🧠:Okay, let me try to work through this problem step by step. So, the problem is about an infinite complex geometric series where the first term z₀ is 1, and the 2013th term z₂₀₁₃ is 1/(2013²⁰¹³). We need to find the sum of all possible sums of this series. Hmm, okay. Let's break this down.First, let's recall the general form of a geometric series. An infinite geometric series has the first term a and common ratio r, right? So, the terms are a, ar, ar², ar³, and so on. The sum S of the series is a / (1 - r), provided that |r| < 1 to ensure convergence. In this problem, the first term z₀ is 1, so a = 1. Therefore, the sum S would be 1 / (1 - r). But since we're dealing with complex numbers, r can be a complex number, and the condition for convergence is still |r| < 1, correct?Now, the 2013th term is given as z₂₀₁₃ = 1/(2013²⁰¹³). Wait, in the standard geometric series, the nth term is arⁿ. But here, the terms are labeled starting from z₀. So, z₀ is the first term, z₁ is the second term, z₂ is the third term, etc. Therefore, the nth term zₙ would be arⁿ. So, z₀ = ar⁰ = a = 1. Then z₁ = ar¹ = r, z₂ = ar² = r², and so on. Therefore, z₂₀₁₃ would be ar²⁰¹³ = r²⁰¹³. But the problem says z₂₀₁₃ = 1/(2013²⁰¹³). Therefore, r²⁰¹³ = 1/(2013²⁰¹³). Therefore, r²⁰¹³ = (1/2013)²⁰¹³. So, taking the 2013th root of both sides, we have r = 1/2013 multiplied by a 2013th root of unity. Because in complex numbers, if xⁿ = yⁿ, then x = y * ω where ω is an nth root of unity.So, that gives r = (1/2013) * ω, where ω is a primitive 2013th root of unity. Wait, but actually, any 2013th root of unity, not necessarily primitive. So, ω can be e^(2πik/2013) for k = 0, 1, 2, ..., 2012. Therefore, all possible values of r are (1/2013) * e^(2πik/2013) for k from 0 to 2012. However, we need to ensure that |r| < 1 for the series to converge. Since |r| = |(1/2013) * ω| = 1/2013, which is less than 1, because 2013 is a positive integer greater than 1. So, 1/2013 is definitely less than 1. Therefore, all these values of r are valid, and the series converges for each possible r.Therefore, each possible r gives a different sum S. The sum S for each r is 1 / (1 - r). Therefore, we need to compute the sum of all possible S, where each S is 1 / (1 - (1/2013) * ω), and ω ranges over all 2013th roots of unity. Therefore, the total sum we need is the sum over k=0 to 2012 of 1 / (1 - (1/2013) * e^(2πik/2013)).Hmm, this seems complicated. How can we compute such a sum? Maybe there's a clever way using complex analysis or roots of unity properties. Let me think.Alternatively, perhaps we can relate this sum to a known formula. Let's denote ω_k = e^(2πik/2013), so that r_k = (1/2013) * ω_k. Then S_k = 1 / (1 - r_k) = 1 / (1 - ω_k / 2013). We need to sum S_k over k from 0 to 2012.So, the sum we need is Σ_{k=0}^{2012} [1 / (1 - ω_k / 2013)]. Let's denote this sum as T = Σ_{k=0}^{2012} [1 / (1 - ω_k / 2013)].Hmm, maybe we can consider this sum T as the sum over all roots ω of the equation ω²⁰¹³ = 1 of the function 1 / (1 - ω / 2013). There is a formula for summing such expressions over roots of unity. For example, sometimes we can use the formula for the sum of 1/(a - ω) over roots ω of unity, which can be connected to derivatives of polynomials or using partial fractions.Alternatively, consider that the denominator 1 - ω / 2013 can be written as (2013 - ω)/2013. Therefore, 1 / (1 - ω / 2013) = 2013 / (2013 - ω). Therefore, T = Σ_{k=0}^{2012} [2013 / (2013 - ω_k)].So, T = 2013 * Σ_{k=0}^{2012} [1 / (2013 - ω_k)].So, the problem reduces to calculating Σ_{k=0}^{2012} [1 / (2013 - ω_k)] where ω_k are the 2013th roots of unity.Hmm, perhaps there is a connection with the roots of the polynomial x²⁰¹³ - 1 = 0. Let's recall that for a polynomial P(x) = (x - ω_0)(x - ω_1)...(x - ω_{2012}), then the sum over 1/(a - ω_k) can be expressed in terms of P'(a)/P(a). Specifically, Σ_{k=0}^{2012} 1/(a - ω_k) = P'(a)/P(a). This is a standard result from complex analysis, where the sum of reciprocals of (a - roots) is equal to the derivative of the polynomial divided by the polynomial evaluated at a.Yes, exactly. So, if P(x) = x²⁰¹³ - 1, then P'(x) = 2013x²⁰¹². Therefore, Σ_{k=0}^{2012} 1/(a - ω_k) = P'(a)/P(a) = (2013a²⁰¹²)/(a²⁰¹³ - 1). Therefore, in our case, a = 2013, so:Σ_{k=0}^{2012} 1/(2013 - ω_k) = (2013*(2013)²⁰¹²)/((2013)²⁰¹³ - 1) = (2013²⁰¹³)/(2013²⁰¹³ - 1).Therefore, T = 2013 * [2013²⁰¹³ / (2013²⁰¹³ - 1)] = 2013²⁰¹⁴ / (2013²⁰¹³ - 1).Simplifying that, we can factor numerator and denominator:2013²⁰¹⁴ / (2013²⁰¹³ - 1) = 2013 * 2013²⁰¹³ / (2013²⁰¹³ - 1) = 2013 / (1 - 1/2013²⁰¹³).Wait, but perhaps we can leave it as 2013²⁰¹⁴ / (2013²⁰¹³ - 1). However, the question is to find the sum of all possible sums of the series. So, each possible r gives a sum S_k, and we need to sum all S_k, which we just found to be T = 2013²⁰¹⁴ / (2013²⁰¹³ - 1).But wait, let me double-check the steps to make sure I didn't make a mistake.First, the key step was recognizing that the sum over 1/(2013 - ω_k) is equal to P'(2013)/P(2013), where P(x) = x²⁰¹³ - 1. Then, P'(x) = 2013x²⁰¹², so P'(2013) = 2013*(2013)²⁰¹² = 2013²⁰¹³. Then, P(2013) = (2013)²⁰¹³ - 1. Therefore, the sum is 2013²⁰¹³ / (2013²⁰¹³ - 1). Therefore, multiplying by 2013 (from T = 2013 * sum), we get T = 2013²⁰¹⁴ / (2013²⁰¹³ - 1). That seems correct.But let me verify with a simpler case. Let's take a smaller exponent instead of 2013, say n=2. Suppose we have z₀=1, z₂=1/2², and find the sum of all possible sums. Following the same steps:r² = (1/2²), so r = ±1/2. The possible sums are 1/(1 - 1/2) = 2 and 1/(1 + 1/2) = 2/3. The sum of these is 2 + 2/3 = 8/3. Using the formula above, T = 2²⁺¹ / (2² - 1) = 2³ / (4 - 1) = 8/3. Which matches. So that works.Another test case: n=1. Wait, n=1 would mean z₁ = 1/1¹ = 1. Then the series would have r¹ = 1, so r=1. But |r|=1, which would make the series divergent. But the problem states it's an infinite series, so convergence is required. So in the problem statement, n=2013 is acceptable since |r|=1/2013 < 1. But for n=1, it's invalid. So maybe n=3. Let's try n=3.If n=3, then z₃=1/3³. So, r³ = 1/27, so r = 1/3 * ω, where ω³=1. So, ω=1, e^(2πi/3), e^(4πi/3). Then the sums are 1/(1 - 1/3) = 3/2, 1/(1 - e^(2πi/3)/3), and 1/(1 - e^(4πi/3)/3). Let's compute those two complex sums.First, 1/(1 - e^(2πi/3)/3). Let me compute denominator: 1 - (e^(2πi/3)/3). Similarly for the other term. Let's compute the sum:3/2 + 1/(1 - e^(2πi/3)/3) + 1/(1 - e^(4πi/3)/3).Using the formula, T = 3⁴ / (3³ -1 ) = 81 / 26 ≈ 3.115. But manually computing:First, 3/2 ≈ 1.5.Now, compute 1/(1 - e^(2πi/3)/3). Let me compute e^(2πi/3) = cos(2π/3) + i sin(2π/3) = -1/2 + i√3/2. So, e^(2πi/3)/3 = (-1/2 + i√3/2)/3 = -1/6 + i√3/6. Therefore, 1 - (-1/6 + i√3/6) = 1 + 1/6 - i√3/6 = 7/6 - i√3/6. The reciprocal is 1 / (7/6 - i√3/6) = 6/(7 - i√3). Multiply numerator and denominator by (7 + i√3):6*(7 + i√3) / (49 + 3) = 6*(7 + i√3)/52 = (42 + 6i√3)/52 = (21 + 3i√3)/26 ≈ (21/26) + i*(3√3)/26 ≈ 0.8077 + i0.299.Similarly, the other term is 1/(1 - e^(4πi/3)/3). e^(4πi/3) is conjugate of e^(2πi/3), so it's -1/2 - i√3/2. Dividing by 3 gives -1/6 - i√3/6. Therefore, 1 - (-1/6 - i√3/6) = 7/6 + i√3/6. The reciprocal is 6/(7 + i√3) = (42 - 6i√3)/52 = (21 - 3i√3)/26 ≈ 0.8077 - i0.299.Adding these two complex numbers: (0.8077 + i0.299) + (0.8077 - i0.299) = 1.6154. Adding the 3/2 (1.5) gives 1.6154 + 1.5 ≈ 3.1154. According to the formula, 81/26 ≈ 3.1154, which matches. So the formula works. Therefore, our steps are correct.Therefore, returning to the original problem, the sum T is 2013²⁰¹⁴ / (2013²⁰¹³ - 1). Let me see if this can be simplified further.Note that 2013²⁰¹⁴ = 2013 * 2013²⁰¹³, so we can write:T = (2013 * 2013²⁰¹³) / (2013²⁰¹³ - 1) = 2013 / (1 - 1/2013²⁰¹³)But that doesn't seem particularly simpler. Alternatively, perhaps factor out 2013²⁰¹³ in the denominator:Denominator: 2013²⁰¹³ - 1 = 2013²⁰¹³(1 - 1/2013²⁰¹³)But when we divide, T = 2013²⁰¹⁴ / (2013²⁰¹³ - 1) = 2013 / (1 - 1/2013²⁰¹³)Hmm, but perhaps this is the simplest form.Alternatively, notice that 2013²⁰¹³ - 1 is a large number, but since the question is about complex series, maybe they expect an answer in terms of 2013 divided by (1 - 1/2013²⁰¹³) or as 2013²⁰¹⁴ / (2013²⁰¹³ - 1). Let me check if that can be simplified more.Alternatively, perhaps we can write it as 2013 + 2013/(2013²⁰¹³ - 1). But that might not be helpful. Alternatively, note that 2013²⁰¹⁴ = 2013*(2013²⁰¹³ - 1) + 2013. Therefore,2013²⁰¹⁴ = 2013*(2013²⁰¹³ - 1) + 2013Therefore,T = [2013*(2013²⁰¹³ - 1) + 2013] / (2013²⁰¹³ - 1) = 2013 + 2013/(2013²⁰¹³ - 1)But this might not be useful. Alternatively, since 2013 and 2013²⁰¹³ - 1 are coprime? Maybe not. But perhaps the answer is simply left as 2013²⁰¹⁴ divided by (2013²⁰¹³ - 1). The problem says "Find the sum of all possible sums of this series." So unless there's a simplification I'm missing, this should be the answer.Alternatively, maybe we can write it as 2013/(1 - (1/2013²⁰¹³)), but that's equivalent. Let me check if these are the same:2013²⁰¹⁴ / (2013²⁰¹³ - 1) = 2013 * 2013²⁰¹³ / (2013²⁰¹³ - 1) = 2013 / (1 - 1/2013²⁰¹³). Yes, that's correct.But which form is more simplified? The problem might prefer the form with exponents as 2013²⁰¹⁴ / (2013²⁰¹³ - 1). Let me check the original problem statement again. It's possible that the answer is simply 2014, but that seems too simplistic. Wait, maybe there's a different approach.Wait, another thought: The sum of all possible sums is the sum over all roots of unity ω of 1/(1 - (ω/2013)). But maybe there's a way to consider that 1/(1 - (ω/2013)) can be expanded as a power series. Wait, but that might not help. Alternatively, consider that when you sum over all roots of unity, certain symmetries cause cancellations or reinforcements.But we already used the formula for summing 1/(a - ω_k) over roots of unity, which gives a clean answer. So unless there's a trick I'm missing, the answer is 2013²⁰¹⁴ / (2013²⁰¹³ - 1).But let me verify once again. Let's redo the steps concisely:1. Given z₀ = 1, z₂₀₁₃ = 1/2013²⁰¹³.2. Since it's a geometric series, zₙ = rⁿ. Therefore, z₂₀₁₃ = r²⁰¹³ = 1/2013²⁰¹³.3. Solving for r: r = (1/2013) * ω, where ω²⁰¹³ = 1. So ω runs over all 2013th roots of unity.4. The sum S for each r is 1/(1 - r).5. Therefore, we need to compute Σ_{ω²⁰¹³=1} [1/(1 - ω/2013)].6. Using the formula for the sum over roots of unity: Σ_{k=0}^{n-1} 1/(a - ω_k) = (n a^{n-1}) / (a^n - 1), where ω_k are the nth roots of unity. Wait, is that the case?Wait, earlier we used that Σ_{k=0}^{n-1} 1/(a - ω_k) = P'(a)/P(a), where P(x) = xⁿ - 1. Then P'(a) = n a^{n-1}, so Σ = n a^{n-1}/(aⁿ - 1). Therefore, in our case, n=2013, a=2013. Therefore, Σ = 2013 * 2013^{2012} / (2013^{2013} - 1) = 2013^{2013}/(2013^{2013} - 1).Then, the total sum T is 2013 multiplied by this, so T = 2013 * 2013^{2013}/(2013^{2013} - 1) = 2013^{2014}/(2013^{2013} - 1). So that confirms our previous result.Therefore, the answer should be 2013²⁰¹⁴ divided by (2013²⁰¹³ - 1). Let me write that as a box:boxed{dfrac{2013^{2014}}{2013^{2013} - 1}}I think that's the correct answer. Let me check once more for any possible miscalculations.Yes, the key steps are:- Identifying that r must be (1/2013) times a 2013th root of unity.- Recognizing that each such r gives a convergent series because |r| = 1/2013 < 1.- Summing 1/(1 - r) over all such r is equivalent to summing 2013/(2013 - ω) over all 2013th roots of unity ω.- Applying the formula for the sum of reciprocals (a - ω_k)⁻¹ over roots ω_k of unity, which gives P'(a)/P(a).- Calculating that sum and multiplying by 2013 to get the final result.Everything checks out. The test cases with smaller exponents also support the formula. Therefore, I'm confident this is the correct answer.Final Answerboxed{dfrac{2013^{2014}}{2013^{2013} - 1}}

✅:Given an infinite complex geometric series with ( z_0 = 1 ) and ( z_{2013} = frac{1}{2013^{2013}} ), we need to find the sum of all possible sums of this series.1. Identify the common ratio ( r ): - The nth term of a geometric series is given by ( z_n = r^n ). - Given ( z_{2013} = frac{1}{2013^{2013}} ), we have ( r^{2013} = frac{1}{2013^{2013}} ). - Solving for ( r ), we get ( r = frac{1}{2013} omega ) where ( omega ) is a 2013th root of unity.2. Sum of the series: - The sum ( S ) of the series for a common ratio ( r ) is ( frac{1}{1 - r} ) provided ( |r| < 1 ). - Here, ( |r| = frac{1}{2013} < 1 ), so the series converges.3. Sum of all possible sums: - We need to sum ( frac{1}{1 - frac{omega}{2013}} ) for all 2013th roots of unity ( omega ). - This translates to finding ( sum_{k=0}^{2012} frac{1}{1 - frac{omega_k}{2013}} ) where ( omega_k ) are the 2013th roots of unity.4. Using polynomial roots: - Consider the polynomial ( P(x) = x^{2013} - 1 ) with roots ( omega_k ). - The sum ( sum_{k=0}^{2012} frac{1}{2013 - omega_k} ) can be found using the derivative of ( P(x) ): [ sum_{k=0}^{2012} frac{1}{2013 - omega_k} = frac{P'(2013)}{P(2013)} = frac{2013 cdot 2013^{2012}}{2013^{2013} - 1} = frac{2013^{2013}}{2013^{2013} - 1} ]5. Final calculation: - Multiplying by 2013 to account for the factor in the denominator: [ sum_{k=0}^{2012} frac{2013}{2013 - omega_k} = 2013 cdot frac{2013^{2013}}{2013^{2013} - 1} = frac{2013^{2014}}{2013^{2013} - 1} ]Thus, the sum of all possible sums of the series is:[boxed{dfrac{2013^{2014}}{2013^{2013} - 1}}]

🔑:1. We are given that ( z_0 + z_1 + z_2 + cdots ) is an infinite geometric series with the first term ( z_0 = 1 ) and ( z_{2013} = frac{1}{2013^{2013}} ). The sum of an infinite geometric series is given by [ S = frac{z_0}{1 - r}, ] where ( r ) is the common ratio. Here, ( z_0 = 1 ), so [ S = frac{1}{1 - r}. ]2. Given ( z_{2013} = 1 cdot r^{2013} = frac{1}{2013^{2013}} ), we find the common ratio ( r ) by solving: [ r^{2013} = frac{1}{2013^{2013}}. ] This implies [ r = left(frac{1}{2013}right)^{frac{2013}{2013}} = frac{1}{2013}. ] However, since there are 2013 possible roots, ( r ) can be any of the 2013 roots of unity multiplied by ( frac{1}{2013} ).3. The possible values for ( r ) are thus: [ r = frac{1}{2013} cdot e^{ifrac{2kpi}{2013}}, ] where ( k = 0, 1, 2, ldots, 2012 ).4. To find the sum of all possible sums of this geometric series, we need to find the sum [ sum_{k=0}^{2012} frac{1}{1 - r_k}, ] where ( r_k = frac{1}{2013} cdot e^{ifrac{2kpi}{2013}} ).5. For each ( r_k ), the expression ( 1 - r_k ) becomes [ 1 - frac{1}{2013} e^{ifrac{2kpi}{2013}}, ] and hence [ frac{1}{1 - r_k} = frac{1}{1 - frac{1}{2013} e^{ifrac{2kpi}{2013}}}. ]6. The roots ( x_n ) of the polynomial ( left(1 - frac{1}{x}right)^{2013} = frac{1}{2013^{2013}} ) can be derived from considering the polynomial: [ left(frac{x - 1}{x}right)^{2013} = frac{1}{2013^{2013}}, ] which simplifies to: [ frac{(x - 1)^{2013}}{x^{2013}} = frac{1}{2013^{2013}}, ] which can be rewritten as: [ (2013x - (2013-1))^{2013} = 2013^{2013}. ]7. Considering the resulting polynomial difference, by Vieta's formulas and polynomial properties, we sum up the roots of the polynomial considering the coefficients.8. Each term corresponding to ( x_k ) as: [ -frac{2013}{frac{1}{2013^{2013}} - 1} = frac{2013 cdot 2013^{2013}}{2013^{2013} - 1} = frac{2013^{2014}}{2013^{2013} - 1}. ] Summarizing the polynomial sums leads us to:9. Conclusively, given all the detailed arithmetic and Vieta’s relationships, the sum of all possible sums of the series is: [ boxed{frac{2013^{2014}}{2013^{2013} - 1}}. ]

🧠:Okay, so I need to find a power of 2 with a natural exponent that ends in three identical digits. Hmm, let's see. Let me start by understanding the problem properly. We're talking about numbers like 2^n where n is a natural number (so positive integers), and the last three digits of this number are all the same. For example, something ending in 111, 222, ..., 999. But wait, 000 is also a possibility, right? Because 000 is three identical digits, but 2^n is never going to end with 000 unless n is such that 2^n is divisible by 1000. But 2^n is only divisible by 2's, so 2^n mod 1000 will never be 000 because 1000 is 2^3 * 5^3, and 5 isn't a factor of 2^n. Therefore, the last three digits can't be 000. So we can eliminate that possibility. So we need 2^n mod 1000 = ddd, where d is from 1 to 9.Alright, so the problem reduces to finding n such that 2^n ≡ ddd mod 1000, where d is a digit from 1 to 9. Now, how do I approach this? Maybe I can compute the last three digits of successive powers of 2 and check if they meet the condition. But since powers of 2 grow exponentially, maybe there's a pattern or cycle in the last three digits that I can exploit.First, let's recall that the last three digits of a number are equivalent to the number modulo 1000. So we need to compute 2^n mod 1000 and see if it's equal to 111, 222, ..., 999.To compute 2^n mod 1000 efficiently, we can use modular exponentiation. Also, Euler's theorem tells us that if two numbers are coprime, then a^φ(n) ≡ 1 mod n. Here, φ(1000) = φ(2^3 * 5^3) = 1000*(1 - 1/2)*(1 - 1/5) = 1000 * 1/2 * 4/5 = 1000 * 2/5 = 400. So φ(1000) = 400. Since 2 and 1000 are not coprime (they share factors 2), Euler's theorem doesn't apply directly. However, perhaps we can find a cycle length for the powers of 2 modulo 1000.Alternatively, we can note that 2^10 = 1024, so 2^10 mod 1000 = 24. Then 2^20 = (2^10)^2 = 24^2 = 576 mod 1000. 2^30 = 576 * 24 = 13,824 mod 1000 = 824. Continuing this way, but this might take a while. Alternatively, we can use the concept of the multiplicative order, but since 2 and 1000 are not coprime, multiplicative order isn't defined. However, perhaps the sequence 2^n mod 1000 will eventually enter a cycle. Let's see.Alternatively, we can break it down using the Chinese Remainder Theorem. Since 1000 = 8 * 125, and 8 and 125 are coprime, we can compute 2^n mod 8 and 2^n mod 125, then combine the results. For the last three digits, combining these two would give us the result.First, let's compute 2^n mod 8. Since 2^3 = 8, for n ≥ 3, 2^n mod 8 = 0. So if n ≥ 3, the last three digits of 2^n will be congruent to 0 mod 8. So the last three digits, if they are three identical digits, must be 000, 222, 444, 666, or 888. But as we saw earlier, 000 is impossible. So possible candidates are 222, 444, 666, 888. Therefore, the last three digits must be one of these. So the problem reduces to finding n such that 2^n ≡ 222, 444, 666, or 888 mod 1000.But wait, let's check if these numbers are possible. For example, 222 mod 8 is 222 - 27*8 = 222 - 216 = 6. But 2^n mod 8 is 0 for n ≥ 3. So 222 mod 8 is 6, which is not 0. Therefore, 222 cannot be the last three digits. Similarly, 444 mod 8 = 444 - 55*8 = 444 - 440 = 4. Not 0. 666 mod 8 = 666 - 83*8 = 666 - 664 = 2. 888 mod 8 = 0. So the only possible candidate among the three identical digits is 888, since 888 mod 8 is 0. So 888 is congruent to 0 mod 8, which matches 2^n mod 8 for n ≥ 3. The others (222, 444, 666) are not divisible by 8, so they cannot be the last three digits of a power of 2. Therefore, the only possible candidate is 888. So we need to find n such that 2^n ≡ 888 mod 1000.So now the problem reduces to solving 2^n ≡ 888 mod 1000. Let's focus on that.Using Chinese Remainder Theorem, since 1000 = 8 * 125, and 8 and 125 are coprime, we can solve 2^n ≡ 888 mod 8 and 2^n ≡ 888 mod 125. However, 2^n ≡ 888 mod 8. Since 888 mod 8 = 0, and as we know, for n ≥ 3, 2^n ≡ 0 mod 8, so that's fine. So the main equation is 2^n ≡ 888 mod 125.So we need to solve 2^n ≡ 888 mod 125. Let's compute 888 mod 125. 125 * 7 = 875, so 888 - 875 = 13. Therefore, 888 ≡ 13 mod 125. So we need to solve 2^n ≡ 13 mod 125.Now, solving 2^n ≡ 13 mod 125. This is a discrete logarithm problem. Let's see how we can approach this. The multiplicative order of 2 modulo 125 might help here. Since 125 is 5^3, and φ(125) = 100. So the multiplicative order of 2 modulo 125 divides φ(125) = 100. Let's check the order.Compute 2^100 ≡ 1 mod 125 by Euler's theorem, since 2 and 125 are coprime. The order must divide 100. Let's check the order.Check 2^20 mod 125. 2^10 = 1024; 1024 mod 125: 125*8 = 1000, so 1024 - 1000 = 24; 24 mod 125 = 24. Then 2^20 = (2^10)^2 = 24^2 = 576 mod 125. 576 - 4*125 = 576 - 500 = 76. So 2^20 ≡ 76 mod 125.2^25 = 2^20 * 2^5 = 76 * 32 mod 125. 76*32 = 2432. 2432 mod 125: 125*19 = 2375, 2432 - 2375 = 57. So 2^25 ≡ 57 mod 125.2^50 = (2^25)^2 = 57^2 = 3249 mod 125. 3249 - 25*125 = 3249 - 3125 = 124. So 2^50 ≡ 124 mod 125.2^100 = (2^50)^2 ≡ 124^2 mod 125. 124^2 = (125 - 1)^2 = 125^2 - 2*125 + 1 ≡ 0 - 0 + 1 ≡ 1 mod 125. So 2^100 ≡ 1 mod 125. Thus, the order of 2 modulo 125 is 100. Therefore, the multiplicative order is 100. So the solutions to 2^n ≡ 13 mod 125 will be of the form n ≡ k mod 100, where k is the discrete logarithm.But finding k such that 2^k ≡ 13 mod 125 is not straightforward. We might need to compute it via methods like baby-step giant-step or Pollard's rho algorithm, but since we're doing this manually, perhaps we can compute powers of 2 modulo 125 until we hit 13.Let me try to compute powers of 2 modulo 125 step by step.Starting from n=1:2^1 = 2 mod 1252^2 = 42^3 = 82^4 = 162^5 = 322^6 = 642^7 = 128 mod 125 = 32^8 = 62^9 = 122^10 = 242^11 = 482^12 = 962^13 = 192 mod 125 = 192 - 125 = 672^14 = 134 mod 125 = 92^15 = 182^16 = 362^17 = 722^18 = 144 mod 125 = 192^19 = 382^20 = 76 (as computed earlier)2^21 = 152 mod 125 = 272^22 = 542^23 = 1082^24 = 216 mod 125 = 216 - 125 = 912^25 = 182 mod 125 = 57 (as computed earlier)2^26 = 1142^27 = 228 mod 125 = 228 - 125 = 1032^28 = 206 mod 125 = 206 - 125 = 812^29 = 162 mod 125 = 162 - 125 = 372^30 = 742^31 = 148 mod 125 = 232^32 = 462^33 = 922^34 = 184 mod 125 = 184 - 125 = 592^35 = 1182^36 = 236 mod 125 = 236 - 125 = 1112^37 = 222 mod 125 = 222 - 125 = 972^38 = 194 mod 125 = 194 - 125 = 692^39 = 138 mod 125 = 13Wait, 2^39 mod 125 = 13? Let's check:Starting from 2^36 ≡ 111 mod 125.2^37 = 111*2 = 222 mod 125 = 222 - 125 = 972^38 = 97*2 = 194 mod 125 = 194 - 125 = 692^39 = 69*2 = 138 mod 125 = 138 - 125 = 13. Yes! So 2^39 ≡ 13 mod 125.Therefore, the solution to 2^n ≡ 13 mod 125 is n ≡ 39 mod 100 (since the order is 100). Therefore, the exponents n that satisfy 2^n ≡ 888 mod 1000 are those congruent to 39 mod 100. But we also need to satisfy 2^n ≡ 0 mod 8, which is already true for n ≥ 3. Since 39 ≥ 3, that's okay.Therefore, the minimal exponent is 39. Let's check if 2^39 ends with 888.But wait, let's verify 2^39 mod 1000. Since 2^10 = 1024, so 2^10 mod 1000 = 242^20 = (2^10)^2 = 24^2 = 576 mod 10002^30 = (2^10)^3 = 24^3 = 24*24=576; 576*24=13,824 mod 1000=8242^40 = (2^10)^4 = 24^4. But since we need 2^39, which is 2^40 / 2. Let's compute 2^40 mod 1000 first.2^10=242^20=5762^30=8242^40= (2^30)*(2^10) = 824*24. Let's compute 824*24:800*24 = 19,20024*24 = 576Total = 19,200 + 576 = 19,776 mod 1000 = 776Therefore, 2^40 ≡ 776 mod 1000. Then 2^39 = 2^40 / 2 ≡ 776 / 2 = 388 mod 1000. Wait, but we thought 2^39 ≡ 888 mod 1000. There's a contradiction here. What did I do wrong?Wait, this discrepancy arises because when we computed using the Chinese Remainder Theorem, we found that 2^39 ≡ 13 mod 125 and 0 mod 8, which should give 888 mod 1000. But when computing directly via powers, 2^39 mod 1000 = 388. That can't be. There must be an error in my calculations.Wait, let's check 2^39 again. Let's compute step by step more carefully.Starting from 2^1 = 22^2 = 42^3 = 82^4 = 162^5 = 322^6 = 642^7 = 1282^8 = 2562^9 = 5122^10 = 1024 → 24 mod 10002^11 = 482^12 = 962^13 = 1922^14 = 3842^15 = 7682^16 = 1,536 → 536 mod 10002^17 = 1,072 → 72 mod 10002^18 = 1442^19 = 2882^20 = 5762^21 = 1,152 → 152 mod 10002^22 = 3042^23 = 6082^24 = 1,216 → 216 mod 10002^25 = 4322^26 = 8642^27 = 1,728 → 728 mod 10002^28 = 1,456 → 456 mod 10002^29 = 9122^30 = 1,824 → 824 mod 10002^31 = 1,648 → 648 mod 10002^32 = 1,296 → 296 mod 10002^33 = 5922^34 = 1,184 → 184 mod 10002^35 = 3682^36 = 7362^37 = 1,472 → 472 mod 10002^38 = 944 mod 10002^39 = 1,888 → 888 mod 1000Ah! There we go. So my previous calculation was wrong. When I divided 776 by 2 to get 2^39, that was incorrect because 2^40 is 776 mod 1000, but 2^40 is 2^39 * 2, so 2^39 = (2^40)/2. But in modular arithmetic, dividing by 2 is multiplying by the inverse of 2 mod 1000. The inverse of 2 mod 1000 is 500, since 2*500=1000≡0 mod 1000. Wait, that's not right. Wait, inverse of 2 mod 1000 is a number x such that 2x ≡1 mod 1000. But 2 and 1000 are not coprime, so 2 doesn't have an inverse mod 1000. Therefore, my mistake was in trying to compute 2^39 as 2^40 / 2 mod 1000, which is invalid because division by 2 isn't defined here. Therefore, my earlier approach using the Chinese Remainder Theorem was correct, but my direct computation was wrong due to an invalid division step.Therefore, the correct way is to compute 2^39 mod 1000 directly step by step. Let's do that again carefully:Starting from 2^10=24 mod 1000.2^20 = (2^10)^2 = 24^2 = 576 mod 10002^30 = (2^10)^3 = 24*576 = 13,824 mod 1000 = 824Now, 2^39 = 2^30 * 2^9. 2^9 = 512 mod 1000. So 824 * 512. Let's compute 824*512:First, 800*512 = 409,60024*512 = 12,288Total = 409,600 + 12,288 = 421,888 mod 1000 = 888. Therefore, 2^39 ≡ 888 mod 1000. There we go! So 2^39 ends with 888. Therefore, the answer is n=39.But let me confirm this once again. Let's compute 2^39 step by step:2^1 = 22^2 = 42^3 = 82^4 = 162^5 = 322^6 = 642^7 = 1282^8 = 2562^9 = 5122^10 = 1024 → 242^11 = 24*2 = 482^12 = 48*2 = 962^13 = 96*2 = 1922^14 = 192*2 = 3842^15 = 384*2 = 7682^16 = 768*2 = 1,536 → 5362^17 = 536*2 = 1,072 → 722^18 = 72*2 = 1442^19 = 144*2 = 2882^20 = 288*2 = 5762^21 = 576*2 = 1,152 → 1522^22 = 152*2 = 3042^23 = 304*2 = 6082^24 = 608*2 = 1,216 → 2162^25 = 216*2 = 4322^26 = 432*2 = 8642^27 = 864*2 = 1,728 → 7282^28 = 728*2 = 1,456 → 4562^29 = 456*2 = 9122^30 = 912*2 = 1,824 → 8242^31 = 824*2 = 1,648 → 6482^32 = 648*2 = 1,296 → 2962^33 = 296*2 = 5922^34 = 592*2 = 1,184 → 1842^35 = 184*2 = 3682^36 = 368*2 = 7362^37 = 736*2 = 1,472 → 4722^38 = 472*2 = 9442^39 = 944*2 = 1,888 → 888 mod 1000. Yes! So 2^39 is indeed 888 mod 1000. Therefore, the exponent n=39 satisfies the condition.Is there a smaller exponent? Let's check n=39 is the first one. Let me check n= less than 39. For example, 2^9=512 ends with 12, not three digits. 2^10=24, 2^20=576, 2^30=824, 2^35=368, 2^36=736, 2^37=472, 2^38=944, 2^39=888. So before 39, there's no power of 2 ending with three identical digits. Hence, n=39 is the smallest such exponent.Therefore, the answer is 2^39, which ends with 888. So the exponent is 39.

✅:To find a power of 2 with a natural exponent that ends in three identical digits, we need (2^n equiv ddd mod 1000) where (d in {1, 2, ldots, 9}). Since (2^n) is divisible by 8 for (n geq 3), the last three digits must be a multiple of 8. The only three-digit numbers with identical digits divisible by 8 are (888). Step-by-Step Solution:1. Modulo 8 Analysis: For (n geq 3), (2^n equiv 0 mod 8), so the last three digits must be (888) (since (888 equiv 0 mod 8)).2. Modulo 125 Analysis: Solve (2^n equiv 888 mod 125). Simplifying, (888 equiv 13 mod 125), so we need (2^n equiv 13 mod 125).3. Finding (n): Compute powers of 2 modulo 125 until we find (2^{39} equiv 13 mod 125). By the Chinese Remainder Theorem, combining with (2^n equiv 0 mod 8), (n = 39) satisfies (2^{39} equiv 888 mod 1000). Verification:- Compute (2^{39} mod 1000) step-by-step: - (2^{10} equiv 24 mod 1000) - (2^{20} equiv 576 mod 1000) - (2^{30} equiv 824 mod 1000) - (2^{39} = 2^{30} cdot 2^9 equiv 824 cdot 512 equiv 888 mod 1000).Answer: The smallest natural exponent is (boxed{39}), and (2^{39}) ends in (888).

🔑:We need to find a power of 2 with a natural exponent that ends in three identical digits.1. Let's denote the number of thousands in 2^k (k in mathbb{N}) through x, and the digit that appears three times at the end of this number through y. Thus, we can write: [ 2^{k} = 1000x + overline{yyy} = 1000x + 111y ]2. From this equation, we infer that 2^k is greater than 1000, hence: [ k geq 10 ]3. Since both 1000x and 2^k are divisible by 8, it follows that: [ 111y mod 8 = 0 quad text{or} quad y mod 8 = 0 ] Therefore, y must be 8 because 111 itself isn't divisible by 8 but 8 times 111 is: [ 111 times y text{ is divisible by } 8 implies y = 8 ]4. So, our equation now is: [ 2^{k} = 1000x + 888 ]5. Simplifying further, we get: [ 2^{k-3} = 125x + frac{888}{8} = 125x + 111 ] Here, since x is a natural number, x must be odd and we have: [ 125x text{ must be congruent to } 125, 375, 625, text{ or } 875 mod 1000 ]6. Thus, sum 125x + 111 leaves us with the following possible endings of 2^{k-3}: [ 236, 486, 736, 986 ]7. Therefore, the task now is to find a natural k geq 10 such that 2^{k-3} ends with one of the four three-digit numbers mentioned.8. Given that the last digits of powers of 2 are cyclic every 20 numbers, we simplify the problem. To end with 6, the exponent of 2 must be divisible by 4: [ k - 3 = 4a implies k = 4a + 3 ] where a is a natural number not less than 2.9. Testing a few values from a = 2 onwards: - For a = 2: [ k - 3 = 4 times 2 = 8 implies k = 11 implies 2^{11} = 2048 quad text{(does not end with 236, 486, 736, or 986)} ] - For a = 3: [ k - 3 = 12 implies k = 15 implies 2^{15} = 32768 quad text{(does not end with 236, 486, 736, or 986)} ] - Continue this process until a = 9: [ k - 3 = 36 implies k = 39 implies 2^{36} = text{ends with 736} ]10. Therefore, 2^{39} satisfies the condition as 2^{39} ends in 888. Conclusion:[boxed{2^{39}}]

🧠:Okay, so I need to figure out if there's an integer k ≥ 2 that is coprime with all numbers in the sequence {a_n}, where a_n = 2^n + 3^n + 6^n + 1. And if such a k exists, find the smallest one. If not, explain why. Alright, let's start breaking this down.First, let me understand what the question is asking. We have this sequence a_n defined for positive integers n, and we need to find a k ≥ 2 such that k shares no common divisors with any term of the sequence. In other words, for every n ≥ 1, gcd(k, a_n) = 1. If such a k exists, find the smallest one; otherwise, show that no such k exists.So, the key here is to check if there's a number k that is coprime with every a_n. To do this, I need to analyze the prime factors of the terms a_n. If there's a prime number that divides infinitely many a_n, then k cannot include that prime. Conversely, if for every prime p ≥ 2, there exists some n where p divides a_n, then there can be no such k. So, the problem reduces to determining whether there exists a prime that does not divide any a_n. If such a prime exists, then k could be that prime, but we need the smallest k, which might be a product of such primes if there are multiple.Alternatively, if all primes divide at least one a_n, then no such k exists. So, maybe the answer is that no such k exists? But I need to check.First, let's compute the first few terms of the sequence to see if we can spot a pattern or identify possible primes.Compute a_1: 2 + 3 + 6 + 1 = 12. So a_1 = 12.a_2: 4 + 9 + 36 + 1 = 50. a_2 = 50.a_3: 8 + 27 + 216 + 1 = 252. a_3 = 252.a_4: 16 + 81 + 1296 + 1 = 1394. a_4 = 1394.a_5: 32 + 243 + 7776 + 1 = 8052. a_5 = 8052.a_6: 64 + 729 + 46656 + 1 = 47450. a_6 = 47450.Wait, so looking at these terms:a_1 = 12 = 2^2 * 3a_2 = 50 = 2 * 5^2a_3 = 252 = 2^2 * 3^2 * 7a_4 = 1394. Let's factorize 1394. 1394 divided by 2 is 697. Then 697: 697 divided by 17 is 41 (17*41=697). So 1394 = 2 * 17 * 41.a_5 = 8052. Let's divide by 2: 8052 / 2 = 4026, /2 again: 2013. 2013: 3*671. 671 is 11*61. So 8052 = 2^2 * 3 * 11 * 61.a_6 = 47450. Divide by 2: 23725. Divided by 5: 4745. Divided by 5 again: 949. 949: 13*73. So 47450 = 2 * 5^2 * 13 * 73.Looking at these factorizations, the primes that appear in the a_n terms are 2, 3, 5, 7, 11, 13, 17, 41, 61, 73. So, primes like 2, 3, 5, 7, 11, etc., are showing up here. Now, if we can find a prime that does not divide any a_n, then that prime would be a candidate for k, or a product of such primes. But if all primes divide some a_n, then there is no such k.But maybe there is a prime that never divides any a_n. Let's think.First, note that k must be coprime with all a_n. So, if 2 divides some a_n, then k cannot be even. Similarly, if 3 divides some a_n, then k cannot be a multiple of 3, and so on.Looking at a_1 = 12, which is divisible by 2 and 3. So, k cannot be divisible by 2 or 3. Then a_2 = 50, which is divisible by 2 and 5. So, k cannot be divisible by 2, 3, or 5. Then a_3 = 252, which is divisible by 2, 3, 7. So, k cannot be divisible by 2, 3, 5, 7. a_4 = 1394, which is 2*17*41. So, k must not be divisible by 17 or 41. But wait, we already know k can't be even, so 2 is already excluded. But 17 and 41: we need to check if 17 divides any a_n. If 17 divides a_4, but does 17 divide any other a_n? Similarly for 41. Hmm.Wait, but the problem states that k must be coprime with all a_n. So, even if 17 divides only a_4, then k cannot include 17. Similarly for 41. So, any prime that divides any a_n is excluded from k. Therefore, the primes that cannot be in k are all primes that divide any a_n.Therefore, if the set of primes dividing some a_n is all primes, then no such k exists. Otherwise, k can be formed by primes not in that set. So, perhaps primes like 19, 23, 29, etc., don't divide any a_n. If such primes exist, then k could be the smallest such prime, or a product. But we need to check.So, first, note that k must be at least 2. Let's check for primes starting from the smallest ones:Already, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, etc.We saw that 2,3,5,7,11,13,17,41,61,73 are already in the factorizations of the first six terms. So, let's check if 19 divides any a_n. Similarly for 23, 29, etc.To check if 19 divides any a_n, we can compute a_n modulo 19 for various n and see if it's 0. If there exists some n where a_n ≡ 0 mod 19, then 19 is excluded. Similarly for other primes.So, let's attempt this. Let's take a prime p and check if there exists n such that 2^n + 3^n + 6^n + 1 ≡ 0 mod p. If yes, then p cannot be part of k.Therefore, we need to check for primes not in the existing factorizations whether such a congruence holds for some n.Alternatively, we can note that 6 = 2*3, so 6^n = (2*3)^n = 2^n * 3^n. So, the term a_n = 2^n + 3^n + (2*3)^n + 1. Maybe there's a way to factor this expression or find a pattern.Alternatively, perhaps we can find a recursive relation or periodicity modulo p.But let's pick a prime, say 19. Let's check for n=1: a_1=12 mod 19=12. Not 0. n=2:50 mod19=50-2*19=50-38=12. Still 12. n=3:252 mod19. 19*13=247, 252-247=5. Not 0. n=4:1394 mod19. Let's compute 1394 divided by 19. 19*70=1330. 1394-1330=64. 64 mod19: 19*3=57, 64-57=7. So 7. Not 0. n=5:8052 mod19. Let's divide 8052 by19. 19*424=8056. 8052-8056=-4, so -4 mod19=15. Not 0. n=6:47450 mod19. 47450 /19=2500 (19*2500=47500). 47450-47500= -50. -50 mod19= (-50 + 57)=7. So 7. Not 0. Hmm. So up to n=6, 19 does not divide any a_n. Maybe check higher n? But this could take a while. Perhaps there's a smarter way.Alternatively, since we're working modulo p, maybe we can find n such that 2^n + 3^n + 6^n ≡ -1 mod p. Let's denote this as 2^n + 3^n + (2*3)^n ≡ -1 mod p. Maybe we can write this as 2^n + 3^n + (2^n * 3^n) ≡ -1 mod p. Let's factor 2^n * 3^n as (2*3)^n, but that's the same as 6^n. Hmm.Alternatively, set x = 2^n, y = 3^n. Then the equation becomes x + y + xy ≡ -1 mod p. Which can be rewritten as x + y + xy + 1 ≡ 0 mod p. Which factors as (x + 1)(y + 1) ≡ 0 mod p. Therefore, (2^n + 1)(3^n + 1) ≡ 0 mod p. Therefore, either 2^n ≡ -1 mod p or 3^n ≡ -1 mod p.Therefore, for a given prime p, there exists n such that (2^n +1)(3^n +1) ≡ 0 mod p if and only if either 2 has even order dividing 2n or 3 has even order dividing 2n, such that 2^n ≡ -1 mod p or 3^n ≡ -1 mod p. Wait, maybe more straightforward: For each prime p, either there exists an n where 2^n ≡ -1 mod p or there exists an n where 3^n ≡ -1 mod p. If either of these is true, then p divides some a_n. If for a prime p, neither 2^n ≡ -1 mod p nor 3^n ≡ -1 mod p for any n, then p does not divide any a_n, and thus p is a candidate for k.Therefore, primes p where neither 2 nor 3 is a quadratic residue modulo p? Wait, no. Wait, if 2^n ≡ -1 mod p, then squaring both sides gives 2^{2n} ≡ 1 mod p. Therefore, the order of 2 modulo p divides 2n but not n. So, the order of 2 modulo p is even, and n is half of that order. Similarly for 3.Therefore, for primes p where the order of 2 modulo p is even, there exists n such that 2^n ≡ -1 mod p. Similarly for 3. So, primes p where both orders of 2 and 3 modulo p are odd, then there does not exist n such that 2^n ≡ -1 or 3^n ≡ -1 mod p. Hence, such primes p would not divide any a_n.Therefore, the problem reduces to finding primes p where the multiplicative orders of 2 and 3 modulo p are both odd. If such primes exist, then k can be the smallest such prime. If not, then no such k exists.So, let's look for primes where both 2 and 3 have odd orders. Let's check small primes not already in the factors of the a_n computed earlier. The primes we saw so are 2,3,5,7,11,13,17,19,23,29,31,37,41,43,47,53,59,61,67,71,73. Wait, we need to check primes starting from 19, since 2,3,5,7,11,13,17, etc., are already in the factors.Take p=19. Let's compute the order of 2 modulo 19. The multiplicative order is the smallest k such that 2^k ≡1 mod19. Since 19 is prime, the multiplicative group has order 18. So possible orders divide 18. Let's check 2^18 ≡1 mod19. Let's compute 2^k mod19:2^1=2, 2^2=4, 2^3=8, 2^4=16, 2^5=32≡13, 2^6=26≡7, 2^7=14, 2^8=28≡9, 2^9=18, 2^10=36≡17, 2^11=34≡15, 2^12=30≡11, 2^13=22≡3, 2^14=6, 2^15=12, 2^16=24≡5, 2^17=10, 2^18=20≡1 mod19. So the order of 2 modulo19 is 18. Which is even. So, 2 has even order mod19. So, there exists n where 2^n≡-1 mod19. For example, n=9: 2^9=512≡18≡-1 mod19. Indeed, 2^9=512. 512 divided by19: 19*26=494, 512-494=18≡-1 mod19. So, n=9, 2^9≡-1 mod19. Therefore, a_9 would be divisible by19. So, 19 divides a_9. Hence, 19 cannot be part of k. So, k cannot be 19.Next prime:23. Let's check orders of 2 and3 modulo23.First, order of2 modulo23. The multiplicative group has order22. So possible orders divide22:1,2,11,22.Compute 2^11 mod23. 2^1=2, 2^2=4, 2^4=16, 2^8=256≡256-11*23=256-253=3, 2^11=2^8*2^2*2^1=3*4*2=24≡1 mod23. So 2^11≡1 mod23. Therefore, the order of2 modulo23 is11, which is odd.Now check order of3 modulo23. Let's compute 3^k mod23.3^1=3, 3^2=9, 3^3=27≡4, 3^4=12, 3^5=36≡13, 3^6=39≡16, 3^7=48≡2, 3^8=6, 3^9=18, 3^10=54≡8, 3^11=24≡1 mod23. So order of3 modulo23 is11, which is odd.Therefore, both orders are odd. So, in this case, there is no n such that 2^n≡-1 mod23 or3^n≡-1 mod23. Because if the order is odd, then the multiplicative order does not divide 2n. Therefore, there is no n where2^n≡-1 mod23 or3^n≡-1 mod23. Therefore, the congruence (2^n +1)(3^n +1)≡0 mod23 never holds. Therefore,23 does not divide any a_n. Therefore,23 is a candidate for k. Since23 is a prime, the smallest such k is23.But wait, we need to check whether this is indeed true, that23 does not divide any a_n. Let's verify for some small n.Compute a_1=12 mod23=12≠0.a_2=50 mod23=50-2*23=4≠0.a_3=252 mod23. 23*10=230, 252-230=22≡-1 mod23. So a_3≡-1 mod23. Not 0.a_4=1394 mod23. Let's compute 1394/23. 23*60=1380, 1394-1380=14≡14 mod23≠0.a_5=8052 mod23. 23*350=8050, 8052-8050=2≡2 mod23≠0.a_6=47450 mod23. 23*2062=47426, 47450-47426=24≡1 mod23≠0.a_7= Let's compute a_7=2^7 +3^7 +6^7 +1. 128 + 2187 + 279936 +1. 128+2187=2315; 2315+279936=282,251; +1=282,252. Compute 282252 mod23.But maybe instead, compute each term mod23:2^7 mod23: 2^7=128≡128-5*23=128-115=13.3^7 mod23: From earlier,3^7≡2 mod23.6^7 mod23:6=2*3, so6^7=2^7*3^7≡13*2=26≡3 mod23.Therefore, a_7 mod23=13 +2 +3 +1=19≡19 mod23≠0.Similarly, a_8: compute 2^8 mod23= (2^7)*2=13*2=26≡3, 3^8= (3^7)*3=2*3=6, 6^8=(6^7)*6=3*6=18. So a_8=3 +6 +18 +1=28≡5 mod23≠0.a_9: 2^9 mod23=3*2=6, 3^9= (3^8)*3=6*3=18, 6^9= (6^8)*6=18*6=108≡108-4*23=108-92=16. So a_9=6 +18 +16 +1=41≡41-1*23=18≡18 mod23≠0.a_10: 2^10 mod23=6*2=12, 3^10= (3^9)*3=18*3=54≡54-2*23=8, 6^10= (6^9)*6=16*6=96≡96-4*23=96-92=4. So a_10=12 +8 +4 +1=25≡2 mod23≠0.Hmm, so up to n=10,23 does not divide a_n. Since we have shown earlier that for prime23, there is no n such that a_n≡0 mod23, because neither2^n ≡-1 mod23 nor3^n≡-1 mod23 is possible, given their orders are11, which is odd. Hence, their powers can't reach-1.Therefore,23 does not divide any a_n, so k=23 is coprime to all a_n.Now, we need to check if there is a smaller k. Since23 is prime, and all smaller primes (2,3,5,7,11,13,17,19) have been shown to divide some a_n, the next possible prime is23. Therefore, the smallest k is23.But wait, let's check if maybe a composite number smaller than23 could work. For example, k=22. But22=2*11. But a_1=12, which shares a factor of2 with22, so gcd(22,12)=2≠1. Similarly, k=21=3*7: a_1=12 shares3, so gcd(21,12)=3≠1. k=19: already in the list, but we saw19 divides a_9. k=17: divides a_4. Similarly, all composites below23 will have prime factors that are already in the forbidden set. Hence, the smallest possible k is23.Therefore, the answer is23.

✅:To determine if there exists an integer ( k geq 2 ) coprime with all terms of the sequence ( {a_n} ), where ( a_n = 2^n + 3^n + 6^n + 1 ), we analyze the prime factors of ( a_n ). If every prime ( p geq 2 ) divides at least one ( a_n ), no such ( k ) exists. Otherwise, the smallest ( k ) is the product of primes not dividing any ( a_n ).1. Factorizing Initial Terms: - ( a_1 = 12 = 2^2 cdot 3 ) - ( a_2 = 50 = 2 cdot 5^2 ) - ( a_3 = 252 = 2^2 cdot 3^2 cdot 7 ) - ( a_4 = 1394 = 2 cdot 17 cdot 41 ) - ( a_5 = 8052 = 2^2 cdot 3 cdot 11 cdot 61 ) - ( a_6 = 47450 = 2 cdot 5^2 cdot 13 cdot 73 ) These factorizations reveal primes ( 2, 3, 5, 7, 11, 13, 17, 41, 61, 73 ), which cannot be part of ( k ).2. Checking Other Primes: For a prime ( p ), ( a_n equiv 0 mod p ) if ( (2^n + 1)(3^n + 1) equiv 0 mod p ). This occurs if either ( 2^n equiv -1 mod p ) or ( 3^n equiv -1 mod p ). 3. Analyzing Orders: - If the multiplicative orders of ( 2 ) and ( 3 ) modulo ( p ) are both odd, neither ( 2^n equiv -1 mod p ) nor ( 3^n equiv -1 mod p ) is possible, implying ( p ) does not divide any ( a_n ).4. Testing ( p = 23 ): - The multiplicative order of ( 2 mod 23 ) is ( 11 ) (odd), and the order of ( 3 mod 23 ) is also ( 11 ) (odd). Hence, ( 2^n notequiv -1 mod 23 ) and ( 3^n notequiv -1 mod 23 ) for all ( n ). - Direct computation confirms ( a_n mod 23 neq 0 ) for ( n leq 10 ), validating that ( 23 ) does not divide any ( a_n ).5. Smaller Composite ( k ): All composites ( < 23 ) include primes ( 2, 3, 5, 7, 11, 13, 17, 19 ), which divide some ( a_n ), ruling them out.Thus, the smallest integer ( k geq 2 ) coprime with all ( a_n ) is:[boxed{23}]

🔑:1. First, we are given the sequence {a_n} defined by:[ a_n = 2^n + 3^n + 6^n + 1 quad text{for} quad n in mathbb{Z}_+ ]2. We need to find the smallest integer k geq 2 such that k and a_n are coprime for all n in mathbb{Z}_+, or show that there is no such k.3. Since k must be coprime with every term in the sequence a_n, it follows that k must not divide a_n for any n. We first seek the smallest such k among the prime numbers.4. Applying Fermat's Little Theorem, if a prime p divides a_n then:[ p mid (2^n + 3^n + 6^n + 1) ] Given this, we know: [ 2^{p-1} equiv 1 pmod{p} ] [ 3^{p-1} equiv 1 pmod{p} ] [ 6^{p-1} equiv 1 pmod{p} ]5. This yields:[ 6^{p-1-r}(2^r + 3^r + 6^r + 1) equiv 0 pmod{p} ] Which simplifies, by Fermat's Little Theorem, to:[ 3^{p-1-r} + 2^{p-1-r} + 1 + 6^{p-1-r} equiv 0 pmod{p} ] Therefore we need to check for divisibility for p considering r = 0, 1, ..., p-2.6. Checking small primes: - For p = 2:[ 2^1 + 3^1 + 6^1 + 1 = 2 + 3 + 6 + 1 = 12 ] -> 2 mid 12. - For p = 3:[ 2^1 + 3^1 + 6^1 + 1 = 2 + 3 + 6 + 1 = 12 ] -> 3 mid 12. - For p = 5:[ 2^2 + 3^2 + 6^2 + 1 = 4 + 9 + 36 + 1 = 50 ] -> 5 mid 50. - For p = 7:[ 2^3 + 3^3 + 6^3 + 1 = 8 + 27 + 216 + 1 = 252 ] -> 7 mid 252. - For p = 13, p = 17, and p = 19, similar divisibility checks show that p divides a_n.7. Finally, we need to check the next prime, p = 23: We need to verify:[ 23 nmid (2^r + 3^r + 6^r + 1) text{ for } r = 0, 1, ..., 11 ] Calculating each individually: - For r = 0:[ 2^0 + 3^0 + 6^0 + 1 = 1 + 1 + 1 + 1 = 4 ] -> 23 nmid 4. - For r = 1:[ 2^1 + 3^1 + 6^1 + 1 = 2 + 3 + 6 + 1 = 12 ] -> 23 nmid 12. - Continue this process up to r = 11...8. Calculations confirm:[ 23 nmid (2^r + 3^r + 6^r + 1) ] for all r as required.Conclusion: The smallest integer k geq 2 such that k and a_n are coprime for all n is:[ boxed{23} ]

🧠:Okay, so I need to figure out the radius of the middle marble when there are five marbles in a conical funnel. Each marble is touching the adjacent ones and also the funnel wall. The smallest has a radius of 8, and the largest is 18. Hmm, this seems like a problem involving similar triangles or maybe some geometric progression. Let me think step by step.First, visualize the setup. A conical funnel, so it's like a cone shape. The marbles are placed inside, each touching the funnel walls and their neighbors. Since they are all in contact with the funnel wall, their positions must be arranged such that their centers lie along the central axis of the funnel. Also, each marble is touching the adjacent ones, so the distance between the centers of two adjacent marbles should equal the sum of their radii.But how does the funnel's cone angle affect this? Maybe the key here is that the marbles are arranged in a way that their points of contact with the funnel form similar triangles. If I consider the cross-section of the cone, it's an isosceles triangle. The marbles will be inscribed circles within this triangle, each touching the sides and the previous circle.Wait, this might relate to the concept of similar triangles in a cone with inscribed spheres. Each sphere touches the cone's sides and the adjacent sphere. In such cases, the radii of the spheres might follow a geometric progression. That is, each subsequent sphere's radius is a constant multiple of the previous one.Let me recall if there's a formula for the radii of spheres stacked in a cone. I think there is a progression factor based on the cone's angle. However, since the cone's angle isn't given here, maybe we can derive the ratio using the given radii of the first and last marbles.Given that there are five marbles, with radii r1=8, r2, r3, r4, r5=18. If they form a geometric sequence, then r5 = r1 * k^(4), where k is the common ratio. Let's check that. If r1=8, then r2=8k, r3=8k^2, r4=8k^3, r5=8k^4=18. Solving for k, we get k^4=18/8=9/4, so k=(9/4)^(1/4). Then the middle marble r3 would be 8*(9/4)^(2/4)=8*(3/2)=12. Wait, is that possible? Let me verify.If k^4=9/4, then k= (9/4)^(1/4) = (3^2/2^2)^(1/4)= (3/2)^(1/2)=sqrt(3/2). So then, r3=8*(sqrt(3/2))^2=8*(3/2)=12. Yes, that works. So the radii would be 8, 8*sqrt(3/2), 12, 12*sqrt(3/2), 18. But wait, 8*(sqrt(3/2))^4=8*(9/4)=18, which matches. So the middle radius is 12. Is that the answer? But wait, does this hold for the conical funnel?But here's the problem: the assumption is that the radii form a geometric progression. But is that valid in a conical funnel? Because in a cone, the relationship might be more complex due to the three-dimensional geometry. Maybe the progression factor isn't just a simple ratio but involves some trigonometric relationship based on the cone's angle. However, since the cone's angle isn't given, perhaps the problem assumes that the radii do form a geometric progression. Alternatively, the ratio can be derived from similar triangles considering the slope of the cone.Let me think again. When a sphere is inside a cone, the radius of the sphere relates to the cone's angle. If we have multiple spheres, each subsequent sphere's radius can be determined based on the cone's geometry. However, without knowing the cone's angle, maybe the problem is set up such that the progression is geometric, given that all marbles touch the funnel walls and each other. Since the cone's angle is constant, the ratio between consecutive radii should be consistent.Alternatively, consider the similar triangles formed by the cone walls and the centers of the marbles. Each time we go from one marble to the next, the distance along the axis increases, and the radius of the marble increases proportionally.Let me model the cone. Let the cone have a half-angle θ. The radius of the cone at a height h from the apex is r = h * tanθ. Now, each marble touches the cone walls, so the radius of the marble must relate to the radius of the cone at the position where the marble is located.Also, the centers of the marbles lie along the central axis of the cone. Let's denote the position of the first marble (radius r1=8) at height h1 from the apex. Then, the radius of the cone at h1 is R1 = h1 tanθ. But the marble of radius 8 is touching the cone walls, so the distance from the center of the marble to the cone wall must be equal to the radius of the marble. However, in a cone, the relationship between the radius of the sphere and its position is a bit more involved.Wait, actually, in three dimensions, the sphere inside the cone will touch the cone along a circle. The radius of the sphere, the half-angle of the cone, and the distance from the apex are related. The formula for the radius r of a sphere inscribed in a cone with half-angle θ, at a distance d from the apex along the axis is r = d * sinθ / (1 + sinθ). Hmm, is that correct?Let me derive that. Imagine the sphere of radius r inside the cone. The center of the sphere is at distance d from the apex. The sphere touches the cone walls. The cone's half-angle is θ. The radius of the cone at distance d is R = d tanθ. The sphere must fit such that the distance from the center of the sphere to the cone wall is equal to the radius of the sphere.Using trigonometry, the distance from the center to the wall is the perpendicular distance. If the cone's half-angle is θ, then the perpendicular distance from the center to the cone wall is d * sinθ - r. Wait, perhaps another approach.Consider a cross-sectional view of the cone and sphere. The cross-section is an isosceles triangle with apex angle 2θ. The sphere appears as a circle with radius r, tangent to both sides of the triangle and the base (if it's the first sphere). The center of the sphere is at a distance h from the apex. The radius of the sphere is related to h and θ.In the cross-section, the distance from the center of the circle to the apex is h. The sides of the triangle make an angle θ with the axis. The center is at (0, h), and the circle is tangent to the sides of the triangle. The equation for the sides of the triangle is y = x tanθ and y = -x tanθ. The distance from the center (0, h) to the side y = x tanθ must be equal to the radius r.The distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a^2 + b^2). The line y = x tanθ can be rewritten as x tanθ - y = 0. So, the distance from (0, h) to this line is |0*tanθ - h| / sqrt(tan²θ + 1) = | -h | / sqrt(tan²θ + 1) = h / secθ = h cosθ. This distance must equal the radius r. Therefore, h cosθ = r. So h = r / cosθ.But in the cross-section, the sphere also has a radius r. Wait, but in the cross-sectional view, the sphere would be a circle of radius r, but in reality, it's a sphere in 3D. However, in cross-section, the radius perpendicular to the plane of the cross-section is the same as the sphere's radius. Hmm, perhaps I need to consider that the sphere touches the cone walls, which in cross-section are the sides of the triangle.But the distance from the center to the side is h cosθ = r. So h = r / cosθ. Therefore, the position of the center is at h = r / cosθ along the axis. However, in the 3D cone, the radius of the cone at height h is R = h tanθ. Substituting h = r / cosθ, we get R = (r / cosθ) tanθ = r (sinθ / cos²θ). But the sphere of radius r must fit into the cone such that the sphere's equator is at radius R - r? Wait, no. Wait, the radius of the cone at the center of the sphere is R = h tanθ = (r / cosθ) tanθ = r sinθ / cos²θ. The sphere of radius r must fit within this radius. But how exactly?Wait, perhaps in cross-section, the sphere of radius r is tangent to the cone walls. The center is at distance h = r / cosθ. The radius of the cone at that height is R = h tanθ = (r / cosθ) * tanθ = r sinθ / cos²θ. The sphere in 3D has a radius r, and the radius of the cone at that point is R. The relationship between R and r is R = r sinθ / cos²θ. Therefore, r = R cos²θ / sinθ.But maybe I'm overcomplicating. The key point is that each subsequent sphere must fit into the remaining space of the cone. Since the problem states that all five marbles touch the funnel walls and each adjacent marble, the ratio between their radii might be consistent.But without knowing the cone's angle θ, perhaps we can relate the radii through a common ratio. Let's suppose that the radii follow a geometric progression r1, r2, r3, r4, r5, with r1=8 and r5=18. Then, as before, r5 = r1 * k^4 => 18 = 8 * k^4 => k = (18/8)^(1/4) = (9/4)^(1/4) = (3/2)^(1/2) = sqrt(3/2). Then, the radii would be 8, 8*sqrt(3/2), 8*(3/2), 8*(3/2)*sqrt(3/2), 18. Simplifying, that's 8, 8*sqrt(1.5), 12, 12*sqrt(1.5), 18. Therefore, the middle radius r3 is 12. But the problem is in a conical funnel, so does this geometric progression hold? If we assume that each subsequent sphere's radius is a constant multiple of the previous, which might be the case if the cone's geometry enforces such a ratio. But how can we be sure?Alternatively, consider that in a cone, the ratio between the radii of consecutive spheres is a constant. This is a known result in some packing problems. For example, in a conical arrangement, if each sphere touches the cone and the adjacent spheres, their radii form a geometric progression. The common ratio depends on the cone's angle. However, since the cone's angle isn't given here, the ratio can be derived from the given radii of the first and last spheres. Since we have five spheres, the ratio k can be found by solving r5 = r1 * k^4. As before, k = (18/8)^(1/4) = (9/4)^(1/4) = (3^2/2^2)^(1/4) = 3^(1/2)/2^(1/2) = sqrt(3/2). Therefore, the radii are 8, 8*sqrt(3/2), 12, 12*sqrt(3/2), 18. So the middle one is 12.But let me verify this with another approach to be sure. Suppose we model the positions of the marbles along the central axis. Let’s denote the centers of the marbles as points along the axis at distances d1, d2, d3, d4, d5 from the apex. Each marble touches the funnel wall, so for each marble i, the radius ri relates to the distance di. From the earlier derivation, ri = di * sinθ / (1 + sinθ). Wait, this comes from some standard formula for a sphere in a cone.Let me check the formula again. The formula for the radius of a sphere inscribed in a cone is r = (h * sinθ) / (1 + sinθ), where h is the height from the apex to the sphere's center. So if we have multiple spheres, each subsequent sphere will have its center at a height h_i, and radius r_i = (h_i * sinθ) / (1 + sinθ). Also, the distance between the centers of two consecutive spheres should be equal to the sum of their radii, because they are touching each other. So, h_{i+1} - h_i = r_i + r_{i+1}.Let me define h_i in terms of r_i. From the formula, h_i = r_i * (1 + sinθ) / sinθ. Then, the distance between h_{i+1} and h_i is:h_{i+1} - h_i = [r_{i+1} * (1 + sinθ)/sinθ] - [r_i * (1 + sinθ)/sinθ] = (r_{i+1} - r_i) * (1 + sinθ)/sinθBut this distance must equal the sum of the radii: r_i + r_{i+1}. Therefore,(r_{i+1} - r_i) * (1 + sinθ)/sinθ = r_i + r_{i+1}Let me rearrange this equation:(r_{i+1} - r_i) * (1 + sinθ) = (r_i + r_{i+1}) * sinθExpand both sides:r_{i+1}(1 + sinθ) - r_i(1 + sinθ) = r_i sinθ + r_{i+1} sinθBring all terms to one side:r_{i+1}(1 + sinθ) - r_{i+1} sinθ - r_i(1 + sinθ) - r_i sinθ = 0Factor terms:r_{i+1}[1 + sinθ - sinθ] - r_i[1 + sinθ + sinθ] = 0Simplify:r_{i+1}(1) - r_i(1 + 2 sinθ) = 0Thus,r_{i+1} = r_i (1 + 2 sinθ)So, the ratio between consecutive radii is (1 + 2 sinθ). Therefore, the radii form a geometric progression with common ratio k = (1 + 2 sinθ). Since this ratio is constant for all consecutive pairs, the radii are indeed a geometric sequence.Given that there are five marbles, from r1=8 to r5=18, we can find k such that r5 = r1 * k^4.Therefore, k^4 = 18/8 = 9/4Thus, k = (9/4)^(1/4) = (3^2/2^2)^(1/4) = 3^(1/2)/2^(1/2) = sqrt(3/2)Therefore, the radii are:r1 = 8r2 = 8 * sqrt(3/2)r3 = 8 * (sqrt(3/2))^2 = 8 * (3/2) = 12r4 = 12 * sqrt(3/2)r5 = 12 * (sqrt(3/2))^2 = 12 * (3/2) = 18Which matches the given radii. Therefore, the middle marble, which is the third one, has radius 12. So the answer is 12.But just to ensure, let me check if the ratio k = sqrt(3/2) is consistent with the formula r_{i+1} = r_i (1 + 2 sinθ). If k = sqrt(3/2) = 1 + 2 sinθ, then sinθ = (sqrt(3/2) - 1)/2 ≈ (1.2247 - 1)/2 ≈ 0.11235. Therefore, θ ≈ arcsin(0.11235) ≈ 6.45 degrees. Is this a reasonable angle for the cone? It seems quite narrow, but since the marbles are in contact with the walls, perhaps that's okay. Alternatively, maybe there's a miscalculation here.Wait, no. From the previous derivation, we had r_{i+1} = r_i (1 + 2 sinθ). So if the common ratio is k, then k = 1 + 2 sinθ. But we found k = sqrt(3/2) ≈ 1.2247. Therefore, sinθ = (k - 1)/2 ≈ (1.2247 - 1)/2 ≈ 0.11235. So θ ≈ 6.45 degrees. That's a very narrow cone. However, the problem doesn't specify the cone's angle, so maybe it's acceptable. The key is that regardless of the cone's angle, the radii must form a geometric progression with ratio k derived from the first and last radii.Therefore, given that the problem doesn't provide the cone's angle, and the fact that the marbles are in contact with each other and the walls, the geometric progression is the valid approach here. Hence, the middle radius is 12.

✅:The radii of the five marbles form a geometric progression due to the consistent conical geometry. Given the smallest radius ( r_1 = 8 ) and largest ( r_5 = 18 ), the common ratio ( k ) satisfies ( 18 = 8k^4 ). Solving:[k^4 = frac{18}{8} = frac{9}{4} implies k = left(frac{9}{4}right)^{1/4} = sqrt{frac{3}{2}}]The radii sequence is:[8, 8sqrt{frac{3}{2}}, 12, 12sqrt{frac{3}{2}}, 18]The middle (third) marble's radius is ( 12 ).Answer: The radius of the middle marble is (boxed{12}).

🔑:To determine the radius of the middle marble when given the radii of the smallest and largest marbles, we can utilize the concept of the geometric mean. The geometric mean of two numbers (a) and (b) is given by:[text{Geometric Mean} = sqrt{a cdot b}]In this problem, we are given:- The radius of the smallest marble, (r_1 = 8),- The radius of the largest marble, (r_5 = 18).Since the marbles follow a geometric sequence in radius, the radius of the middle marble, (r_3) (the third marble in a sequence of five), should be the geometric mean of the radii of the smallest and the largest marbles. Therefore, we need to find:[r_3 = sqrt{r_1 cdot r_5}]Substituting the values, we get:[r_3 = sqrt{8 cdot 18}]First, calculate the product inside the square root:[8 cdot 18 = 144]Now, take the square root of 144:[r_3 = sqrt{144} = 12]Hence, the radius of the middle marble is:[boxed{12}]

🧠:Okay, so I need to find the ratio of the surface area and volume of a sphere to the total surface area and volume of a circumscribed cone that has an equilateral axial section. Hmm, let me start by understanding the problem step by step.First, let's parse the question. A cone circumscribes a sphere, which means the sphere is tangent to the cone's sides and base. The axial section of the cone is an equilateral triangle. An axial section is the cross-section through the axis of the cone, right? So if that cross-section is an equilateral triangle, then all sides of that triangle are equal. That should help in relating the dimensions of the cone to the sphere.Let me visualize this. If the axial section is equilateral, then when we slice the cone along its axis, we get an equilateral triangle. The sphere inscribed in the cone will touch the base of the cone and the sides. The sphere's center should be along the axis of the cone. Since the cone is circumscribed around the sphere, the sphere is perfectly snug inside the cone. So the radius of the sphere should relate to the dimensions of the cone. Let me denote the radius of the sphere as r. Then, the cone must have a height h and a base radius R. The goal is to express h and R in terms of r, then compute the surface areas and volumes of the sphere and cone, and finally find their ratios.First, let's figure out the relationship between the sphere's radius and the cone's dimensions. Since the axial section is equilateral, the triangle has all sides equal. Let me consider the axial section: it's an equilateral triangle with side length equal to the slant height of the cone? Wait, no. Wait, in the axial section, the triangle's base is the diameter of the cone's base, which is 2R, and the two other sides are the slant heights of the cone. But if the axial section is equilateral, then all sides of the triangle are equal. So 2R = slant height. But the slant height (l) of a cone is given by l = sqrt(R² + h²). So if 2R = sqrt(R² + h²), then we can solve for h in terms of R.Let me write that equation down: 2R = sqrt(R² + h²). Squaring both sides: 4R² = R² + h² => h² = 3R² => h = R*sqrt(3). So the height of the cone is R times sqrt(3), and the slant height is 2R. That's because the axial section is an equilateral triangle, so each side is 2R.Wait, but in reality, the height of the cone is h, the base radius is R, and the slant height is sqrt(R² + h²). But here, the axial section is an equilateral triangle, so the sides of the triangle (the two equal sides) must be equal to the base. Therefore, the triangle has all sides equal, so the base is equal to the other two sides. Therefore, the base of the triangle (which is 2R) must equal the slant height (sqrt(R² + h²)). Therefore, as before, 2R = sqrt(R² + h²), leading to h = R*sqrt(3). So that gives us the relationship between h and R for the cone.Now, we need to relate this to the sphere inscribed in the cone. The sphere touches the base of the cone and its lateral surface. The center of the sphere is along the axis of the cone, at a distance equal to the sphere's radius from the base. So the center is at height r from the base. Then, the sphere must also be tangent to the lateral sides of the cone. The distance from the center of the sphere to the lateral surface of the cone must be equal to r.To find the relationship between the sphere's radius r and the cone's dimensions R and h, we can use the formula for the distance from a point to a plane. The lateral surface of the cone can be represented as a plane in the axial cross-section. Let's model the axial cross-section, which is an equilateral triangle with side length 2R. The cone's apex is at the top of this triangle, and the base is at the bottom.Let me place this coordinate system with the origin at the center of the sphere. Wait, no. Let me set up coordinates with the base of the cone at the origin (0,0), and the apex at (0, h). Then, the center of the sphere is at (0, r), since the sphere is tangent to the base, so the distance from the center to the base is r. The sphere has radius r, so its equation is x² + (y - r)^2 = r². The lateral edges of the cone in this cross-section are lines from the apex (0, h) to the base points (R, 0) and (-R, 0). So the right lateral edge is the line from (0, h) to (R, 0).The equation of the right lateral edge can be found using the two points (0, h) and (R, 0). The slope is (0 - h)/(R - 0) = -h/R. So the equation is y = (-h/R)x + h.The distance from the center of the sphere (0, r) to this line must be equal to the radius of the sphere, which is r. The formula for the distance from a point (x0, y0) to the line ax + by + c = 0 is |ax0 + by0 + c| / sqrt(a² + b²). Let's write the equation of the lateral edge in standard form. Starting from y = (-h/R)x + h, we can rearrange to (h/R)x + y - h = 0. So a = h/R, b = 1, c = -h.Plugging into the distance formula: |(h/R)(0) + 1*(r) - h| / sqrt((h/R)^2 + 1^2) = |r - h| / sqrt( (h²/R²) + 1 ). This distance must equal r.So:|r - h| / sqrt( (h²/R²) + 1 ) = rSince h > r (the center is at r, and the apex is at h, which is higher), so h - r is positive. Therefore, we can drop the absolute value:(h - r) / sqrt( (h²/R²) + 1 ) = rLet me square both sides to eliminate the square root:(h - r)^2 / ( (h²/R²) + 1 ) = r²Multiply both sides by the denominator:(h - r)^2 = r² * ( (h²/R²) + 1 )Expand the left side:h² - 2hr + r² = r²*(h²/R² + 1)Divide both sides by r² (assuming r ≠ 0, which it isn't):(h² - 2hr + r²)/r² = h²/R² + 1Simplify left side:(h²/r² - 2h/r + 1) = h²/R² + 1Subtract 1 from both sides:h²/r² - 2h/r = h²/R²Bring all terms to left side:h²/r² - 2h/r - h²/R² = 0Factor h² terms:h²(1/r² - 1/R²) - 2h/r = 0Hmm, not sure if factoring helps here. Maybe express in terms of h and R. Recall from earlier that in the cone with equilateral axial section, h = R*sqrt(3). So we can substitute h = R*sqrt(3) into the equation. Let's do that.Given h = R*sqrt(3), substitute into the equation:h²/r² - 2h/r - h²/R² = 0Replace h with R*sqrt(3):(R²*3)/r² - 2*(R*sqrt(3))/r - (R²*3)/R² = 0Simplify each term:3R²/r² - 2R*sqrt(3)/r - 3 = 0Let me divide all terms by R² to make it in terms of (1/r²) and (1/r):3/r² - 2*sqrt(3)/(R r) - 3/R² = 0Wait, but this seems messy. Maybe instead, since h = R*sqrt(3), we can express R in terms of h: R = h / sqrt(3). Then substitute R into the equation. Let's try that.Let me start again. We have h = R*sqrt(3) from the axial section being equilateral. So R = h / sqrt(3). Then, substituting R into the previous equation.Original equation after substituting h = R*sqrt(3):3R²/r² - 2R*sqrt(3)/r - 3 = 0Express R in terms of h: R = h / sqrt(3). Wait, but h is already R*sqrt(3). Maybe this substitution is not helpful. Alternatively, perhaps we can express everything in terms of R.Wait, if h = R*sqrt(3), then let's plug that into the equation:3R²/r² - 2R*sqrt(3)/r - 3 = 0Let me divide all terms by 3:R²/r² - (2R*sqrt(3))/(3r) - 1 = 0Let me let x = R/r. Then, the equation becomes:x² - (2*sqrt(3)/3)x - 1 = 0This is a quadratic equation in x. Let's solve for x:x = [ (2*sqrt(3)/3) ± sqrt( (2*sqrt(3)/3)^2 + 4 ) ] / 2Calculate discriminant:( (2*sqrt(3)/3)^2 + 4 ) = (4*3)/9 + 4 = (12/9) + 4 = (4/3) + 4 = (4/3 + 12/3) = 16/3So sqrt(16/3) = 4/sqrt(3) = (4*sqrt(3))/3Thus,x = [ (2*sqrt(3)/3) ± (4*sqrt(3)/3) ] / 2We have two solutions:x = [ (2*sqrt(3)/3 + 4*sqrt(3)/3 ) ] / 2 = (6*sqrt(3)/3)/2 = (2*sqrt(3))/2 = sqrt(3)Orx = [ (2*sqrt(3)/3 - 4*sqrt(3)/3 ) ] / 2 = (-2*sqrt(3)/3)/2 = (-sqrt(3)/3 )Since x = R/r must be positive, we discard the negative solution. So x = sqrt(3). Therefore, R/r = sqrt(3) => R = r*sqrt(3)But we also know that h = R*sqrt(3) = r*sqrt(3)*sqrt(3) = r*3Therefore, the height of the cone is h = 3r and the base radius is R = r*sqrt(3).So now we have the cone's dimensions in terms of the sphere's radius. Great!Now, let's compute the surface area and volume of the sphere and the cone.First, the sphere:Surface area of sphere, A_sphere = 4πr²Volume of sphere, V_sphere = (4/3)πr³Now, the cone. The total surface area of the cone includes the base and the lateral surface area.Total surface area of cone, A_cone = πR² + πR*l, where l is the slant height. We know from earlier that slant height l = 2R (since the axial section is equilateral, so each side is 2R). Wait, hold on. Wait, in the axial section, the triangle is equilateral with sides of length 2R. But the slant height is equal to the side length of the triangle, which is 2R. Wait, but earlier we found that slant height l = 2R, but h = R*sqrt(3). However, we just found that h = 3r and R = r*sqrt(3). Let's check slant height in terms of R.Wait, if h = R*sqrt(3), then slant height l = sqrt(R² + h²) = sqrt(R² + 3R²) = sqrt(4R²) = 2R. So yes, slant height is 2R, which is equal to the side length of the equilateral triangle. So that's consistent.Therefore, the slant height l = 2R.But R = r*sqrt(3), so l = 2r*sqrt(3)Therefore, total surface area of cone:A_cone = πR² + πR*l = π*(r*sqrt(3))² + π*(r*sqrt(3))*(2r*sqrt(3))Compute each term:First term: π*(3r²) = 3πr²Second term: π*(r*sqrt(3))*(2r*sqrt(3)) = π*2r²*3 = 6πr²So total A_cone = 3πr² + 6πr² = 9πr²Volume of cone:V_cone = (1/3)πR²h = (1/3)π*(r*sqrt(3))²*(3r)Compute:(r*sqrt(3))² = 3r²So V_cone = (1/3)π*3r²*3r = (1/3)*π*9r³ = 3πr³Therefore, the ratios are:Surface area ratio: A_sphere / A_cone = (4πr²) / (9πr²) = 4/9Volume ratio: V_sphere / V_cone = (4/3 πr³) / (3πr³) = (4/3)/3 = 4/9Wait, both ratios are 4/9? That's interesting. Let me verify the calculations.Surface area:Sphere: 4πr²Cone: 9πr²So ratio 4/9. That's correct.Volume:Sphere: (4/3)πr³Cone: 3πr³So (4/3)/3 = 4/9. That's also correct.Hmm, so both the surface area and volume ratios are 4:9. That seems surprising but mathematically consistent given the relationships we derived. Let me double-check the cone's total surface area and volume.Cone's total surface area is base area (πR²) plus lateral surface area (πRl). We found R = r√3, l = 2R = 2r√3. So:Base area: π*(r√3)^2 = π*3r²Lateral surface area: π*(r√3)*(2r√3) = π*2r²*3 = 6πr²Total: 3πr² + 6πr² = 9πr². Correct.Volume: (1/3)πR²h. R²h = (3r²)*(3r) = 9r³, so (1/3)*π*9r³ = 3πr³. Correct.Sphere's surface area and volume are standard formulas.Therefore, the ratios are both 4/9. So the answer is the ratio of surface areas is 4:9 and the ratio of volumes is also 4:9.But wait, the problem says "respectively to the total surface area and volume of a cone". So the question is asking for (surface area of sphere)/(surface area of cone) and (volume of sphere)/(volume of cone). So both ratios are 4/9. Therefore, the answer is 4/9 for both. But let me confirm once again if there was any miscalculation.Alternatively, maybe the cone is only the lateral surface area? The problem says "total surface area", which usually includes the base. If it were lateral surface area, the base would be excluded. But in the problem statement, it's specified as "total surface area", so including the base. Therefore, our calculation is correct.Hence, both ratios are 4:9.Final AnswerThe ratios of the surface area and volume of the sphere to those of the cone are both boxed{dfrac{4}{9}}.

✅:To find the ratio of the surface area and volume of a sphere to the total surface area and volume of a cone that circumscribes it with an equilateral axial section, we start by analyzing the relationships between the dimensions of the cone and the sphere.1. Equilateral Axial Section: The axial section of the cone is an equilateral triangle, meaning all sides are equal. For a cone with base radius ( R ) and height ( h ), the slant height ( l ) is equal to ( 2R ). Using the Pythagorean theorem, we find: [ l = sqrt{R^2 + h^2} = 2R implies h = Rsqrt{3} ]2. Sphere Inscribed in the Cone: The sphere is tangent to the base and the lateral surface of the cone. The radius ( r ) of the sphere relates to the cone's dimensions. By analyzing the distance from the sphere's center to the cone's lateral surface, we derive: [ R = rsqrt{3} quad text{and} quad h = 3r ]3. Surface Area and Volume Calculations: - Sphere: - Surface area: ( 4pi r^2 ) - Volume: ( frac{4}{3}pi r^3 ) - Cone: - Total surface area: ( pi R^2 + pi R l = pi (rsqrt{3})^2 + pi (rsqrt{3})(2rsqrt{3}) = 3pi r^2 + 6pi r^2 = 9pi r^2 ) - Volume: ( frac{1}{3}pi R^2 h = frac{1}{3}pi (rsqrt{3})^2 (3r) = 3pi r^3 )4. Ratios: - Surface area ratio: ( frac{4pi r^2}{9pi r^2} = frac{4}{9} ) - Volume ratio: ( frac{frac{4}{3}pi r^3}{3pi r^3} = frac{4}{9} )Thus, the ratios of the surface area and volume of the sphere to those of the cone are both boxed{dfrac{4}{9}}.

🔑:1. Calculate the Volume of the Sphere: - Given the radius ( r ) of the sphere, the volume ( V_{text{ш}} ) of the sphere is calculated using the formula: [ V_{text{ш}} = frac{4}{3} pi r^3 ]2. Calculate the Volume of the Cone: - Given the cone is described with a base radius ( R ) and height ( H ), the volume ( V_{text{κ}} ) of the cone is calculated using the formula: [ V_{text{κ}} = frac{1}{3} pi R^2 H ] As shown in figure 11.81.3. Relate Dimensions Using an Equilateral Triangle: - Since the equilateral triangle is an axial section of the cone, we denote the side of the equilateral triangle as ( a ). - The radius ( r ) of the sphere can be related to ( a ) as: [ r = frac{a sqrt{3}}{6} ] - Solving for ( a ): [ a = frac{6r}{sqrt{3}} = 2sqrt{3} r ]4. Compute Base Radius and Height of the Cone: - The radius ( R ) of the base of the cone is: [ R = frac{a}{2} = sqrt{3} r ] - The height ( H ) of the cone, using Pythagorean Theorem in an equilateral triangle, is: [ H = sqrt{a^2 - left(frac{a}{2}right)^2} = sqrt{(2sqrt{3}r)^2 - left(sqrt{3}rright)^2} = sqrt{12r^2 - 3r^2} = 3r ]5. Determine the Volume Ratio: - Substituting the values into the volumes: [ frac{V_{text{ш}}}{V_{text{κ}}} = frac{frac{4}{3} pi r^3}{frac{1}{3} pi (sqrt{3}r)^2 (3r)} = frac{4 pi r^3}{3 pi 3 r^3} = frac{4}{9} ]6. Calculate the Surface Area of the Sphere: - The surface area ( S_{text{ш}} ) of the sphere is: [ S_{text{ш}} = 4 pi r^2 ]7. Calculate the Surface Area of the Cone: - The surface area ( S_{text{κ}} ) of the cone includes the lateral surface area and base surface area: [ S_{text{κ}} = S_{text{бок}} + S_{text{осн}} = pi R cdot l + pi R^2 ] Where ( l ) is the slant height of the cone, which is ( l = a ) because of the equilateral triangle properties. Hence: [ S_{text{κ}} = pi (sqrt{3}r) cdot (2sqrt{3}r) + pi (sqrt{3}r)^2 = 6pi r^2 + 3pi r^2 = 9 pi r^2 ]8. Determine the Surface Area Ratio: - Substituting the values into the surface areas: [ frac{S_{text{ш}}}{S_{text{κ}}} = frac{4 pi r^2}{9 pi r^2} = frac{4}{9} ] Conclusion:[boxed{frac{V_{text{ш}}}{V_{text{κ}}} = frac{frac{4}{3} pi r^3}{frac{1}{3} pi (sqrt{3} r)^2 cdot 3r} = frac{4}{9} quad text{и} quad frac{S_{text{ш}}}{S_{text{κ}}} = frac{4 pi r^2}{9 pi r^2} = frac{4}{9}}]

🧠:Okay, so I need to find the zeros and local extrema of the function y = f(x) = x² - {x}² + x + |x² - 1|. Hmm, let's start by understanding each part of this function. First, the function has three main components: x², -{x}², x, and the absolute value term |x² - 1|. The fractional part {x} is defined as x - floor(x), right? So {x} is always between 0 (inclusive) and 1 (exclusive). That means {x}² will be between 0 and 1 as well. Since the function involves the fractional part {x}, the behavior of the function might change at each integer. Therefore, it's probably a good idea to analyze the function on each interval [n, n+1) where n is an integer. That way, {x} can be expressed as x - n in each interval, simplifying the expression. Also, the absolute value term |x² - 1| will have different expressions depending on whether x² - 1 is non-negative or negative. So x² - 1 ≥ 0 when |x| ≥ 1, and x² - 1 < 0 when |x| < 1. Therefore, the absolute value term can be split into two cases: when x is in (-1, 1), the term becomes -(x² - 1) = -x² + 1, and otherwise, it's x² - 1. But since the function is defined piecewise due to the fractional part, I need to combine these two piecewise definitions. Let me try to structure this step by step.First, let's consider different intervals based on the integer part of x. Let x be in [n, n+1) where n is an integer. Then {x} = x - n. So substituting into the function:f(x) = x² - (x - n)² + x + |x² - 1|Let's expand the (x - n)² term:(x - n)² = x² - 2nx + n²So substituting back:f(x) = x² - (x² - 2nx + n²) + x + |x² - 1| Simplify the terms:x² - x² + 2nx - n² + x + |x² - 1| So that simplifies to:(2n + 1)x - n² + |x² - 1|So, on each interval [n, n+1), the function f(x) is a linear function in x (since |x² - 1| is either quadratic or linear, but wait, |x² - 1| is still quadratic. Wait, hold on, but the function f(x) would be (2n + 1)x - n² + |x² - 1|. So the absolute value term complicates things because it's not linear. So even within each interval [n, n+1), the function f(x) is not purely linear but has a quadratic component from |x² - 1|. Therefore, I need to analyze f(x) on each interval [n, n+1) but also considering the points where x² - 1 changes sign, which are at x = -1 and x = 1. Therefore, if the interval [n, n+1) includes x = -1 or x = 1, then the expression for |x² - 1| will change within that interval. So first, let's list all possible intervals where both the fractional part {x} and the absolute value term can be expressed without change. That is, for integer intervals [n, n+1), but also split at x = -1 and x = 1 if necessary. Let's consider different integer intervals:Case 1: n ≤ x < n+1, where n is an integer.But depending on n, the interval [n, n+1) may include x = -1 or x = 1. For example:- If n = -2: interval [-2, -1). Here, x is less than -1, so x² -1 ≥ 0, so |x² -1| = x² -1.- If n = -1: interval [-1, 0). Here, x is between -1 and 0. So x² -1: since x is between -1 and 0, x² is between 0 and 1, so x² -1 is negative. Hence, |x² -1| = -x² +1.- If n = 0: interval [0, 1). Here, x is between 0 and 1. Similarly, x² -1 is negative, so |x² -1| = -x² +1.- If n = 1: interval [1, 2). Here, x is ≥1, so x² -1 ≥0, so |x² -1| = x² -1.Similarly, for n ≥2, x is ≥2, so x² -1 is positive, and for n ≤-2, x is ≤-2, so x² -1 is positive.So, the absolute value term |x² -1| can be split as follows:- For x in (-∞, -1) ∪ [1, ∞): |x² -1| = x² -1- For x in [-1,1): |x² -1| = -x² +1Therefore, combining the two piecewise definitions (fractional part and absolute value), the intervals to consider are:1. x in [n, n+1) where n ≤ -22. x in [-1, 0)3. x in [0,1)4. x in [1,2)5. x in [n, n+1) where n ≥2Wait, but for n ≤ -2, the interval [n, n+1) is entirely in (-∞, -1), so |x² -1| = x² -1. For n = -1, the interval is [-1,0), which is in [-1,1), so |x² -1| = -x² +1. For n=0, same as above. For n=1, [1,2) is in [1, ∞), so |x² -1| = x² -1. Similarly for n ≥2.Therefore, perhaps the function can be written piecewise as:For x in [n, n+1):- If n ≤ -1: |x² -1| = x² -1 if n ≤ -2, but for n = -1, x is in [-1,0), so |x² -1| = -x² +1.Wait, so perhaps better to split into the following intervals:1. x < -12. -1 ≤ x < 03. 0 ≤ x <14. 1 ≤ x <25. x ≥2But actually, for x < -1, x is in some interval [n, n+1) where n ≤ -2. Similarly, x ≥2 is [n, n+1) with n ≥2. Then for -1 ≤ x <0, n = -1; 0 ≤ x <1, n=0; 1 ≤ x <2, n=1. So combining both the fractional part and the absolute value.Therefore, let's handle each interval one by one.First interval: x < -1 (n ≤ -2)Here, {x} = x - n, where n = floor(x). But since x is in [n, n+1) with n ≤ -2, which is less than -1. Then:f(x) = (2n + 1)x - n² + |x² -1|But since x < -1, |x² -1| = x² -1. Therefore:f(x) = (2n + 1)x - n² + x² -1Simplify:x² + (2n +1)x - n² -1So this is a quadratic function in x. But n is the integer part of x, which is floor(x). Since x is in [n, n+1), n is fixed for each interval. However, since n ≤ -2, each interval [n, n+1) with n ≤ -2 will have a different quadratic function. Hmm, this seems complicated. Maybe it's better to write the function as a quadratic in each interval [n, n+1) with n ≤ -2. For example, take n = -2, then x ∈ [-2, -1):f(x) = x² - {x}² + x + |x² -1|Here, {x} = x - (-2) = x +2. So {x}² = (x +2)²So substituting:f(x) = x² - (x +2)^2 + x + |x² -1|Compute (x +2)^2 = x² +4x +4. Therefore:f(x) = x² - (x² +4x +4) + x + |x² -1|Simplify:x² -x² -4x -4 +x + |x² -1|Which is:(-3x -4) + |x² -1|But since x is in [-2, -1), x < -1, so |x² -1| = x² -1. Therefore:f(x) = -3x -4 + x² -1 = x² -3x -5So on [-2, -1), f(x) is a quadratic function x² -3x -5.Similarly, for n = -3, x ∈ [-3, -2):{x} = x +3. Then {x}² = (x +3)^2 = x² +6x +9Thus:f(x) = x² - (x² +6x +9) + x + |x² -1| = x² -x² -6x -9 +x + |x² -1| = (-5x -9) + |x² -1|Again, since x < -1, |x² -1| =x² -1, so:f(x) = -5x -9 +x² -1 = x² -5x -10So in general, for x ∈ [n, n+1) where n ≤ -2:{x} = x -n = x - n (since n is negative, x is in [n, n+1), so {x} = x -n)Therefore, {x}² = (x -n)^2 = x² -2nx +n²Therefore, f(x) = x² - (x² -2nx +n²) +x + |x² -1| = 2nx -n² +x + |x² -1| = (2n +1)x -n² + |x² -1|But since x ∈ [n, n+1) and n ≤ -2, x < -1, so |x² -1| =x² -1. Therefore:f(x) = (2n +1)x -n² +x² -1 = x² + (2n +1)x -n² -1Thus, in each interval [n, n+1) with n ≤ -2, the function is a quadratic function: x² + (2n +1)x -n² -1.Similarly, for x ∈ [-1, 0):Here, n = -1. So {x} = x - (-1) = x +1. So {x}² = (x +1)^2Thus:f(x) = x² - (x +1)^2 +x + |x² -1|Compute (x +1)^2 =x² +2x +1. Therefore:f(x) =x² - (x² +2x +1) +x + |x² -1| = x² -x² -2x -1 +x + |x² -1| = (-x -1) + |x² -1|Since x ∈ [-1, 0), then x is in [-1,0). Therefore, x² is between 0 and 1, so x² -1 is negative. Thus, |x² -1| = -x² +1. Therefore:f(x) = -x -1 -x² +1 = -x² -xSo on [-1,0), f(x) = -x² -x.For x ∈ [0,1):n =0, so {x} =x -0 =x. Therefore {x}² =x². Thus:f(x) =x² -x² +x + |x² -1| =0 +x + |x² -1|Since x ∈ [0,1), x² -1 <0, so |x² -1|= -x² +1. Therefore:f(x) =x + (-x² +1) = -x² +x +1So on [0,1), f(x) is -x² +x +1.For x ∈ [1,2):n =1, so {x} =x -1. Therefore {x}² = (x -1)^2.Thus:f(x) =x² - (x -1)^2 +x + |x² -1|Compute (x -1)^2 =x² -2x +1. Therefore:f(x) =x² - (x² -2x +1) +x + |x² -1| = x² -x² +2x -1 +x + |x² -1| = 3x -1 + |x² -1|Since x ∈ [1,2), x ≥1, so |x² -1| =x² -1. Therefore:f(x) =3x -1 +x² -1 =x² +3x -2So on [1,2), f(x) =x² +3x -2.For x ≥2, n ≥2. Then similar to the case n ≤ -2, for each interval [n, n+1), n ≥2:{x} =x -n. So {x}² =(x -n)^2. Therefore:f(x) =x² - (x -n)^2 +x + |x² -1|Expanding (x -n)^2:x² -2nx +n². Therefore:f(x) =x² -x² +2nx -n² +x + |x² -1| = (2n +1)x -n² + |x² -1|Since x ∈ [n, n+1) with n ≥2, x ≥2, so |x² -1| =x² -1. Therefore:f(x) = (2n +1)x -n² +x² -1 = x² + (2n +1)x -n² -1So in each interval [n, n+1) with n ≥2, f(x) is quadratic: x² + (2n +1)x -n² -1.Okay, so now we have expressions for f(x) on different intervals. Let me summarize:1. For x < -1 (n ≤ -2): f(x) =x² + (2n +1)x -n² -1 (quadratic in each interval [n, n+1))2. For x ∈ [-1,0): f(x) = -x² -x3. For x ∈ [0,1): f(x) = -x² +x +14. For x ∈ [1,2): f(x) =x² +3x -25. For x ≥2 (n ≥2): f(x) =x² + (2n +1)x -n² -1 (quadratic in each interval [n, n+1))Now, the task is to find zeros and local extrema.Starting with zeros: solve f(x) =0 in each interval.Also, for local extrema, we need to check where the derivative is zero (critical points), and also endpoints of intervals where the function might have extrema.But since the function is piecewise defined, we need to check each interval separately, compute derivatives where differentiable, and check critical points within each interval as well as continuity at the endpoints.First, let's handle each interval.1. x < -1 (n ≤ -2): quadratic function f(x) =x² + (2n +1)x -n² -1.Each interval [n, n+1) has its own quadratic function. Let's first check if these quadratics can have zeros in their respective intervals.For x ∈ [n, n+1), n ≤ -2.The quadratic is f(x) =x² + (2n +1)x -n² -1.To find zeros, set x² + (2n +1)x -n² -1 =0.Solve for x:x = [ - (2n +1) ± sqrt( (2n +1)^2 +4(n² +1) ) ] /2Compute discriminant D:D = (2n +1)^2 +4(n² +1) =4n² +4n +1 +4n² +4 =8n² +4n +5Since 8n² +4n +5 is always positive (quadratic in n with a=8 >0, discriminant 16 - 160 = -144 <0), so two real roots.But need to check if these roots lie within [n, n+1).But since n ≤ -2, and quadratic is x² + (2n +1)x -n² -1.Alternatively, perhaps it's better to check for specific n. For example, take n = -2:x ∈ [-2, -1), f(x) =x² -3x -5.Solve x² -3x -5 =0. The roots are [3 ± sqrt(9 +20)]/2 = [3 ± sqrt(29)]/2 ≈ [3 ±5.385]/2. So positive root ≈ (3 +5.385)/2 ≈4.193, negative root ≈(3 -5.385)/2≈-1.193. So in [-2,-1), the root is ≈-1.193, which is in [-2,-1). Therefore, in this interval, there is a zero at x≈-1.193.Similarly, for n = -3, interval [-3,-2):f(x) =x² -5x -10.Solve x² -5x -10 =0. Roots: [5 ± sqrt(25 +40)]/2 = [5 ±sqrt(65)]/2 ≈[5 ±8.062]/2. Positive root ≈6.531, negative root≈(5 -8.062)/2≈-1.531. So in [-3,-2), the root is≈-1.531, which is not in [-3,-2). Wait, that's a problem.Wait, wait, when n=-3, the quadratic is x² + (2*(-3)+1)x - (-3)^2 -1 =x² + (-6 +1)x -9 -1 =x² -5x -10. So the roots are [5 ±sqrt(25 +40)]/2≈[5 ±8.062]/2. So ≈6.531 and ≈-1.531. But the interval is [-3,-2). The roots ≈-1.531 is in [-2, -1), not in [-3,-2). So in [-3,-2), the quadratic x² -5x -10 doesn't cross zero. Wait, but then when n=-3, f(x)=x² -5x -10 on [-3,-2). Let's evaluate f(-3)= (-3)^2 -5*(-3) -10=9 +15 -10=14. f(-2)= (-2)^2 -5*(-2) -10=4 +10 -10=4. So f(x) at both endpoints is 14 and 4, both positive. So f(x) is always positive in [-3,-2). Therefore, no zeros here.Similarly, for n=-4, interval [-4,-3):Quadratic f(x)=x² + (2*(-4)+1)x - (-4)^2 -1 =x² -7x -16 -1 =x² -7x -17.Roots: [7 ±sqrt(49 +68)]/2 = [7 ±sqrt(117)]/2≈[7 ±10.816]/2. So roots≈(7+10.816)/2≈8.908 and≈(7-10.816)/2≈-1.908. Again, the negative root is≈-1.908, which is in [-2,-1), but the interval here is [-4,-3). So in [-4,-3), f(-4)=16 +28 -17=27, f(-3)=9 +21 -17=13. Both positive. So no zeros in this interval.Therefore, for n ≤ -2, except n=-2, the quadratics do not cross zero in their intervals. Only for n=-2, x ∈[-2,-1), the quadratic x² -3x -5 has a root at≈-1.193, which is in [-2,-1). So there is one zero in this interval.So the zeros from this part are x≈-1.193, but exact value is needed.Wait, but we can compute exact roots.For n=-2, quadratic is x² -3x -5=0. The exact roots are [3 ±sqrt(9 +20)]/2 = [3 ±sqrt(29)]/2. So the root in [-2,-1) is [3 -sqrt(29)]/2. Compute sqrt(29)≈5.385, so [3 -5.385]/2≈(-2.385)/2≈-1.1925. So exact root is (3 -sqrt(29))/2. So that's one zero at x=(3 -sqrt(29))/2 ≈-1.1925.Now, moving to the next interval: x ∈[-1,0). Here, f(x)= -x² -x. Set -x² -x =0. Factor: -x(x +1)=0. Solutions x=0 or x=-1. But the interval is [-1,0). So x=-1 is included in the interval, but at x=-1, since the interval is [-1,0), x=-1 is included here. So f(-1) = -(-1)^2 -(-1) = -1 +1=0. Similarly, at x=0, f(0) would be -0 -0=0, but x=0 is not in this interval, it's in [0,1). So zeros at x=-1 and x=0? Wait, but x=0 is not in [-1,0). So only x=-1. But let's check.Wait, f(-1)=0, which is part of this interval. Then in the open interval (-1,0), is there another zero? Let's see. Solve -x² -x =0. Solutions x=0 and x=-1. But in (-1,0), only x=-1 is the left endpoint. So the zeros here are x=-1 and x=0? Wait, but x=0 is not in this interval. Wait, confusion here.Wait, in the interval [-1,0), the function f(x)= -x² -x. So setting -x² -x=0, we get x=0 or x=-1. So x=-1 is in the interval, x=0 is not. Therefore, the only zero in [-1,0) is at x=-1. But since x=-1 is the endpoint, but since the interval is [-1,0), x=-1 is included. So x=-1 is a zero.Wait, but let's check continuity. For example, approaching x=-1 from the left (from n=-2 interval), we have the function x² -3x -5. At x=-1, f(-1) = (-1)^2 -3*(-1) -5 =1 +3 -5= -1. But from the right (x approaching -1 from [-1,0)), f(-1)=0. So there's a discontinuity at x=-1. Therefore, the function has a jump discontinuity at x=-1. Therefore, f(-1) is defined as the right-hand limit, which is 0. But in reality, the fractional part function {x} is continuous from the right. Wait, actually, the function {x} is continuous on [n, n+1), but at x=n, {x} approaches 0 from the right. Wait, but when x is approaching -1 from the left, in the interval [-2,-1), {x} =x +2, so as x approaches -1 from the left, {x} approaches -1 +2=1. But {x} is defined as x - floor(x). For x approaching -1 from the left (just below -1), floor(x)= -2, so {x}=x - (-2)=x +2. So approaching -1 from the left, {x} approaches 1. But approaching -1 from the right (just above -1), floor(x)= -1, so {x}=x - (-1)=x +1, which approaches 0. Therefore, {x} has a jump discontinuity at each integer, from 1 to 0. Therefore, the function f(x) will also have jump discontinuities at each integer. Therefore, at x=-1, the left limit of f(x) is f(-1^-)=lim_{x→-1^-} x² - {x}² +x + |x² -1|. Let's compute that.From the left (x approaching -1 from [-2,-1)):f(x)=x² - (x +2)^2 +x + |x² -1|At x approaching -1:x² =1, (x +2)^2=1, so x² - (x +2)^2 =1 -1=0. Then x + |x² -1|: x approaches -1, |x² -1|=|1 -1|=0. Therefore, f(-1^-)=0 + (-1) +0= -1.From the right (x approaching -1 from [-1,0)):f(x)= -x² -x. So f(-1)= -(-1)^2 -(-1)= -1 +1=0.Therefore, there's a jump discontinuity at x=-1. So the function is not continuous there, so x=-1 is a zero from the right, but the left limit is -1. Therefore, x=-1 is a zero (from the right side).Moving on to the interval [0,1): f(x)= -x² +x +1.Set -x² +x +1=0. Multiply by -1: x² -x -1=0. Solutions: [1 ±sqrt(1 +4)]/2 = [1 ±sqrt(5)]/2≈[1 ±2.236]/2. So roots≈1.618 and≈-0.618. The interval is [0,1). So the positive root≈1.618 is outside the interval, and the negative root≈-0.618 is also outside. Therefore, no zeros in [0,1). Wait, but let's check at endpoints.At x=0: f(0)= -0 +0 +1=1≠0.At x=1: approaching from the left, f(1^-)= -1 +1 +1=1≠0.So no zeros in [0,1).Next interval [1,2): f(x)=x² +3x -2.Set x² +3x -2=0. Solutions: [-3 ±sqrt(9 +8)]/2 = [-3 ±sqrt(17)]/2. sqrt(17)≈4.123. So roots≈(-3 +4.123)/2≈0.561 and≈(-3 -4.123)/2≈-3.561. The interval [1,2), so the positive root≈0.561 is outside, and the negative root is also outside. Therefore, no zeros in [1,2). Check endpoints:At x=1: f(1)=1 +3 -2=2≠0.At x=2: approaching from the left, f(2^-)=4 +6 -2=8≠0.So no zeros here.For x ≥2 (n ≥2): quadratic function f(x)=x² + (2n +1)x -n² -1.Similar to the n ≤-2 case, but here x is in [n, n+1), n ≥2. Let's check if these quadratics have zeros in their intervals.For example, take n=2: interval [2,3). f(x)=x² +5x -4 -1=x² +5x -5.Solve x² +5x -5=0. Roots: [-5 ±sqrt(25 +20)]/2 = [-5 ±sqrt(45)]/2≈[-5 ±6.708]/2. Positive root≈(1.708)/2≈0.854, negative root≈-11.708/2≈-5.854. Both roots are outside [2,3). At x=2: f(2)=4 +10 -5=9. At x=3: f(3)=9 +15 -5=19. So function is positive here, no zeros.Similarly for n=3: interval [3,4). f(x)=x² +7x -9 -1=x² +7x -10.Roots: [-7 ±sqrt(49 +40)]/2 = [-7 ±sqrt(89)]/2≈[-7 ±9.434]/2. Positive root≈(2.434)/2≈1.217, negative root≈-16.434/2≈-8.217. Both outside [3,4). f(3)=9 +21 -10=20, f(4)=16 +28 -10=34. Positive, so no zeros.Thus, for n ≥2, the quadratics do not cross zero in [n, n+1). So no zeros in x ≥2.Therefore, compiling all zeros found:- x=(3 -sqrt(29))/2≈-1.1925 in [-2,-1)- x=-1 in [-1,0)But wait, at x=0, what's f(0)? For x in [0,1), f(0)= -0 +0 +1=1≠0. So no zero at x=0.Wait, but in the interval [-1,0), the function is f(x)= -x² -x, which is zero at x=-1 and x=0. But x=0 is not in the interval. However, when we move to the next interval [0,1), f(x)= -x² +x +1, which at x=0 is 1≠0. Therefore, the only zeros are x=(3 -sqrt(29))/2 and x=-1.Wait, but what about the interval [1,2)? The quadratic there has no zeros. And in x ≥2, no zeros.But wait, let's check if there's any zero in other intervals.Wait, maybe in the interval [n, n+1) with n=1, [1,2), f(x)=x² +3x -2. We found roots≈0.561 and≈-3.561, which are outside [1,2). So no zeros.Similarly, in [0,1), no zeros.So total zeros are x=-1 and x=(3 -sqrt(29))/2≈-1.1925. Wait, but x=-1 is also part of the interval [-1,0). So two zeros? Wait, but x=(3 -sqrt(29))/2 is≈-1.1925, which is in [-2,-1), and x=-1 is in [-1,0). So two zeros.But we need to check if there are any more zeros.Wait, in the interval [-1,0), the function is f(x)= -x² -x. We found it's zero at x=-1 and x=0. But x=0 is not in the interval. However, at x=0, moving to the next interval [0,1), f(0)=1≠0. So only x=-1 is a zero here.Therefore, total zeros are at x=-1 and x=(3 -sqrt(29))/2. Wait, is that correct?Wait, but when solving in the interval [-2,-1), we have a zero at (3 -sqrt(29))/2≈-1.1925. Then in [-1,0), a zero at x=-1. But x=-1.1925 is less than -1, so it's in the interval [-2,-1). Therefore, total two zeros.Wait, but let's verify:At x=(3 -sqrt(29))/2≈-1.1925, which is in [-2,-1). Therefore, there is a zero here. And at x=-1, which is the left endpoint of [-1,0), also a zero. So two zeros in total.Are there any more zeros?In other intervals, we saw no zeros. So zeros are at x=-1 and x=(3 -sqrt(29))/2.Now, moving to local extrema.Local extrema occur where the derivative is zero (critical points) or at endpoints of intervals (since the function may have corners or jumps there). However, due to the piecewise nature, we need to check each interval separately.Again, let's go interval by interval.1. For x < -1 (n ≤ -2): Each interval [n, n+1) has f(x)=x² + (2n +1)x -n² -1. This is a quadratic function in x. The derivative is f’(x)=2x +2n +1. Setting derivative to zero: 2x +2n +1=0 →x=-(2n +1)/2.We need to check if this critical point lies within [n, n+1). Let's solve for x:x=-(2n +1)/2. Since n ≤ -2, let's check for n=-2:x= -(-4 +1)/2= -(-3)/2=1.5. But n=-2 interval is [-2,-1), x=1.5 is not in [-2,-1). So no critical point here.For n=-3:x= -(-6 +1)/2= -(-5)/2=2.5. Again, not in [-3,-2).Similarly, for any n ≤ -2, x=-(2n +1)/2.Since n is integer ≤-2, let's compute x:Let n =k where k ≤ -2. Then x=-(2k +1)/2.But since k ≤ -2, 2k ≤-4, so 2k +1 ≤-3. Therefore, -(2k +1)/2 ≥3/2=1.5. So x=1.5 or higher. But x is in [k, k+1) where k ≤-2. So x is negative, but the critical point x=1.5 is positive. Therefore, none of these critical points lie in the intervals x < -1. Therefore, on each interval [n, n+1) with n ≤ -2, the function is a quadratic opening upwards (since coefficient of x² is 1>0), so it has a minimum at x=-(2n +1)/2, but since this x is not in [n, n+1), the extrema in these intervals occur at the endpoints.Therefore, in each interval [n, n+1), n ≤-2, the minimum is at the left endpoint x=n (since the quadratic opens upwards and the vertex is to the right of the interval), and maximum at the right endpoint x=n+1. Wait, but wait, if the vertex is to the right of the interval, then on the interval [n, n+1), the function is increasing if the vertex is to the left, or decreasing if the vertex is to the right. Wait, the quadratic is f(x)=x² + (2n +1)x -n² -1. Since the coefficient of x² is positive, the parabola opens upwards. The vertex is at x= - (2n +1)/(2*1)= -(2n +1)/2. For n ≤ -2, 2n +1 ≤-4 +1=-3, so x=vertex= positive number, as before. Since the vertex is at x>0, which is way to the right of the interval [n, n+1) where x < -1. Therefore, on the interval [n, n+1), the function is decreasing towards the vertex, but since the vertex is to the right of the interval, the function is decreasing on the entire interval? Wait, no. Wait, the vertex is at x=-(2n +1)/2. For example, take n=-2: vertex at x= -(2*(-2)+1)/2= -(-4 +1)/2=3/2=1.5. So for n=-2, the interval is [-2,-1), and the vertex is at x=1.5. Therefore, the function is decreasing on (-∞,1.5) and increasing on (1.5, ∞). Therefore, on the interval [-2,-1), which is entirely to the left of the vertex, the function is decreasing. Similarly, for n=-3, the vertex is at x=-(2*(-3)+1)/2= -(-6 +1)/2=5/2=2.5. So on interval [-3,-2), which is to the left of the vertex, the function is decreasing. Therefore, in each interval [n, n+1) with n ≤-2, the function is decreasing. Therefore, the maximum in the interval is at the left endpoint x=n, and the minimum at the right endpoint x=n+1.But wait, since it's decreasing, the maximum is at x=n and the minimum at x=n+1. However, since these intervals are closed on the left and open on the right, the minimum is not attained but approaches as x→(n+1)^-.But for the purposes of local extrema, since the function is decreasing on each interval [n, n+1), n ≤-2, there are no local extrema within the intervals, only at the endpoints. However, endpoints are shared between intervals, but due to the jump discontinuities, the function may have local maxima or minima at these points.For example, at x=-2, approaching from the left (from n=-3 interval), but wait x=-2 is in [-2,-1) interval. Wait, actually, each integer is the left endpoint of its interval. So for x=n, the function is defined on [n, n+1). So at x=n, the function has a value, and the left limit (from [n-1,n)) is different due to the jump.Therefore, when considering local extrema, we need to check if there is a local maximum or minimum at these integer points.For example, take x=-2. In the interval [-2,-1), f(x)=x² -3x -5. The left limit as x approaches -2 from the left (from [-3,-2)) is f(-2^-)= (-2)^2 -5*(-2) -10=4 +10 -10=4. The right limit at x=-2 is f(-2)= (-2)^2 -3*(-2) -5=4 +6 -5=5. So the function jumps from 4 to 5 at x=-2. Therefore, there's a local minimum at x=-2?But actually, since the function is decreasing on [-2,-1), and the value at x=-2 is 5, and it decreases to the right, then x=-2 is a local maximum?Wait, in the interval [-2,-1), the function is decreasing from x=-2 to x=-1. So f(-2)=5 and approaches f(-1^-)= (-1)^2 -3*(-1) -5=1 +3 -5=-1. So the function decreases from 5 to -1. Therefore, at x=-2, since the function jumps from 4 (left limit) to 5 (right limit), and then decreases. Therefore, x=-2 is a local maximum because the function decreases after it. Similarly, at x=-1, from the left side (interval [-2,-1)), the function approaches -1, but from the right side (interval [-1,0)), it jumps to f(-1)=0. Therefore, there's a local minimum at x=-1.Similarly, at each integer x=n ≤-2, the function jumps from the left limit (lower) to a higher value at x=n, and then decreases. Therefore, each x=n (n ≤-2) is a local maximum.But let's verify with n=-2:Left limit at x=-2 is 4, f(-2)=5, which is higher, then decreases to -1. So yes, x=-2 is a local maximum.Similarly, at x=-3: left limit is f(-3^-) from [-4,-3): let's compute for n=-4, interval [-4,-3): f(x)=x² -7x -17. At x=-3, f(-3)=9 +21 -17=13. Left limit approaching x=-3 from the left (n=-4 interval): f(-3^-)=x approaching -3 from [-4,-3), f(x)=x² -7x -17. f(-3)=9 +21 -17=13. Wait, but x=-3 is not in [-4,-3). Wait, maybe I need to re-express.Wait, for n=-3: interval is [-3,-2). So x=-3 is in this interval. Wait, no, n=-3 corresponds to x ∈[-3,-2). Therefore, x=-3 is the left endpoint of this interval.Wait, this is confusing. Let's clarify:For n ≤-2, each interval [n, n+1). For example:- n=-2: interval [-2,-1)- n=-3: interval [-3,-2)- n=-4: interval [-4,-3)Etc.Therefore, for n=-2, the interval is [-2,-1), so x=-2 is included here. The function in this interval is x² -3x -5. At x=-2, f(-2)=4 +6 -5=5. The left limit as x approaches -2 from the left (from interval [-3,-2)):In interval [-3,-2), the function is x² -5x -10. So lim_{x→-2^-} f(x)= (-2)^2 -5*(-2) -10=4 +10 -10=4. Therefore, f(-2)=5, which is higher than the left limit 4. Then, in the interval [-2,-1), the function decreases from 5 to -1. Therefore, x=-2 is a local maximum.Similarly, at x=-3: in the interval [-3,-2), the function is x² -5x -10. At x=-3: f(-3)=9 +15 -10=14. The left limit as x approaches -3 from the left (from [-4,-3)): f(x)=x² -7x -17. lim_{x→-3^-} f(x)=9 +21 -17=13. So f(-3)=14, which is higher than the left limit 13. Then, in [-3,-2), the function decreases from 14 to f(-2^-)=4. So x=-3 is a local maximum.Therefore, each integer x=n ≤-1 is a local maximum because the function jumps up at x=n and then decreases.Wait, but for x=-1, what happens?At x=-1, from the left (interval [-2,-1)), the function approaches f(-1^-)= -1. From the right, f(-1)=0. So there's a jump from -1 to 0. Then, in the interval [-1,0), the function f(x)=-x² -x. Its derivative is f’(x)=-2x -1. Setting to zero: -2x -1=0 →x=-0.5. So critical point at x=-0.5 in [-1,0).Therefore, in [-1,0), the function is a downward opening parabola (since coefficient of x² is -1), with vertex at x=-0.5. Therefore, the function increases from x=-1 to x=-0.5, reaching a maximum, then decreases towards x=0.Therefore, in [-1,0), the function f(x)=-x² -x has a local maximum at x=-0.5. Let's compute f(-0.5)= -0.25 -(-0.5)= -0.25 +0.5=0.25.So in [-1,0), the function increases from f(-1)=0 to f(-0.5)=0.25, then decreases to f(0^-)=0. Therefore, x=-0.5 is a local maximum. Also, at x=-1, since the left limit is -1 and the function jumps to 0, then increases, so x=-1 is a local minimum.Wait, but at x=-1, the left limit is -1 and the right value is 0. So since it's lower on the left, but the function jumps up, so x=-1 is not a local minimum because the function isn't defined to the left of -1 in this piece. Wait, but actually, for x < -1, the function continues. So approaching x=-1 from the left, the limit is -1, but the function at x=-1 is 0. So from the left side, it's approaching -1, but at x=-1, it's 0. Therefore, x=-1 is a local minimum? Because if you consider a small neighborhood around x=-1, to the left it approaches -1, at x=-1 it's 0, and to the right it increases to 0.25. So the lowest point around x=-1 is the left limit, but since x=-1 is defined as 0, which is higher, perhaps x=-1 is not a local minimum. This is a bit tricky due to the discontinuity.Generally, local minima require the function value at the point to be less than or equal to nearby points. However, due to the jump, at x=-1, the function is 0, which is higher than the left limit (-1) and lower than the right neighborhood (which increases to 0.25). Therefore, x=-1 is not a local minimum because there are points immediately to the left (in the previous interval) with lower values, but since those points are in a different interval, depending on the definition, sometimes endpoints can be considered for local extrema only if they are higher or lower than their immediate neighborhood. In this case, x=-1 has a jump from -1 to 0, so it's a discontinuity. In some contexts, a jump discontinuity can create a local maximum or minimum if the function value at the point is higher or lower than both sides. Here, at x=-1, the left limit is -1, the right limit is 0. So the function value at x=-1 is 0, which is higher than the left limit but lower than the right side which increases to 0.25. So compared to the immediate right, it's lower, but compared to the immediate left, it's higher. Therefore, x=-1 is not a local extremum in the strict sense, but in some contexts, it might be considered a local minimum from the right. However, typically, local extrema require the point to be in the domain, and to be higher or lower than all neighboring points in the domain. Since the domain includes x=-1, and to the left of x=-1 is another interval, but in the immediate right neighborhood (within [-1,0)), the function increases from 0 to 0.25, so x=-1 is a local minimum on the interval [-1,0) if we consider it as a closed interval. But since it's half-open, [-1,0), x=-1 is the left endpoint. The function at x=-1 is 0, and to the right, it increases. Therefore, on the interval [-1,0), x=-1 is a local minimum.But this depends on the definition. In many calculus textbooks, local extrema are defined with respect to points in the domain, so if x=-1 is part of the domain, and there exists a neighborhood around x=-1 where f(x=-1) is the minimum or maximum, then it could be considered. However, since to the left of x=-1, the function is not part of the same piece, but technically, the domain is all real numbers, just defined piecewise. Therefore, for the entire domain, x=-1 is a point where the function has a jump from -1 to 0. Therefore, in the entire domain, x=-1 is not a local minimum or maximum because in any neighborhood around x=-1, there are points both higher and lower than f(-1)=0. Specifically, immediately to the left, the function approaches -1 (lower), and to the right, it increases to 0.25 (higher). Therefore, x=-1 is not a local extremum.But within the interval [-1,0), x=-1 is the left endpoint, and since the function increases from 0 to 0.25, then x=-1 is a local minimum on this interval. However, in the global sense, it's not a local minimum.This is a bit ambiguous, so perhaps we should proceed carefully.Similarly, at x=0, moving from interval [-1,0) to [0,1), the function jumps from f(0^-)=0 to f(0)=1. So there's a jump from 0 to 1. Then in [0,1), the function f(x)=-x² +x +1. Its derivative is f’(x)=-2x +1. Setting to zero: -2x +1=0 →x=0.5. So critical point at x=0.5 in [0,1).Evaluating f(0.5)= -0.25 +0.5 +1=1.25. So the function increases from f(0)=1 to f(0.5)=1.25, then decreases to f(1^-)=-1 +1 +1=1. Therefore, in [0,1), there's a local maximum at x=0.5.At x=0, since the left limit is 0 and the right value is1, which is higher, so x=0 is not a local extremum globally, but within [0,1), it's the left endpoint, and the function increases from 1 to 1.25, so x=0 is a local minimum on this interval.Similarly, at x=1, moving from [0,1) to [1,2): f(1^-)=1, and f(1)=1 +3 -2=2. So jump from1 to2. In [1,2), the function f(x)=x² +3x -2, which has derivative f’(x)=2x +3. Setting to zero:2x +3=0→x=-1.5, which is outside [1,2). Therefore, in [1,2), the function is increasing (since derivative 2x +3 is positive for x ≥1: at x=1, derivative=5>0). Therefore, function increases from f(1)=2 to f(2^-)=4 +6 -2=8. Therefore, on [1,2), the function is increasing, so no local extrema within the interval, but at x=1, the left limit is1 and the right value is2, so x=1 is a local minimum if we consider the jump.But similar to x=-1 and x=0, x=1 has a jump from1 to2, so in the entire domain, it's not a local extremum because immediately to the left, the function approaches1, and to the right, it's increasing. But within [1,2), x=1 is the left endpoint, and since the function increases, x=1 is a local minimum on that interval.For x ≥2, each interval [n, n+1), n ≥2: function f(x)=x² + (2n +1)x -n² -1. The derivative is f’(x)=2x +2n +1. Setting to zero:2x +2n +1=0→x=-(2n +1)/2. Since n ≥2, x is negative, which is outside [n, n+1). Therefore, no critical points in these intervals. The function is quadratic opening upwards, so increasing on [n, n+1) since the vertex is to the left. Therefore, on each interval [n, n+1), n ≥2, the function is increasing, so minima at x=n and maxima approaching x=n+1. Therefore, each x=n, n ≥2, is a local minimum.But let's check for n=2, interval [2,3). The derivative f’(x)=2x +5. At x=2, derivative=4 +5=9>0, so function is increasing. Therefore, local minimum at x=2, value f(2)=4 +10 -5=9, and approaches f(3^-)=9 +15 -5=19.But since the function is increasing, at x=2, it's a local minimum.Similarly, at x=3, the next interval's function is increasing, so x=3 is a local minimum, etc.Now, compiling all local extrema:For x < -1 (n ≤-2):Each integer x=n ≤-2 is a local maximum (as function jumps up and then decreases).In the interval [-1,0):Local maximum at x=-0.5 (critical point), and no local minima within the interval except at endpoints, but x=-1 is not a local minimum globally.In [0,1):Local maximum at x=0.5.In [1,2):No local extrema within the interval, but x=1 is a local minimum on the interval [1,2).For x ≥2:Each integer x=n ≥2 is a local minimum.Additionally, we should check if the critical points found are indeed within their intervals.In [-1,0), the critical point x=-0.5 is within the interval.In [0,1), the critical point x=0.5 is within the interval.In other intervals, there are no critical points.Therefore, the local extrema are:- Local maxima at each x=n ≤-2 (integers), x=-0.5, x=0.5.- Local minima at x=n ≥2 (integers), and possibly at x=1.But need to verify:For x=-0.5 in [-1,0), it's a local maximum.For x=0.5 in [0,1), it's a local maximum.At x=1, since the function jumps from1 to2, and then increases, x=1 is a local minimum.Similarly, at each x=n ≥2, it's a local minimum.At x=-2, -3, etc., they are local maxima.But also, at x=0, what's the behavior? Moving from [-1,0) to [0,1), the function jumps from0 to1. So x=0 is not a local extremum because the left limit is0, the function at x=0 is1, and then it increases to1.25. So x=0 is a local minimum within [0,1), but globally, since the left limit is lower, it's not a local minimum.Similarly, at x=1, the function jumps from1 to2, then increases, so x=1 is a local minimum within [1,2), but globally, since the left limit is1, which is lower than the right side, it's not a local minimum.But the problem statement says "local extrema". Local extrema are defined with respect to the entire domain. Therefore, to determine if these points are local extrema, we need to check if there's a neighborhood around the point where the function value is maximum or minimum.For example, x=-0.5 in [-1,0). In a small neighborhood around x=-0.5 within the domain, f(x) has a maximum there, so it's a local maximum.Similarly, x=0.5 in [0,1) is a local maximum.For x=n ≤-2, in a small neighborhood around x=n, the function jumps up at x=n and then decreases. Therefore, x=n is a local maximum.For x=n ≥2, in a small neighborhood around x=n, the function is increasing to the right, so x=n is a local minimum.At x=1, in a small neighborhood around x=1, the function jumps up from1 to2 and then increases. Therefore, x=1 is not a local minimum because the left limit is lower. However, if we only consider the right neighborhood, x=1 is a local minimum. But since the left neighborhood is in a different interval, strictly speaking, x=1 is not a local minimum.Similarly, at x=-1, the function jumps from -1 to0 and then increases. Therefore, x=-1 is not a local minimum or maximum.Therefore, the local extrema are:- Local maxima at x=-0.5, x=0.5, and at each integer x=n ≤-2.- Local minima at each integer x=n ≥2.Additionally, need to check if there are any other local extrema in the intervals.In the interval [-2,-1), the function is decreasing, so no local extrema within the interval.In the interval [1,2), the function is increasing, so no local extrema within.In the intervals [n, n+1) for n ≤-3 and n ≥3, similarly, the function is decreasing or increasing without critical points.Therefore, the complete list of local extrema:Local maxima:- x=-0.5 (value 0.25)- x=0.5 (value 1.25)- x=n for each integer n ≤-2 (value f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1)Local minima:- x=n for each integer n ≥2 (value f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1)Wait, compute f(n) for n ≤-2:f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.Similarly, for n ≥2:f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.So for example, at n=-2:f(-2)= (-2)^2 + (-2) -1=4 -2 -1=1.Wait, but earlier, in the interval [-2,-1), the function at x=-2 is5. Wait, inconsistency here. What's wrong?Wait, earlier, in the interval [-2,-1), f(x)=x² -3x -5. At x=-2, f(-2)=4 +6 -5=5. But according to the general formula for f(n)=n² +n -1, plugging n=-2:4 + (-2) -1=1. This discrepancy suggests an error.Wait, no. The expression for f(x) in interval [n, n+1) is x² + (2n +1)x -n² -1. So at x=n, f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1. But in the specific case of n=-2, we had f(x)=x² -3x -5. At x=-2, f(-2)=4 +6 -5=5. But according to n=-2 in the general formula: (-2)^2 + (-2) -1=4 -2 -1=1. This contradicts. Therefore, there must be a mistake in the general formula.Wait, let's re-examine the earlier steps.For x ∈ [n, n+1), n ≤-2:f(x)=x² + (2n +1)x -n² -1.At x=n, f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But in the specific case of n=-2, f(x)=x² -3x -5. At x=-2:(-2)^2 -3*(-2) -5=4 +6 -5=5.But according to n² +n -1: (-2)^2 + (-2) -1=4 -2 -1=1. Not matching. Therefore, there's an error in the general formula.Wait, where did that come from?Original derivation:For x ∈ [n, n+1), n ≤-2:f(x)=x² - {x}² +x + |x² -1|Since {x}=x -n, {x}²=(x -n)^2=x² -2nx +n².Therefore:f(x)=x² - (x² -2nx +n²) +x + |x² -1|Simplify:x² -x² +2nx -n² +x + |x² -1| = (2n +1)x -n² + |x² -1|But for x ∈ [n, n+1), n ≤-2, x < -1, so |x² -1|=x² -1.Thus, f(x)= (2n +1)x -n² +x² -1 =x² + (2n +1)x -n² -1.But when we compute f(n), substituting x=n:n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But in the specific case n=-2:f(x)=x² -3x -5.At x=-2: 4 +6 -5=5.But according to n² +n -1: 4 -2 -1=1. This discrepancy suggests a mistake in the general formula.Ah! Wait, the |x² -1| term at x=n. For n ≤-2, x=n is in [n, n+1), and since n ≤-2, x=n ≤-2 < -1, so |x² -1|=x² -1. So substituting x=n into f(x):f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But in the specific case of n=-2, substituting x=-2 into f(x)=x² -3x -5:f(-2)=4 +6 -5=5.But according to the general formula, f(-2)= (-2)^2 + (-2) -1=4 -2 -1=1. This is inconsistent. Therefore, the error must be in the derivation.Wait, let's re-express the function for n=-2.For x ∈ [-2,-1), n=-2, {x}=x - (-2)=x +2. So {x}²=(x +2)^2.Therefore, f(x)=x² - (x +2)^2 +x + |x² -1|.Expand (x +2)^2= x² +4x +4.So f(x)=x² -x² -4x -4 +x + |x² -1|= -3x -4 + |x² -1|.Since x ∈ [-2,-1), |x² -1|=x² -1.Therefore, f(x)= -3x -4 +x² -1= x² -3x -5. At x=-2:(-2)^2 -3*(-2) -5=4 +6 -5=5. Correct.But according to the general formula derived earlier: f(x)=x² + (2n +1)x -n² -1 for x ∈ [n, n+1), n ≤-2. For n=-2:x² + (2*(-2)+1)x - (-2)^2 -1= x² + (-4 +1)x -4 -1= x² -3x -5. Correct. So f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But for n=-2: n² +n -1=4 + (-2) -1=1. However, the actual f(n)=5. Therefore, the general formula for f(n) is incorrect. This suggests that the general formula is derived under the assumption that x is in [n, n+1), but when x=n, the |x² -1| term must be computed correctly.Wait, for x=n, which is in [n, n+1), and n ≤-2, so x=n ≤-2 < -1, so |x² -1|=x² -1. Therefore, f(n)=x² + (2n +1)x -n² -1 at x=n.Plugging x=n:f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But in reality, when we compute f(n) for n=-2, we get 5, which is different from n² +n -1=1.Therefore, there's a mistake here. The error is that when x=n, the term |x² -1|=x² -1, which is correct, but when substituting into the expression f(x)=x² + (2n +1)x -n² -1, this is accurate.Wait, but for n=-2:f(n)= (-2)^2 + (2*(-2)+1)*(-2) - (-2)^2 -1=4 + (-3)*(-2) -4 -1=4 +6 -4 -1=5. Yes, that works. So the general formula is correct. The mistake was in the manual calculation earlier. So f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.For n=-2: 4 + (-2) -1=1, but using the expanded form:2n² +n -n² -1=2*4 + (-2) -4 -1=8 -2 -4 -1=1. Wait, no, this is incorrect. Wait, 2n² +n -n² -1=2n² -n² +n -1=n² +n -1.But when n=-2: n² +n -1=4 + (-2) -1=1, but actual f(-2)=5. So why the discrepancy?Wait, the function in the interval [n, n+1) is f(x)=x² + (2n +1)x -n² -1.So at x=n, f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But in the specific case when n=-2, substituting into this formula: (-2)^2 + (-2) -1=4 -2 -1=1, but in reality, f(-2)=5. Therefore, there's an inconsistency. This suggests that the general formula is incorrect.Wait, let's re-derive f(x) for x ∈ [n, n+1), n ≤-2:f(x)=x² - {x}^2 +x + |x² -1|.Since {x}=x -n, {x}^2=(x -n)^2.Thus, f(x)=x² - (x -n)^2 +x + |x² -1|.Expanding (x -n)^2:x² -2nx +n².Therefore:f(x)=x² -x² +2nx -n² +x + |x² -1|= (2n +1)x -n² + |x² -1|.For x ∈ [n, n+1), n ≤-2, |x² -1|=x² -1.Therefore, f(x)= (2n +1)x -n² +x² -1= x² + (2n +1)x -n² -1.So at x=n:f(n)=n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But for n=-2:f(n)= (-2)^2 + (2*(-2) +1)*(-2) - (-2)^2 -1=4 + (-3)*(-2) -4 -1=4 +6 -4 -1=5. Which matches the specific calculation.So the general formula gives f(n)=n² +n -1, but when n=-2:n² +n -1=4 + (-2) -1=1, but actual calculation gives5. Therefore, the general formula is wrong.Wait, no. Wait, confusion between variables. When we write f(n)=n² + (2n +1)n -n² -1, it's correctly evaluated as:n² + (2n +1)n -n² -1= n² +2n² +n -n² -1=2n² +n -1.But when n=-2:2*(-2)^2 + (-2) -1=2*4 + (-2) -1=8 -2 -1=5. Correct.Therefore, the mistake was in simplifying 2n² +n -n² -1 to n² +n -1. Which is incorrect.Wait, original expression:n² + (2n +1)n -n² -1= n² +2n² +n -n² -1=2n² +n -1.Ah! So the correct general formula is f(n)=2n² +n -1.Earlier, incorrect simplification:n² + (2n +1)n -n² -1=2n² +n -n² -1=n² +n -1.But that's wrong. Correct is:n² + (2n +1)n -n² -1= n² +2n² +n -n² -1= (n² +2n² -n²) +n -1=2n² +n -1.Therefore, the correct general formula for f(n) where x=n is 2n² +n -1.Therefore, previous mistake was in simplifying.Therefore, for n=-2:2*(-2)^2 + (-2) -1=8 -2 -1=5, which matches.Similarly, for n=-3:2*(-3)^2 + (-3) -1=18 -3 -1=14, which matches f(-3)=14 as calculated earlier.Therefore, the general formula for f(n) is 2n² +n -1.Therefore, the local maxima at x=n ≤-2 have values f(n)=2n² +n -1.Similarly, for local minima at x=n ≥2, the value is f(n)=2n² +n -1.Therefore, correcting the earlier mistake, the local maxima at x=n ≤-2 are indeed f(n)=2n² +n -1, and local minima at x=n ≥2 are also f(n)=2n² +n -1.Therefore, compiling all local extrema:Local maxima:- x=-0.5 (value 0.25)- x=0.5 (value 1.25)- x=n for each integer n ≤-2 (value 2n² +n -1)Local minima:- x=n for each integer n ≥2 (value 2n² +n -1)Additionally, checking x=1: since in the interval [1,2), the function is increasing, and the value at x=1 is2. The previous interval [0,1) ends at f(1^-)=1. Therefore, there's a jump from1 to2. Hence, x=1 is not a local minimum globally but could be considered a local minimum on the interval [1,2). However, since the function is increasing, the minimum on the interval is at x=1, but in the global sense, since the left limit is lower, it's not a local minimum.Therefore, the final list of local extrema:Local maxima:- x=-0.5: f(-0.5)=0.25- x=0.5: f(0.5)=1.25- x=n for integers n ≤-2: f(n)=2n² +n -1Local minima:- x=n for integers n ≥2: f(n)=2n² +n -1This seems consistent.To summarize:Zeros: x=-1 and x=(3 -sqrt(29))/2≈-1.1925.Local maxima:- At x=-0.5, value 0.25- At x=0.5, value 1.25- At each integer n ≤-2, value 2n² +n -1Local minima:- At each integer n ≥2, value 2n² +n -1Thus, these are the zeros and local extrema of the function.

✅:Zeros and Local Extrema of the Function ( y = f(x) = x^2 - {x}^2 + x + |x^2 - 1| ) Zeros1. ( x = -1 ) - Located in the interval ([-1, 0)), where ( f(x) = -x^2 - x ). - Direct substitution: ( f(-1) = -(-1)^2 - (-1) = -1 + 1 = 0 ).2. ( x = frac{3 - sqrt{29}}{2} approx -1.192 ) - Located in the interval ([-2, -1)), where ( f(x) = x^2 - 3x - 5 ). - Root of ( x^2 - 3x - 5 = 0 ): ( x = frac{3 pm sqrt{29}}{2} ). - The valid root in ([-2, -1)) is ( x = frac{3 - sqrt{29}}{2} ). Local Extrema# Local Maxima1. ( x = -0.5 ) - In ([-1, 0)), ( f(x) = -x^2 - x ). - Critical point: ( x = -0.5 ). - Value: ( f(-0.5) = -(-0.5)^2 - (-0.5) = 0.25 ).2. ( x = 0.5 ) - In ([0, 1)), ( f(x) = -x^2 + x + 1 ). - Critical point: ( x = 0.5 ). - Value: ( f(0.5) = -(0.5)^2 + 0.5 + 1 = 1.25 ).3. Integers ( n leq -2 ) - At ( x = n ), value: ( f(n) = 2n^2 + n - 1 ). - Example: For ( n = -2 ), ( f(-2) = 5 ).# Local Minima1. Integers ( n geq 2 ) - At ( x = n ), value: ( f(n) = 2n^2 + n - 1 ). - Example: For ( n = 2 ), ( f(2) = 9 ).---Final Answer- Zeros: ( boxed{x = -1} ) and ( boxed{x = frac{3 - sqrt{29}}{2}} )- Local Maxima: ( boxed{x = -0.5} ), ( boxed{x = 0.5} ), and integers ( boxed{n leq -2} )- Local Minima: Integers ( boxed{n geq 2} )Values at Local Extrema:- For ( n leq -2 ): ( f(n) = 2n^2 + n - 1 )- For ( n geq 2 ): ( f(n) = 2n^2 + n - 1 )- At ( x = -0.5 ): ( f(-0.5) = 0.25 )- At ( x = 0.5 ): ( f(0.5) = 1.25 )

🔑:Given the function:[ y = f(x) = x^2 - {x}^2 + x + left| x^2 - 1 right| ]where {x} denotes the fractional part of x.# Step 1: Determining the expression of ( f(x) )When k leq x < k+1 (where k is an integer), the fractional part ({x} = x - k). Thus:[f(x) = x^2 - (x - k)^2 + x + left| x^2 - 1 right|]Expanding the square:[(x - k)^2 = x^2 - 2kx + k^2]So,[f(x) = x^2 - (x^2 - 2kx + k^2) + x + left| x^2 - 1 right| = x^2 - x^2 + 2kx - k^2 + x + left| x^2 - 1 right|]which simplifies to:[f(x) = 2kx + x - k^2 + left| x^2 - 1 right|]or:[f(x) = (2k+1)x - k^2 + left| x^2 - 1 right|]Since |x^2 - 1| can be x^2 - 1 or -x^2 + 1:- For ( x^2 geq 1 ): (left| x^2 - 1 right| = x^2 - 1)- For ( x^2 < 1 ): (left| x^2 - 1 right| = -x^2 + 1)# Step 2: Evaluating at (k) and checking continuityAt ( x = k ):- The substitution value of ( f(x) ) is (left| k^2 - 1 right| + k^2 + k).- The left-hand limit is (left| k^2 - 1 right| + k^2 + k - 1).Thus, ( f(x) ) jumps up by 1 at each integer ( k ).# Step 3: Solving for specific intervalsa) ( k geq 1 ): ( x geq 1 )For ( k geq 1 ):[f(x) = left| x^2 - 1 right| + (2k+1)x - k^2]Since ( x geq 1 ):[|x^2 - 1| = x^2 - 1]Therefore,[f(x) = x^2 - 1 + (2k+1)x - k^2]Completing the square for the quadratic term:[(2k+1)x = (x + k + frac{1}{2})^2 - k^2 - frac{1}{4}]Hence,[f(x) = (x + k + frac{1}{2})^2 - k^2 - frac{5}{4}]Given ( x + k + frac{1}{2} > 0 ), ( f(x) ) is strictly increasing in ([k, k+1)). Combining the jumps at integer points, ( f(x) ) is strictly increasing for ( x geq 1 ). Since ( f(1) = 2 ), there are no zeros for ( x geq 1 ).b) ( k = 0 ): ( 0 leq x < 1 )[f(x) = left| x^2 - 1 right| + x]Since ( x^2 < 1 ):[|x^2 - 1| = -x^2 + 1]Therefore,[f(x) = 0 + x - x^2 + 1 = 1 - (x - 0.5)^2 + 1 = frac{5}{4} - (x - 0.5)^2]At ( x = 0.5 ),[f(0.5) = frac{5}{4}]This is a local maximum point with no zero in the interval since:[|x - 0.5| leq 0.5 implies f(x) geq 1]c) ( k = -1 ): ( -1 leq x < 0 )[f(x) = left| x^2 - 1 right| + (2k + 1)x - k^2]For ( -1 leq x < 0 ):[|x^2 - 1| = -x^2 + 1]Hence,[f(x) = 1 - x^2 + (2(-1) + 1)x - (-1)^2 = 1 - x^2 - x - 1 = frac{1}{4} - (x + frac{1}{2})^2]Maximum at ( x = -0.5 ), zero at ( x = -1 ):[x_{1}^{*} = -1]d) ( k < -1 ): ( x < -1 )For ( k leq x < k+1 ):[f(x) = (2k+1)x - k^2 + left| x^2 - 1 right|]Since ( x < -1 ):[|x^2 - 1| = x^2 - 1]Thus,[f(x) = x^2 - 1 + (2k + 1)x - k^2]Completing the square:[f(x) = (x - k - frac{1}{2})^2 - 2k^2 - k - frac{5}{4}]Since ( x < k + frac{1}{2} ):[f(x) text{ strictly decreasing}]Consider zeros in periods spanning several intervals:For ( k = -2 ):[(x + frac{3}{2})^2 - frac{29}{4}=0 rightarrow x = frac{3 - sqrt{29}}{2} = -1.193] Conclusion:The identified zeroes and local extremum points are as follows:Local maximum:- x_0 = frac{1}{2}- x_1 = -frac{1}{2}- x_k = -k (k = 2, 3, ldots) Zero:- x = -1- x = -1.193Therefore: [boxed{begin{aligned}&text{Local Maximum Points:} &frac{1}{2}, -frac{1}{2}, -k, (k = 2, 3, ldots) &text{Local Minimum Points:} &text{none} &text{Zero Points:} &-1, -1.193end{aligned}}]