and (m < n), we get:[sum_{n=1}^{100} sum_{m=n}^{100} frac{1}{m} + sum_{m=1}^{100} sum_{n=m+1}^{100} frac{1}{n}]The first part simplifies to 100, and the second part simplifies to (100 - H_{100}). Combining these, we get:[100 + (100 - H_{100}) = 200 - H_{100}]Adding (a_1) (which is (H_{100})) to this result gives:[H_{100} + (200 - H_{100}) = 200]Thus, the value of (a_1 + a_1^2 + a_2^2 + ldots + a_{100}^2) is (boxed{200}).

🔑:To solve the given problem, we'll break it down into clear steps and compute the values one by one.Given:[sum_{k=1}^{100} a_k + sum_{k=1}^{100} a_k^2]where [a_k = sum_{j=k}^{100} frac{1}{j}]1. Understand the nature of ( a_k ): Each ( a_k ) represents a partial sum of the harmonic series: [ a_k = frac{1}{k} + frac{1}{k+1} + cdots + frac{1}{100} ]2. Compute the total ( sum_{k=1}^{100} a_k ): By definition, the sum ( sum_{k=1}^{100} a_k ) will repeatedly add terms from the harmonic series. In particular, each ( frac{1}{j} ) (for ( j = 1, 2, ldots, 100 )) will appear ( j ) times. Specifically: [ sum_{k=1}^{100} a_k = sum_{k=1}^{100} left( sum_{j=k}^{100} frac{1}{j} right) ] Reversing the order of summation, we get: [ sum_{k=1}^{100} a_k = sum_{j=1}^{100} sum_{k=1}^{j} frac{1}{j} ] Simplifying further: [ sum_{k=1}^{100} a_k = sum_{j=1}^{100} frac{j}{j} = sum_{j=1}^{100} 1 = 100 ]3. Compute the total ( sum_{k=1}^{100} a_k^2 ): Each ( a_k ) represents the sum of harmonic terms, and squaring each term requires considering each squared term and cross product. We first focus on the square terms: [ a_k^2 = left( sum_{j=k}^{100} frac{1}{j} right)^2 ] By expanding the squared sum, every ( frac{1}{j^2} ) appears ( j ) times for each ( k leq j ), and each mixed term ( frac{2}{j cdot i} ) appears when summing over all possible ( k leq i ).4. Handle self terms ( sum_{k=1}^{100} frac{1}{k^2} ): Sums over the squared terms: [ sum_{k=1}^{100} frac{k}{k^2} = sum_{k=1}^{100} frac{1}{k} ] Therefore: [ sum_{k=1}^{100} frac{1}{k^2} = a_1 ]5. Handle mixed terms: Each term of the form ( frac{2}{i cdot j} ) appears for each combination where ( k leq j ). Summing these accounts for ( (i-1) times frac{2}{i} = 2 - frac{2}{i} ), then summing this over all ( i ): [ sum_{j < i} (2 cdot frac{1}{ij}) = 2 - frac{2}{i} ] Total sum for all values of ( i ): [ 2n - 2a_1 = 200 - 2a_1 ]Combine all the terms:[a_1 + sum_{k=1}^{100} a_k^2 = a_1 + a_1 + 200 - 2a_1 = 200]Therefore, the final result is:[boxed{200}]

🧠:Okay, let me try to figure out how to solve this geometry problem. Hmm, so we have triangle ABC with angles at B being 30 degrees and angle A being 90 degrees. So that makes it a right-angled triangle at A, with angle B = 30°, which would mean angle C must be 60°, since the angles in a triangle add up to 180°. So, right-angled at A, with angles 30°, 60°, 90°. That tells me the sides are in the ratio 1 : √3 : 2. But I need to figure out the actual lengths or maybe use coordinates?The problem mentions point K on side AC, points L and M on side BC such that KL = KM, with L lying on segment BM. We need to find the length of segment LM, given that AK = 4, BL = 31, and MC = 3. First, let me sketch the triangle to visualize. Let me label the triangle: right-angled at A, angle at B is 30°, so side opposite to 30° is the shortest side, which would be BC if angle A is 90°, wait no. Wait, in triangle ABC, angle A is 90°, angle B is 30°, so angle C is 60°. Then, side opposite angle B (30°) is AC, and side opposite angle C (60°) is AB. The hypotenuse is BC. Wait, in a right-angled triangle, the side opposite the right angle is the hypotenuse. Wait, angle A is 90°, so BC is the hypotenuse. Then, side opposite 30° (angle B) is AC, and side opposite 60° (angle C) is AB. So, the sides are in the ratio AC : AB : BC = 1 : √3 : 2. So, if we let AC = x, then AB = x√3, and BC = 2x. Hmm, but we might need actual lengths. Maybe assign coordinates?Let me place the triangle in coordinate system. Let’s put point A at (0,0), since it's the right angle. Then, since angle B is 30°, and angle C is 60°, let's decide which sides to assign. Let me suppose side AB is along the x-axis and AC is along the y-axis. Wait, but if angle B is 30°, then side AC is opposite angle B, which is 30°, so AC should be the shortest side. Let's confirm:In triangle ABC, angle at A is 90°, angle at B is 30°, angle at C is 60°. Therefore, sides:- Opposite angle A (90°): BC (hypotenuse)- Opposite angle B (30°): AC- Opposite angle C (60°): ABSo, hypotenuse BC is the longest side. Then AC is the shortest side (opposite 30°), and AB is the middle length (opposite 60°). So, the sides are in the ratio AC : AB : BC = 1 : √3 : 2.So, if we let AC = 1 unit, then AB = √3 units, BC = 2 units. But since we need to use actual lengths given (AK = 4, BL = 31, MC = 3), we need to relate these to the triangle. Maybe coordinate geometry would help here.Let me set up coordinates with point A at (0,0). Let’s let AB be along the x-axis and AC along the y-axis. So, point B is at (b, 0), point C is at (0, c). Since angle at B is 30°, which is at point B. Wait, no: angle at B is 30°, so the angle between sides AB and BC is 30°. Hmm, maybe coordinate system is a bit tricky here. Wait, perhaps another approach. Wait, angle at B is 30°, and since the triangle is right-angled at A, we can compute the sides.Let’s denote:- AB = adjacent side to angle B (30°)- AC = opposite side to angle B (30°)- BC = hypotenuseIn a right-angled triangle, the tangent of angle B is opposite over adjacent. So tan(30°) = AC / AB. Since tan(30°) = 1/√3, so AC / AB = 1/√3, meaning AC = AB / √3. Then, using Pythagoras theorem: AB² + AC² = BC². Substitute AC = AB / √3:AB² + (AB² / 3) = BC² => (4/3)AB² = BC² => BC = (2/√3)AB.But maybe instead of keeping variables, assign specific coordinates. Let me set AB as the side adjacent to 30°, so point B is along the x-axis. Let’s set point A at (0,0). Let’s let AB = x√3 (since in a 30-60-90 triangle, sides opposite 30°, 60°, 90° are 1, √3, 2, so maybe if AB is opposite 60°, it's x√3, AC opposite 30° is x, and BC is 2x). So, coordinates:- A: (0,0)- B: (x√3, 0) since AB is along x-axis- C: (0, x) since AC is along y-axis with length xThen BC would be hypotenuse from (x√3, 0) to (0, x). Let me compute BC's length:Distance between B and C: √[(x√3 - 0)^2 + (0 - x)^2] = √[3x² + x²] = √(4x²) = 2x. Which matches the 30-60-90 triangle ratios.So, in this coordinate system:- AB = x√3- AC = x- BC = 2xNow, point K is on AC. Since AC is from (0,0) to (0,x), point K must be somewhere along the y-axis. Given that AK = 4, so from point A (0,0) moving up 4 units along AC, so point K is at (0,4). But wait, AC has length x, so AK = 4 implies that x must be at least 4. But we don't know x yet. Wait, but maybe we need to find x first?Wait, but we also have points L and M on BC. BL = 31 and MC = 3. Since BC has length 2x, then BL + LM + MC = 2x. Given BL = 31, MC = 3, so LM = 2x - 31 - 3 = 2x - 34. But we need to find LM, so we need to find x.Alternatively, maybe we can find x through other given information. However, maybe we need to use the condition that KL = KM. So points L and M are on BC such that KL = KM. Also, point L lies on segment BM. Wait, BM is part of BC? Wait, BC is a side, and M is a point on BC such that MC = 3, so M is 3 units away from C. Similarly, BL = 31, so L is 31 units away from B on BC. But BC has length 2x, so BL + LM + MC = 2x => 31 + LM + 3 = 2x => LM = 2x - 34. But we need to find LM, so if we can find x, then we can compute LM.But how do we find x? Maybe using the coordinates and the condition KL = KM.So let me proceed step by step.First, set up the coordinates with A at (0,0), B at (x√3, 0), C at (0, x). Then point K is on AC, with AK = 4, so since AC is from (0,0) to (0,x), K is at (0,4). Therefore, x must be greater than 4 (since K is on AC, which has length x). So, point K is at (0,4).Points L and M are on BC. Let's parameterize BC. The coordinates of BC: from B (x√3, 0) to C (0, x). So parametric equations for BC can be written as:A point P on BC can be expressed as (x√3 - t(x√3)/ (2x), 0 + t(x)/ (2x)) where t is the distance from B to P. Wait, maybe better to use a parameter s, 0 ≤ s ≤ 2x, such that when s = 0, we are at B, and s = 2x at C. Then coordinates would be:x-coordinate: x√3 - (s / (2x)) * x√3 = x√3 - (s√3)/2y-coordinate: 0 + (s / (2x)) * x = s/2Wait, that might be a way. Alternatively, since BC is from (x√3, 0) to (0, x), the parametric equations can be written as:x = x√3 - (x√3 / (2x)) * t = x√3 - ( (√3)/2 ) ty = 0 + (x / (2x)) * t = (1/2) twhere t ranges from 0 to 2x. So when t = 0, we are at B (x√3, 0), and when t = 2x, we are at C (0, x). Then, for a point at distance s from B along BC, t = s, and coordinates would be:x = x√3 - ( (√3)/2 ) sy = (1/2) sTherefore, points L and M can be located using BL = 31 and MC = 3. Since BL is the distance from B to L along BC, so s = 31. Therefore, coordinates of L are:x_L = x√3 - ( (√3)/2 ) * 31y_L = (1/2) * 31 = 31/2Similarly, MC = 3, which is the distance from M to C along BC. Since the total length of BC is 2x, then BM = 2x - 3. Wait, no. If MC = 3, then the distance from M to C is 3, so the distance from B to M is BC - MC = 2x - 3. Therefore, the parameter t for point M would be t = 2x - 3. Therefore, coordinates of M are:x_M = x√3 - ( (√3)/2 ) * (2x - 3) = x√3 - √3 x + (3√3)/2 = (3√3)/2Wait, that seems odd. Wait, x_M = x√3 - (√3/2)(2x - 3) = x√3 - (√3 x - (3√3)/2 ) = x√3 - √3 x + (3√3)/2 = (3√3)/2. Similarly, y_M = (1/2)(2x - 3) = x - 3/2.Wait, but x_M is (3√3)/2, which is a constant? That can't be right, unless x is fixed. Wait, this suggests that x is a variable here, but perhaps we have to find x such that KL = KM. Because coordinates of M depend on x, so we need to set KL = KM and solve for x, then compute LM.But let's check. Let's first write coordinates for points K, L, M.Point K: (0,4)Point L: (x√3 - ( (√3)/2 ) * 31, 31/2 )Point M: ( (3√3)/2, x - 3/2 )Now, KL = KM. So distance from K to L equals distance from K to M.Compute KL^2:= [x√3 - ( (√3)/2 * 31 ) - 0]^2 + [31/2 - 4]^2= [x√3 - (31√3)/2]^2 + [(31/2 - 8/2)]^2= [ (2x√3 - 31√3)/2 ]^2 + (23/2)^2= [ (√3(2x - 31))/2 ]^2 + (529/4 )= (3(2x - 31)^2)/4 + 529/4Similarly, compute KM^2:= [ (3√3)/2 - 0 ]^2 + [x - 3/2 - 4]^2= ( (9 * 3)/4 ) + [x - 11/2]^2= (27/4) + (x - 11/2)^2Set KL^2 = KM^2:(3(2x - 31)^2)/4 + 529/4 = 27/4 + (x - 11/2)^2Multiply both sides by 4 to eliminate denominators:3(2x - 31)^2 + 529 = 27 + 4(x - 11/2)^2Expand both sides:Left side: 3(4x² - 124x + 961) + 529 = 12x² - 372x + 2883 + 529 = 12x² - 372x + 3412Right side: 27 + 4(x² - 11x + 121/4) = 27 + 4x² - 44x + 121 = 4x² - 44x + 148Set left = right:12x² - 372x + 3412 = 4x² - 44x + 148Subtract right side from left:8x² - 328x + 3264 = 0Divide equation by 8:x² - 41x + 408 = 0Solve quadratic equation:x = [41 ± √(41² - 4*1*408)] / 2Compute discriminant:41² = 16814*1*408 = 1632Discriminant = 1681 - 1632 = 49So sqrt(49) = 7Thus, x = [41 ± 7]/2Possible solutions:x = (41 + 7)/2 = 48/2 = 24x = (41 - 7)/2 = 34/2 = 17But x must be greater than 4 (since AK = 4 and AC = x), and also, since BL = 31 and BC = 2x, then BC must be greater than BL + MC = 31 + 3 = 34, so 2x > 34 => x > 17. Therefore, x = 24 is valid, while x = 17 is invalid (since x must be >17). So x = 24.Therefore, BC = 2x = 48. Then LM = BC - BL - MC = 48 - 31 - 3 = 14. Wait, but hold on. Wait, LM is between L and M on BC. Since BL = 31, and MC = 3, then BM = BC - MC = 48 - 3 = 45. So point M is 45 units from B. Then point L is 31 units from B, so LM = BM - BL = 45 - 31 = 14. So LM = 14. But wait, the problem didn't directly give BC; we found BC through solving the equation KL = KM. So according to this, LM = 14. Let me check if this is consistent with the KL = KM condition.But wait, let's verify with x = 24. Let me compute coordinates:Point K: (0,4)Point L: x_L = 24√3 - (√3/2)*31 = √3(24 - 15.5) = √3(8.5) = 8.5√3y_L = 31/2 = 15.5Point M: x_M = (3√3)/2 ≈ 2.598..., but wait earlier calculation gave x_M = (3√3)/2 regardless of x? Wait, that can't be right. Wait, earlier when we derived coordinates for M:x_M = x√3 - (√3/2)(2x - 3) = x√3 - x√3 + (3√3)/2 = (3√3)/2But if x =24, then x_M = (3√3)/2 ≈ 2.598, but BC is from B (24√3, 0) to C (0,24). So point M should be closer to C. But if x_M is (3√3)/2, which is approximately 2.598, but the x-coordinate of point B is 24√3 ≈ 41.569, so moving from B to C, the x-coordinate decreases from 41.569 to 0, and y-coordinate increases from 0 to 24. So point M with x-coordinate ≈2.598 would be near point C. But if MC = 3, then the distance from M to C is 3. Let's compute the coordinates:Point M: ( (3√3)/2, x - 3/2 ) where x =24, so y_M =24 - 3/2 =22.5. So point M is at ( (3√3)/2, 22.5 ). Let me compute the distance from M to C (0,24):Distance MC = √[ ( (3√3)/2 - 0 )^2 + (22.5 -24)^2 ] = √[ (27/4) + (2.25) ] = √[6.75 + 2.25] = √9 =3. Correct. So that's good.Similarly, point L is at (8.5√3, 15.5). Let me check the distance BL:Coordinates of B: (24√3, 0). Distance from B to L:√[ (24√3 -8.5√3 )^2 + (0 -15.5)^2 ] = √[ (15.5√3)^2 + (-15.5)^2 ] = √[ (15.5)^2 *3 + (15.5)^2 ] = √[15.5^2 (3 +1)] = √[15.5^2 *4] =15.5*2=31. Correct.Now, check KL = KM. Compute KL and KM.First, point K: (0,4)Point L: (8.5√3,15.5)Compute KL:√[ (8.5√3 -0)^2 + (15.5 -4)^2 ] = √[ (72.25*3) + (11.5)^2 ] = √[216.75 + 132.25] = √[349] ≈18.68Point M: ( (3√3)/2,22.5 )Compute KM:√[ ( (3√3)/2 -0 )^2 + (22.5 -4)^2 ] = √[ (27/4) + (18.5)^2 ] = √[6.75 +342.25] = √[349] ≈18.68Yes, so KL = KM = √349. So that's correct.Therefore, LM =14. So the answer should be 14. But let me check again with BC=48, BL=31, MC=3, so LM=48-31-3=14. Yep.But just to make sure, is there another way this could be approached without coordinates? Maybe using triangle properties or trigonometry?Alternatively, since triangle ABC is right-angled at A with angles 30°, 60°, 90°, sides in ratio 1:√3:2. But since we found x=24, AC=24, which is opposite 30°, so BC=48. Then BL=31, LM=14, MC=3, which adds up to 48.Alternatively, since KL=KM, triangle KML is isoceles with KL=KM. Therefore, point K is equidistant from L and M, so it lies on the perpendicular bisector of LM. Since L and M are on BC, the perpendicular bisector of LM should pass through K. But since K is on AC, perhaps we can use properties of perpendicular bisectors?But the coordinate method worked, giving LM=14. The quadratic gave x=24, which makes BC=48, LM=14. All checks out. So I think the answer is 14.Final AnswerThe length of segment ( LM ) is (boxed{14}).

✅:In triangle (ABC), we know (angle B = 30^circ) and (angle A = 90^circ), making (angle C = 60^circ). This forms a 30-60-90 triangle with sides in the ratio (1 : sqrt{3} : 2). We are given (AK = 4), (BL = 31), and (MC = 3), and need to find the length of segment (LM) where (KL = KM).1. Coordinate System Setup: - Place (A) at ((0, 0)), (B) at ((xsqrt{3}, 0)), and (C) at ((0, x)). - Point (K) is on (AC) such that (AK = 4), so (K) is at ((0, 4)).2. Parameterizing Points on (BC): - Points (L) and (M) are on (BC). Using the distance from (B) and (C): - (L) is 31 units from (B), coordinates: ((xsqrt{3} - frac{31sqrt{3}}{2}, frac{31}{2})). - (M) is 3 units from (C), coordinates: ((frac{3sqrt{3}}{2}, x - frac{3}{2})).3. Using Distance Formula: - Calculate distances (KL) and (KM): - (KL^2 = left(xsqrt{3} - frac{31sqrt{3}}{2}right)^2 + left(frac{31}{2} - 4right)^2) - (KM^2 = left(frac{3sqrt{3}}{2}right)^2 + left(x - frac{11}{2}right)^2)4. Solving for (x): - Set (KL^2 = KM^2) and solve the quadratic equation: - (8x^2 - 328x + 3264 = 0) - Solutions: (x = 24) (valid) and (x = 17) (invalid as (x > 17)).5. Finding (LM): - With (x = 24), (BC = 2x = 48). - (LM = BC - BL - MC = 48 - 31 - 3 = 14).The length of segment (LM) is (boxed{14}).

🔑:1. Identify Given Information: - In the triangle ( triangle ABC ), we are given: [ angle B = 30^circ quad text{and} quad angle A = 90^circ ] - On side ( AC ), a point ( K ) is marked with ( AK = 4 ). - Points ( L ) and ( M ) are marked on side ( BC ) such that ( L ) lies on the segment ( BM ) and ( KL = KM ). - Additionally, ( BL = 31 ) and ( MC = 3 ).2. Relation in Triangle ( triangle ABC ): - Since triangle ( triangle ABC ) is a right-angled triangle with (angle B = 30^circ) and (angle A = 90^circ), it implies (angle C = 60^circ). - The side opposite the (30^circ) angle (side ( AB )) is half the hypotenuse. Thus, [ AB = frac{1}{2} BC ]3. Calculate (AC): - Given (AK = 4), then (KC = AC - AK = AC - 4).4. Analyze right triangle ( triangle CKH ) (drop a perpendicular from ( K ) to ( BC ) at point ( H )): - (angle CKH ) is (60^circ) since ( angle C = 60^circ ). - ( triangle CKH ) forms another (30^circ-60^circ-90^circ) triangle. - In such triangles, [ CK = 2 times CH = 2 times (CM + MH) = 2 times (3+x) ] - Substituting this into our earlier equation, [ KC = 6 + 2x ]5. Determine segment ( BC ): - From the given information, [ BC = BL + LM + MC = 31 + LM + 3 ] - We also know, [ BC = 2 times AC = 2 times (4 + (6 + 2x)) = 20 + 4x ]6. Equate expressions for ( BC ): [ 31 + LM + 3 = 20 + 4x ] Simplifying, [ 34 + LM = 20 + 4x ] [ LM = 4x - 20 + 34 ] [ LM = 4x - 9 - text{(Since if LM was some other value)} 4x - 14 = LM. ]7. Solve for ( x ) and then find ( LM ): - Using condition ( 31 + 2x + 3 = 2(10) + 6=26), [ 34=26 + 4+x x=4, 34+4=40. 34=31 +4--> 34+LM=normal setting 34+ positions - calculate with 34+7. LM = 2 ×x=28=29, = 14.not necessarily calculate element only answer8. Final Answer: Analysis is true and the intersection is correct as solving does not mean we should alter steps but the sketch mechanism is corrected with points and hence the final segment is calculated Conclusion:[boxed{14}] piecing both proving values normal.

🧠:Okay, so I need to figure out the minimum number of colors required to color the wires between 20 telephones such that no two wires connected to the same telephone have the same color. Each telephone can have at most 3 wires connected to it, and each wire connects two telephones. Also, each pair of telephones is connected by at most one wire. Let me try to break this down step by step.First, this seems related to graph theory. Telephones can be thought of as vertices, and wires as edges. The problem is then asking for the edge chromatic number of this graph, which is the minimum number of colors needed to color the edges so that no two adjacent edges (edges sharing a common vertex) have the same color. Given that each telephone (vertex) has at most 3 wires (edges) connected to it, the graph is 3-regular or less. Wait, actually, no. The problem states that each telephone is connected by at most 3 wires. So, the maximum degree of any vertex is 3. Therefore, the graph has a maximum degree Δ = 3. In graph theory, there's a theorem called Vizing's Theorem which states that any simple graph can be edge-colored with either Δ or Δ + 1 colors. So, for a graph with maximum degree 3, the edge chromatic number is either 3 or 4. Vizing's Theorem tells us that it's class 1 (Δ colors) or class 2 (Δ + 1 colors). But the question is, is this particular graph class 1 or class 2? If it's class 1, then 3 colors would suffice. If it's class 2, then we need 4 colors. So, how do we determine which class this graph falls into?I remember that a graph is class 1 if it is possible to color its edges with Δ colors. Factors that might make a graph class 2 include having an odd number of vertices in a regular graph. Wait, but this graph isn't necessarily regular. Each vertex can have degree up to 3, but not necessarily exactly 3. However, in this problem, there are 20 telephones (vertices) connected with wires (edges) such that each wire connects two telephones. Let me check the total number of edges.But wait, the problem doesn't specify how many wires there are in total. It just says there are 20 telephones connected with wires, each wire connects two telephones, each pair connected by at most one wire, and each telephone has at most 3 wires. So, the total number of edges can vary. But the question is about the minimum number of colors required regardless of how the wires are arranged, as long as they meet the given conditions. Wait, no, actually, the problem is about coloring the wires given that the graph is already formed under these constraints. So, we need to find the edge chromatic number for any such graph (with maximum degree 3), and find the minimal number of colors that would work for all such graphs.Wait, no. The problem is not about all possible such graphs, but about a specific graph: 20 telephones connected with wires, each pair connected by at most one wire, each telephone connected by at most 3 wires. So, it's a specific graph with maximum degree 3, but the exact structure isn't specified. So, we need to determine the minimal number of colors that can color any such graph, given the constraints. So, since the graph could be any graph with maximum degree 3, we need to consider the worst-case scenario, i.e., the graph that requires the most colors. Therefore, the answer would be the edge chromatic number for the class of graphs with maximum degree 3, which by Vizing's theorem is 4.But wait, maybe the graph is not simple? No, the problem states each pair is connected by at most one wire, so it's a simple graph. Therefore, according to Vizing's theorem, the edge chromatic number is either 3 or 4. However, to determine the minimal number of colors needed such that any graph meeting the problem's constraints can be colored, we need to use 4 colors, since some graphs with maximum degree 3 might require 4 colors.But is there a way to confirm whether all such graphs with maximum degree 3 on 20 vertices can be edge-colored with 3 colors? Or is there a possibility that some of them require 4?I recall that a connected graph is class 2 (i.e., requires Δ + 1 colors) if it is regular with an odd number of vertices. But here, we have 20 vertices, which is even. However, the graph might not be regular. If the graph is 3-regular and connected, then if it has an even number of vertices, it might still be class 1. Wait, no. For example, a 3-regular graph with an even number of vertices could still be class 2. For instance, the Petersen graph is 3-regular, has 10 vertices, and is class 2.But the Petersen graph has 10 vertices. In our problem, there are 20 telephones. However, the Petersen graph is a famous example of a 3-regular class 2 graph. Therefore, if our graph can contain a Petersen graph as a component, then even with 20 vertices, we could have a subgraph that requires 4 colors. Therefore, in such a case, the entire graph would require 4 colors.Wait, but the Petersen graph has 10 vertices. If we have two copies of the Petersen graph, that would give us 20 vertices, each 3-regular. Each copy would require 4 colors, so the whole graph would require 4 colors. Therefore, in that case, the minimal number of colors needed would be 4.Therefore, even though some 3-regular graphs can be 3-edge-colored (class 1), others cannot. Therefore, since the problem allows for the possibility of such a graph (like two Petersen graphs), the minimal number of colors required to cover all possible cases is 4.However, let's make sure. The problem states "each telephone is connected by at most 3 wires". So, the graph could have maximum degree 3, but not necessarily regular. So, the presence of some vertices with degree less than 3 might allow the entire graph to be edge-colored with 3 colors. Wait, but even if some vertices have lower degrees, if there's a component with a 3-regular graph that is class 2 (like the Petersen graph), then that component would still require 4 colors, regardless of the rest of the graph.Therefore, since the problem allows for the possibility that the graph includes a class 2 component, then the minimal number of colors needed is 4. Therefore, the answer should be 4.But let me check some references. Vizing's theorem states that any graph with maximum degree Δ is edge-colorable with Δ or Δ + 1 colors. So, for Δ = 3, it's 3 or 4. Therefore, the edge chromatic number is 4 if the graph is class 2. So, if the graph in question is class 2, then the answer is 4.But how do we know whether the given graph is class 1 or 2? If the graph is a simple graph with maximum degree 3, but not necessarily connected or regular. If it contains a class 2 component, then the entire graph would require 4 colors. Therefore, to determine the minimal number of colors needed in the worst case, we have to assume the graph could be class 2, hence requiring 4 colors.Therefore, the answer is 4.Wait, but let me think again. Suppose we have a graph where all components are class 1. Then 3 colors would suffice. However, the problem is asking for the minimal number of colors needed such that any such graph (with the given constraints) can be colored. Since there exists graphs (like the Petersen graph) that require 4 colors, even with 20 vertices, we must use 4 colors to cover all possibilities.Therefore, the minimum number of colors required is 4.Alternatively, maybe the structure of the graph (with 20 vertices) allows for some properties that prevent it from being class 2? For example, if the graph is bipartite, then it is class 1. But the problem doesn't state that the graph is bipartite. So, if the graph can be arbitrary, with maximum degree 3, then we need to consider the possibility of it being class 2.Another thought: for a 3-regular graph, if it has a perfect matching, then maybe it can be decomposed into a 2-regular graph and a 1-regular graph, each of which can be edge-colored with 2 and 1 colors respectively. But that might not necessarily work. Alternatively, if the graph is 3-edge-colorable, that would mean it's class 1.But the key point is that there exist 3-regular graphs that are not 3-edge-colorable (class 2), such as the Petersen graph. Therefore, if the telephones are connected in such a way that the graph is similar to the Petersen graph, which is a 3-regular class 2 graph, then 4 colors would be necessary.Since the problem doesn't specify any additional constraints on the connections, we must assume the worst-case scenario where such a class 2 graph is possible, hence requiring 4 colors.Therefore, the minimal number of colors needed is 4.

✅:To determine the minimum number of colors required to color the wires such that no two wires connected to the same telephone share the same color, we model the problem using graph theory. Here, telephones are vertices, wires are edges, and the constraints translate to a simple graph with a maximum degree of 3 (each vertex has at most 3 edges).Key Analysis:1. Vizing's Theorem: Any simple graph with maximum degree ( Delta ) can be edge-colored with ( Delta ) or ( Delta + 1 ) colors. For ( Delta = 3 ), the edge chromatic number is either 3 or 4.2. Class 1 vs. Class 2 Graphs: A graph is class 1 if it can be edge-colored with ( Delta ) colors, and class 2 if it requires ( Delta + 1 ). Notably, some 3-regular graphs (e.g., the Petersen graph) are class 2 and require 4 colors.3. Applicability to the Problem: Since the problem allows for arbitrary connections (with maximum degree 3), the graph could contain class 2 components (e.g., a Petersen graph subgraph). This necessitates 4 colors to accommodate the worst-case scenario.Conclusion: While certain 3-regular graphs can be 3-edge-colored, the existence of class 2 graphs (like the Petersen graph) within the given constraints means the minimal number of colors required to guarantee proper coloring for any valid configuration is 4.Answer: boxed{4}

🔑:1. Interpret the problem: - We have 20 telephone machines, represented as vertices in a graph. - Each telephone machine can connect to at most 3 other machines (vertices of degree at most 3). - There is at most one connecting line between any two telephone machines (edges). - We need to color these lines such that lines from any single machine have different colors. - The task is to determine the minimum number of colors required to achieve this.2. Graph interpretation: - Let's represent each telephone machine as a vertex of a graph. - If two machines are connected by a line, there is an edge between the corresponding vertices. - The graph has 20 vertices, and each vertex can have a degree of at most 3.3. Coloring edges: - We need to color the edges such that no two edges incident to the same vertex have the same color. - This requirement implies a need for a proper edge coloring of the graph.4. Upper and Lower Bounds: - Upper Bound: By Vizing's theorem, a simple graph of maximum degree Delta needs either Delta or Delta + 1 colors for a proper edge coloring. Here, Delta = 3, so we may need either 3 or 4 colors. - Lower Bound: Since each vertex can connect to 3 other vertices, we can't guarantee that 3 colors are sufficient without verifying it on this specific graph.5. Detailed Solution: - Consider the constraints: - There are 20 vertices. - Each vertex (telephone machine) connects to at most 3 other vertices. - To satisfy the requirement that lines from any one telephone machine have different colors, we inspect a few specific structures (cycles and chains): - Cycle: Any cycle in this graph must respect the degree constraints. Since every vertex has a degree of 3, the cycles that occur in the maximum set would dictate coloring complexity. - Chains: Chains can be simplified, but in combination with cycles, distinguishable cases complete the complexity. - We start by analyzing if 3 colors suffice: - Since we know each vertex is involved in at most 3 edges, a degree-3 vertex coloring problem suggests testing a 3-color strategy. - However, if cycles' lengths aren't even, at least one cycle will demand an extra color. The extra handling guarantees the possibility of up to Delta + 1 = 4 colors. - Step-by-Step Process Verification: - Step 1: Find the longest cycle. - Step 2: Color its edges using either 3 colors if possible, otherwise involve a 4th color for odd lengths. - Step 3: Remove the cycle and all dangling edges (edges leading out from vertices of the cycle). Analyze remaining parts for simplifiable sections. - Step 4: Continually break down the new longest cycle until none remains. - Step 5: After breaking cycles, inspect chains or isolated vertices: - Ensure new colors if it yields three edges ending at a vertex requiring a non-redundant color. - Analysis for Eccentric case: A vertex with 3 edges can't consistently get colored if neighbors share required remaining only color distinct but overlapping in next analysis chains.6. Conclusion: - From comprehensive step reductions and infallible combinations of needing distinct colors: - Four-color solution reliably accommodates the constraints: [ boxed{4} ]

🧠:Okay, so I need to find all pairs of strictly positive integers (a, b) that satisfy the equation ab + 2 = a³ + 2b. Hmm, let's see. Let me start by trying to rearrange the equation to make it easier to handle. Maybe I can get all the terms involving b on one side and the terms involving a on the other. Let's try that.Starting with the original equation:ab + 2 = a³ + 2bIf I subtract ab and 2b from both sides, I get:2 = a³ - ab - 2b + abWait, that doesn't make sense. Let me check that again. Wait, no, maybe I should subtract ab and 2b from both sides properly. Let's rearrange the terms:ab - 2b = a³ - 2So, factoring out b from the left side:b(a - 2) = a³ - 2Ah, that's better. So now, we can express b in terms of a:b = (a³ - 2)/(a - 2)Since a and b are strictly positive integers, the denominator (a - 2) must divide the numerator (a³ - 2) exactly, and the result must be a positive integer. Also, since a is a positive integer and a - 2 is in the denominator, a cannot be 2 because that would make the denominator zero. So a ≠ 2. But since a must be strictly positive, possible values of a start from 1, 3, 4, etc.Now, let's analyze this expression for b. Let's compute (a³ - 2)/(a - 2). To do this, maybe I can perform polynomial division or factor the numerator. Let's try polynomial division. Divide a³ - 2 by a - 2.Using polynomial long division:Divide a³ by a to get a². Multiply (a - 2) by a² to get a³ - 2a². Subtract that from a³ - 2:(a³ - 2) - (a³ - 2a²) = 2a² - 2Now, divide 2a² by a to get 2a. Multiply (a - 2) by 2a to get 2a² - 4a. Subtract that from 2a² - 2:(2a² - 2) - (2a² - 4a) = 4a - 2Next, divide 4a by a to get 4. Multiply (a - 2) by 4 to get 4a - 8. Subtract that from 4a - 2:(4a - 2) - (4a - 8) = 6So, the division gives us a² + 2a + 4 with a remainder of 6. Therefore:(a³ - 2)/(a - 2) = a² + 2a + 4 + 6/(a - 2)Since b must be an integer, the remainder 6/(a - 2) must also be an integer. Therefore, (a - 2) must divide 6. So, a - 2 is a positive divisor of 6 (since a > 2, otherwise if a < 2, the denominator would be negative or zero, but since a is a strictly positive integer, and a ≠ 2, let's check possible a values:Wait, a must be strictly positive integer, so possible values for a are 1, 3, 4, 5, etc. But when a is 1, then a - 2 = -1, which divides 6 as well. So maybe we need to consider both positive and negative divisors of 6, but since a is a positive integer, a - 2 can be positive or negative. Let's consider all divisors of 6, positive and negative.The divisors of 6 are: ±1, ±2, ±3, ±6.So, a - 2 must be one of these values. Therefore, a = 2 + d, where d is a divisor of 6. Let's list all possible d:d ∈ { -6, -3, -2, -1, 1, 2, 3, 6 }Therefore, possible a values are:a = 2 + (-6) = -4 (invalid, since a must be positive)a = 2 + (-3) = -1 (invalid)a = 2 + (-2) = 0 (invalid)a = 2 + (-1) = 1 (valid)a = 2 + 1 = 3 (valid)a = 2 + 2 = 4 (valid)a = 2 + 3 = 5 (valid)a = 2 + 6 = 8 (valid)So possible a values are 1, 3, 4, 5, 8. Let's check each of these and compute b to see if it's a positive integer.Starting with a = 1:b = (1³ - 2)/(1 - 2) = (1 - 2)/(-1) = (-1)/(-1) = 1. So b = 1. Check if this satisfies the original equation:1*1 + 2 = 1 + 2 = 3. On the other side, 1³ + 2*1 = 1 + 2 = 3. So yes, (1,1) is a solution.Next, a = 3:b = (3³ - 2)/(3 - 2) = (27 - 2)/1 = 25/1 = 25. So b = 25. Check the equation:3*25 + 2 = 75 + 2 = 77. On the other side, 3³ + 2*25 = 27 + 50 = 77. That works. So (3,25) is a solution.Next, a = 4:b = (4³ - 2)/(4 - 2) = (64 - 2)/2 = 62/2 = 31. So b = 31. Check the equation:4*31 + 2 = 124 + 2 = 126. On the other side, 4³ + 2*31 = 64 + 62 = 126. Correct. So (4,31) is a solution.a = 5:b = (5³ - 2)/(5 - 2) = (125 - 2)/3 = 123/3 = 41. So b = 41. Check:5*41 + 2 = 205 + 2 = 207. On the other side, 5³ + 2*41 = 125 + 82 = 207. Correct. So (5,41) is a solution.a = 8:b = (8³ - 2)/(8 - 2) = (512 - 2)/6 = 510/6 = 85. So b = 85. Check:8*85 + 2 = 680 + 2 = 682. Other side: 8³ + 2*85 = 512 + 170 = 682. Correct. So (8,85) is a solution.Now, let's check if there are any other possible a values. Wait, when a - 2 is a negative divisor of 6, which gives a = 2 + d where d is a negative divisor. But when d is negative, a could be 1, 0, etc., but since a must be strictly positive, only a = 1 is valid in that case (since a = 2 + (-1) =1). We already checked a =1. The other negative divisors lead to a ≤0, which are invalid. So those are all possible a's. Therefore, the solutions are (1,1), (3,25), (4,31), (5,41), (8,85).Wait, but wait. Let me confirm if these are all the possible divisors. For example, is there a divisor of 6 that we missed? Let's see: divisors of 6 are ±1, ±2, ±3, ±6, as I listed before, so we considered all possible d's.Alternatively, maybe there's another way to approach the problem. Let me try substituting small a values just to make sure.Starting with a =1:Original equation: 1*b +2 =1 +2b → b +2 =1 +2b → Subtract b: 2 =1 +b → b=1. Which matches.a=2: Not allowed since denominator becomes zero. So skip.a=3: as before, gives b=25.a=4: gives b=31.a=5: b=41.a=6: Let's compute b. Using the formula:b = (6³ -2)/(6-2) = (216 -2)/4 =214/4=53.5. Not an integer. So not valid.a=7: b=(343 -2)/5=341/5=68.2. Not integer.a=8: 85, which is integer.a=9: b=(729 -2)/7=727/7≈103.857. Not integer.So seems like after a=8, next a's give non-integer b's. So that's consistent with our earlier conclusion.Therefore, the solutions are the ones we found. So final answer should be these pairs: (1,1), (3,25), (4,31), (5,41), (8,85).Wait, but wait a second. Let me check a=0. Even though a must be strictly positive, just to confirm. a=0 would make the original equation: 0*b +2=0 +2b → 2=2b →b=1. But a=0 is invalid, so that's not considered. Similarly, negative a's: like a=-1, but since a must be strictly positive, we don't consider those.Another way to check: suppose we rearrange the original equation as:ab - 2b = a³ - 2b(a - 2) = a³ -2So, since a and b are positive integers, a must be greater than 2 or less than 2. If a=1, which is less than 2, then a-2=-1, so b=(1 -2)/(-1)=1, which is positive. If a=3,4,5,8 as before, which are greater than 2, leading to positive denominators. So all possibilities considered.Therefore, the conclusion is correct. All solutions are the five pairs mentioned above. Wait, but wait, actually in our list, we have five solutions? Let's count:(1,1), (3,25), (4,31), (5,41), (8,85). That's five pairs. Wait, but let me check once more for a=6:As before, a=6 gives b=(216-2)/4=214/4=53.5, not integer. So invalid. a=7 gives 341/5=68.2, invalid. a=8 gives 510/6=85, which is valid. a=9: 727/7=103.857 invalid. a=10: (1000-2)/8=998/8=124.75 invalid. So no more solutions beyond a=8.Therefore, the five pairs are indeed the only solutions.Wait, but wait. Let me check a=1 again. Original equation: 1*1 +2=1+2=3. On the other side: 1³ +2*1=1+2=3. Correct.a=3: 3*25+2=75+2=77; 27 +50=77. Correct.a=4:4*31+2=124+2=126;64+62=126. Correct.a=5:5*41+2=205+2=207;125+82=207. Correct.a=8:8*85+2=680+2=682;512+170=682. Correct.Therefore, all these pairs are valid. Hence, the solution set is correct.Final AnswerThe solutions are (boxed{(1, 1)}), (boxed{(3, 25)}), (boxed{(4, 31)}), (boxed{(5, 41)}), and (boxed{(8, 85)}).

✅:To determine all pairs of strictly positive integers ((a, b)) that satisfy the equation (ab + 2 = a^3 + 2b), we start by rearranging the equation:[ab - 2b = a^3 - 2]Factoring out (b) on the left side, we get:[b(a - 2) = a^3 - 2]Solving for (b), we have:[b = frac{a^3 - 2}{a - 2}]For (b) to be a positive integer, (frac{a^3 - 2}{a - 2}) must be an integer. This requires that (a - 2) divides (a^3 - 2). Using polynomial division, we find:[frac{a^3 - 2}{a - 2} = a^2 + 2a + 4 + frac{6}{a - 2}]For the remainder (frac{6}{a - 2}) to be an integer, (a - 2) must be a divisor of 6. The divisors of 6 are (pm 1, pm 2, pm 3, pm 6). Thus, (a - 2) can be one of these values, leading to possible values of (a):- (a - 2 = -1 implies a = 1)- (a - 2 = 1 implies a = 3)- (a - 2 = 2 implies a = 4)- (a - 2 = 3 implies a = 5)- (a - 2 = 6 implies a = 8)We check each of these values of (a) to find corresponding (b):- For (a = 1): (b = frac{1^3 - 2}{1 - 2} = frac{-1}{-1} = 1)- For (a = 3): (b = frac{3^3 - 2}{3 - 2} = frac{25}{1} = 25)- For (a = 4): (b = frac{4^3 - 2}{4 - 2} = frac{62}{2} = 31)- For (a = 5): (b = frac{5^3 - 2}{5 - 2} = frac{123}{3} = 41)- For (a = 8): (b = frac{8^3 - 2}{8 - 2} = frac{510}{6} = 85)Each pair ((a, b)) is verified to satisfy the original equation. Therefore, the solutions are:[boxed{(1, 1)}, boxed{(3, 25)}, boxed{(4, 31)}, boxed{(5, 41)}, boxed{(8, 85)}]

🔑:We are given the equation: [ ab + 2 = a^3 + 2b ]To start solving for the pairs ((a, b)), we first rearrange terms to isolate (b):1. Move all the terms involving (b) to one side:[ab + 2 = a^3 + 2b implies ab - 2b = a^3 - 2]2. Factor out (b) on the left-hand side:[b(a - 2) = a^3 - 2]From this, we see that (a - 2) must be a divisor of (a^3 - 2), and:[ b = frac{a^3 - 2}{a - 2}]Next, we need to determine when (a - 2) divides (a^3 - 2). To simplify (a^3 - 2), we use polynomial long division or the remainder theorem:3. Simplify the expression:[a^3 - 2 = (a - 2)left(frac{a^3-2}{a-2}right) + R]4. Using polynomial long division:Divide (a^3 - 2) by (a - 2):[a^3 - 2 = (a - 2)(a^2 + 2a + 4) + 6]This confirms:[a - 2 mid a^3 - 2Rightarrow a - 2 mid 6]Thus (a - 2) must be a divisor of 6. Therefore the possible values are:[a - 2 in {-1, 1, 2, 3, 6}]This leads to:[a in {1, 3, 4, 5, 8}]Now we calculate (b) for each (a):- When (a = 1):[b = frac{1^3 - 2}{1 - 2} = frac{-1}{-1} = 1]Hence, ((a, b) = (1, 1))- When (a = 3):[b = frac{3^3 - 2}{3 - 2} = frac{27 - 2}{1} = 25]Hence, ((a, b) = (3, 25))- When (a = 4):[b = frac{4^3 - 2}{4 - 2} = frac{64 - 2}{2} = 31]Hence, ((a, b) = (4, 31))- When (a = 5):[b = frac{5^3 - 2}{5 - 2} = frac{125 - 2}{3} = 41]Hence, ((a, b) = (5, 41))- When (a = 8):[b = frac{8^3 - 2}{8 - 2} = frac{512 - 2}{6} = 85]Hence, ((a, b) = (8, 85)) Conclusion:The pairs of strictly positive integers ((a, b)) that satisfy the equation (ab + 2 = a^3 + 2b) are:[boxed{(1, 1), (3, 25), (4, 31), (5, 41), (8, 85)}]

🧠:Okay, let me try to figure this out step by step. So, Sasha lives in apartment number 333 in entrance 10, but Peter needs to know which floor to go to. The building has nine floors. Each floor has the same number of apartments, and the apartment numbers start from one. Hmm, alright.First, I need to determine how many apartments there are per floor in each entrance. Since the building has nine floors and each entrance presumably has apartments on all floors, the total number of apartments in one entrance would be the number of apartments per floor multiplied by nine. Then, apartment numbers are consecutive starting from 1 in each entrance. So, if Sasha is in entrance 10, that entrance has apartments numbered starting from 1 up to some maximum number, with each floor contributing equally to that total.Wait, but entrance 10 is just one entrance. So, apartment numbers in entrance 10 start at 1, and each floor in entrance 10 has a certain number of apartments. The key is to find out how many apartments are on each floor in entrance 10. Then, since apartment 333 is in entrance 10, we can divide 333 by the number of apartments per floor to find the floor number, right?But how do we find the number of apartments per floor? Let's denote the number of apartments per floor in each entrance as x. Then, each entrance has 9 floors, so total apartments per entrance would be 9x. Therefore, apartment numbers in entrance 10 would range from (9x * 9) * (10 - 1) + 1 to 9x * 10. Wait, that might not be correct. Wait, maybe each entrance is independent, so entrance 1 has apartments 1 to 9x, entrance 2 has 9x + 1 to 18x, and so on. But the problem says apartment numbers start from one in each entrance. Wait, that might not be standard. Wait, the problem says "apartment numbers in the building start from one." Hmm, maybe the entire building's apartments are numbered starting from 1, but each entrance has a block of apartments. Wait, but Sasha is in entrance 10, apartment 333. So, perhaps each entrance has apartments numbered starting from 1, but that can't be because the building's apartments start from one. Wait, the problem says "apartment numbers in the building start from one." So, maybe each entrance is a separate block where apartment numbers start from 1 again? Wait, that's conflicting. If the entire building's apartments start from one, then entrance 10's apartments would be part of the continuous numbering. But Sasha said she is in apartment No. 333 in entrance 10. So, perhaps the building has multiple entrances, each with their own apartment numbering. Wait, that's possible. For example, in some buildings, each entrance has apartments numbered independently. So entrance 1 has apartments 1, 2, 3..., entrance 2 has apartments 1, 2, 3..., etc. But the problem says "apartment numbers in the building start from one," which might mean that all apartments in the building are numbered consecutively from 1, regardless of entrance. But then, if that's the case, entrance 10's apartments would be part of a continuous sequence. But the problem states that Sasha is in apartment No. 333 in entrance 10. This is a bit confusing.Wait, perhaps the problem is structured such that each entrance has apartments numbered starting from 1. So, entrance 1 has apartments 1, 2, 3..., entrance 2 has apartments 1, 2, 3..., etc. Then, Sasha is in entrance 10, apartment 333. However, the building has nine floors. So, each entrance is a separate vertical section of the building, each with nine floors. Each floor in an entrance has the same number of apartments. So, the total number of apartments in one entrance would be floors * apartments per floor. So, entrance 10's apartment numbers are 1 to 9x, where x is apartments per floor. Then, apartment 333 is within entrance 10. Therefore, we can calculate x by seeing how many apartments are in entrance 10. But since entrance 10 has apartment numbers up to 9x, and apartment 333 is in entrance 10, that means 9x must be at least 333. So, 9x ≥ 333 → x ≥ 37. So, apartments per floor would be at least 37. But that can't be right because that would make each floor in entrance 10 have 37 apartments, which seems high. Wait, perhaps I'm approaching this wrong.Alternatively, maybe the entire building is divided into entrances, each entrance having multiple floors, and each floor in an entrance has a certain number of apartments. The apartment numbers are sequential within each entrance. So, entrance 1 has apartments 1 to N, entrance 2 has N+1 to 2N, etc., but the problem states that apartment numbers start from one in each entrance. Wait, no, the problem says "apartment numbers in the building start from one." So, maybe the entire building's apartments are numbered from 1 onwards, and each entrance is a subset of that numbering. So, entrance 1 has apartments 1 to K, entrance 2 has K+1 to 2K, etc. Then, Sasha is in entrance 10, which would have apartments from (10-1)*K +1 to 10*K. But she is in apartment 333, so 9*K +1 ≤ 333 ≤10*K. Then, solving for K: 9K +1 ≤333 ⇒ 9K ≤332 ⇒ K ≤332/9≈36.888. So K=36. Then, 10*K=360. So entrance 10 would have apartments from 325 to 360. But 333 is within that range. Then, each entrance has 36 apartments. But the building has nine floors, so each entrance must have 36 apartments spread over nine floors. Therefore, each floor in an entrance has 36/9=4 apartments. Therefore, each floor has 4 apartments. Therefore, apartment 333 in entrance 10 is calculated as follows: entrance 10 starts at apartment (10-1)*36 +1= 324 +1=325. So apartment 325 is the first apartment in entrance 10, on floor 1. Then, each floor has 4 apartments, so floor 1: 325-328, floor 2:329-332, floor3:333-336. Therefore, apartment 333 is on floor 3. Therefore, Peter should go to floor 3. Wait, but let's check this.Wait, if each entrance has 36 apartments (4 per floor *9 floors), then entrance 1:1-36, entrance2:37-72,... entrance10:325-360. So apartment 333 is in entrance10. Then, within entrance10, the apartments are 325-360. Apartment 333 is 333-325+1=9th apartment in entrance10. Since each floor has 4 apartments, 9 divided by4 is 2.25, so floor 3. Because floor1:1-4 (325-328), floor2:5-8 (329-332), floor3:9-12 (333-336). So yes, apartment 333 is on floor3.But wait, is the assumption correct that each entrance has 36 apartments? Because 9 floors *4 apartments per floor=36. But how did we get 4 apartments per floor? Because K=36 per entrance, divided by9 floors=4 per floor. But how did we get K=36? Because entrance10 starts at apartment 9*K +1, and 9*K +1 ≤333 ≤10*K. Solving 9K +1 ≤333 ≤10K. So from the upper bound:333 ≤10K → K≥33.3. From the lower bound:9K +1 ≤333 →9K ≤332→ K≤36.888. So K can be 34,35,36. Wait, but K must be integer. So possible K values are 34,35,36. Let's check each.If K=34: entrance10 would start at 9*34 +1=306 +1=307. So entrance10:307-340. Then apartment333 is in entrance10. Then, within entrance10, apartment333 is 333-307 +1=27th apartment. Each floor has 34/9≈3.77 apartments, which is not possible. So K must be a multiple of 9? Because each entrance has 9 floors, so total apartments per entrance must be 9*x, where x is apartments per floor. Therefore, K=9x. So K must be divisible by9. So from previous K=36.888, the maximum integer K divisible by9 less than or equal to36.888 is36. So K=36. Then, apartments per entrance=36, which is9 floors *4 apartments per floor. Therefore, each floor has4 apartments. Then entrance10 starts at9*36 +1=325. So entrance10 apartments are325-360. Apartment333 is the9th apartment in entrance10. 9 divided by4 apartments per floor gives2 full floors (8 apartments) and1 remaining, so floor3. Therefore, floor3.Alternatively, maybe the problem assumes that all entrances have the same number of apartments, and the total apartments per entrance is9 floors * apartments per floor. So total apartments per entrance=9x. The entire building's apartments are numbered starting from1, so entrance1 has1-9x, entrance2 has9x+1-18x, etc. So entrance10 would be from9x*9 +1 to9x*10. Wait, entrance1:1-9x, entrance2:9x+1-18x, ..., entrance10:9x*9 +1 to9x*10. Wait, no, entrance10 would be (10-1)*9x +1 to10*9x. So entrance10 starts at81x +1. But Sasha is in entrance10, apartment333. So81x +1 ≤333 ≤90x. So solving forx:81x +1 ≤333 →81x ≤332 →x ≤332/81≈4.098And 333 ≤90x →x≥333/90≈3.7So x must be an integer between4.098 and3.7, so x=4.So x=4 apartments per floor. Then, entrance10 starts at81*4 +1=324 +1=325. So entrance10 has apartments325 to360. Apartment333 is in entrance10. Then, within entrance10, apartment333 is333-325 +1=9th apartment. Each floor has4 apartments. So floor1:1-4 (325-328), floor2:5-8 (329-332), floor3:9-12 (333-336). Therefore, apartment333 is on floor3.Yes, this seems consistent. Therefore, the answer is the3rd floor.

✅:To determine the correct floor for apartment 333 in entrance 10 of a 9-floor building where each floor has an equal number of apartments:1. Assume each entrance has (9x) apartments (9 floors × (x) apartments per floor).2. Entrance 10 starts at apartment ((10-1) times 9x + 1 = 81x + 1).3. Solve for (x) such that (81x + 1 leq 333 leq 90x): - Lower bound: (81x leq 332 Rightarrow x leq 4.098). - Upper bound: (333 leq 90x Rightarrow x geq 3.7). - Thus, (x = 4) (integer value).4. Entrance 10 apartments: (81 times 4 + 1 = 325) to (90 times 4 = 360).5. Apartment 333 is the 9th apartment in entrance 10 ((333 - 325 + 1 = 9)).6. Floor calculation: With 4 apartments per floor: - Floor 1: 1-4 (325-328) - Floor 2: 5-8 (329-332) - Floor 3: 9-12 (333-336).Answer: Peter should go to the 3rd floor.

🔑:1. Determine the constraints on the number of apartments per floor: - If the building has up to three apartments per floor: [ text{Total number of apartments in 10 entrances} = 10 times 9 times 3 = 270 ] Since apartment number 333 exceeds 270, there cannot be only three apartments per floor.2. Determine the possibility with at least five apartments per floor: - If the building has at least five apartments per floor: [ text{Total number of apartments in 9 entrances} = 9 times 9 times 5 = 405 ] Since 405 is much larger than 333, it contradicts the scenario of having 5 or more apartments per floor in a 9-story building with 10 entrances.3. Evaluate the scenario with four apartments per floor: - If each floor has four apartments: [ text{Total number of apartments in 9 entrances} = 9 times 9 times 4 = 324 ] This implies that the numbering starts from 325 in the 10th entrance.4. Compute the apartment numbers in the 10th entrance: - Here, the apartments in the 10th entrance start at: [ 325 ]5. Identify the floor of apartment number 333: - Apartments in the 10th entrance, by floor: - 1st floor: 325, 326, 327, 328 - 2nd floor: 329, 330, 331, 332 - 3rd floor: 333, 334, 335, 3366. Conclusion: - Since apartment number 333 is on the 3rd floor, Petya needs to go to the 3rd floor.[boxed{text{On the 3rd floor}}]