🧠:[<think>Okay, let me try to figure out this problem. So, we have 1000 lamps and 1000 switches. Each switch controls all lamps that are multiples of its own number. Initially, all lamps are on. Then we pull the switches numbered 2, 3, and 5. We need to find out how many lamps are still on after that.First, let me make sure I understand the setup correctly. Each switch corresponds to a number from 1 to 1000. Switch number n controls all lamps that are multiples of n. So, for example, switch 2 controls lamps 2, 4, 6, 8, ..., up to 1000. Similarly, switch 3 controls lamps 3, 6, 9, 12, ..., and so on.Initially, all lamps are on. Then we pull switches 2, 3, and 5. Pulling a switch toggles the state of the lamps it controls. So if a lamp is on, pulling the switch turns it off, and vice versa. Since all lamps start on, after pulling switch 2, all even-numbered lamps (multiples of 2) will be turned off. Then pulling switch 3 will toggle the state of all multiples of 3, turning them off if they were on, or on if they were off. Similarly, pulling switch 5 will toggle all multiples of 5.But wait, the problem says "pull the 2, 3, 5 switches". Does that mean pulling each of them once? So each switch is pulled once, and each pull affects all the lamps it controls. So the order might not matter because toggling is commutative? Hmm, maybe. Let me think.Each lamp's final state depends on how many times it was toggled. Since we start with all lamps on, if a lamp is toggled an even number of times, it will end up on (since each toggle switches it twice: off then on, or on then off). Wait, no. If a lamp is toggled once, it turns off. If toggled again, it turns back on. So the number of times it's toggled determines its final state: even number of toggles means it's on, odd number means it's off.So the key is to find for each lamp, how many switches that affect it have been pulled. The switches pulled are 2, 3, 5. So a lamp number L will be toggled once for each switch s in {2, 3, 5} such that s divides L. So, for each L from 1 to 1000, count the number of divisors among 2, 3, 5. If the count is even, the lamp remains on; if odd, it's off.Therefore, we need to count the number of lamps L where the number of divisors from {2, 3, 5} is even. Wait, no. Wait, no. Wait, if a lamp is toggled an odd number of times, it's off. If even, it's on. But initially, all lamps are on. Each toggle inverts the state. So the number of toggles is equal to the number of switches that affect it (i.e., divisors in {2, 3, 5}). So the final state is on if the number of such divisors is even, and off if odd.But we need the number of lamps that are still on, which is the number of L in 1..1000 where the number of divisors in {2, 3, 5} is even. So we need to compute the size of the set { L | L ∈ [1,1000], the number of s ∈ {2,3,5} that divide L is even }.Alternatively, we can model this using inclusion-exclusion principles. Let's see.First, note that each time we pull a switch, all multiples of that number are toggled. So pulling switch 2 toggles all even lamps, pulling switch 3 toggles multiples of 3, and pulling 5 toggles multiples of 5. So each lamp L is toggled once for each s ∈ {2,3,5} that divides L.Therefore, the total number of toggles for lamp L is equal to the number of elements in {2,3,5} that divide L. Let's denote this number as t(L). Then, the lamp L is on if t(L) is even, and off if t(L) is odd.Therefore, we need to count the number of L in 1..1000 such that t(L) is even.Alternatively, since t(L) is the number of divisors among 2,3,5, perhaps we can model this using inclusion-exclusion or Möbius inversion.But maybe a better approach is to consider the possible subsets of {2,3,5} that divide L.Each lamp can be associated with the set S = { s ∈ {2,3,5} | s divides L }.Then, the size of S is t(L), and we need |S| even.So, we need the number of L where |S| is even. That is, the number of L not divisible by any of 2,3,5, or divisible by exactly two of them.Wait, no. Wait, the subsets with even size: size 0, 2. So, lamps that are not divisible by 2, 3, or 5 (size 0), or divisible by exactly two of them (size 2). The ones divisible by all three would have size 3 (odd), so they are off, and ones divisible by exactly one would have size 1 (odd), so also off.Therefore, the lamps that remain on are those not divisible by 2,3,5 or divisible by exactly two of them.Therefore, we need to compute:Number of lamps not divisible by 2,3,5: this is the count of numbers from 1 to 1000 not divisible by 2, 3, or 5.Plus the number of lamps divisible by exactly two of the numbers 2,3,5.So, total_lamps_on = (numbers not divisible by 2,3,5) + (numbers divisible by exactly two of 2,3,5).Alternatively, perhaps using inclusion-exclusion:First, compute the numbers not divisible by 2,3,5. That is, total numbers minus numbers divisible by 2,3, or 5.But we need to add the numbers divisible by exactly two of them. Wait, but let's see:Wait, let's consider each case.Let me denote:A: numbers divisible by 2B: numbers divisible by 3C: numbers divisible by 5We need | (A ∪ B ∪ C)^c | + | (A∩B C) ∪ (A∩C B) ∪ (B∩C A) |.That is, the numbers not in A, B, or C, plus the numbers in exactly two of A, B, C.So total_lamps_on = |not (A ∪ B ∪ C)| + |exactly two of A,B,C|.Alternatively, we can compute this as:Total numbers - |A ∪ B ∪ C| + |exactly two of A,B,C|. Wait, no. Wait:Wait, |not (A ∪ B ∪ C)| = total - |A ∪ B ∪ C|and |exactly two of A,B,C| = |A∩B| + |A∩C| + |B∩C| - 3|A∩B∩C|.But perhaps it's easier to compute directly.Alternatively, using the principle that the number of elements with even number of properties is (Total + inclusion-exclusion sum)/2.Wait, there is a formula in combinatorics where if you have properties and you want to count the number of elements that satisfy an even number of properties, it's (Total + weighted inclusion-exclusion)/2. Let me recall.Yes, for each element, let f(L) be the number of properties it satisfies. Then sum_{L} (-1)^{f(L)} = sum_{k} (-1)^k * N_k, where N_k is the number of elements with exactly k properties. But if we want the number of elements with even k, it's (Total + sum_{L} (-1)^{f(L)}) / 2.Alternatively, from generating functions, the number of elements with even number of properties is (Total + sum_{S} (-1)^{|S|} |A_S| ) / 2, where A_S is the intersection over S.Wait, maybe that's a stretch, but let me check.Suppose we have three sets A, B, C. The number of elements in exactly even number of sets is:|A| + |B| + |C| - 2|A∩B| - 2|A∩C| - 2|B∩C| + 4|A∩B∩C| ?Wait, perhaps not. Let's consider the standard inclusion-exclusion.Alternatively, the number of elements in exactly two sets is |A∩B| + |A∩C| + |B∩C| - 3|A∩B∩C|.And the number of elements in exactly zero sets is Total - |A ∪ B ∪ C|.Therefore, total_lamps_on = (Total - |A ∪ B ∪ C|) + (|A∩B| + |A∩C| + |B∩C| - 3|A∩B∩C|).So let me compute each term:First, Total = 1000.Compute |A|: numbers divisible by 2: floor(1000/2) = 500|B|: floor(1000/3) = 333|C|: floor(1000/5) = 200|A ∩ B|: numbers divisible by LCM(2,3)=6: floor(1000/6) ≈ 166|A ∩ C|: divisible by LCM(2,5)=10: floor(1000/10)=100|B ∩ C|: divisible by LCM(3,5)=15: floor(1000/15)≈66|A ∩ B ∩ C|: divisible by LCM(2,3,5)=30: floor(1000/30)=33Now, |A ∪ B ∪ C| = |A| + |B| + |C| - |A∩B| - |A∩C| - |B∩C| + |A∩B∩C|Plugging numbers:500 + 333 + 200 - 166 - 100 - 66 + 33Compute step by step:500 + 333 = 833833 + 200 = 10331033 - 166 = 867867 - 100 = 767767 - 66 = 701701 + 33 = 734So |A ∪ B ∪ C| = 734Therefore, numbers not in A, B, C: Total - 734 = 1000 - 734 = 266Now, numbers in exactly two sets:|A∩B| + |A∩C| + |B∩C| - 3|A∩B∩C| = 166 + 100 + 66 - 3*33Calculate:166 + 100 = 266266 + 66 = 3323*33 = 99332 - 99 = 233Therefore, numbers in exactly two sets: 233Hence, total_lamps_on = 266 + 233 = 499Wait, 266 + 233 is 499? Let me check:266 + 200 = 466466 + 33 = 499. Yes.So according to this calculation, the number of lamps still on is 499.But wait, let me verify this with another approach to be sure.Alternative approach: For each lamp L from 1 to 1000, determine how many of 2,3,5 divide L. If even number, count it.Alternatively, we can model this using the principle of inclusion-exclusion for even numbers.Alternatively, consider that each lamp's toggle count is t(L) = number of switches in {2,3,5} that divide L.We can compute the number of L with t(L) even.This is equal to (Total + sum_{L=1}^{1000} (-1)^{t(L)} ) / 2Because for each L, (-1)^{t(L)} is 1 if t(L) is even, -1 if odd. So sum_{L} (-1)^{t(L)} = (number of even t(L) - number of odd t(L)) = 2*number_even - TotalHence, number_even = (Total + sum_L (-1)^{t(L)} ) / 2So compute sum_L (-1)^{t(L)}.But t(L) is the number of primes dividing L among {2,3,5}. Wait, but actually, t(L) counts the number of elements in {2,3,5} that divide L, which is similar to the number of distinct prime factors, except 2,3,5 are primes, but L could be divisible by higher powers. However, t(L) is just the count of how many of 2,3,5 divide L, regardless of multiplicity.So for example, L=6 is divisible by 2 and 3, so t(L)=2. L=12 is divisible by 2 and 3, t(L)=2. L=30 is divisible by 2,3,5, t(L)=3.Therefore, (-1)^{t(L)} depends only on whether L is divisible by each of 2,3,5.This is similar to the multiplicative function over square-free divisors.In fact, sum_{L=1}^{N} (-1)^{t(L)} is equal to sum_{L=1}^{N} μ(L) where μ is the Möbius function? Wait, not exactly. The Möbius function is (-1)^k where k is the number of prime factors, but only for square-free numbers. Here, t(L) is the number of primes in {2,3,5} that divide L, regardless of their exponents.But perhaps we can use the principle similar to generating functions.Let me think of generating function for the sum.Each number L can be associated with the exponents of 2,3,5 in its prime factorization, but since we are only concerned with divisibility by 2,3,5, we can model each number as either being divisible or not by each of 2,3,5.So, for each number L, let’s define three independent properties: whether it's divisible by 2, by 3, or by 5. Then t(L) is the count of these properties that hold for L. Then, sum_{L=1}^{1000} (-1)^{t(L)} = sum_{L=1}^{1000} product_{s ∈ {2,3,5}} (-1)^{1_{s divides L}}}.This is equivalent to product_{s ∈ {2,3,5}} (1 + (-1)^{1_{s divides L}}). Wait, no.Wait, actually, the sum over L of (-1)^{t(L)} can be expressed as the product over s ∈ {2,3,5} of (1 + (-1) * floor(N/s))). Wait, maybe not.Alternatively, the sum can be calculated using inclusion-exclusion. Let's see.Let’s note that (-1)^{t(L)} is equal to (-1)^{number of s ∈ {2,3,5} dividing L} = product_{s | L, s ∈ {2,3,5}} (-1).So, for each L, the term is the product over s ∈ {2,3,5} of (-1)^{1_{s divides L}}}.Therefore, sum_{L=1}^{1000} (-1)^{t(L)} = sum_{L=1}^{1000} product_{s ∈ {2,3,5}} (-1)^{1_{s divides L}}}.This can be expanded as:sum_{L=1}^{1000} [1 - 2*1_{2 divides L}] * [1 - 2*1_{3 divides L}] * [1 - 2*1_{5 divides L}]Wait, because (-1)^{1_{s divides L}} = 1 - 2*1_{s divides L} since if s divides L, it's -1, otherwise 1.But perhaps expanding this product would give a way to compute the sum.Alternatively, since we have multiplicative functions, maybe we can use the formula:sum_{L=1}^N (-1)^{t(L)} = sum_{d | 2*3*5} μ(d) * floor(N/d)}.Wait, not sure. Let's think.Alternatively, using the principle that:sum_{L=1}^N (-1)^{t(L)} = sum_{L=1}^N product_{s ∈ {2,3,5}} (-1)^{1_{s divides L}} = product_{s ∈ {2,3,5}} (1 + (-1) * floor(N/s)) + ... )Wait, perhaps it's the product over s ∈ {2,3,5} of (1 - 2*floor(N/s)). Wait, that might not be correct.Wait, actually, when we have multiplicative functions over inclusion of divisors, the generating function is multiplicative. So for each prime, we can model the contribution. However, since we're dealing with fixed primes 2,3,5, and considering divisibility by them, perhaps we can model each number as a combination of being divisible or not by each of 2,3,5.So, the total sum can be calculated as follows:Each number can be in one of 8 categories (divisible by none, 2, 3, 5, 2&3, 2&5, 3&5, 2&3&5). For each category, we can compute (-1)^{t(L)}, where t(L) is the number of divisors from {2,3,5} that divide L. Then multiply by the number of elements in each category.But this is exactly what we did earlier with inclusion-exclusion. Let's see:The sum would be:Number not divisible by 2,3,5: 266 terms, each contributing (-1)^0 = 1. So total 266.Numbers divisible by exactly one of 2,3,5: each contributes (-1)^1 = -1.Numbers divisible by exactly two: contribute (-1)^2 = 1.Numbers divisible by all three: contribute (-1)^3 = -1.So sum = (266 * 1) + (N1 * (-1)) + (233 * 1) + (N3 * (-1)) where N1 is numbers divisible by exactly one, N3 is numbers divisible by all three.But from earlier:Numbers not divisible by any: 266Numbers divisible by exactly two: 233Numbers divisible by all three: |A ∩ B ∩ C| = 33Therefore, numbers divisible by exactly one: Total - (none + exactly two + all three) = 1000 - (266 + 233 + 33) = 1000 - 532 = 468So sum = 266*1 + 468*(-1) + 233*1 + 33*(-1) = 266 - 468 + 233 - 33Calculate step by step:266 - 468 = -202-202 + 233 = 3131 - 33 = -2Therefore, sum_{L=1}^{1000} (-1)^{t(L)} = -2Then, number of lamps on is (Total + sum) / 2 = (1000 + (-2)) / 2 = 998 / 2 = 499So that confirms the previous result of 499.Therefore, the number of lamps still on is 499.But just to be thorough, let me check with a small example.Suppose N=10, switches 2,3,5.Compute manually.Lamps 1-10.Initially all on.After pulling 2,3,5.Let's see:Lamp 1: not divisible by 2,3,5. Toggles: 0. Stays on.Lamp 2: divisible by 2. Toggles:1. Off.Lamp 3: divisible by 3. Toggles:1. Off.Lamp 4: divisible by 2. Toggles:1. Off.Lamp 5: divisible by 5. Toggles:1. Off.Lamp 6: divisible by 2 and 3. Toggles:2. On.Lamp 7: none. On.Lamp 8: divisible by 2. Toggles:1. Off.Lamp 9: divisible by 3. Toggles:1. Off.Lamp 10: divisible by 2 and 5. Toggles:2. On.So lamps on: 1,6,7,10. Total 4.Compute via formula:Total=10|A|=5 (div by 2: 2,4,6,8,10)|B|=3 (3,6,9)|C|=2 (5,10)|A∩B|=1 (6)|A∩C|=1 (10)|B∩C|=0|A∩B∩C|=0|A ∪ B ∪ C|=5+3+2 -1 -1 -0 +0=8Numbers not in A,B,C:10 -8=2 (1,7)Numbers exactly two: |A∩B| + |A∩C| + |B∩C| - 3|A∩B∩C| =1+1+0 -0=2 (6,10)Total lamps on:2+2=4. Which matches manual count.So formula works here.Similarly, applying the other formula:sum_{L=1}^{10} (-1)^{t(L)} = sum over L=1 to 10:L=1: (-1)^0=1L=2: (-1)^1=-1L=3: (-1)^1=-1L=4: (-1)^1=-1L=5: (-1)^1=-1L=6: (-1)^2=1L=7: (-1)^0=1L=8: (-1)^1=-1L=9: (-1)^1=-1L=10: (-1)^2=1Sum: 1 -1 -1 -1 -1 +1 +1 -1 -1 +1 = (1 +1 +1 +1) - (1 +1 +1 +1 +1) ) Wait:Wait, compute step by step:1 (L1)-1 (L2) total 0-1 (L3) total -1-1 (L4) total -2-1 (L5) total -3+1 (L6) total -2+1 (L7) total -1-1 (L8) total -2-1 (L9) total -3+1 (L10) total -2Sum is -2Then number of lamps on is (10 + (-2))/2=8/2=4. Which matches.Therefore, the formula works.Thus, applying it to N=1000 gives 499.So, after pulling switches 2, 3, and 5, the number of lamps still on is 499.Final Answerboxed{499}</think>To determine the number of lamps that remain on after pulling switches 2, 3, and 5, we need to consider how many times each lamp is toggled. A lamp will be on if it is toggled an even number of times and off if toggled an odd number of times. We use the principle of inclusion-exclusion to calculate the number of lamps that are not divisible by 2, 3, or 5 (toggled 0 times) and the number of lamps divisible by exactly two of 2, 3, or 5 (toggled 2 times, which is even).1. Calculate the number of lamps divisible by 2, 3, or 5: - (|A|) (divisible by 2): ( leftlfloor frac{1000}{2} rightrfloor = 500 ) - (|B|) (divisible by 3): ( leftlfloor frac{1000}{3} rightrfloor = 333 ) - (|C|) (divisible by 5): ( leftlfloor frac{1000}{5} rightrfloor = 200 ) - (|A cap B|) (divisible by 6): ( leftlfloor frac{1000}{6} rightrfloor = 166 ) - (|A cap C|) (divisible by 10): ( leftlfloor frac{1000}{10} rightrfloor = 100 ) - (|B cap C|) (divisible by 15): ( leftlfloor frac{1000}{15} rightrfloor = 66 ) - (|A cap B cap C|) (divisible by 30): ( leftlfloor frac{1000}{30} rightrfloor = 33 )2. Calculate the number of lamps in the union of sets A, B, and C: [ |A cup B cup C| = 500 + 333 + 200 - 166 - 100 - 66 + 33 = 734 ]3. Calculate the number of lamps not divisible by 2, 3, or 5: [ 1000 - 734 = 266 ]4. Calculate the number of lamps divisible by exactly two of 2, 3, or 5: [ (166 + 100 + 66 - 3 times 33) = 233 ]5. Total number of lamps still on: [ 266 + 233 = 499 ]Thus, the number of lamps that are still on is (boxed{499}).,<think>Okay, let me try to figure out this problem. So, there are 1000 lamps and 1000 switches. Each switch controls all the lamps that are multiples of its own number. Initially, all lamps are on. Then we pull the switches numbered 2, 3, and 5. We need to find out how many lamps are still on after pulling these switches.First, I need to understand how the switches work. Each switch corresponds to a number from 1 to 1000. When you pull a switch, it toggles all the lamps that are multiples of that switch's number. So, for example, switch 2 controls lamps 2, 4, 6, 8, etc. If a lamp is on, pulling its controlling switch will turn it off, and vice versa. But here, all lamps start on, and we only pull switches 2, 3, and 5. So, each time we pull a switch, it's flipping the state of the lamps it controls.Wait, so initially, all lamps are on. Then pulling switch 2 will turn off all even-numbered lamps (2, 4, 6, ..., 1000). Then pulling switch 3 will turn off all lamps that are multiples of 3 (3, 6, 9, ..., 999). But wait, hold on, actually, when you pull a switch, it toggles the lamps. So if a lamp was on, it turns off, and if it was off, it turns on. Since we start with all lamps on, the first switch pull (switch 2) will turn them off. Then pulling switch 3 will toggle the multiples of 3. But some of those might have been already turned off by switch 2. So, for example, lamp 6 is a multiple of both 2 and 3. So first, lamp 6 is turned off by switch 2, then when we pull switch 3, lamp 6 is toggled again, turning it back on. Then pulling switch 5 will toggle multiples of 5. So lamp 10, 15, 20, etc., will be toggled. But again, some of these might have been toggled multiple times.Therefore, the state of each lamp depends on how many times it was toggled. If it's toggled an even number of times, it remains on (since it started on). If it's toggled an odd number of times, it ends up off. Wait, actually, starting from on, each toggle switches the state. So even number of toggles would bring it back to on, and odd number to off. Therefore, the lamps that are still on are those that are toggled an even number of times (including zero times). But since we only pulled three switches: 2, 3, and 5, the only toggles that happened are for these three switches.Therefore, the number of times a lamp is toggled is equal to the number of switches that control it. For example, lamp number 1 is not a multiple of 2, 3, or 5, so it's never toggled and remains on. Lamp 2 is a multiple of 2, so it's toggled once (off). Lamp 6 is a multiple of 2 and 3, so it's toggled twice (off then on). Lamp 30 is a multiple of 2, 3, and 5, so it's toggled three times (off, on, off). So the key is to find all lamps that are not multiples of 2, 3, or 5, or are multiples of an even number of these switches.But actually, no. Wait, if a lamp is a multiple of two of the switches, say 2 and 3, it's toggled twice, which brings it back to on. If it's a multiple of all three, it's toggled three times, ending up off. So to summarize, the lamps that remain on are those that are not multiples of 2, 3, or 5, plus those that are multiples of exactly two of them (since two toggles cancel each other out), and subtracting those that are multiples of all three (since three toggles result in off). Hmm, actually, this seems like an inclusion-exclusion problem.Alternatively, using the principle of inclusion-exclusion, we can calculate the number of lamps turned off by any of the switches 2, 3, or 5, and subtract that from the total 1000. Wait, but since the lamps can be toggled multiple times, the inclusion-exclusion might be more complicated. Let me think.But actually, maybe not. Because the final state depends on the number of times a lamp is toggled. So, each lamp's final state is on if it's toggled even number of times, off if odd. So, for each lamp, we need to count the number of divisors it has among 2, 3, 5. Since each switch corresponds to a divisor. So a lamp number n will be toggled once for each divisor in {2,3,5} that divides n. Therefore, the number of toggles is equal to the number of elements in {2,3,5} that divide n.So if n is divisible by none of 2,3,5, it remains on (0 toggles, even). If n is divisible by exactly one of them, it's toggled once (off). If divisible by exactly two, toggled twice (on). If divisible by all three, toggled three times (off). Therefore, the lamps that remain on are those numbers from 1 to 1000 that are either not divisible by 2, 3, or 5, or divisible by exactly two of them.Therefore, to compute the number of lamps still on, we need to compute:Number of lamps not divisible by 2, 3, or 5 + Number of lamps divisible by exactly two of 2, 3, 5.Let me verify that logic. For example, take a number divisible by 2 and 3 but not 5. It will be toggled twice (once by 2, once by 3), so it's on. Similarly, a number divisible by 2 and 5 but not 3 is toggled twice, hence on. Similarly for 3 and 5. But a number divisible by all three is toggled three times, hence off. And numbers divisible by exactly one are toggled once, hence off. So yes, that seems right.Therefore, total lamps on = (numbers not divisible by 2,3,5) + (numbers divisible by exactly two of 2,3,5).So now, let's compute these two quantities.First, the number of lamps not divisible by 2,3, or 5. This is equivalent to 1000 minus the number of lamps divisible by 2,3, or 5. But since we need the count of numbers not divisible by any, we can use inclusion-exclusion.Alternatively, maybe directly compute it using inclusion-exclusion.The formula for numbers not divisible by 2, 3, or 5 is equal to:Total numbers - (numbers divisible by 2 + numbers divisible by 3 + numbers divisible by 5) + (numbers divisible by 2 and 3 + numbers divisible by 2 and 5 + numbers divisible by 3 and 5) - (numbers divisible by 2,3,5).Which is:1000 - (floor(1000/2) + floor(1000/3) + floor(1000/5)) + (floor(1000/6) + floor(1000/10) + floor(1000/15)) - floor(1000/30)Compute each term:floor(1000/2) = 500floor(1000/3) = 333 (since 3*333=999, 1000 - 999=1)floor(1000/5) = 200floor(1000/6) = 166 (since 6*166=996)floor(1000/10) = 100floor(1000/15) = 66 (15*66=990)floor(1000/30) = 33 (30*33=990)So plugging in:1000 - (500 + 333 + 200) + (166 + 100 + 66) - 33Compute step by step:First term: 1000Subtract: 500 + 333 + 200 = 1033So 1000 - 1033 = -33Then add: 166 + 100 + 66 = 332So -33 + 332 = 299Then subtract 33: 299 - 33 = 266Therefore, there are 266 numbers between 1 and 1000 not divisible by 2, 3, or 5.Now, the second part: numbers divisible by exactly two of 2, 3, 5.This can be calculated by taking numbers divisible by each pair of numbers (2&3, 2&5, 3&5) and subtracting those divisible by all three (since those would have been counted in all three pairs).So, for each pair:Numbers divisible by 2 and 3: floor(1000/6) = 166Numbers divisible by 2 and 5: floor(1000/10) = 100Numbers divisible by 3 and 5: floor(1000/15) = 66Sum these: 166 + 100 + 66 = 332Now subtract 3 times the numbers divisible by all three (since each of the three pairs includes the numbers divisible by all three). Wait, actually, each of the three pair counts includes numbers divisible by all three. So to get numbers divisible by exactly two, we need to subtract 3 times the numbers divisible by all three. Wait, no. Let me think.Wait, if a number is divisible by all three (2,3,5), it was counted in each of the three pair counts (2&3, 2&5, 3&5). Therefore, each such number is included three times in the total pair counts. But we want numbers divisible by exactly two, so we need to subtract 3 times the count of numbers divisible by all three, then add back 0 (since we're not dealing with higher overlaps). Wait, perhaps not. Let me check with inclusion-exclusion.Alternatively, numbers divisible by exactly two of 2,3,5 is equal to:(numbers divisible by 2 and 3) + (numbers divisible by 2 and 5) + (numbers divisible by 3 and 5) - 3*(numbers divisible by 2,3,5)Wait, because numbers divisible by all three are included in each of the three pair counts. So each of those numbers is subtracted three times. Wait, perhaps not. Wait, actually, if you have a number divisible by all three, it's counted in each pair. So in the sum of the three pairs, such numbers are counted three times. But we only want numbers divisible by exactly two, so we need to subtract twice the numbers divisible by all three. Wait, here's the formula:The count of numbers divisible by exactly two is equal to the sum of the counts of each pair minus 3 times the count of numbers divisible by all three. Wait, no.Wait, let's consider:Let’s denote:A = numbers divisible by 2 and 3B = numbers divisible by 2 and 5C = numbers divisible by 3 and 5Each of A, B, C includes numbers divisible by all three (2,3,5). Let’s denote D = numbers divisible by all three.Then the numbers divisible by exactly two are (A - D) + (B - D) + (C - D) = A + B + C - 3DTherefore, exactly two: A + B + C - 3DSo in our case:A = floor(1000/6) = 166B = floor(1000/10) = 100C = floor(1000/15) = 66D = floor(1000/30) = 33Therefore, exactly two = 166 + 100 + 66 - 3*33 = 332 - 99 = 233Therefore, the number of lamps divisible by exactly two of 2,3,5 is 233.Therefore, total lamps on = 266 (not divisible by any) + 233 (divisible by exactly two) = 266 + 233 = 499.Wait, but let me confirm this calculation.First, numbers not divisible by 2,3,5: 266Numbers divisible by exactly two: 233Total: 266 + 233 = 499But let's check with an alternative approach. Alternatively, since each lamp's state is on if it's toggled even number of times (0, 2 times). So, total lamps on = number of numbers not divisible by 2,3,5 (0 toggles) + numbers divisible by exactly two of them (2 toggles). Therefore, 266 + 233 = 499.But let me verify with a small example. Let's take a smaller case, say up to 30 lamps.In the case of 30 lamps, how many would be on after pulling 2,3,5.Compute numbers not divisible by 2,3,5: numbers like 1,7,11,13,17,19,23,29. That's 8 numbers.Numbers divisible by exactly two: numbers divisible by 2 and 3 (6,12,18,24,30), but subtract those divisible by all three (30). So 5 -1=4. Similarly, numbers divisible by 2 and 5 (10,20,30) minus 30: 3-1=2. Numbers divisible by 3 and 5 (15,30) minus 30: 2-1=1. So total exactly two: 4+2+1=7.Total on: 8 +7=15. Now, 30 - 15=15 lamps off. Let's check manually.But maybe this is time-consuming. Alternatively, according to our formula: for N=30,Numbers not divisible by 2,3,5:30 - floor(30/2) - floor(30/3) - floor(30/5) + floor(30/6) + floor(30/10) + floor(30/15) - floor(30/30)=30 -15 -10 -6 +5 +3 +2 -1=30 -31 +10 -1=30 -31= -1 +10=9 -1=8. Which matches.Numbers divisible by exactly two: A+B+C -3D=5+3+2 -3*1=10 -3=7. Which also matches. So total on:8+7=15. Which would be correct. Let's count the actual numbers:Numbers from 1-30:Ones not divisible by 2,3,5:1,7,11,13,17,19,23,29 (8 numbers)Ones divisible by exactly two:- Divisible by 2 and 3 but not 5:6,12,18,24 (excluding 30)- Divisible by 2 and 5 but not 3:10,20 (excluding 30)- Divisible by 3 and 5 but not 2:15 (excluding 30)Total:4+2+1=7So total on:8+7=15. Now, total lamps 30, 15 on and 15 off. So that seems correct.Therefore, our formula seems to work for N=30. Hence, probably correct.Therefore, back to N=1000, the answer is 499. Wait, but 266 +233=499. But 266+233=499. 499 is less than 500. But let me check whether the inclusion-exclusion was done correctly.Wait, in the numbers not divisible by 2,3,5, we calculated 266. Then numbers divisible by exactly two:233. So 266+233=499.But let me cross-verify with another approach.Alternatively, the number of lamps on can be calculated using the principle that a lamp is on if it has an even number of toggles (0 or 2). So, each lamp is toggled once for each switch that divides it. Since we have three switches (2,3,5), the number of divisors among these is 0,1,2,3.Thus, the lamps on are those with 0 or 2 divisors. Therefore, the total number is the number of numbers not divisible by 2,3,5 (0 divisors) plus the number of numbers divisible by exactly two of them (2 divisors). So, that's what we calculated.Alternatively, we can model each number as a vector indicating divisibility by 2,3,5. For example, each number can be represented by a triplet (a,b,c) where a=1 if divisible by 2, 0 otherwise; b=1 if divisible by 3, 0 otherwise; c=1 if divisible by 5, 0 otherwise. Then, the number of toggles is a + b + c. We need a + b + c to be even. So, the number of such vectors where a + b + c is even.Each position (a,b,c) corresponds to a unique number if we consider the least common multiple. But in reality, numbers can have multiple prime factors, so this might not directly correspond. However, since 2,3,5 are primes, the triplet (a,b,c) corresponds to the exponents in the prime factorization (either 0 or 1 for each prime). Therefore, each number can be uniquely represented by the set of primes it's divisible by among 2,3,5. Therefore, the count of numbers with a certain combination can be calculated via inclusion-exclusion.Alternatively, the number of numbers where a + b + c is even is equal to the total numbers minus the numbers where a + b + c is odd. So, total numbers is 1000. The numbers where a + b + c is even is 1000 - numbers where a + b + c is odd.But in our previous calculation, numbers where a + b + c is even is 266 + 233 = 499, and numbers where odd is 1000 - 499 = 501. But let's see if this makes sense.Wait, if we have three sets, the number of elements in exactly even number of sets can be calculated by inclusion-exclusion. But this might be complicated. Alternatively, using generating functions.Alternatively, for each number from 1 to 1000, assign a variable indicating whether it's divisible by 2, 3, or 5. Then, the total number of numbers where the count is even is 1/2*(Total + (-1)^{0}*Product terms). Wait, this is using the principle from combinatorics where the number of elements with even number of properties is 1/2[N + (1 - 1)^{k}], but maybe more precise.Wait, perhaps using the inclusion-exclusion formula for even number of sets. The formula is:Number of elements in even number of sets = (Sum_{i} |A_i| - Sum_{i<j} |A_i ∩ A_j| + Sum_{i<j<k} |A_i ∩ A_j ∩ A_k| - ...)/2But this is for symmetric cases where all intersections are similar, but in our case, the sets are distinct.Alternatively, for each number, define x_i as 1 if divisible by the i-th prime (2,3,5), else 0. Then, we want the sum over x1 + x2 + x3 ≡ 0 mod 2. The number of such numbers is equal to:(Total numbers + numbers with even number of properties - numbers with odd number of properties)/2. Wait, actually, more precisely:Number of elements with even number of properties = (Total + (-1)^{k} * something)/2. Wait, perhaps not. Maybe using Möbius inversion.Alternatively, use the generating function. The generating function for the counts would be:(1 + y)^{3}, where each term corresponds to including or excluding each set. The coefficient of y^m gives the number of ways to have m sets. But in our case, the sets are overlapping.Wait, maybe not directly applicable.Alternatively, using the principle that the number of elements in exactly m sets is C(3, m) * something. But again, overlaps complicate this.Alternatively, think of each number n from 1 to 1000, and compute the parity of the number of divisors among 2,3,5. The total number with even parity is 266 +233 =499. Since that's what we calculated. Let's check with another approach.Alternatively, the number of numbers not divisible by 2,3,5 is 266. The number of numbers divisible by exactly two is 233. So total is 266 +233=499. So that's the answer. But just to make sure, let's check another way.Total number of lamps:1000Number of lamps toggled once (divisible by exactly one of 2,3,5): ?Number of lamps toggled three times (divisible by all three):33Therefore, total lamps off would be numbers toggled once + numbers toggled three times. Which is (numbers divisible by exactly one) + (numbers divisible by all three). Then lamps on is 1000 - (exactly one + exactly three).But we can compute exactly one:Numbers divisible by exactly one of 2,3,5 = (numbers divisible by 2 only + numbers divisible by 3 only + numbers divisible by 5 only)To compute this:Numbers divisible by 2 only = numbers divisible by 2 - numbers divisible by 2 and 3 - numbers divisible by 2 and 5 + numbers divisible by all threeWait, using inclusion-exclusion again.Similarly,Numbers divisible by 2 only: floor(1000/2) - floor(1000/6) - floor(1000/10) + floor(1000/30) =500 -166 -100 +33=500 -266 +33=500 -266=234; 234 +33=267Wait, that can't be. Let me compute step by step:Numbers divisible by 2:500Subtract numbers divisible by 2 and 3 (166) and numbers divisible by 2 and 5 (100). So 500 -166 -100=234But then we have subtracted too much because numbers divisible by all three (2,3,5) were subtracted twice, so we need to add them back once. So 234 +33=267Therefore, numbers divisible by 2 only:267Similarly, numbers divisible by 3 only: floor(1000/3) - floor(1000/6) - floor(1000/15) + floor(1000/30)=333 -166 -66 +33=333 -232=101 +33=134Wait:Numbers divisible by 3:333Subtract numbers divisible by 3 and 2 (166) and numbers divisible by 3 and 5 (66):333 -166 -66=101Then add back numbers divisible by all three (33):101 +33=134Similarly, numbers divisible by 5 only: floor(1000/5) - floor(1000/10) - floor(1000/15) + floor(1000/30)=200 -100 -66 +33=200 -166=34 +33=67Therefore, numbers divisible by exactly one:267 +134 +67=468Numbers divisible by exactly three:33Therefore, total lamps off:468 +33=501Therefore, lamps on:1000 -501=499Which matches our previous result.Therefore, the number of lamps still on is 499.But let me verify once more.Another approach: each lamp n is on if μ(n) is... Wait, maybe not. Alternatively, using Möbius function, but I think that complicates things. Alternatively, think of each number's toggle count.But given that we have verified with two different methods (direct inclusion-exclusion for numbers not divisible and exactly two, and another approach calculating exactly one and exactly three) both leading to 499, I think this is correct.Therefore, the answer should be 499. So, the number of lamps still on after pulling switches 2,3,5 is 499.Final Answerboxed{499}</think>Initially, all 1000 lamps are on. When a switch is pulled, it toggles the state of all lamps whose numbers are multiples of the switch's number. We need to determine the number of lamps that remain on after pulling switches 2, 3, and 5.1. Understanding the Problem: - Each switch toggles the state of lamps that are multiples of its number. - A lamp remains on if it is toggled an even number of times (including zero times) and turns off if toggled an odd number of times.2. Key Observations: - Lamps not divisible by 2, 3, or 5 remain on (0 toggles). - Lamps divisible by exactly two of 2, 3, or 5 are toggled twice (even), so they remain on. - Lamps divisible by all three (2, 3, 5) are toggled three times (odd), turning off.3. Inclusion-Exclusion Principle: - Numbers not divisible by 2, 3, or 5: [ 1000 - left( leftlfloor frac{1000}{2} rightrfloor + leftlfloor frac{1000}{3} rightrfloor + leftlfloor frac{1000}{5} rightrfloor right) + left( leftlfloor frac{1000}{6} rightrfloor + leftlfloor frac{1000}{10} rightrfloor + leftlfloor frac{1000}{15} rightrfloor right) - leftlfloor frac{1000}{30} rightrfloor ] Calculating each term: [ 1000 - (500 + 333 + 200) + (166 + 100 + 66) - 33 = 266 ] - Numbers divisible by exactly two of 2, 3, 5: [ left( leftlfloor frac{1000}{6} rightrfloor + leftlfloor frac{1000}{10} rightrfloor + leftlfloor frac{1000}{15} rightrfloor right) - 3 times leftlfloor frac{1000}{30} rightrfloor ] Calculating each term: [ (166 + 100 + 66) - 3 times 33 = 233 ]4. Total Lamps Remaining On: - Sum of lamps not divisible by 2, 3, or 5 and lamps divisible by exactly two of them: [ 266 + 233 = 499 ]Thus, the number of lamps that are still on is (boxed{499}).]

🔑:9. 499 Only the numbers that are not divisible by any of 2,3,5, or are exactly divisible by two of them, have their lights on.In 1 to 1000, there are 500 numbers divisible by 2, 333 numbers divisible by 3, 200 numbers divisible by 5, 166 numbers divisible by 2 times 3, 100 numbers divisible by 2 times 5, 66 numbers divisible by 3 times 5, and 33 numbers divisible by 2 times 3 times 5.Therefore, the numbers divisible by 2 and 3 but not by 5 are 166-33=133; the numbers divisible by 2 and 5 but not by 3 are 100-33=67; the numbers divisible by 3 and 5 but not by 2 are 66-33=33. The numbers not divisible by any of 2,3,5 are 1000-(500+ 333+200-166-100-66+33)=266.Thus, there are a total of 133+67+33+266=499 lights that are on.

🧠:[<think>Okay, so I need to find the value of cos(2π/5) + cos(4π/5). Hmm, let me think. I remember that sometimes when you have sums of cosines, you can use trigonometric identities to simplify them. Maybe there's a formula for the sum of two cosines. Let me recall... Oh, right! The sum-to-product identities might help here. The formula for the sum of two cosines is:cos A + cos B = 2 cos[(A + B)/2] cos[(A - B)/2]Let me try applying that. Let A be 2π/5 and B be 4π/5. Then:cos(2π/5) + cos(4π/5) = 2 cos[(2π/5 + 4π/5)/2] cos[(2π/5 - 4π/5)/2]Calculating the arguments inside the cosines:First term inside the first cosine: (2π/5 + 4π/5)/2 = (6π/5)/2 = 3π/5Second term inside the second cosine: (2π/5 - 4π/5)/2 = (-2π/5)/2 = -π/5But cosine is even, so cos(-π/5) = cos(π/5). So this simplifies to:2 cos(3π/5) cos(π/5)Hmm, okay. So now I have 2 cos(3π/5) cos(π/5). Let me compute each of these terms.First, I know that cos(π/5) is a known value. I remember that cos(π/5) is equal to (1 + √5)/4 multiplied by 2, or maybe it's (√5 + 1)/4? Wait, let's recall the exact value. Actually, cos(π/5) is (1 + √5)/4 * 2? Wait, no. Let me think. Maybe I can derive it using some identity.Alternatively, I recall that in a regular pentagon, the cosine of 36 degrees (which is π/5 radians) is (1 + √5)/4 multiplied by 2. Wait, let me verify.Alternatively, using the formula for cosine of 36 degrees:cos(36°) = (1 + √5)/4 × 2, which is (1 + √5)/4 × 2 = (1 + √5)/2. Wait, that can't be, because the maximum value of cosine is 1. Wait, no. Wait, (1 + √5)/2 is approximately (1 + 2.236)/2 ≈ 1.618/2 ≈ 0.809, which is approximately cos(36°), which is about 0.8090. So yes, cos(π/5) = (1 + √5)/4 × 2? Wait, actually, (1 + √5)/4 × 2 is (1 + √5)/2. So cos(π/5) = (√5 + 1)/4 × 2? Wait, maybe I got confused.Alternatively, let's use the identity for cos(36°). Let me convert π/5 to degrees: π/5 radians is 36 degrees. So, cos(36°). There's a formula for that. Let's consider a regular pentagon. In a regular pentagon, the central angles are 72°, so maybe there are some triangles we can use. Alternatively, using the double-angle formula.Alternatively, use the identity that cos(5θ) = 16 cos^5θ - 20 cos^3θ + 5 cosθ. If we take θ = π/5, then 5θ = π, so cos(π) = -1 = 16 cos^5(π/5) - 20 cos^3(π/5) + 5 cos(π/5). Let me set x = cos(π/5). Then:16x^5 - 20x^3 + 5x + 1 = 0Hmm, that's a quintic equation, which might be difficult to solve. But maybe we can factor it. Let me try plugging x = (√5 + 1)/4 into the equation. Wait, but maybe there's a quadratic factor. Alternatively, note that for θ = π/5, 5θ = π, so 2θ = 2π/5, 3θ = 3π/5. Wait, maybe there's another approach.Alternatively, consider a 36-72-72 triangle. If we have an isosceles triangle with angles 36°, 72°, 72°, then the ratio of sides can be related to the golden ratio. Let me recall that in such a triangle, the ratio of the longer side to the shorter side is the golden ratio φ = (1 + √5)/2. Maybe that can help.Alternatively, use the formula for cos(2π/5) and cos(4π/5). Wait, but the problem is to compute their sum.Wait, let me get back to the original expression after applying the sum-to-product formula: 2 cos(3π/5) cos(π/5). Let's compute each term.First, cos(3π/5). 3π/5 is 108 degrees. The cosine of 108 degrees. Hmm, but 108 degrees is 180 - 72 degrees, so cos(108°) = -cos(72°). Therefore, cos(3π/5) = -cos(2π/5). Wait, but that's circular because cos(2π/5) is part of the original expression. Maybe that's not helpful.Alternatively, express cos(3π/5) in terms of known cosines. Wait, 3π/5 is π - 2π/5, so cos(3π/5) = -cos(2π/5). So then:2 cos(3π/5) cos(π/5) = 2 (-cos(2π/5)) cos(π/5) = -2 cos(2π/5) cos(π/5)But that seems like going in circles. So maybe the sum-to-product identity isn't helpful here. Let me think of another approach.Alternatively, use complex numbers. Let me recall Euler's formula: e^(iθ) = cosθ + i sinθ. Then, perhaps, summing the real parts.But the expression is cos(2π/5) + cos(4π/5). Let me note that 4π/5 is equal to π - π/5, so cos(4π/5) = -cos(π/5). Wait, no. Wait, cos(π - θ) = -cosθ. So cos(4π/5) = cos(π - π/5) = -cos(π/5). So then, cos(2π/5) + cos(4π/5) = cos(2π/5) - cos(π/5). Hmm, is that right? Wait, cos(4π/5) = -cos(π/5), so substituting:cos(2π/5) + (-cos(π/5)) = cos(2π/5) - cos(π/5)Hmm, but how does that help? Maybe I can compute this difference.Alternatively, use another identity. For the difference of cosines:cos A - cos B = -2 sin[(A + B)/2] sin[(A - B)/2]So, applying that to cos(2π/5) - cos(π/5):cos(2π/5) - cos(π/5) = -2 sin[(2π/5 + π/5)/2] sin[(2π/5 - π/5)/2]Calculating the arguments:First term: (3π/5)/2 = 3π/10Second term: (π/5)/2 = π/10Therefore:= -2 sin(3π/10) sin(π/10)Hmm, now I need to compute sin(3π/10) and sin(π/10). Let's see. 3π/10 is 54 degrees, and π/10 is 18 degrees. The sine of 54 degrees and 18 degrees.I recall that sin(54°) is (√5 + 1)/4 multiplied by 2? Wait, let's recall exact values. From trigonometric tables:sin(18°) = (√5 - 1)/4 * 2? Wait, let me think.Alternatively, use the formula for sin(3θ). Let's take θ = 18°, so 3θ = 54°, or θ = π/10 radians. Let me try to compute sin(3π/10).Alternatively, use the identity for sin(3θ) = 3 sinθ - 4 sin^3θ. Let θ = π/10, so sin(3π/10) = 3 sin(π/10) - 4 sin^3(π/10). But this might lead to a cubic equation. Alternatively, note that sin(3π/10) = cos(π/2 - 3π/10) = cos(π/5). Wait, π/2 is 5π/10, so 5π/10 - 3π/10 = 2π/10 = π/5. So sin(3π/10) = cos(π/5). Similarly, sin(π/10) = cos(2π/5). Wait, because sin(π/10) = cos(π/2 - π/10) = cos(4π/10) = cos(2π/5). Therefore, sin(3π/10) = cos(π/5) and sin(π/10) = cos(2π/5).So substituting back into the expression:-2 sin(3π/10) sin(π/10) = -2 cos(π/5) cos(2π/5)But that's equal to -2 cos(π/5) cos(2π/5). Wait, but this seems to lead us back to a similar expression as before. So perhaps this approach isn't helpful. Let's try another idea.Alternatively, recall that the sum of cosines can be related to roots of unity. The angles 2π/5 and 4π/5 correspond to two of the fifth roots of unity. Specifically, the sum of all fifth roots of unity is zero, and the real parts are the cosines. The fifth roots of unity are 1, e^(2πi/5), e^(4πi/5), e^(6πi/5), e^(8πi/5). The sum is 1 + e^(2πi/5) + e^(4πi/5) + e^(6πi/5) + e^(8πi/5) = 0. Taking the real parts:1 + cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = 0But cos(6π/5) = cos(π + π/5) = -cos(π/5) and cos(8π/5) = cos(2π - 2π/5) = cos(2π/5). Wait, let me check:6π/5 is π + π/5, so cos(6π/5) = -cos(π/5), sin(6π/5) = -sin(π/5)Similarly, 8π/5 is 2π - 2π/5, so cos(8π/5) = cos(2π/5), sin(8π/5) = -sin(2π/5)Therefore, the sum becomes:1 + cos(2π/5) + cos(4π/5) - cos(π/5) + cos(2π/5) = 0Wait, let me verify that step by step. The fifth roots of unity are at angles 0, 2π/5, 4π/5, 6π/5, 8π/5. Their cosines are:cos(0) = 1,cos(2π/5),cos(4π/5),cos(6π/5) = cos(π + π/5) = -cos(π/5),cos(8π/5) = cos(2π - 2π/5) = cos(2π/5).Therefore, summing the real parts:1 + cos(2π/5) + cos(4π/5) - cos(π/5) + cos(2π/5) = 0Wait, so combining like terms:1 + 2 cos(2π/5) + cos(4π/5) - cos(π/5) = 0But our original problem is to compute cos(2π/5) + cos(4π/5). Let me denote S = cos(2π/5) + cos(4π/5). Then, the equation becomes:1 + 2 cos(2π/5) + S - cos(π/5) = 0But that introduces another term, cos(π/5). Maybe we can find another equation involving S and cos(π/5). Alternatively, use another identity.Alternatively, note that the sum S = cos(2π/5) + cos(4π/5) and we can relate this to other trigonometric values. Let me consider the identity for the sum of cos(kθ) for k from 1 to n, but maybe that's overcomplicating.Alternatively, recall that in a regular pentagon, the diagonal over the side is the golden ratio φ = (1 + √5)/2. Perhaps there is a relationship here.Alternatively, consider using the identity for sin(5θ) and relate it to sinθ, but maybe that's too involved.Wait, another approach: Let’s consider the equation we got from the fifth roots of unity:1 + 2 cos(2π/5) + cos(4π/5) - cos(π/5) = 0But S = cos(2π/5) + cos(4π/5). Then the equation becomes:1 + cos(2π/5) + S - cos(π/5) = 0But this is still one equation with two variables, S and cos(π/5). Maybe another equation can relate these.Alternatively, we can use the identity that cos(4π/5) = -cos(π - 4π/5) = -cos(π/5). Wait, no: cos(π - θ) = -cosθ. So cos(4π/5) = cos(π - π/5) = -cos(π/5). Wait, that's correct. So cos(4π/5) = -cos(π/5). Therefore, substituting into S:S = cos(2π/5) + (-cos(π/5)) = cos(2π/5) - cos(π/5)So S = cos(2π/5) - cos(π/5)Hmm, so now we can substitute that back into the equation from the roots of unity:1 + 2 cos(2π/5) + S - cos(π/5) = 0But since S = cos(2π/5) - cos(π/5), substitute S:1 + 2 cos(2π/5) + cos(2π/5) - cos(π/5) - cos(π/5) = 0Simplify:1 + 3 cos(2π/5) - 2 cos(π/5) = 0Hmm, this seems more complicated. Maybe we need another equation relating cos(2π/5) and cos(π/5).Alternatively, use the double-angle formula. For example, cos(2π/5) can be written in terms of cos(π/5). The double-angle formula is:cos(2θ) = 2 cos²θ - 1Let θ = π/5. Then:cos(2π/5) = 2 cos²(π/5) - 1Therefore, substituting into S:S = [2 cos²(π/5) - 1] - cos(π/5) = 2 cos²(π/5) - cos(π/5) - 1So S is expressed in terms of cos(π/5). Let me denote x = cos(π/5). Then:S = 2x² - x - 1Now, if I can find another equation involving x, I can solve for x and then compute S.From the equation we had earlier:1 + 3 cos(2π/5) - 2 cos(π/5) = 0But cos(2π/5) is 2x² - 1, so substitute:1 + 3(2x² - 1) - 2x = 0Simplify:1 + 6x² - 3 - 2x = 0Which becomes:6x² - 2x - 2 = 0Divide both sides by 2:3x² - x - 1 = 0So, 3x² - x - 1 = 0. Solving for x:x = [1 ± √(1 + 12)] / 6 = [1 ± √13]/6Wait, but this can’t be right because we know that cos(π/5) is a positive value approximately 0.8090, and [1 + √13]/6 ≈ (1 + 3.605)/6 ≈ 4.605/6 ≈ 0.767, which is close but not exactly 0.8090. The other root is negative, which we can disregard. But this suggests a discrepancy. That must mean there was an error in deriving the equation.Wait, let's check the steps again. We had:From the roots of unity equation:1 + 2 cos(2π/5) + cos(4π/5) - cos(π/5) = 0But cos(4π/5) = -cos(π/5), so substituting:1 + 2 cos(2π/5) - cos(π/5) - cos(π/5) = 0Which is:1 + 2 cos(2π/5) - 2 cos(π/5) = 0Then we expressed cos(2π/5) as 2x² - 1 where x = cos(π/5):1 + 2(2x² - 1) - 2x = 0Expanding:1 + 4x² - 2 - 2x = 0Simplifies to:4x² - 2x - 1 = 0Ah! There was a mistake in the previous calculation. I had incorrectly written 3(2x² - 1) instead of 2(2x² - 1). Let's correct that.So, starting again:From the roots of unity:1 + 2 cos(2π/5) + cos(4π/5) - cos(π/5) = 0But cos(4π/5) = -cos(π/5), so substituting:1 + 2 cos(2π/5) - cos(π/5) - cos(π/5) = 0Which is:1 + 2 cos(2π/5) - 2 cos(π/5) = 0Now, expressing cos(2π/5) as 2x² - 1, where x = cos(π/5):1 + 2(2x² - 1) - 2x = 0Calculating:1 + 4x² - 2 - 2x = 0Simplifies to:4x² - 2x - 1 = 0Now, solving for x:x = [2 ± √(4 + 16)] / 8 = [2 ± √20]/8 = [2 ± 2√5]/8 = [1 ± √5]/4Since x = cos(π/5) is positive, we take the positive root:x = [1 + √5]/4 × 2? Wait, no. Wait:Wait, x = [1 + √5]/4? Wait, the quadratic equation gives x = [1 ± √5]/4. Wait, no:Wait, the quadratic equation is 4x² - 2x - 1 = 0.Using the quadratic formula:x = [2 ± √(4 + 16)] / 8 = [2 ± √20]/8 = [2 ± 2√5]/8 = [1 ± √5]/4So, x = [1 + √5]/4 or x = [1 - √5]/4. Since x is cos(π/5), which is approximately 0.8090, we take the positive root:x = [1 + √5]/4 ≈ [1 + 2.236]/4 ≈ 3.236/4 ≈ 0.809, which matches. Therefore, cos(π/5) = (1 + √5)/4 × 2? Wait, (1 + √5)/4 is approximately 0.809, but actually (1 + √5)/4 is 0.809/2 ≈ 0.4045. Wait, no. Wait, √5 ≈ 2.236, so 1 + √5 ≈ 3.236, divided by 4 is ≈ 0.809. Yes, so cos(π/5) = (1 + √5)/4 × 2? Wait, no. Wait, (1 + √5)/4 is ≈ 0.809, but cos(π/5) ≈ 0.8090. Therefore, cos(π/5) = (√5 + 1)/4 × 2? Wait, no. Wait, actually, cos(π/5) is (1 + √5)/4 multiplied by 2? Let me verify:Wait, (1 + √5)/4 is approximately (1 + 2.236)/4 ≈ 3.236/4 ≈ 0.809, which is exactly cos(36°). Therefore, cos(π/5) = (1 + √5)/4 × 2? Wait, no. Wait, (1 + √5)/4 ≈ 0.809, so cos(π/5) = (1 + √5)/4 × 2? No, actually, cos(π/5) = (1 + √5)/4 × 2? Wait, this is confusing. Wait, (1 + √5)/4 is approximately 0.809, which is exactly the value of cos(π/5). So cos(π/5) = (√5 + 1)/4 × 2? Wait, no, because (√5 + 1)/4 × 2 would be (√5 + 1)/2 ≈ (2.236 + 1)/2 ≈ 1.618/2 ≈ 0.809, which is correct. So cos(π/5) = (√5 + 1)/4 × 2? Wait, perhaps the confusion arises from different forms. In reality, cos(π/5) is (1 + √5)/4 multiplied by 2? No, actually, (√5 + 1)/4 is approximately 0.809, which is the correct value. Wait, but (√5 + 1)/4 is approximately (2.236 + 1)/4 ≈ 3.236/4 ≈ 0.809, so cos(π/5) = (√5 + 1)/4? No, that would be 0.809, which is correct. Wait, but 0.809 is (√5 + 1)/4? Wait, no:Wait, let's compute (√5 + 1)/4:√5 ≈ 2.236, so √5 + 1 ≈ 3.236; divided by 4 gives ≈ 0.809. So yes, cos(π/5) = (√5 + 1)/4? Wait, but 0.809 is indeed equal to (√5 + 1)/4. Wait, no! Wait, (√5 + 1)/4 ≈ 0.809, but (√5 + 1) is approximately 3.236, so 3.236 divided by 4 is approximately 0.809. So cos(π/5) is equal to (√5 + 1)/4? No, that can't be because (√5 + 1)/4 is approximately 0.809, which is exactly the value of cos(π/5). Wait, but according to the quadratic equation solution, x = [1 + √5]/4? No, in the quadratic equation, we had:4x² - 2x - 1 = 0Solutions:x = [2 ± √(4 + 16)] / 8 = [2 ± √20]/8 = [2 ± 2√5]/8 = [1 ± √5]/4So positive solution is x = [1 + √5]/4 ≈ 0.809, which is correct. Therefore, cos(π/5) = (1 + √5)/4? Wait, no, (1 + √5)/4 ≈ 3.236/4 ≈ 0.809, which matches cos(π/5). Therefore, yes, cos(π/5) = (√5 + 1)/4 × 2? No, it's (√5 + 1)/4 is already 0.809, so cos(π/5) = (1 + √5)/4. Wait, but that conflicts with my previous thought. Wait, no, maybe I confused it with another value.Wait, let me check with calculator: cos(π/5) = cos(36°) ≈ 0.8090. (1 + √5)/4 ≈ (1 + 2.236)/4 ≈ 3.236/4 ≈ 0.809. So yes, exactly. Therefore, cos(π/5) = (1 + √5)/4. Wait, but (1 + √5)/4 is approximately 0.809, which is correct. So the exact value is (1 + √5)/4. Wait, but often I see cos(36°) expressed as (1 + √5)/4 × 2? Hmm, maybe different forms. Let me confirm.Wait, actually, the exact value of cos(π/5) is (1 + √5)/4 multiplied by 2. Wait, let's compute (1 + √5)/2: that's approximately (1 + 2.236)/2 ≈ 1.618, which is the golden ratio φ, which is approximately 1.618. But cos(π/5) is approximately 0.809, which is φ/2. So yes, cos(π/5) = φ/2 = (1 + √5)/4 × 2. Wait, but φ is (1 + √5)/2, so φ/2 is (1 + √5)/4. So cos(π/5) = φ/2 = (1 + √5)/4. That's correct.Therefore, cos(π/5) = (√5 + 1)/4.Wait, no, (√5 + 1)/4 is equal to approximately 0.809, which matches. Therefore, cos(π/5) = (√5 + 1)/4.Wait, but 0.809 is (√5 + 1)/4? Let's compute (√5 ≈ 2.236):(2.236 + 1)/4 = 3.236/4 ≈ 0.809, yes. So, cos(π/5) = (√5 + 1)/4.Therefore, x = (√5 + 1)/4. Now, going back to S = 2x² - x - 1. Let's compute S.First, compute x²:x² = [(√5 + 1)/4]^2 = (5 + 2√5 + 1)/16 = (6 + 2√5)/16 = (3 + √5)/8Therefore, 2x² = 2*(3 + √5)/8 = (3 + √5)/4Then, subtract x: (3 + √5)/4 - (√5 + 1)/4 = [3 + √5 - √5 -1]/4 = (2)/4 = 1/2Then subtract 1: 1/2 - 1 = -1/2Therefore, S = -1/2Wait, what? That can't be right. Let me check the steps again.Given S = 2x² - x - 1Compute x = (√5 + 1)/4x² = [(√5 + 1)/4]^2 = (5 + 2√5 + 1)/16 = (6 + 2√5)/16 = (3 + √5)/8Multiply by 2: 2x² = (3 + √5)/4Subtract x: (3 + √5)/4 - (√5 + 1)/4 = (3 + √5 - √5 -1)/4 = (2)/4 = 1/2Subtract 1: 1/2 - 1 = -1/2So S = -1/2But wait, S was defined as cos(2π/5) + cos(4π/5). But according to this calculation, it's -1/2. However, let's check with approximate values:cos(2π/5) ≈ cos(72°) ≈ 0.3090cos(4π/5) ≈ cos(144°) ≈ -0.8090Adding them: 0.3090 + (-0.8090) ≈ -0.5, which is -1/2. So yes, the exact value is -1/2. But the question is to find cos(2π/5) + cos(4π/5), which is S = -1/2.But wait, the problem didn't specify to compute the absolute value or anything. So according to this, the answer is -1/2. But let me confirm with the approximate calculation:cos(72°) ≈ 0.3090cos(144°) ≈ -0.8090Sum ≈ 0.3090 - 0.8090 ≈ -0.5, which is exactly -1/2. So the exact value is -1/2.Therefore, despite the complex journey through various identities and equations, the answer is -1/2. However, let me cross-verify with another method.Another approach:Let’s consider the identity for sum of cosines at angles that are supplementary.Wait, but 2π/5 and 4π/5 are not supplementary. However, note that 2π/5 + 4π/5 = 6π/5, which is greater than π, so not supplementary.Alternatively, use the identity that the sum of cos(kθ) for k=1 to n, but that might not apply here.Alternatively, recall that in the roots of unity sum, we had:1 + 2 cos(2π/5) + 2 cos(4π/5) = 0Wait, no. Wait, original sum of real parts was:1 + cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = 0But cos(6π/5) = cos(π + π/5) = -cos(π/5)cos(8π/5) = cos(2π - 2π/5) = cos(2π/5)Therefore:1 + cos(2π/5) + cos(4π/5) - cos(π/5) + cos(2π/5) = 0So 1 + 2 cos(2π/5) + cos(4π/5) - cos(π/5) = 0But cos(4π/5) = -cos(π/5). Wait, no:Wait, cos(4π/5) = cos(π - π/5) = -cos(π/5)So substituting that:1 + 2 cos(2π/5) - cos(π/5) - cos(π/5) = 0Which is 1 + 2 cos(2π/5) - 2 cos(π/5) = 0Which was the equation we used earlier.But we found that S = cos(2π/5) + cos(4π/5) = -1/2But this contradicts the equation if we plug S = -1/2 into the equation:1 + 2 cos(2π/5) - 2 cos(π/5) = 0But S = cos(2π/5) + cos(4π/5) = cos(2π/5) - cos(π/5) = -1/2So, cos(2π/5) - cos(π/5) = -1/2Therefore, rearranged, cos(2π/5) = cos(π/5) - 1/2Then substitute back into the equation:1 + 2(cos(π/5) - 1/2) - 2 cos(π/5) = 0Simplify:1 + 2 cos(π/5) - 1 - 2 cos(π/5) = 0Which is 0 = 0, which is consistent. Therefore, there's no contradiction. Thus, the value of S is indeed -1/2.But wait, when I computed using approximate values, cos(2π/5) + cos(4π/5) ≈ 0.3090 + (-0.8090) ≈ -0.5, which is exactly -1/2. Therefore, despite the negative result, the answer is -1/2.But let me check again, since the problem is stated as "Find the value of cos(2π/5) + cos(4π/5)". According to our calculations, this is equal to -1/2. But maybe I made a mistake in the sign somewhere.Wait, cos(4π/5) is cos(π - π/5) = -cos(π/5). So cos(4π/5) = -cos(π/5). Therefore, cos(2π/5) + cos(4π/5) = cos(2π/5) - cos(π/5). But when I computed numerically:cos(2π/5) ≈ 0.3090cos(π/5) ≈ 0.8090Therefore, 0.3090 - 0.8090 ≈ -0.5. So the sum is indeed -0.5, which is -1/2.Therefore, despite the initial surprise, the answer is -1/2.But wait, the problem says "Find the value of cos(2π/5) + cos(4π/5)." So according to all these derivations and numerical checks, the answer is -1/2.But why did the sum-to-product formula give us 2 cos(3π/5) cos(π/5) which is 2*(-cos(2π/5))*cos(π/5) = -2 cos(2π/5) cos(π/5). But if the answer is -1/2, then:-2 cos(2π/5) cos(π/5) = -1/2Therefore, cos(2π/5) cos(π/5) = 1/4Is this true? Let me check with approximate values:cos(2π/5) ≈ 0.3090cos(π/5) ≈ 0.8090Their product ≈ 0.3090 * 0.8090 ≈ 0.25, which is 1/4. So yes, it's 1/4. Therefore, -2 * 1/4 = -1/2. So all the identities are consistent.Therefore, the answer is indeed -1/2.Final AnswerThe value of cos frac{2 pi}{5}+cos frac{4 pi}{5} is boxed{-dfrac{1}{2}}.</think>To find the value of (cos frac{2 pi}{5} + cos frac{4 pi}{5}), we can use trigonometric identities and properties of roots of unity.1. Using the Sum-to-Product Identity: [ cos frac{2 pi}{5} + cos frac{4 pi}{5} = 2 cos left( frac{frac{2 pi}{5} + frac{4 pi}{5}}{2} right) cos left( frac{frac{2 pi}{5} - frac{4 pi}{5}}{2} right) ] Simplifying the arguments: [ = 2 cos left( frac{3 pi}{5} right) cos left( -frac{pi}{5} right) ] Since (cos(-theta) = cos theta): [ = 2 cos left( frac{3 pi}{5} right) cos left( frac{pi}{5} right) ] Knowing (cos left( frac{3 pi}{5} right) = -cos left( frac{2 pi}{5} right)): [ = -2 cos left( frac{2 pi}{5} right) cos left( frac{pi}{5} right) ]2. Using Roots of Unity: The sum of the real parts of the fifth roots of unity is zero: [ 1 + cos frac{2 pi}{5} + cos frac{4 pi}{5} + cos frac{6 pi}{5} + cos frac{8 pi}{5} = 0 ] Using (cos frac{6 pi}{5} = -cos frac{pi}{5}) and (cos frac{8 pi}{5} = cos frac{2 pi}{5}): [ 1 + cos frac{2 pi}{5} + cos frac{4 pi}{5} - cos frac{pi}{5} + cos frac{2 pi}{5} = 0 ] Simplifying: [ 1 + 2 cos frac{2 pi}{5} + cos frac{4 pi}{5} - cos frac{pi}{5} = 0 ] Knowing (cos frac{4 pi}{5} = -cos frac{pi}{5}): [ 1 + 2 cos frac{2 pi}{5} - 2 cos frac{pi}{5} = 0 ] Let (S = cos frac{2 pi}{5} + cos frac{4 pi}{5}). Substituting (S): [ S = cos frac{2 pi}{5} - cos frac{pi}{5} ] Using the quadratic equation derived from the roots of unity and solving, we find (S = -frac{1}{2}).3. Numerical Verification: Approximate values: [ cos frac{2 pi}{5} approx 0.3090, quad cos frac{4 pi}{5} approx -0.8090 ] Sum: [ 0.3090 + (-0.8090) approx -0.5 = -frac{1}{2} ]Thus, the value of (cos frac{2 pi}{5} + cos frac{4 pi}{5}) is (boxed{-dfrac{1}{2}}).,<think>Okay, so I need to find the value of cos(2π/5) + cos(4π/5). Hmm, I remember that there are some trigonometric identities that relate sums of cosines. Maybe I can use the sum-to-product formulas? Let me recall... The sum of cosines formula is cos A + cos B = 2 cos[(A+B)/2] cos[(A-B)/2]. Let me try applying that here.Let me set A = 2π/5 and B = 4π/5. Then, (A + B)/2 = (6π/5)/2 = 3π/5, and (A - B)/2 = (-2π/5)/2 = -π/5. So, plugging into the formula, cos(2π/5) + cos(4π/5) = 2 cos(3π/5) cos(-π/5). But cos(-π/5) is the same as cos(π/5), since cosine is an even function. So this simplifies to 2 cos(3π/5) cos(π/5).Hmm, but I'm not sure if this helps. Maybe I need to find the numerical values of these cosines? Let me recall that 3π/5 is 108 degrees, and π/5 is 36 degrees. I know that cos(36°) is related to the golden ratio. Let me recall that cos(36°) = (1 + √5)/4 * 2 = (1 + √5)/4 * 2? Wait, no, let me check.Actually, cos(36°) is equal to (1 + √5)/4 multiplied by 2. Wait, maybe I should derive it. Let's consider a regular pentagon. The internal angles are 108°, and the central angles are 72°. The cosine of 36° comes up in relationships in a pentagon. There's a formula for cos(36°) which is (1 + √5)/4 multiplied by 2. Let me check:Alternatively, using the identity that cos(36°) = sin(54°). But perhaps another approach. Let's use the double-angle formula. Let me set θ = 36°, so 2θ = 72°, and 3θ = 108°, which is the angle we have in our problem.Wait, maybe I can express cos(3π/5) in terms of cos(π/5). Since 3π/5 is π - 2π/5, and cos(π - x) = -cos(x). Therefore, cos(3π/5) = cos(π - 2π/5) = -cos(2π/5). So substituting back into the expression we had earlier:2 cos(3π/5) cos(π/5) = 2 (-cos(2π/5)) cos(π/5) = -2 cos(2π/5) cos(π/5)But wait, that gives us cos(2π/5) + cos(4π/5) = -2 cos(2π/5) cos(π/5). Hmm, but this seems like we're going in circles. Let me see.Alternatively, perhaps instead of using the sum-to-product formula, I can consider using complex numbers. Let me recall Euler's formula: e^(iθ) = cosθ + i sinθ. So, maybe if I consider the sum of e^(i2π/5) and e^(i4π/5), then take the real part. Let's try that.Let me denote z = e^(i2π/5) + e^(i4π/5). Then, the real part of z is cos(2π/5) + cos(4π/5), which is exactly what we need. So if I can compute Re(z), that would be the answer. Let's compute z first.Note that e^(i2π/5) and e^(i4π/5) are two of the fifth roots of unity, excluding 1. The sum of all fifth roots of unity is 0. So, 1 + e^(i2π/5) + e^(i4π/5) + e^(i6π/5) + e^(i8π/5) = 0. Therefore, e^(i2π/5) + e^(i4π/5) + e^(i6π/5) + e^(i8π/5) = -1. But e^(i6π/5) is e^(i(2π - 4π/5)) = e^(-i4π/5), and e^(i8π/5) is e^(i(2π - 2π/5)) = e^(-i2π/5). So, the sum can be written as e^(i2π/5) + e^(i4π/5) + e^(-i4π/5) + e^(-i2π/5) = -1. But these terms are just 2[cos(2π/5) + cos(4π/5)] = -1. Therefore, cos(2π/5) + cos(4π/5) = -1/2.Wait, is that possible? Let me check. If that's the case, then the answer would be -1/2. But let me verify this.Alternatively, let's compute the sum step by step.Suppose S = cos(2π/5) + cos(4π/5). Let me compute numerical values. Let's convert radians to degrees: 2π/5 is 72°, 4π/5 is 144°, so cos(72°) + cos(144°). Let me recall that cos(72°) ≈ 0.3090, and cos(144°) ≈ -0.8090. Adding them together: 0.3090 + (-0.8090) ≈ -0.5. So, -0.5 is -1/2. Therefore, the value is indeed -1/2. So maybe the answer is -1/2. But how does that align with the previous steps?Wait, in the complex roots approach, we had that sum of the roots equals zero, so excluding 1, the sum of the other four roots is -1. Then, grouping them in conjugate pairs: e^(i2π/5) + e^(-i2π/5) + e^(i4π/5) + e^(-i4π/5) = 2[cos(2π/5) + cos(4π/5)] = -1, so indeed S = -1/2. Therefore, the answer is -1/2. But I need to make sure that this is correct.But let me try another approach. Let's use the identity for sum of cosines with angles in arithmetic progression, but here we have two terms. Alternatively, maybe express these cosines in terms of known values.I remember that in a regular pentagon, the diagonal over the side is the golden ratio φ = (1 + √5)/2. Perhaps there's a relationship here. Let me recall that cos(2π/5) and cos(4π/5) can be expressed in terms of radicals.From trigonometric exact values, I know that cos(36°) = (1 + √5)/4 * 2, which is (1 + √5)/4 * 2? Wait, let's compute it properly.Let me recall that cos(36°) = (1 + √5)/4 * 2, but actually, exact value is cos(π/5) = (1 + √5)/4 * 2, which is (√5 + 1)/4 * 2 = (√5 + 1)/2. Wait, no. Wait, cos(36°) is equal to (1 + √5)/4 multiplied by 2. Wait, let me check this.Let me recall that in a regular pentagon, cos(72°) is equal to (sqrt(5)-1)/4 multiplied by 2. Let me use the exact value.We can derive the exact value of cos(2π/5) as follows. Let θ = 2π/5, so 5θ = 2π. Then, cos(5θ) = cos(2π) = 1. Using the multiple-angle formula for cosine: cos(5θ) = 16 cos^5θ - 20 cos^3θ + 5 cosθ = 1. Let me set x = cosθ. Then, 16x^5 - 20x^3 + 5x - 1 = 0. But since θ = 2π/5, which is 72°, and cos(72°) is a root of this equation. However, solving a fifth-degree equation is complicated. Alternatively, we can factor it.We know that x = 1 is a root (since cos0 = 1), so we can factor out (x - 1). Let me perform polynomial division.Divide 16x^5 - 20x^3 + 5x - 1 by (x - 1). Using synthetic division:Coefficients: 16, 0, -20, 0, 5, -1Root at x=1:Bring down 16Multiply by 1: 16Add to next coefficient: 0 +16=16Multiply by 1:16Add to next coefficient: -20+16=-4Multiply by1:-4Add to next coefficient:0 +(-4)=-4Multiply by1:-4Add to next coefficient:5 +(-4)=1Multiply by1:1Add to last coefficient: -1 +1=0Therefore, the polynomial factors as (x -1)(16x^4 +16x^3 -4x^2 -4x +1)=0So, the quartic equation 16x^4 +16x^3 -4x^2 -4x +1=0. Now, we need to solve this quartic equation. Maybe factor it further.Let me try to factor this quartic. Let me attempt to factor it into quadratics. Suppose it factors as (ax^2 + bx + c)(dx^2 + ex + f) = 0.Multiplying out: adx^4 + (ae + bd)x^3 + (af + be + cd)x^2 + (bf + ce)x + cf = 0Comparing coefficients:ad =16ae + bd =16af + be + cd = -4bf + ce = -4cf=1Assuming a,d are integers. Since ad=16, possible pairs (a,d)=(4,4),(8,2),(16,1), etc. Let's try a=4, d=4.Then:ad=16ae + bd=4e +4b=16 ⇒ e + b=4af + be + cd=4f + be +4c= -4bf + ce= bf + ce= -4cf=1. Since c and f are integers, possible c=1, f=1 or c=-1, f=-1. Let's try c=1, f=1.Then:From cf=1, c=1, f=1.From e + b=4.From bf + ce= b*1 + e*1= b + e= -4. But we already have e + b=4. So, 4 = -4? Contradiction. So, invalid.Try c=-1, f=-1.Then, cf=(-1)(-1)=1. Good.From e + b=4.From bf + ce = b*(-1) + e*(-1) = -b -e = -4 ⇒ b + e=4. Which is same as before. So, same equation. Therefore, no solution here. So perhaps a different a,d.Try a=8, d=2.Then:ad=16.ae + bd=8e +2b=16 ⇒ 4e + b=8.af + be + cd=8f + be +2c= -4.bf + ce= bf + ce= -4.cf=1. So c and f are either 1,1 or -1,-1.Let's try c=1, f=1:Then,From 4e + b=8.From 8f + be +2c=8(1) + be +2(1)=8 + be +2=10 + be= -4 ⇒ be= -14.From bf + ce= b(1) + e(1)=b + e= -4.But 4e + b=8 and b + e= -4. Let's solve these equations.From b + e= -4 ⇒ b= -4 -e.Substitute into 4e + b=8:4e + (-4 -e)=8 ⇒ 3e -4=8 ⇒3e=12 ⇒e=4.Then, b= -4 -4= -8.Check if be= (-8)(4)= -32, but we need be= -14. Doesn't work.So, invalid.Try c=-1, f=-1.Then,From 4e + b=8.From 8f + be +2c=8(-1) + be +2(-1)= -8 + be -2= be -10= -4 ⇒ be=6.From bf + ce= b(-1) + e(-1)= -b -e= -4 ⇒ b + e=4.But 4e + b=8 and b + e=4.From b + e=4 ⇒ b=4 - e.Substitute into 4e + b=8:4e + (4 -e)=8 ⇒3e +4=8 ⇒3e=4 ⇒e=4/3. Not integer. Disregard.Thus, maybe a different a,d.Try a=2, d=8.Similarly, ad=16.ae + bd=2e +8b=16 ⇒ e +4b=8.af + be + cd=2f + be +8c= -4.bf + ce= bf + ce= -4.cf=1. Again, c=1,f=1 or c=-1,f=-1.First, c=1,f=1.From e +4b=8.From 2f + be +8c=2(1) + be +8(1)=2 + be +8=10 + be= -4 ⇒ be= -14.From bf + ce= b(1) + e(1)=b + e= -4.But we have e +4b=8 and b + e= -4. Let's solve:From b + e= -4 ⇒ e= -4 -b.Substitute into e +4b=8:(-4 -b) +4b=8 ⇒3b -4=8 ⇒3b=12 ⇒b=4. Then e= -4 -4= -8.Check be=4*(-8)= -32 ≠ -14. Not good.c=-1,f=-1:From e +4b=8.From 2f + be +8c=2(-1) + be +8(-1)= -2 + be -8= be -10= -4 ⇒ be=6.From bf + ce= b(-1) + e(-1)= -b -e= -4 ⇒b + e=4.Now, we have e +4b=8 and e +b=4. Subtract: (e +4b) - (e +b)=3b=4 ⇒b=4/3. Not integer. Disregard.Hmm, maybe another approach. Let's try to factor the quartic equation 16x^4 +16x^3 -4x^2 -4x +1=0.Alternatively, perhaps substitute y = x + something. Let me try to perform a substitution y = x + a to eliminate the cubic term. But this might be complicated. Alternatively, notice that the quartic might be a quadratic in terms of x^2. Let me check.16x^4 +16x^3 -4x^2 -4x +1. Hmm, not symmetric, but maybe group terms:(16x^4 +16x^3) + (-4x^2 -4x) +1 = 16x^3(x +1) -4x(x +1) +1. Hmm, factor (x +1):= (x +1)(16x^3 -4x) +1. But then that's (x +1)(16x^3 -4x) +1. Not sure.Alternatively, factor differently:16x^4 -4x^2 +1 +16x^3 -4x. That's (16x^4 -4x^2 +1) + (16x^3 -4x). Not obvious.Alternatively, maybe use substitution t = x + 1/(4x). Not sure. Alternatively, maybe check for rational roots. The quartic equation 16x^4 +16x^3 -4x^2 -4x +1=0. By rational root theorem, possible roots are ±1, ±1/2, ±1/4, ±1/8, ±1/16. Let me test x=1: 16 +16 -4 -4 +1=25 ≠0. x=-1:16 -16 -4 +4 +1=1≠0. x=1/2:16*(1/16) +16*(1/8) -4*(1/4) -4*(1/2)+1=1 +2 -1 -2 +1=1≠0. x=-1/2:16*(1/16) +16*(-1/8) -4*(1/4) -4*(-1/2)+1=1 -2 -1 +2 +1=1≠0. x=1/4:16*(1/256)+16*(1/64)-4*(1/16)-4*(1/4)+1≈0.0625 +0.25 -0.25 -1 +1=0.0625≠0. So no rational roots. Hmm.Alternatively, perhaps use the quadratic formula in some substitution. Let me divide the equation by x^2: 16x^2 +16x -4 -4/x +1/x^2=0. Let me set t = x + 1/x. Then, x^2 +1/x^2 = t^2 - 2. So:16(x^2 +1/x^2) +16(x -1/x) -4=0.But x^2 +1/x^2 = t^2 -2, and x -1/x = z (let me denote z = x -1/x). Then, the equation becomes 16(t^2 -2) +16z -4=0. Hmm, but this introduces both t and z. Maybe not helpful.Alternatively, maybe set u = x +1/x. Wait, but with the cubic term. Hmm, this seems messy. Maybe this approach is not the best.Alternatively, since this quartic equation is related to the cosine values, perhaps using trigonometric identities is a better approach.Alternatively, let me recall that cos(2π/5) and cos(4π/5) can be written in terms of radicals. Let me check known values.I recall that:cos(π/5) = (1 + √5)/4 * 2 = (√5 +1)/4 *2 = (√5 +1)/2 * (1/2). Wait, perhaps not. Wait, let me look it up mentally. The exact value of cos(36°) is (1 + √5)/4 * 2. Let me calculate:cos(36°) = (1 + √5)/4 * 2 = (1 + √5)/2 * (1/2). Wait, maybe not. Let me recall that in a regular pentagon, the diagonal is φ times the side, where φ is the golden ratio (1 + √5)/2. From the properties of the pentagon, we can derive that cos(36°) = (1 + √5)/4 * 2.Alternatively, consider a 36-54-90 triangle. Hmm, perhaps using the double-angle formula.Let θ = 36°, so 2θ = 72°, 3θ=108°, etc. Let me use the double-angle formula:cos(2θ) = 2cos²θ -1. So cos(72°) = 2cos²(36°) -1. But we need expressions for cos(72°) and cos(144°). Wait, cos(144°) is cos(180° -36°)= -cos(36°). Wait, no. cos(180° -x)= -cos(x). So cos(144°)=cos(180° -36°)= -cos(36°). Wait, but 144° is 2*72°, so perhaps cos(144°)=2cos²(72°)-1. Hmm, let's see.But perhaps using multiple-angle formulas:cos(72°) = 2cos²(36°)-1.Similarly, cos(36°) = 1 - 2sin²(18°). But this might not help.Alternatively, let me use the identity for cos(2π/5) + cos(4π/5). Let me denote S = cos(2π/5) + cos(4π/5). Then, perhaps compute 2S cos(2π/5) and use product-to-sum identities?Alternatively, multiply S by 2 and see:2S = 2cos(2π/5) + 2cos(4π/5). Wait, but this is the same as the original problem. Not helpful.Alternatively, consider the sum S and relate it to known angles.Alternatively, since 4π/5 = 2π/5 * 2, so maybe express cos(4π/5) in terms of cos(2π/5). Using double-angle formula:cos(4π/5) = 2cos²(2π/5) -1. So S = cos(2π/5) + 2cos²(2π/5) -1. Let me denote x = cos(2π/5). Then, S = x + 2x² -1.But we need to express S in terms of x. However, we need another equation relating x. Since we know from the earlier equation that 16x^4 -20x^2 +5x + ... Wait, no, this seems complicated.Alternatively, from the previous approach with complex numbers, we saw that the sum S is -1/2. But let me confirm with approximate values. cos(72°) ≈ 0.3090, cos(144°)≈-0.8090. Adding them: 0.3090 -0.8090 ≈ -0.5. Which is exactly -1/2. Therefore, the answer is indeed -1/2.But wait, the problem is asking for the value of cos(2π/5) + cos(4π/5). From the numerical check, it's approximately -0.5, which is -1/2. And the complex roots approach gave us that the sum is -1/2. Therefore, the answer is -1/2.But let me check once again through another method. Let's use the identity that the sum of cos(kθ) for k=1 to n can be expressed in terms of sine functions. But here we have only two terms. Alternatively, use geometric series.Alternatively, remember that in the complex plane, the sum of the real parts of the roots of unity (excluding 1) equals -1. As we used before. Therefore, the sum of cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5) = -1. But cos(6π/5) = cos(π + π/5) = -cos(π/5), and cos(8π/5)=cos(2π - 2π/5)=cos(2π/5). Wait, no. Wait, 6π/5 is π + π/5, so cos(6π/5) = -cos(π/5), and similarly, cos(8π/5)=cos(2π - 2π/5)=cos(2π/5). Wait, no, 8π/5 is 2π - 2π/5, so cos(8π/5)=cos(2π/5). Similarly, cos(6π/5)=cos(π + π/5)= -cos(π/5). Therefore, the sum becomes:cos(2π/5) + cos(4π/5) + (-cos(π/5)) + cos(2π/5) = -1. Wait, this can't be right. Wait, original sum is cos(2π/5) + cos(4π/5) + cos(6π/5) + cos(8π/5). But 6π/5 is equivalent to -4π/5 (since 6π/5 = 2π -4π/5), so cos(6π/5)=cos(-4π/5)=cos(4π/5). Similarly, 8π/5=2π -2π/5, so cos(8π/5)=cos(-2π/5)=cos(2π/5). Therefore, the sum is cos(2π/5) + cos(4π/5) + cos(4π/5) + cos(2π/5) = 2[cos(2π/5) + cos(4π/5)] = -1. Therefore, cos(2π/5) + cos(4π/5) = -1/2. Exactly as before.Therefore, the answer is -1/2. So I think that settles it. The value is -1/2. Hence, the answer is boxed{-dfrac{1}{2}}.Final Answerboxed{-dfrac{1}{2}}</think>To find the value of (cos frac{2 pi}{5} + cos frac{4 pi}{5}), we can use the properties of the roots of unity. The fifth roots of unity are given by (e^{i frac{2kpi}{5}}) for (k = 0, 1, 2, 3, 4). The sum of all fifth roots of unity is zero:[1 + e^{i frac{2pi}{5}} + e^{i frac{4pi}{5}} + e^{i frac{6pi}{5}} + e^{i frac{8pi}{5}} = 0]Excluding the root (1), we have:[e^{i frac{2pi}{5}} + e^{i frac{4pi}{5}} + e^{i frac{6pi}{5}} + e^{i frac{8pi}{5}} = -1]Noting that (e^{i frac{6pi}{5}} = e^{-i frac{4pi}{5}}) and (e^{i frac{8pi}{5}} = e^{-i frac{2pi}{5}}), we can rewrite the sum using cosine terms (since the imaginary parts cancel out):[e^{i frac{2pi}{5}} + e^{-i frac{2pi}{5}} + e^{i frac{4pi}{5}} + e^{-i frac{4pi}{5}} = 2 left( cos frac{2pi}{5} + cos frac{4pi}{5} right)]Thus, we have:[2 left( cos frac{2pi}{5} + cos frac{4pi}{5} right) = -1]Solving for the sum, we get:[cos frac{2pi}{5} + cos frac{4pi}{5} = -frac{1}{2}]Therefore, the value is (boxed{-dfrac{1}{2}}).]

🔑:Analysis from the perspective of basic methods and construction methods.Solution 1 (Reverse use of sum-to-product formulas)cos frac{2 pi}{5}+cos frac{4 pi}{5}=2 cos frac{pi}{5} cos frac{3 pi}{5}=-2 cos frac{pi}{5} cos frac{2 pi}{5},Multiplying the numerator and denominator by 4 sin frac{pi}{5}, and using the double-angle formula twice, we get the value as -frac{1}{2}.Solution 2 (Solving by constructing a conjugate expression)Let x=cos frac{2 pi}{5}+cos frac{4 pi}{5}, y=cos frac{2 pi}{5}-cos frac{4 pi}{5}, x y=cos ^{2} frac{2 pi}{5}-cos ^{2} frac{4 pi}{5}= frac{1}{2}left(1+cos frac{4 pi}{5}right)-frac{1}{2}left(1+cos frac{8 pi}{5}right)=frac{1}{2}left(cos frac{4 pi}{5}-cos frac{2 pi}{5}right)=-frac{1}{2} y, canceling y(y>0) gives x=-frac{1}{2}.Solution 3 (Constructing an equation by self-substitution)x=cos frac{2 pi}{5}+cos frac{4 pi}{5}=cos frac{2 pi}{5}-cos frac{pi}{5}<0, squaring gives x^{2}=cos ^{2} frac{2 pi}{5}+cos ^{2} frac{4 pi}{5}+2 cos frac{2 pi}{5} cos frac{4 pi}{5}=frac{1}{2}left(1+cos frac{4 pi}{5}right)+frac{1}{2}left(1+cos frac{8 pi}{5}right)+2 frac{sin frac{4 pi}{5} sin frac{8 pi}{5}}{2 sin frac{2 pi}{5} cdot 2 sin frac{4 pi}{5}}=frac{1}{2} +frac{1}{2}left(cos frac{4 pi}{5}+cos frac{8 pi}{5}right)=frac{1}{2}+frac{1}{2}left(cos frac{4 pi}{5}+cos frac{2 pi}{5}right)=frac{1}{2}+frac{x}{2}, thus x^{2}=frac{1}{2}+ frac{1}{2} x, solving gives x=-frac{1}{2}.Solution 4 (Constructing a similar equation)Let cos x+cos 2 x=cos frac{2 pi}{5}+cos frac{4 pi}{5}, then cos x=cos frac{2 pi}{5}, cos frac{4 pi}{5} simultaneously satisfy this similar equation. Using the double-angle formula, we get the quadratic equation 2 cos ^{2} x+cos x-left(1+cos frac{2 pi}{5}+cos frac{4 pi}{5}right)=0, which indicates that cos frac{2 pi}{5}, cos frac{4 pi}{5} are the roots of the equation 2 y^{2}+y-left(1+cos frac{2 pi}{5}+cos frac{4 pi}{5}right)=0, and are all the roots. By the relationship between roots and coefficients, we get cos frac{2 pi}{5}+cos frac{4 pi}{5}=-frac{1}{2}.

🧠:[<think>Okay, let's tackle this problem step by step. There are four propositions given, and I need to determine which ones are true. The options are A, B, C, D, each combining different propositions. Let me go through each proposition one by one.First, proposition ①: "Complementary angles are congruent." Hmm. Complementary angles are two angles that add up to 90 degrees. But does that mean they have to be congruent? Let me think. For example, if one angle is 30 degrees and the other is 60 degrees, they are complementary because 30 + 60 = 90, but they are not congruent. So, complementary angles don't necessarily have to be equal. Therefore, proposition ① is false. So, any option that includes ① (which is option C: ①②③) can be eliminated. That leaves options A, B, D. But wait, A includes ① and ④. Since ① is false, A is out. So remaining options are B and D.Moving on to proposition ②: "The bisectors of a pair of adjacent supplementary angles are perpendicular to each other." Let's parse this. Adjacent supplementary angles are two angles that are next to each other and their measures add up to 180 degrees. So, if you have two angles, say, angle A and angle B, which are adjacent and supplementary, each of them is 90 degrees? Wait, no. Wait, adjacent supplementary angles form a linear pair, which means they are adjacent and their non-common sides form a straight line. But their measures add up to 180 degrees. So, for example, angle A could be 120 degrees and angle B 60 degrees. Then, the bisectors of each angle would split them into two equal parts.So, if angle A is 120 degrees, its bisector would create two 60-degree angles. Similarly, angle B is 60 degrees, so its bisector would create two 30-degree angles. Now, are the bisectors perpendicular? Let's visualize this. If the original angles are adjacent and supplementary, their sum is 180 degrees. Let me denote the two angles as x and 180 - x degrees. The bisector of angle x divides it into two angles of x/2 each, and the bisector of angle 180 - x divides it into two angles of (180 - x)/2 each. The angle between the two bisectors would be the sum of x/2 and (180 - x)/2. Wait, no. Wait, the bisectors themselves are lines that split the original angles. Let me think more carefully.Suppose we have two adjacent supplementary angles forming a straight line. Let me draw this mentally: two angles, let's say angle AOB and angle BOC, with point O being the vertex, and points A, B, C lying on a straight line. So angle AOB + angle BOC = 180 degrees. Now, the bisector of angle AOB would split it into two equal parts, each of measure (AOB)/2. Similarly, the bisector of angle BOC would split it into two equal parts, each of measure (BOC)/2. Now, what's the angle between these two bisectors?Since the original angles are adjacent and supplementary, AOB + BOC = 180 degrees. Let’s denote angle AOB as θ, so angle BOC is 180 - θ. The bisector of angle AOB divides it into θ/2 each, and the bisector of angle BOC divides it into (180 - θ)/2 each. Now, if we consider the angle between the two bisectors, how would that be calculated?The angle between the two bisectors would be equal to the angle between their respective directions. Since they are bisecting angles θ and 180 - θ, which are adjacent, the angle between the bisectors would be θ/2 + (180 - θ)/2. Let me compute that: θ/2 + (180 - θ)/2 = (θ + 180 - θ)/2 = 180/2 = 90 degrees. Wait, that's 90 degrees. So regardless of the value of θ, the angle between the bisectors is always 90 degrees. That means the bisectors are perpendicular to each other. So proposition ② is actually true.Wait, but earlier I thought of specific angles like 120 and 60. Let's test that. If θ is 120, then the bisectors would split them into 60 and 30 degrees. The angle between the bisectors would be 60 + 30 = 90 degrees. Yes, that's right. Similarly, if θ is 150, the bisectors would split into 75 and 15 degrees, which also sum to 90. So regardless, the angle between the bisectors is 90 degrees. Therefore, proposition ② is true. Hmm, so I was wrong earlier when I thought ② might be false. Therefore, proposition ② is actually true. Therefore, options that include ② are C and D. But we had already eliminated C because it includes ①. So D includes ②, ③, ④.But previously, we thought that since ① is false, A and C are out, leaving B and D. But if ② is true, then D is a candidate. So let's re-examine.Wait, hold on. Initially, I thought ① was false, which eliminates options C and A. Then proposition ② is actually true, so among the remaining options B (③④) and D (②③④), since ② is true, D includes an additional true proposition. So we need to check ③ and ④.Proposition ③: "Through a point outside a line, there is one and only one line parallel to this line." This is one of Euclid's axioms, the parallel postulate. In Euclidean geometry, this is true. However, in non-Euclidean geometries, this isn't the case. But I think the question is assuming we are working in Euclidean geometry, so ③ is true.Proposition ④: "Among all line segments connecting a point outside a line to points on the line, the perpendicular segment is the shortest." This is a well-known theorem in geometry. The distance from a point to a line is defined as the length of the perpendicular segment, which is indeed the shortest among all such segments. So ④ is true.Therefore, propositions ②, ③, ④ are all true. Therefore, the correct answer is D: ②③④.Wait, but initially, I thought ② was false, but after working through the math, it turned out to be true. So, the key was to carefully analyze proposition ②, which upon examination, is actually true. Therefore, the answer is D.Final Answerboxed{D}</think>Let's analyze each proposition step by step:1. Proposition ①: "Complementary angles are congruent." Complementary angles add up to 90 degrees, but they do not necessarily have to be congruent. For example, 30° and 60° are complementary but not congruent. Therefore, proposition ① is false.2. Proposition ②: "The bisectors of a pair of adjacent supplementary angles are perpendicular to each other." Adjacent supplementary angles sum to 180 degrees. If each angle is bisected, the angle between the bisectors is calculated as follows: Let the angles be ( theta ) and ( 180^circ - theta ). The bisectors split these angles into ( theta/2 ) and ( (180^circ - theta)/2 ). The angle between the bisectors is ( theta/2 + (180^circ - theta)/2 = 90^circ ). Hence, proposition ② is true.3. Proposition ③: "Through a point outside a line, there is one and only one line parallel to this line." This is Euclid's parallel postulate, which is true in Euclidean geometry. Therefore, proposition ③ is true.4. Proposition ④: "Among all line segments connecting a point outside a line to points on the line, the perpendicular segment is the shortest." This is a well-known theorem in geometry, stating the perpendicular segment is the shortest distance. Hence, proposition ④ is true.After evaluating all propositions, propositions ②, ③, and ④ are true. Therefore, the correct answer is:boxed{D},<think>Okay, so I need to figure out which of these four propositions are true. The options are given as ①, ②, ③, and ④, and the answer choices combine different numbers. Let me go through each proposition one by one.Starting with ①: "Complementary angles are congruent." Hmm. Complementary angles add up to 90 degrees, right? So if two angles are complementary, their measures sum to 90. But does that mean they have to be congruent? Let me think. For example, if one angle is 30 degrees and the other is 60 degrees, they are complementary but not congruent. So ① is not necessarily true. So ① is false. That means any answer choice that includes ① (like A or C) is incorrect. So we can rule out options A and C. Now, between B and D.Next, ②: "The bisectors of a pair of adjacent supplementary angles are perpendicular to each other." Let me visualize this. Adjacent supplementary angles add up to 180 degrees and share a common side. If we bisect each of these angles, then the bisectors should form some angle. Let's say the two angles are 180 degrees total. Let each be x and 180 - x. Then their bisectors would split them into x/2 and (180 - x)/2. The angle between the bisectors would be x/2 + (180 - x)/2. Wait, no. Let me think again. If the two original angles are adjacent, forming a linear pair, then their bisectors would each split their respective angles. The angle between the two bisectors would be the sum of half of each original angle. Since the original angles add up to 180, each bisected angle is half of that. So, each bisected angle is x/2 and (180 - x)/2. So the angle between the bisectors is x/2 + (180 - x)/2 = (x + 180 - x)/2 = 180/2 = 90 degrees. So the angle between the bisectors is 90 degrees, which means they are perpendicular. So ② is true. Wait, so ② is true? But if ② is true, then answer D includes ②, ③, ④. But in the answer choices, B is ③④ and D is ②③④. So if ② is true, then D is possible. But I thought ① is false, so C is out. So maybe D is correct. But let me confirm.Wait, let me double-check ②. Suppose two adjacent supplementary angles: let's say 100 degrees and 80 degrees. Then their bisectors would split them into 50 degrees and 40 degrees. So the angle between the bisectors would be 50 + 40 = 90 degrees. Hmm, that works. Another example: 150 and 30. Bisectors split into 75 and 15, sum is 90. So yes, regardless of the original angles, the bisectors would always add up to 90 degrees. Therefore, the angle between the bisectors is 90 degrees, so they are perpendicular. Therefore, ② is true. Okay, so ② is true.Now, moving to ③: "Through a point outside a line, there is one and only one line parallel to this line." This sounds like the parallel postulate from Euclidean geometry. In Euclidean geometry, this is indeed true. However, in other geometries like hyperbolic, there are infinitely many lines through a point that don't intersect the given line. But assuming we're working in standard Euclidean geometry (which is typical unless stated otherwise), this is true. So ③ is true.Then, ④: "Among all line segments connecting a point outside a line to points on the line, the perpendicular segment is the shortest." This is a standard theorem in geometry. The shortest distance from a point to a line is the perpendicular distance. So all other segments from the point to the line are longer than the perpendicular segment. Therefore, ④ is true.So propositions ②, ③, and ④ are all true, which corresponds to option D. But let me just confirm once again if ② is indeed true. Let's take another example. Suppose two adjacent angles: both 90 degrees each. Then their bisectors would split them into 45 degrees each. The angle between the bisectors would be 45 + 45 = 90 degrees. Still perpendicular. If one angle is 180 and the other is 0, but adjacent angles can't be 0 and 180, they must be supplementary and adjacent, so each has to be between 0 and 180, not inclusive. Wait, no, adjacent supplementary angles can be 180 degrees together, but each can be any measure as long as they add up to 180 and are adjacent. So even if one is acute and the other is obtuse, their bisectors would still split them into two angles adding to 90. So yes, regardless of the measures, the angle between the bisectors is 90 degrees. Therefore, ② is indeed true.Therefore, the correct answer is D: ②③④. So among the options given, D is the correct choice.Final Answerboxed{D}</think>To determine which propositions are true, we analyze each one step by step:1. Proposition ①: "Complementary angles are congruent." - Complementary angles add up to 90 degrees, but they do not necessarily have to be congruent. For example, 30° and 60° are complementary but not congruent. - Conclusion: False.2. Proposition ②: "The bisectors of a pair of adjacent supplementary angles are perpendicular to each other." - Adjacent supplementary angles sum to 180°. If each angle is bisected, the resulting angles are half of the original angles. The angle between the bisectors is the sum of these halves, which is ( frac{x}{2} + frac{180^circ - x}{2} = 90^circ ). Thus, the bisectors are perpendicular. - Conclusion: True.3. Proposition ③: "Through a point outside a line, there is one and only one line parallel to this line." - This is the parallel postulate in Euclidean geometry, which is true. - Conclusion: True.4. Proposition ④: "Among all line segments connecting a point outside a line to points on the line, the perpendicular segment is the shortest." - This is a standard theorem in geometry, stating the perpendicular segment is the shortest distance. - Conclusion: True.Propositions ②, ③, and ④ are true. Therefore, the correct answer is:boxed{D}]

🔑:Let's analyze each proposition step by step:① Complementary angles are congruent;- Complementary angles are defined as two angles whose sum is 90^circ. The statement that complementary angles are congruent (meaning they have the same measure) is not universally true. For example, a 30^circ angle and a 60^circ angle are complementary but not congruent.- Therefore, this proposition is false.② The bisectors of a pair of adjacent supplementary angles are perpendicular to each other;- Supplementary angles are two angles whose sum is 180^circ. If you bisect each of these angles, you create two pairs of angles that each sum to 90^circ. Since the bisectors divide the angles into two equal parts, the angles formed between the bisectors are also supplementary, and because they sum to 90^circ, the bisectors are perpendicular.- This proposition is true.③ Through a point outside a line, there is one and only one line parallel to this line;- This statement is a fundamental principle of Euclidean geometry, known as the Parallel Postulate. It asserts that, given a line and a point not on the line, there is exactly one line through the point that does not intersect the original line, meaning it is parallel.- This proposition is true.④ Among all line segments connecting a point outside a line to points on the line, the perpendicular segment is the shortest;- This is a principle derived from the Pythagorean theorem. For any point not on a line, the shortest distance to the line is the length of the perpendicular segment from the point to the line. Any other segment from the point to the line will form the hypotenuse of a right-angled triangle, which is always longer than the legs.- This proposition is true.Given the analysis:- Proposition ① is false.- Propositions ②, ③, and ④ are true.Therefore, the correct answer is boxed{D}.

🧠:[<think>Okay, let's see. I need to find the limit as x approaches 0 of [3^(x+1) - 3] divided by ln(1 + x*sqrt(1 + x*e^x)). Hmm. Limits can sometimes be tricky, especially when dealing with exponentials and logarithms. Let me start by analyzing both the numerator and the denominator separately to see if I can simplify them or apply standard limit rules like L'Hôpital's Rule if necessary.First, the numerator: 3^(x+1) - 3. Let's rewrite 3^(x+1) as 3*3^x. So the numerator becomes 3*3^x - 3, which factors out a 3, giving 3*(3^x - 1). That might be helpful because I know that the limit of (a^x - 1)/x as x approaches 0 is ln(a). So maybe the numerator can be approximated as 3*(x*ln(3)) when x is near 0. Let me check that. If x approaches 0, then 3^x ≈ 1 + x*ln(3) by the Taylor series expansion. Therefore, 3^x - 1 ≈ x*ln(3), so the numerator 3*(3^x - 1) ≈ 3*x*ln(3). So that's the numerator's approximation for small x.Now the denominator: ln(1 + x*sqrt(1 + x*e^x)). Let me focus on simplifying the argument of the natural logarithm first. Let's denote the inner expression as A = x*sqrt(1 + x*e^x). So the denominator is ln(1 + A). For small x, A should be small, so ln(1 + A) ≈ A - A^2/2 + A^3/3 - ... But since x is approaching 0, A will be approaching 0 as well, so we can approximate ln(1 + A) ≈ A. So maybe I can approximate the denominator as just A, which is x*sqrt(1 + x*e^x). But maybe I need to consider higher-order terms? Let's see.But before that, let me check if both the numerator and denominator approach 0. If they do, then maybe I can apply L'Hôpital's Rule. Let's evaluate the limit as x approaches 0.First, numerator: When x approaches 0, 3^(0+1) - 3 = 3 - 3 = 0. So numerator is 0.Denominator: ln(1 + 0*sqrt(1 + 0*e^0)) = ln(1 + 0) = ln(1) = 0. So we have 0/0, which is indeterminate. Therefore, L'Hôpital's Rule can be applied. That means the limit is equal to the limit of the derivative of the numerator divided by the derivative of the denominator.Alternatively, maybe expanding both numerator and denominator using Taylor series would be easier. Let's try both approaches.First, let's try the Taylor series expansion approach.Starting with the numerator: 3*(3^x - 1). As mentioned before, 3^x = e^{x*ln(3)}, so expanding that as a Taylor series around x=0: e^{x*ln(3)} ≈ 1 + x*ln(3) + (x^2*(ln(3))^2)/2 + ... Therefore, 3^x - 1 ≈ x*ln(3) + (x^2*(ln(3))^2)/2 + ..., so multiplying by 3 gives the numerator ≈ 3*x*ln(3) + (3*x^2*(ln(3))^2)/2 + ... So the leading term is 3*ln(3)*x.Now the denominator: ln(1 + x*sqrt(1 + x*e^x)). Let's first compute sqrt(1 + x*e^x). Let me denote B = x*e^x. Then sqrt(1 + B) can be expanded as 1 + (1/2)B - (1/8)B^2 + ... for small B. Since x approaches 0, B = x*e^x ≈ x*(1 + x + x^2/2 + ...) ≈ x + x^2 + x^3/2 + ... So B is approximately x for small x. Then sqrt(1 + B) ≈ 1 + (1/2)B - (1/8)B^2 + ... ≈ 1 + (1/2)(x + x^2) - (1/8)x^2 + ... ≈ 1 + (1/2)x + (1/2)x^2 - (1/8)x^2 + ... = 1 + (1/2)x + (3/8)x^2 + ... So up to the first order, sqrt(1 + x*e^x) ≈ 1 + (1/2)x. Therefore, multiplying by x, we get x*sqrt(1 + x*e^x) ≈ x*(1 + (1/2)x) = x + (1/2)x^2.Therefore, the argument inside the logarithm is approximately x + (1/2)x^2. So ln(1 + x + (1/2)x^2). Let's denote C = x + (1/2)x^2. Then ln(1 + C) ≈ C - C^2/2 + C^3/3 - ... For small C (since x approaches 0), we can approximate ln(1 + C) ≈ C - C^2/2. Let's compute that:C = x + (1/2)x^2C^2 = (x + (1/2)x^2)^2 = x^2 + x^3 + (1/4)x^4 ≈ x^2 (since higher powers are negligible for small x)Therefore, ln(1 + C) ≈ C - C^2/2 ≈ (x + (1/2)x^2) - (x^2)/2 = x + (1/2)x^2 - (1/2)x^2 = x. So up to the first order, the denominator is approximately x. So denominator ≈ x.Putting numerator and denominator together: numerator ≈ 3*ln(3)*x, denominator ≈ x, so the limit as x approaches 0 is (3*ln(3)*x)/x = 3*ln(3). Therefore, the limit should be 3 ln 3.Wait, but let's verify this with L'Hôpital's Rule to make sure.First, applying L'Hôpital's Rule: take the derivative of the numerator and denominator.Numerator: d/dx [3^(x+1) - 3] = 3^(x+1) * ln(3). Because derivative of 3^u is 3^u * ln(3) * u', where u = x + 1, so u' = 1. So derivative is 3^(x+1)*ln(3). At x = 0, this is 3^(1)*ln(3) = 3 ln(3).Denominator: d/dx [ln(1 + x*sqrt(1 + x e^x))]. Let's compute this derivative.Let me denote D = 1 + x*sqrt(1 + x e^x). Then the derivative is (1/D) * derivative of D.Derivative of D is derivative of x*sqrt(1 + x e^x). Let's compute that:Let me denote f(x) = x and g(x) = sqrt(1 + x e^x). Then the derivative is f'(x)g(x) + f(x)g'(x).f'(x) = 1.g(x) = sqrt(1 + x e^x) = (1 + x e^x)^(1/2). So g'(x) = (1/2)(1 + x e^x)^(-1/2) * derivative of (1 + x e^x).Derivative of (1 + x e^x) is e^x + x e^x. Therefore, g'(x) = (1/2)(1 + x e^x)^(-1/2)*(e^x + x e^x).Putting it all together, derivative of D is:1 * sqrt(1 + x e^x) + x * [ (1/2)(1 + x e^x)^(-1/2)*(e^x + x e^x) ]At x = 0:First term: sqrt(1 + 0*e^0) = sqrt(1) = 1.Second term: 0 * [ ... ] = 0.Therefore, derivative of D at x=0 is 1 + 0 = 1. Hence, derivative of the denominator is (1/D) * derivative of D, but at x=0, D = 1 + 0*sqrt(...) = 1. Therefore, the derivative of the denominator at x=0 is (1/1)*1 = 1.Therefore, by L'Hôpital's Rule, the limit is numerator derivative / denominator derivative at x=0, which is 3 ln(3)/1 = 3 ln(3). So this confirms the earlier result from the Taylor series expansion.Therefore, the answer should be 3 ln 3. To present it in boxed notation, that's boxed{3 ln 3}.Final Answerboxed{3 ln 3}</think>To find the limit of the function:lim _{x rightarrow 0} frac{3^{x+1}-3}{ln left(1+x sqrt{1+x e^{x}}right)}we start by analyzing the numerator and denominator separately.Numerator Analysis:The numerator is (3^{x+1} - 3). Rewriting (3^{x+1}) as (3 cdot 3^x), we get:3^{x+1} - 3 = 3(3^x - 1)Using the Taylor series expansion for (3^x) around (x = 0), we have:3^x approx 1 + x ln 3 + frac{(x ln 3)^2}{2} + cdotsThus,3^x - 1 approx x ln 3So, the numerator approximates to:3(3^x - 1) approx 3x ln 3Denominator Analysis:The denominator is (ln(1 + x sqrt{1 + x e^x})). Let (B = x e^x). For small (x), (B approx x). Then,sqrt{1 + x e^x} approx 1 + frac{1}{2}xThus,x sqrt{1 + x e^x} approx x left(1 + frac{1}{2}xright) = x + frac{1}{2}x^2The argument inside the logarithm is approximately (x + frac{1}{2}x^2). Using the Taylor series expansion for (ln(1 + C)) where (C = x + frac{1}{2}x^2), we get:ln(1 + x + frac{1}{2}x^2) approx xApplying L'Hôpital's Rule:Since both the numerator and denominator approach 0 as (x to 0), we apply L'Hôpital's Rule. The derivatives are:- Numerator derivative: (frac{d}{dx} [3^{x+1} - 3] = 3^{x+1} ln 3). At (x = 0), this is (3 ln 3).- Denominator derivative: (frac{d}{dx} ln(1 + x sqrt{1 + x e^x})). Using the chain rule, the derivative at (x = 0) is 1.Thus, the limit is:frac{3 ln 3}{1} = 3 ln 3The final answer is:boxed{3 ln 3},<think>Okay, so I have this limit problem to solve: the limit as x approaches 0 of [3^(x+1) - 3] divided by the natural logarithm of [1 + x times the square root of (1 + x e^x)]. Hmm, let's see. I need to calculate this limit. Alright, let me start by understanding each part of the function and then see how they behave as x approaches 0. First, the numerator is 3^(x+1) minus 3. Let me rewrite 3^(x+1) as 3*3^x because 3^(x+1) is the same as 3^x multiplied by 3^1. So that would make the numerator 3*3^x - 3, which factors out a 3, so 3*(3^x - 1). So the numerator simplifies to 3*(3^x - 1). Then, the denominator is ln[1 + x*sqrt(1 + x e^x)]. Alright, so as x approaches 0, both the numerator and the denominator approach 0. Let's check that. For the numerator: when x approaches 0, 3^x approaches 1, so 3*(1 - 1) = 0. For the denominator: inside the log, we have 1 + x*sqrt(1 + x e^x). When x approaches 0, x e^x approaches 0*1 = 0, so sqrt(1 + 0) is 1. Therefore, inside the log, we have 1 + x*1 = 1 + x. Then ln(1 + x) as x approaches 0 is ln(1) which is 0. So yes, both numerator and denominator approach 0, so we have a 0/0 indeterminate form. That means we can apply L'Hospital's Rule here. But before jumping into L'Hospital's, maybe there's a way to expand these expressions using Taylor series or use known limits. Let me think. For the numerator, 3^x can be expressed as e^{x ln 3}, so 3^x - 1 is e^{x ln 3} - 1. The expansion of e^y around y=0 is 1 + y + y^2/2! + ..., so e^{x ln 3} - 1 ≈ x ln 3 + (x ln 3)^2 / 2 + ... So the numerator is 3*(x ln 3 + higher order terms). So approximately 3 x ln 3 for small x.For the denominator, ln(1 + something) when that something is small can be approximated by that something minus (something)^2/2 + ... So if we let the argument inside the log be 1 + z, where z = x*sqrt(1 + x e^x). So first, let's approximate z. Let's look at sqrt(1 + x e^x). Again, when x is near 0, e^x is approximately 1 + x + x^2/2, so x e^x ≈ x*(1 + x + x^2/2) ≈ x + x^2 + x^3/2. Then, inside the square root, we have 1 + x e^x ≈ 1 + x + x^2 + x^3/2. The square root of 1 + a small term is approximately 1 + (small term)/2 - (small term)^2/8 + ... So sqrt(1 + x e^x) ≈ 1 + (x + x^2)/2 - (x + x^2)^2 / 8. Let me compute that:First, expand the square root up to x^2 terms. Let me denote the small term as h = x e^x ≈ x + x^2 + x^3/2. Then sqrt(1 + h) ≈ 1 + h/2 - h^2/8. So substituting h ≈ x + x^2:sqrt(1 + h) ≈ 1 + (x + x^2)/2 - (x + x^2)^2 /8. Let's compute (x + x^2)^2: that's x^2 + 2x^3 + x^4. So then:sqrt(1 + h) ≈ 1 + x/2 + x^2/2 - (x^2)/8 + higher order terms. Wait, let's do the calculation step by step:First term: 1 + h/2 = 1 + (x + x^2)/2 = 1 + x/2 + x^2/2.Second term: subtract h^2 /8. h^2 = (x + x^2)^2 = x^2 + 2x^3 + x^4. So h^2 /8 = x^2/8 + x^3/4 + x^4/8. Therefore, subtracting that:1 + x/2 + x^2/2 - x^2/8 - x^3/4 - x^4/8. Combining like terms:Constant term: 1x term: x/2x^2 terms: x^2/2 - x^2/8 = (4x^2 - x^2)/8 = 3x^2/8x^3 term: -x^3/4x^4 term: -x^4/8So up to x^2, sqrt(1 + h) ≈ 1 + x/2 + 3x^2/8.Therefore, sqrt(1 + x e^x) ≈ 1 + x/2 + 3x^2/8. Then multiplying by x:z = x*sqrt(1 + x e^x) ≈ x*(1 + x/2 + 3x^2/8) = x + x^2/2 + 3x^3/8.Then, the denominator is ln(1 + z) where z ≈ x + x^2/2. Since z approaches 0 as x approaches 0, we can approximate ln(1 + z) ≈ z - z^2/2 + z^3/3 - ... So let's compute up to x^2 terms:z ≈ x + x^2/2, so z^2 ≈ (x + x^2/2)^2 = x^2 + x^3 + x^4/4. Therefore:ln(1 + z) ≈ z - z^2/2 ≈ (x + x^2/2) - (x^2)/2 = x + x^2/2 - x^2/2 = x. Wait, that cancels out the x^2 term. Hmm, so is it just x? Let me check again.Wait, z is approximately x + x^2/2 + 3x^3/8, but maybe we need to consider higher terms? Wait, perhaps when we compute z, the x^3 term might not be necessary if we are only going up to x^2 in the denominator.Wait, let me re-express:If we take sqrt(1 + x e^x) ≈ 1 + x/2 + 3x^2/8, then multiplying by x gives z = x + x^2/2 + 3x^3/8. Then ln(1 + z) ≈ z - z^2/2 + z^3/3 - ...So let's compute z:z = x + x^2/2 + 3x^3/8z^2 = (x + x^2/2 + 3x^3/8)^2 = x^2 + x^3 + (3x^4/4 + ...). Up to x^3 terms, z^2 ≈ x^2 + x^3.z^3 = x^3 + ... So up to x^3 terms, z^3 ≈ x^3.Therefore, ln(1 + z) ≈ z - z^2/2 + z^3/3 ≈ [x + x^2/2 + 3x^3/8] - [x^2 + x^3]/2 + x^3/3.Let's compute term by term:First term: x + x^2/2 + 3x^3/8Second term: subtract (x^2 + x^3)/2: -x^2/2 - x^3/2Third term: add x^3/3Combine all terms:x + (x^2/2 - x^2/2) + (3x^3/8 - x^3/2 + x^3/3)Simplify:x + 0 + x^3*(3/8 - 1/2 + 1/3). Let's compute the coefficients:3/8 - 1/2 + 1/3 = 3/8 - 4/8 + 1/3 = (-1/8) + 1/3 = (-3/24 + 8/24) = 5/24So ln(1 + z) ≈ x + (5/24) x^3. Therefore, the denominator is approximately x + (5/24) x^3.But since we are taking the limit as x approaches 0, the dominant term is x. So perhaps up to the first order, the denominator is approximately x. But if we need a better approximation to compute the limit, since both numerator and denominator approach 0, maybe we need to consider higher order terms?Wait, the numerator is 3*(3^x -1). Let's expand 3^x as e^{x ln3} ≈ 1 + x ln3 + (x ln3)^2 / 2 + (x ln3)^3 / 6 + ... So 3^x -1 ≈ x ln3 + (x ln3)^2 / 2 + (x ln3)^3 / 6. Multiply by 3: numerator ≈ 3x ln3 + 3(x ln3)^2 / 2 + 3(x ln3)^3 / 6 + ...Denominator: as above, approximated as x + (5/24)x^3.So if we write the numerator as 3x ln3 + (3 (ln3)^2 / 2)x^2 + ... and denominator as x + (5/24)x^3, then if we divide numerator by denominator:[3x ln3 + (3 (ln3)^2 / 2)x^2 + ...] / [x + (5/24)x^3] = [3 ln3 + (3 (ln3)^2 / 2)x + ...] / [1 + (5/24)x^2 + ...]As x approaches 0, the denominator [1 + (5/24)x^2 + ...] approaches 1, so the entire expression approaches 3 ln3. Therefore, the limit would be 3 ln3. Wait, but let's check if the first-order terms are sufficient. Since both numerator and denominator are approaching zero like x, so their ratio should approach 3 ln3 / 1 = 3 ln3. So is that the answer?Alternatively, maybe I can use L'Hospital's Rule once. Let's try that.Given that both numerator and denominator approach 0, we can apply L'Hospital's Rule, which says that the limit is equal to the limit of the derivatives of numerator and denominator.So first, compute derivative of numerator: d/dx [3^{x+1} - 3] = 3^{x+1} * ln3. Because derivative of 3^{x+1} is 3^{x+1} * ln3 * derivative of (x+1), which is 1. So derivative is 3^{x+1} ln3.Derivative of denominator: d/dx [ln(1 + x sqrt(1 + x e^x})]. Let's denote the inside of the log as f(x) = 1 + x sqrt(1 + x e^x). Then derivative of ln(f(x)) is f'(x)/f(x).Compute f'(x): derivative of 1 is 0, plus derivative of x sqrt(1 + x e^x). Use product rule: derivative of x times sqrt(1 + x e^x) plus x times derivative of sqrt(1 + x e^x).So:f'(x) = sqrt(1 + x e^x) + x * [ (1/(2 sqrt(1 + x e^x))) * (e^x + x e^x) ]Wait, derivative of sqrt(u) is (1/(2 sqrt(u))) * u'. Here, u = 1 + x e^x, so u' = e^x + x e^x. Therefore:f'(x) = sqrt(1 + x e^x) + x * [ (e^x + x e^x) / (2 sqrt(1 + x e^x)) ) ].Simplify this:= sqrt(u) + x * (u' ) / (2 sqrt(u)) where u = 1 + x e^x.So, f'(x) = sqrt(u) + [x u'] / (2 sqrt(u)).Combine terms:= [2 u + x u'] / (2 sqrt(u)).But maybe it's better to leave it as is for now.So, then the derivative of the denominator at x is [f'(x)] / f(x).Therefore, applying L'Hospital's Rule, the limit becomes:lim_{x→0} [3^{x+1} ln3] / [f'(x)/f(x)].Wait, no. Wait, the derivative of the numerator is 3^{x+1} ln3, and derivative of denominator is f'(x)/f(x). Wait, no: actually, L'Hospital's Rule says that the limit of [numerator]/[denominator] is equal to the limit of [numerator'] / [denominator'].Therefore, the original limit is equal to:lim_{x→0} [3^{x+1} ln3] / [f'(x)].But wait, denominator was ln(f(x)), so derivative is f'(x)/f(x), so when applying L'Hospital's, the derivative of the denominator is [f'(x)/f(x)]. Therefore, the ratio after L'Hospital's is [3^{x+1} ln3] / [f'(x)/f(x)] = [3^{x+1} ln3 * f(x)] / f'(x).But since f(x) = 1 + x sqrt(1 + x e^x), and as x approaches 0, f(x) approaches 1 + 0 = 1. Similarly, f'(x) as x approaches 0:Let's compute f'(0). Let's compute u = 1 + x e^x. At x=0, u = 1 + 0 =1, u' = e^0 + 0*e^0 =1.So f'(0) = sqrt(u) + x * (u')/(2 sqrt(u)) evaluated at x=0.sqrt(u) = sqrt(1) =1. The second term is x*(1)/(2*1) = x/2, which at x=0 is 0. Therefore, f'(0)=1 +0=1.Therefore, the limit becomes [3^{0+1} ln3 * f(0)] / f'(0) = [3*1 * ln3 * 1] / 1 = 3 ln3. So that's the same result as before.So whether I use the Taylor series approximation or L'Hospital's Rule once, I get the same answer: 3 ln3.Wait, but just to be thorough, let me verify if applying L'Hospital's was valid here. The original limit is 0/0, so L'Hospital's can be applied once, and after applying, we get a new limit. Let's check that when x approaches 0, the new numerator is 3^{x+1} ln3, which approaches 3*1*ln3 = 3 ln3. The new denominator is f'(x)/f(x). f'(x) as x approaches 0 is 1, and f(x) approaches 1, so the denominator approaches 1/1=1. Therefore, the limit is 3 ln3 /1 =3 ln3.Therefore, the limit is 3 ln3. So the answer is 3 ln3, which can be written as 3 natural logarithm of 3. Therefore, boxed, it would be boxed{3 ln 3}.But let me just cross-verify once again using another approach to be safe. Suppose I use substitution. Let x approach 0. Let me plug in x=0 into numerator and denominator:Numerator: 3^{0+1} -3=3-3=0.Denominator: ln(1 +0*sqrt(...))=ln1=0. So 0/0, so indeterminate. Therefore, L'Hospital's applies.Alternatively, perhaps using substitution variables. Let me set t =x as x approaches 0.Wait, not sure. Alternatively, using the standard limit lim_{x→0} (a^x -1)/x = ln a.In the numerator: 3^{x+1} -3=3(3^x -1). So numerator is 3*(3^x -1). So the numerator is 3*(3^x -1). The denominator is ln(1 +x*sqrt(1 +x e^x)}.So, if I write the expression as:[3*(3^x -1)] / [ln(1 +x*sqrt(1 +x e^x))]Divide numerator and denominator by x:3*( (3^x -1)/x ) / [ (ln(1 +x*sqrt(1 +x e^x)) ) /x ]Now, as x approaches 0, (3^x -1)/x approaches ln3, by the standard limit. Similarly, the denominator part: [ln(1 +x*sqrt(1 +x e^x})]/x. Let me set z = x*sqrt(1 +x e^x). Then, as x→0, z≈x*1= x. So [ln(1 + z)] / z * (z/x). Then, as x→0, z/x = sqrt(1 +x e^x} ≈1. Therefore, [ln(1 + z)/z] * [z/x] ≈1*1=1. Wait, let's see:Let me write [ln(1 + z)] /x = [ln(1 + z)/z] * [z/x].Now, z =x*sqrt(1 +x e^x), so z/x = sqrt(1 +x e^x). As x→0, sqrt(1 +x e^x) approaches 1. Therefore, [ln(1 + z)/z] approaches 1 as z→0. Therefore, the entire denominator part [ln(1 + z)] /x approaches 1*1=1. Therefore, the denominator divided by x approaches1. Therefore, the original expression is 3*ln3 /1=3 ln3. So same result.Therefore, this confirms that the limit is indeed 3 ln3. So the answer is boxed{3 ln 3}.Final Answerboxed{3 ln 3}</think>To calculate the limit:lim _{x rightarrow 0} frac{3^{x+1}-3}{ln left(1+x sqrt{1+x e^{x}}right)}we start by recognizing that both the numerator and the denominator approach 0 as (x) approaches 0, resulting in a (0/0) indeterminate form. This allows us to use L'Hospital's Rule.First, we rewrite the numerator:[3^{x+1} - 3 = 3 cdot 3^x - 3 = 3(3^x - 1)]As (x rightarrow 0), (3^x - 1) can be approximated using the Taylor series expansion (e^{x ln 3} - 1 approx x ln 3). Thus, the numerator approximates to (3 cdot x ln 3).For the denominator, we consider the argument inside the natural logarithm:[ln left(1 + x sqrt{1 + x e^x}right)]As (x rightarrow 0), (x e^x approx x), so (sqrt{1 + x e^x} approx 1 + frac{x}{2}). Multiplying by (x), we get:[x sqrt{1 + x e^x} approx x left(1 + frac{x}{2}right) = x + frac{x^2}{2}]Using the approximation (ln(1 + z) approx z) for small (z), the denominator approximates to (x).Applying L'Hospital's Rule, we differentiate the numerator and the denominator. The derivative of the numerator (3(3^x - 1)) is:[3 cdot 3^x ln 3]Evaluated at (x = 0), this is (3 ln 3).For the derivative of the denominator (ln(1 + x sqrt{1 + x e^x})), we use the chain rule. Let (f(x) = 1 + x sqrt{1 + x e^x}). The derivative of (ln(f(x))) is:[frac{f'(x)}{f(x)}]At (x = 0), (f(x) approx 1) and (f'(x)) simplifies to 1, so the derivative of the denominator is 1.Thus, applying L'Hospital's Rule, the limit becomes:[frac{3 ln 3}{1} = 3 ln 3]Therefore, the final answer is:[boxed{3 ln 3}]]

🔑:## Solutionbegin{aligned}& lim _{x rightarrow 0} frac{3^{x+1}-3}{ln left(1+x sqrt{1+x e^{x}}right)}=lim _{x rightarrow 0} frac{3left(3^{x}-1right)}{ln left(1+x sqrt{1+x e^{x}}right)}= & =lim _{x rightarrow 0} frac{3left(left(e^{ln 3}right)^{x}-1right)}{ln left(1+x sqrt{1+x e^{x}}right)}=lim _{x rightarrow 0} frac{3left(e^{x ln 3}-1right)}{ln left(1+x sqrt{1+x e^{x}}right)}=end{aligned}Using the substitution of equivalent infinitesimals:begin{aligned}& e^{x ln 3}-1 sim x ln 3, text { as } x rightarrow 0(x ln 3 rightarrow 0) & ln left(1+x sqrt{1+x e^{x}}right) sim x sqrt{1+x e^{x}}, text { as } & ln rightarrow 0left(x sqrt{1+x e^{x}} rightarrow 0right)end{aligned}We get:=lim _{x rightarrow 0} frac{3 x ln 3}{x sqrt{1+x e^{x}}}=lim _{x rightarrow 0} frac{3 ln 3}{sqrt{1+x e^{x}}}==frac{3 ln 3}{sqrt{1+0 cdot e^{0}}}=frac{3 ln 3}{1}=3 ln 3## Problem Kuznetsov Limits 16-26

🧠:[<think>Okay, so I need to find the value of a + b + c for the parabola y = ax² + bx + c that passes through the points (-1, 12), (0, 5), and (2, -3). Let me start by recalling that a parabola in this form has three coefficients: a, b, and c. Since we have three points that lie on the parabola, each of these points should satisfy the equation when plugged in. That means I can set up three equations with the three unknowns (a, b, c) and solve the system. Let me start with the first point, (-1, 12). If I plug x = -1 and y = 12 into the equation, it should hold true. So substituting those in:12 = a*(-1)² + b*(-1) + cCalculating the squares and multiplications:12 = a*(1) + b*(-1) + c 12 = a - b + cOkay, that's my first equation: a - b + c = 12. Next, the point (0, 5). Let me plug x = 0 and y = 5 into the equation:5 = a*(0)² + b*(0) + c 5 = 0 + 0 + c 5 = cOh, that's nice and simple. So c is 5. That immediately gives me one of the variables. So I can substitute c = 5 into the other equations to reduce the number of variables.Now, the third point is (2, -3). Let's plug x = 2 and y = -3 into the equation:-3 = a*(2)² + b*(2) + c -3 = a*4 + 2b + cAgain, substitute c = 5 into this:-3 = 4a + 2b + 5Let me subtract 5 from both sides to simplify:-3 -5 = 4a + 2b -8 = 4a + 2bI can divide both sides of this equation by 2 to simplify further:-4 = 2a + bSo that's the second equation: 2a + b = -4Now, let's go back to the first equation we had from the point (-1, 12). Remember that was:a - b + c = 12But since we already found c = 5, substitute that in:a - b + 5 = 12Subtract 5 from both sides:a - b = 7So now we have two equations:1) a - b = 7 2) 2a + b = -4Hmm, so we can solve this system of equations to find a and b. Let's add the two equations together to eliminate b. If I add equation 1 and equation 2:(a - b) + (2a + b) = 7 + (-4) a - b + 2a + b = 3 3a = 3Wait, that simplifies to 3a = 3? Let me check:Left side: a + 2a is 3a, and -b + b is 0. So 3a = 3. Then a = 1.Okay, so a is 1. Now, substitute a = 1 back into one of the equations to find b. Let's use equation 1: a - b = 7.So 1 - b = 7 Subtract 1 from both sides: -b = 6 Multiply both sides by -1: b = -6Wait, hold on. If a is 1, then substituting into equation 1:1 - b = 7 So -b = 7 -1 -b = 6 Therefore, b = -6.But let's check this with equation 2 as well to make sure. Equation 2: 2a + b = -4Substitute a = 1 and b = -6:2*(1) + (-6) = 2 -6 = -4. Which matches. So that's correct.So we have a = 1, b = -6, and c = 5.Therefore, the equation of the parabola is y = x² -6x + 5.But the question is asking for a + b + c. Let me compute that:a + b + c = 1 + (-6) + 5 = 1 -6 +5 = (1 +5) -6 = 6 -6 = 0.Wait, that gives me 0? But the answer choices are (A) -4, (B)-2, (C)0, (D)1, (E)2. So 0 is option C.But let me double-check my work to make sure I didn't make a mistake. Let me verify all three points with the equation y = x² -6x +5.First point (-1,12):Left side: y = 12 Right side: (-1)^2 -6*(-1) +5 = 1 +6 +5 = 12. Correct.Second point (0,5):Left side: y =5 Right side: 0 -0 +5 =5. Correct.Third point (2, -3):Left side: y = -3 Right side: 2² -6*(2) +5 =4 -12 +5 = (4 +5) -12 =9 -12 = -3. Correct.So all three points satisfy the equation. Therefore, the coefficients are indeed a=1, b=-6, c=5. So a + b + c =0. So the answer is C.But wait, just to make sure I didn't miscalculate a + b + c. Let me compute again:1 (a) + (-6) (b) +5 (c) =1 -6 +5. 1 -6 is -5, -5 +5 is 0. Yep, that's 0.Hmm. So why is option C 0? That's one of the answer choices. The options given are (A) -4, (B)-2, (C)0, (D)1, (E)2. So 0 is option C. Therefore, the answer must be C.But just to check again, maybe I miscalculated the equations?First equation from (-1,12):a - b + c =12 Since c=5, a - b =7. Correct.Second equation from (2, -3):4a +2b +5 = -3 4a +2b = -8 Divide by 2: 2a +b = -4. Correct.Then solving the system:a - b =7 2a +b =-4 Adding them: 3a =3 => a=1. Then b= -6. Correct. Therefore, all steps are correct. So the answer is C) 0.But wait, the problem says "the value of a + b + c". Wait, but in the equation y = ax² +bx +c, when x=1, y =a(1)^2 +b(1) +c =a +b +c. So actually, the value of a +b +c is the y-coordinate when x=1. So maybe there's a shortcut here. If the problem is asking for a +b +c, that's the same as evaluating the parabola at x=1. So if I can find y when x=1, that's a +b +c. Maybe that's a quicker way. Let me check.Given that the parabola passes through the three points, maybe I can check if (1, ?) is known? But the given points are (-1,12), (0,5), (2,-3). So (1, y) is not given. But if I can compute y at x=1 using the equation we found, which is y = x² -6x +5. So plugging x=1: 1 -6 +5 =0. So y=0. Therefore, a +b +c=0. Which is consistent. So that's another way to see it. But since we already found a, b, c, both methods give the same result. So yeah, the answer is 0, which is C.Therefore, I think the answer is C) 0.Final Answerboxed{C}</think>Given the parabola ( y = ax^2 + bx + c ) passing through the points ((-1, 12)), ((0, 5)), and ((2, -3)), we need to find the value of (a + b + c).1. Substituting the point ((0, 5)) into the equation: [ 5 = a(0)^2 + b(0) + c implies c = 5 ]2. Substituting the point ((-1, 12)) into the equation: [ 12 = a(-1)^2 + b(-1) + c implies 12 = a - b + 5 implies a - b = 7 ]3. Substituting the point ((2, -3)) into the equation: [ -3 = a(2)^2 + b(2) + 5 implies -3 = 4a + 2b + 5 implies 4a + 2b = -8 implies 2a + b = -4 ]4. Solving the system of equations: [ begin{cases} a - b = 7 2a + b = -4 end{cases} ] Adding the two equations: [ 3a = 3 implies a = 1 ] Substituting (a = 1) into (a - b = 7): [ 1 - b = 7 implies b = -6 ]5. Calculating (a + b + c): [ a + b + c = 1 + (-6) + 5 = 0 ]Thus, the value of (a + b + c) is (boxed{C}).,<think>Alright, let's see. I need to find the value of a + b + c for the parabola y = ax² + bx + c that passes through the points (-1, 12), (0, 5), and (2, -3). Hmm, okay. So since it's a parabola, it's a quadratic equation, and I have three points. That should be enough to determine the coefficients a, b, and c. Once I find those, I just add them up to get the answer. Let me start by plugging in each point into the equation.First, the point (0, 5). If x is 0, then the equation becomes y = a*0² + b*0 + c, which simplifies to y = c. Since the point is (0, 5), that means c = 5. Okay, so that's one variable down. So now the equation is y = ax² + bx + 5.Next, let's use the point (-1, 12). Plugging x = -1 into the equation, we get y = a*(-1)² + b*(-1) + 5. That simplifies to y = a*1 - b + 5, so y = a - b + 5. But we know when x is -1, y is 12. Therefore, 12 = a - b + 5. If I subtract 5 from both sides, that gives me 7 = a - b. So that's my first equation: a - b = 7. Got that.Now the third point is (2, -3). Plugging x = 2 into the equation y = ax² + bx + 5, we get y = a*(2)² + b*(2) + 5. That simplifies to 4a + 2b + 5. Since the point is (2, -3), that means -3 = 4a + 2b + 5. Subtracting 5 from both sides, we get -8 = 4a + 2b. Let me write that as 4a + 2b = -8. Hmm, maybe simplify that equation by dividing both sides by 2: 2a + b = -4. So now we have two equations:1. a - b = 72. 2a + b = -4Now, I can solve these two equations simultaneously. Let's add them together to eliminate b. If I add the left sides: (a - b) + (2a + b) = 3a. The right sides: 7 + (-4) = 3. So 3a = 3, which means a = 1. Okay, so a is 1. Then plugging back into the first equation: 1 - b = 7. Subtract 1 from both sides: -b = 6, so b = -6. Wait, so a is 1, b is -6, and c is 5. Then a + b + c would be 1 + (-6) + 5. Let's compute that: 1 -6 is -5, plus 5 is 0. So 0? Hmm, but let me check to make sure I didn't make any mistakes.First, let's verify the equations with a = 1 and b = -6. First equation: a - b = 1 - (-6) = 1 + 6 = 7. That's correct. Second equation: 2a + b = 2*1 + (-6) = 2 -6 = -4. Correct. So the values of a and b check out. Then c is 5 as we found earlier.Let me double-check the original points. Let's plug in x = -1 into the equation y = x² -6x +5. So (-1)^2 is 1, -6*(-1) is +6, so 1 +6 +5 = 12. Which matches the point (-1, 12). Good.Then x = 0: 0 +0 +5 =5. Correct. Then x =2: 2² is 4, -6*2 is -12, so 4 -12 +5 = -3. That matches (2, -3). Perfect. So all three points are satisfied. Therefore, a + b + c is 1 + (-6) +5 = 0. So the answer should be C) 0. But wait, the options given are A)-4, B)-2, C)0, D)1, E)2. So 0 is option C. Hmm. Wait, but the problem is phrased as "the value of a + b + c". But in the quadratic equation, a + b + c would be when x =1, because y = a(1)^2 + b(1) + c = a + b + c. So they could also be asking for the value of the function at x=1. Let me check that. If x=1, then y=1 -6 +5=0. So indeed, a + b + c is the value at x=1, which is 0. So that's correct. So I think the answer is C)0. Hmm. But why is the problem giving options? Wait, the options are A)-4, B)-2, C)0, D)1, E)2, so C is 0. So I think that's correct. Let me just check again.Wait, let me recap. When they say a + b + c, that's the sum of the coefficients. But since c is the constant term, which is when x=0. But in the equation, when x=1, y = a + b + c, which is exactly the sum of the coefficients. So yeah, if you plug x=1 into the equation, you get a + b + c. So maybe another way to approach this problem is to realize that if you plug x=1 into the equation, you can get the answer directly. But since we don't have the point (1, something), we can't do that directly. But since we found the equation, we can compute y when x=1, which is a + b + c. But since we already found a, b, and c, then adding them gives the same result. So either way, the answer is 0. So that's option C. Therefore, I think the answer is C.But wait, just to make sure, maybe I made a mistake in solving the equations. Let's check again. So first equation from (-1,12): a - b =7. Second equation from (2,-3):2a + b = -4. Adding them: 3a =3, so a=1. Then b= a -7=1 -7=-6. So a=1, b=-6, c=5. So sum is 1 + (-6) +5=0. Correct. Yeah, that's right. So the answer is 0. So option C. Final Answerboxed{C}</think>Given the parabola ( y = ax^2 + bx + c ) passing through the points ((-1, 12)), ((0, 5)), and ((2, -3)), we need to find the value of (a + b + c).1. Using the point ((0, 5)): [ y = ax^2 + bx + c implies 5 = a(0)^2 + b(0) + c implies c = 5 ]2. Using the point ((-1, 12)): [ 12 = a(-1)^2 + b(-1) + 5 implies 12 = a - b + 5 implies a - b = 7 quad text{(Equation 1)} ]3. Using the point ((2, -3)): [ -3 = a(2)^2 + b(2) + 5 implies -3 = 4a + 2b + 5 implies 4a + 2b = -8 implies 2a + b = -4 quad text{(Equation 2)} ]4. Solving the system of equations: - Adding Equation 1 and Equation 2: [ (a - b) + (2a + b) = 7 + (-4) implies 3a = 3 implies a = 1 ] - Substituting (a = 1) into Equation 1: [ 1 - b = 7 implies -b = 6 implies b = -6 ]5. Summing the coefficients (a), (b), and (c): [ a + b + c = 1 + (-6) + 5 = 0 ]Thus, the value of (a + b + c) is (boxed{C}).]

🔑:Substituting in the (x, y) pairs gives the following system of equations:[a-b+c=12][c=5][4a+2b+c=-3]We know c=5, so plugging this in reduces the system to two variables:[a-b=7][4a+2b=-8]Dividing the second equation by 2 gives 2a+b=-4, which can be added to the first equation to get 3a=3, or a=1. So the solution set is (1, -6, 5), and the sum is boxed{0textbf{ (C)}}.

🧠:[<think>Okay, let's try to solve this cryptarithmetic puzzle. The problem is in Chinese, but the translation says that each Chinese character represents a unique digit, and different characters represent different digits. There are two equations:1. 数字花园 + 探秘 = 20152. 探秘 (as a two-digit number) + 1 + 2 + 3 + ... + 10 = 花园 (as a two-digit number)We need to find the four-digit number 数字花园.First, let's parse the equations. The first equation is a four-digit number (数字花园) plus a two-digit number (探秘) equals 2015. The second equation says that the two-digit number 探秘 plus the sum from 1 to 10 equals the two-digit number 花园.Let me note down the sum from 1 to 10 first. That's a known value. The sum of numbers from 1 to n is given by n(n+1)/2. For n=10, that's 10*11/2 = 55. So the second equation simplifies to:探秘 + 55 = 花园But 探秘 and 花园 are two-digit numbers. Let me denote them as TM and HY for simplicity. So TM + 55 = HY.But HY is the last two digits of the four-digit number 数字花园 (which is SZHY, where S, Z, H, Y are digits). So HY is a two-digit number, and TM is another two-digit number. Also, since all characters represent different digits, the characters in 探秘 (T and M) must be different from each other and from those in 数字花园 (S, Z, H, Y).Now, let's note that in the first equation: SZHY + TM = 2015.Since SZHY is a four-digit number and 2015 is a four-digit number, SZHY must be less than 2015 because adding a positive two-digit number (TM) to it gives 2015. So SZHY is a four-digit number starting with either 1 or 2. But since 2015 starts with 2, if SZHY starts with 2, then adding even a two-digit number would make it more than 2015. Wait, but 2015 - TM = SZHY. So SZHY must be 2015 minus a two-digit number. Therefore, SZHY is between 2015 - 99 = 1916 and 2015 - 10 = 2005. Therefore, the first two digits of SZHY (数字) must be 19 or 20. But 数字花园 is a four-digit number. If 数字 is 20, then the four-digit number would be 20xx, but 2015 - 20xx = TM, which is a two-digit number. Let's check:If SZHY is 20ab, then 20ab + TM = 2015. So TM = 2015 - 20ab. For example, if SZHY is 2000, then TM would be 15. If SZHY is 2010, TM would be 5. But TM is a two-digit number, so 10 ≤ TM ≤ 99. Therefore, 2015 - SZHY must be between 10 and 99. Therefore, SZHY must be between 2015 - 99 = 1916 and 2015 - 10 = 2005. Therefore, SZHY is a four-digit number from 1916 to 2005 inclusive.Therefore, the first two digits (数字) are either 19 or 20. However, if it's 20, then the third digit (花) is 0 or 1. Wait, 2005 is the upper limit. Let's see. For example, 2005 + 10 = 2015, so TM=10. But in the second equation, TM + 55 = HY. So HY = 10 + 55 = 65. Then 花园 is 65, so 花 is 6 and 园 is 5. Then SZHY would be 2005, but 花园 is 65, which would mean 花 is 6, but in 2005, the third digit is 0, which contradicts 花 being 6. So that's not possible. Therefore, the first two digits might be 19.Wait, let's check. If SZHY is between 1916 and 1999, then the thousands digit is 1, and the hundreds digit is 9. So 数字 is 19. Then 数字花园 is 19花园, where 花 is the hundreds digit (third digit) and 园 is the units digit. Then adding TM (探秘) to 19花园 gives 2015.So 19花园 + TM = 2015. Let's write this as 1900 + 100*花 + 10*园 + 园 + TM = 2015. Wait, no, 花园 is the last two digits. Wait, no, 花园 is a two-digit number, which is the last two digits of the four-digit number 数字花园. So 数字花园 is a four-digit number where 数字 are the first two digits (19) and 花园 are the last two digits. So the number is 19花园, i.e., 1000*1 + 100*9 + 10*花 + 园. So 19花园 in digits is 1 9 H Y, where H is 花 and Y is 园.So the first equation becomes 1900 + 10H + Y + TM = 2015. Therefore, 10H + Y + TM = 2015 - 1900 = 115. So 10H + Y + TM = 115.From the second equation, TM + 55 = HY. So HY is a two-digit number, so H is the tens digit and Y is the units digit. Therefore, HY = 10H + Y. Therefore, from the second equation: TM + 55 = 10H + Y.But from the first equation, we have 10H + Y + TM = 115. Let's substitute TM from the second equation into the first equation. From the second equation, TM = 10H + Y - 55. Substitute into the first equation:10H + Y + (10H + Y - 55) = 115Simplify:10H + Y + 10H + Y - 55 = 11520H + 2Y - 55 = 11520H + 2Y = 115 + 55 = 170Divide both sides by 2:10H + Y = 85But 10H + Y is HY, the two-digit number 花园. Therefore, HY = 85. So 花园 is 85, meaning H = 8 and Y = 5.Then from the second equation, TM + 55 = 85. Therefore, TM = 85 - 55 = 30. So TM is 30, which is 探秘. So 探 is 3 and 秘 is 0.Wait, but let's check the digits. The four-digit number is 19花园 = 1985. Then 探秘 is 30. So the digits are:数字花园: 1, 9, 8, 5探秘: 3, 0So all digits must be unique. Checking: 1, 9, 8, 5, 3, 0. These are all different. So that seems okay.Now check the first equation: 1985 + 30 = 2015. 1985 + 30 is 2015. Correct.Second equation: 30 + 55 = 85. 30 + 55 is 85. Correct.So all conditions are satisfied. Therefore, 数字花园 is 1985.Wait, but let me check if there are other possibilities. Suppose HY = 85, H=8, Y=5. Then TM=30. Then 探 is 3, 秘 is 0. Are there any other possible HY that satisfy 10H + Y = 85? No, because H and Y are digits, so H=8, Y=5 is the only solution.But let me confirm if the substitution is correct.From the second equation: TM + 55 = HY => TM = HY - 55.From the first equation: HY + TM = 115.Substituting TM: HY + (HY - 55) = 115 => 2HY - 55 = 115 => 2HY = 170 => HY = 85.Yes, that's correct. So HY must be 85, so H=8, Y=5. Then TM = 85 -55=30.Therefore, the four-digit number is 19花园 = 1985. Let's check all digits:1 (数), 9 (字), 8 (花), 5 (园), 3 (探), 0 (秘). All unique. Yes, 0 is allowed as a digit as long as it's not the leading digit. Since 秘 is the second digit in TM=30, which is okay because leading digit 3 is fine. So 0 is allowed here.Therefore, the answer should be 1985.But wait, let me check again:数字花园 is 1985, which is 1 9 8 5.探秘 is 30, which is 3 0.Adding 1985 + 30 = 2015. Correct.Sum from 1 to 10 is 55. 30 + 55 = 85 = 花园. Correct.All digits are unique: 1,9,8,5,3,0. No duplicates. Yes.Therefore, 数字花园 = 1985.Final Answerboxed{1985}</think>Given the problem, we need to find the four-digit number 数字花园 such that:1. 数字花园 + 探秘 = 20152. 探秘 (as a two-digit number) + 1 + 2 + 3 + ... + 10 = 花园 (as a two-digit number)First, we calculate the sum from 1 to 10, which is 55. The second equation simplifies to:[overline{text{探秘}} + 55 = overline{text{花园}}]Let ( overline{text{探秘}} = TM ) and ( overline{text{花园}} = HY ). Then:[TM + 55 = HY]From the first equation, we have:[数字花园 + TM = 2015]Let 数字花园 be the four-digit number SZHY. Since SZHY + TM = 2015, SZHY must be between 1916 and 2005. Therefore, the first two digits (数字) are either 19 or 20. Given the range, we consider SZHY as 19花园 (19HY). Substituting into the first equation:[1900 + 10H + Y + TM = 2015]Simplifying, we get:[10H + Y + TM = 115]From the second equation, substituting ( TM = HY - 55 ):[10H + Y + (10H + Y - 55) = 115]Simplifying:[20H + 2Y - 55 = 115][20H + 2Y = 170][10H + Y = 85]Thus, ( HY = 85 ), meaning ( H = 8 ) and ( Y = 5 ). Substituting back, we find ( TM = 85 - 55 = 30 ). Therefore, 探秘 is 30, where 探 is 3 and 秘 is 0.Checking the digits:- 数字花园: 1 (数), 9 (字), 8 (花), 5 (园)- 探秘: 3 (探), 0 (秘)All digits are unique. Verifying the equations:1. ( 1985 + 30 = 2015 )2. ( 30 + 55 = 85 )Both equations are satisfied. Therefore, the four-digit number 数字花园 is:[boxed{1985}],<think>Alright, let's tackle this problem. Hmm, it's a cryptarithmetic puzzle where Chinese characters represent digits. The two equations given are:1. 数字花园 (a four-digit number) + 探秘 (a two-digit number) = 20152. 探秘 (the two-digit number) + 1+2+3+...+10 = 花园 (a two-digit number)First, I need to figure out what digits each Chinese character represents. All the same characters represent the same digits, and different characters represent different digits. Let's break it down step by step.Starting with the second equation: 探秘 + sum(1 to 10) = 花园. The sum from 1 to 10 is 55, right? So that equation simplifies to:探秘 + 55 = 花园Since both 探秘 and 花园 are two-digit numbers, let's denote 探秘 as TM (where T is 探 and M is 秘) and 花园 as HY (H is 花 and Y is 园). So, TM + 55 = HY. But wait, HY is a two-digit number. However, adding 55 to a two-digit number could result in a three-digit number if TM is 45 or higher. But since HY is a two-digit number, TM must be such that TM + 55 is still a two-digit number. That means TM must be less than or equal to 44 because 45 + 55 = 100, which is three digits. Therefore, TM is between 00 and 44. But since it's a two-digit number, TM is between 10 and 44 (assuming leading zeros are not allowed). Wait, but in cryptarithmetic puzzles, leading zeros are usually not allowed, so TM is between 10 and 44. Therefore, HY would be between 65 (10 + 55) and 99 (44 + 55). So HY is between 65 and 99. But HY is a two-digit number, so that's okay.But let's check this again. If HY is a two-digit number, then TM + 55 must be less than 100. Therefore, TM < 45. So TM is between 10 and 44. Then HY is between 65 and 99. So HY is a two-digit number between 65 and 99. That's an important point.Now, moving to the first equation: 数字花园 (a four-digit number) + TM = 2015. Let's denote 数字花园 as SZHY, where S is 数, Z is 字, H is 花, and Y is 园. So SZHY + TM = 2015.Since SZHY is a four-digit number and 2015 is also a four-digit number, adding a two-digit number (TM) to it. Let's write this as:SZHY (a four-digit number) + TM (a two-digit number) = 2015So, SZHY = 2015 - TMBut from the second equation, we know that HY = TM + 55. Therefore, HY is the last two digits of SZHY. So, SZHY is a four-digit number ending with HY, which is equal to TM + 55. Also, since SZHY = 2015 - TM, we can substitute HY with (TM + 55). Let me try to write this out.Let’s represent SZHY as 1000*S + 100*Z + 10*H + Y. Similarly, TM is 10*T + M. Then:1000*S + 100*Z + 10*H + Y + 10*T + M = 2015But from the second equation, 10*T + M + 55 = 10*H + YSo we have two equations:1. 1000S + 100Z + 10H + Y + 10T + M = 20152. 10T + M + 55 = 10H + YLet me substitute the second equation into the first. From equation 2: 10H + Y = 10T + M + 55. Let's plug this into equation 1.Equation 1 becomes: 1000S + 100Z + (10H + Y) + 10T + M = 2015But (10H + Y) = (10T + M + 55), so substituting:1000S + 100Z + (10T + M + 55) + 10T + M = 2015Simplify:1000S + 100Z + 10T + M + 55 + 10T + M = 2015Combine like terms:1000S + 100Z + (10T + 10T) + (M + M) + 55 = 2015Which is:1000S + 100Z + 20T + 2M + 55 = 2015Subtract 55 from both sides:1000S + 100Z + 20T + 2M = 2015 - 55 = 1960So:1000S + 100Z + 20T + 2M = 1960Divide both sides by 2 to simplify:500S + 50Z + 10T + M = 980So now we have:500S + 50Z + 10T + M = 980Hmm, this equation. Let's note that S, Z, T, M are digits (0-9), with S ≠ 0 (since it's a four-digit number), T ≠ 0 (since TM is a two-digit number). Also, all characters represent different digits.But let's analyze this equation: 500S + 50Z + 10T + M = 980Since 500S is the largest term here. Let's see possible values of S. Since 500S ≤ 980, S must be 1, because 500*2 = 1000 which is more than 980. So S = 1.So substituting S = 1:500*1 + 50Z + 10T + M = 980Which simplifies to:500 + 50Z + 10T + M = 980Subtract 500:50Z + 10T + M = 480So now we have:50Z + 10T + M = 480Again, Z, T, M are digits (0-9), different from each other and different from S (which is 1). Also, since Z is part of the four-digit number SZHY, Z can be 0-9 except 1. Similarly, T is the first digit of TM, so T cannot be 0 (since it's a two-digit number). Therefore, T is 2-9, but different from S=1. Also, all characters are different: S, Z, H, Y, T, M are all distinct digits.So 50Z + 10T + M = 480Let’s note that 50Z must be a multiple of 50. Let's see possible values of Z. 50Z ≤ 480, so Z ≤ 9 (since 50*9=450). But let's compute possible Z values:Z can be from 0 to 9, but 50Z must be ≤480. So maximum Z is 9 (50*9=450). Then 10T + M = 480 - 50Z.So 10T + M = 480 - 50ZSince 10T + M is a two-digit number (TM), which must be between 10 and 44 as established earlier. So:10 ≤ 10T + M ≤ 44But 10T + M = 480 - 50ZTherefore:10 ≤ 480 - 50Z ≤ 44Let's solve for Z.First inequality: 480 - 50Z ≥ 10480 - 10 ≥ 50Z470 ≥ 50ZZ ≤ 470 / 50 = 9.4, so Z ≤ 9 (since Z is integer). Which is already true.Second inequality: 480 - 50Z ≤ 44480 - 44 ≤ 50Z436 ≤ 50ZZ ≥ 436 / 50 = 8.72Since Z is an integer, Z ≥ 9So combining both inequalities, Z must be 9.So Z = 9.Then, 10T + M = 480 - 50*9 = 480 - 450 = 30So 10T + M = 30Therefore, TM (探秘) is 30.But TM is a two-digit number. So T is 3 and M is 0. So 探=3, 秘=0.But check if digits are unique. So far, S=1, Z=9, T=3, M=0. All different. Good.Now, from the second equation, HY (花园) = TM + 55 = 30 + 55 = 85. Therefore, H=8 and Y=5.Check uniqueness: H=8, Y=5. So digits used so far: 1,9,3,0,8,5. All unique. Good.Now, let's reconstruct the four-digit number SZHY: S=1, Z=9, H=8, Y=5. So 数字花园 is 1985.Check the first equation: 1985 + 30 = 2015. Yes, that's correct. 1985 + 30 = 2015.Check the second equation: 30 + 55 = 85. Correct. 30 + 55 = 85.Now, check all digits: 数=1, 字=9, 花=8, 园=5, 探=3, 秘=0. All different digits. No overlaps. So this satisfies the problem's conditions.Therefore, the four-digit number 数字花园 is 1985.Final AnswerThe four-digit number 数字花园 is boxed{1985}.</think>Given the problem, we need to find the four-digit number 数字花园 (represented as SZHY) such that:1. 数字花园 + 探秘 = 20152. 探秘 + 1 + 2 + 3 + cdots + 10 = 花园First, we note that the sum from 1 to 10 is 55. Therefore, the second equation simplifies to:[overline{text{探秘}} + 55 = overline{text{花园}}]Let 探秘 be represented as TM (a two-digit number) and 花园 as HY (another two-digit number). This gives:[TM + 55 = HY]Since TM must be a two-digit number, TM ranges from 10 to 44 (as adding 55 to TM must result in another two-digit number HY). Next, we consider the first equation:[SZHY + TM = 2015]Substituting HY from the second equation into the first equation, we express SZHY as:[SZHY = 2015 - TM]We denote SZHY as 1000S + 100Z + 10H + Y and TM as 10T + M. Substituting HY = TM + 55 into the equation, we get:[1000S + 100Z + 10H + Y + 10T + M = 2015]Using the second equation, we substitute HY (10H + Y) with TM + 55:[1000S + 100Z + (10T + M + 55) + 10T + M = 2015]Simplifying this, we get:[1000S + 100Z + 20T + 2M + 55 = 2015]Subtracting 55 from both sides:[1000S + 100Z + 20T + 2M = 1960]Dividing by 2:[500S + 50Z + 10T + M = 980]Since S must be 1 (as 500*2 = 1000 exceeds 980), we substitute S = 1:[500 + 50Z + 10T + M = 980]Subtracting 500:[50Z + 10T + M = 480]Solving for Z, we find Z = 9 (as 50*9 = 450), leading to:[10T + M = 30]Thus, TM = 30 (T = 3, M = 0). Using the second equation, HY = 30 + 55 = 85 (H = 8, Y = 5).Finally, reconstructing SZHY with S = 1, Z = 9, H = 8, Y = 5, we get the four-digit number 1985. Checking the equations confirms the solution:- 1985 + 30 = 2015- 30 + 55 = 85All digits are unique and satisfy the conditions. The four-digit number 数字花园 is boxed{1985}.]

🔑:5. (10 points) In the following two equations, the same Chinese characters represent the same digits, and different Chinese characters represent different digits:数字花园 + 探秘 =2015, overline{text { 探秘 }}+1+2+3+cdots+10=overline{text { 花园 }}Then the four-digit number 数学花园 = qquad .【Solution】Solution: According to the analysis, let x=overline{text { 数学 }}, y=overline{text { 花园 }} quad mathrm{z}=overline{text { 探秘 }}Then: 数字花园 +overline{text { 探秘 }}=100 x+y+z=2015 cdots (1)overline{text { 探秘 }}+1+2+3+cdots+10=z+1+2+3+cdots+10=yz+frac{(1+10) times 10}{2}=yz+55=y cdots(2)Solving (1) and (2) together:left{begin{array}{l}100 x+y+z=2015 z+55=yend{array}right.We get: 50 x+y=1035begin{array}{l}because 100>z=y-55 geqslant 10 therefore 65 leqslant y<100 quad 65 leqslant 1035-50 x<100 935<50 x leqslant 970end{array}18.7<x leqslant 19.4 and because x is an integerbegin{array}{l}therefore x=19 y=1035-50 times 19=85end{array}Therefore, the answer is: 1985