🧠:Okay, so I need to prove that by pressing 19 switches, we can turn on at least eight lamps. Let me start by understanding the problem setup.We have 70 switches and 15 lamps. Each lamp is connected to 35 switches. Also, no two switches are connected to the same set of lamps. That means each switch has a unique combination of lamps it's connected to. When you press a switch, it toggles all the lamps connected to it. All lamps start off, so pressing a switch will turn some lamps on and others off, depending on their previous state.The goal is to show that there's a subset of 19 switches that, when pressed, result in at least eight lamps being on. Hmm. This feels like a combinatorial problem, maybe involving linear algebra over a finite field or some probabilistic method. Let me think.Since each switch is connected to 35 lamps, and each lamp is connected to 35 switches. Wait, each lamp has 35 switches connected to it? Let me confirm: the problem says each lamp is connected to 35 switches. So each lamp has 35 switches that can toggle it. And each switch is connected to 35 lamps. Wait, but 15 lamps in total. So if each lamp is connected to 35 switches, how does that work? Each switch is connected to 35 lamps, but there are only 15 lamps. Wait, that can't be. Wait, maybe I misread.Wait, the problem says: "Each lamp is connected to 35 switches." So each lamp is connected to 35 switches. Since there are 15 lamps, each connected to 35 switches, and each switch is connected to some number of lamps. Let's see: each switch is connected to 35 lamps? Wait, but there are only 15 lamps. That can't be. Wait, no. Wait, no, the problem says "each lamp is connected to 35 switches." So each lamp has 35 switches connected to it. So each switch can be connected to multiple lamps. How many lamps is each switch connected to? It doesn't say. Wait, that's important. Wait, the problem says: "No two switches are connected to the same set of lamps." So each switch has a unique subset of lamps. But each lamp is connected to 35 switches.So each lamp has 35 switches connected to it. Therefore, for each lamp, there are 35 switches that can toggle it. Since there are 15 lamps, each with 35 switches, the total number of connections is 15*35=525. On the other hand, each switch is connected to some number of lamps. Let’s denote that each switch is connected to k lamps. Then total number of connections is also 70*k. Therefore, 70*k = 525, so k=525/70=7.5. Wait, that can't be. 525 divided by 70 is 7.5, which is not an integer. That's a problem.Wait, this suggests a contradiction. Because if each of the 15 lamps is connected to 35 switches, the total number of connections is 15*35=525. Each switch is connected to k lamps, so total connections is 70*k. Therefore, 70*k=525, so k=525/70=7.5. But switches can't be connected to half a lamp. Therefore, this must be a mistake. So either the problem is stated incorrectly, or I misinterpret it.Wait, let's check the original problem again. "Each lamp is connected to 35 switches. No two switches are connected to the same set of lamps. Pressing a switch changes the state of all connected lamps (turns on the off ones and vice versa). Initially, all lamps are off. Prove that it is possible to press some 19 switches in such a way that at least eight lamps are turned on."Wait, perhaps the problem says each switch is connected to 35 lamps? That would make sense. Let me see. If each switch is connected to 35 lamps, but there are only 15 lamps. So each switch is connected to all 15 lamps? But 35 is larger than 15. So that can't be. Wait, no. Wait, 35 is the number of lamps each switch is connected to. But there are only 15 lamps. Therefore, that's impossible. Therefore, there's a problem here.Wait, maybe I misread. Wait, original problem: "Each lamp is connected to 35 switches." So each lamp has 35 switches connected to it. Then total connections are 15*35=525. Each switch is connected to some number of lamps. Let’s call that number k. Then total connections would be 70*k. Therefore, 70*k=525. So k=525/70=7.5. Hmm, which is 7.5. But that's impossible. So there's a contradiction here. Therefore, the problem must have been misstated. Maybe each lamp is connected to 36 switches? 15*36=540, 540 divided by 70 is 540/70=7.714… Still not integer. Wait, maybe the numbers are different? Wait, original problem: 70 switches, 15 lamps. Each lamp connected to 35 switches. So, 15*35=525 connections. 525 divided by 70 is 7.5. Hmm. Therefore, maybe the problem has an error? But the user is asking to prove it. So perhaps I misinterpret the problem.Wait, perhaps "each lamp is connected to 35 switches" means that each lamp can be toggled by 35 different switches. So each lamp has 35 switches that affect it. Then, for each lamp, the number of switches connected to it is 35. Then, the total number of connections is 15*35=525. Each switch is connected to some number of lamps. If there are 70 switches, then the average number of lamps per switch is 525/70=7.5. But each switch must be connected to an integer number of lamps. So unless some switches are connected to 7 lamps and others to 8, but the problem says "No two switches are connected to the same set of lamps." Therefore, all switches have distinct subsets. Therefore, each switch is connected to a unique subset of the 15 lamps. Therefore, the number of possible subsets is 2^15, which is 32768. So we have 70 distinct subsets. Now, if each lamp is connected to 35 switches, then for each lamp, 35 of the 70 switches include that lamp in their subset. Therefore, each lamp is in exactly 35 switches. Therefore, the total number of connections is 15*35=525. Then, the average number of lamps per switch is 525/70=7.5. So, since each switch is connected to a unique subset, and each switch is connected to 7 or 8 lamps? Because 7.5 is the average. Therefore, maybe half the switches are connected to 7 lamps and half to 8? But 70 is even, so 35 switches with 7 and 35 with 8. Then total connections would be 35*7 +35*8=245+280=525. Yes, that works. Therefore, each switch is connected to either 7 or 8 lamps, with 35 switches each. Therefore, okay, that resolves the confusion. So each lamp is in 35 switches, each switch is in 7 or 8 lamps, with exactly 35 switches having 7 lamps and 35 having 8. Got it. So that makes sense.So now, the problem is: we have 70 switches, each connected to a unique subset of the 15 lamps, with each lamp in 35 switches. Each switch is connected to 7 or 8 lamps. All lamps start off. Pressing a switch toggles all its connected lamps. We need to show that there exists a subset of 19 switches such that, after pressing them, at least 8 lamps are on.Hmm. So, how to approach this? Maybe linear algebra over GF(2). Each lamp's state is a vector in GF(2)^15, and each switch corresponds to a vector in GF(2)^15, where the entries are 1 if the switch is connected to the lamp, 0 otherwise. Then, pressing switches corresponds to adding their vectors (mod 2). We need to find a combination of 19 vectors (switch presses) such that the sum has at least 8 ones (lamps on). Alternatively, the weight of the sum vector is at least 8.Alternatively, think in terms of covering the lamps: we need that the union (with possible overlaps) of the pressed switches' lamp sets toggles the lamps an odd number of times, resulting in them being on. Wait, since each press toggles the lamp. So, if a lamp is toggled an odd number of times, it's on; even, off. So, we need that for at least 8 lamps, the number of pressed switches connected to them is odd.Therefore, the problem reduces to selecting 19 switches such that at least 8 lamps are included in an odd number of the selected switches.Hmm. So how can we guarantee that? Maybe using probabilistic method? If we randomly select 19 switches, what's the expected number of lamps that are toggled an odd number of times? Then, show that with positive probability, this number is at least 8.Alternatively, use combinatorial arguments. Let me think.First, note that each lamp is connected to 35 switches. So, for each lamp, the number of switches connected to it is 35. So, if we consider selecting a subset of switches, then for each lamp, the probability that it's toggled an odd number of times depends on the number of selected switches connected to it.But since we have to pick exactly 19 switches, it's a combinatorial problem. Let me consider linear algebra. Since the problem is over GF(2), perhaps the dimension is 15, and we have 70 vectors. The question is whether the span of these vectors has a vector of weight at least 8, and whether such a vector can be expressed as a combination of 19 vectors. But I'm not sure. Maybe another approach.Alternatively, think of the problem in terms of set systems. Each switch is a subset of the 15 lamps, each of size 7 or 8, with all subsets distinct. Each lamp is in 35 subsets. We need to select 19 subsets such that the symmetric difference (since toggling is addition mod 2) results in a set of size at least 8. Wait, the symmetric difference of the pressed switches would correspond to the lamps that are toggled an odd number of times. So we need the symmetric difference of 19 subsets to have size at least 8.Alternatively, the problem is similar to covering the lamps with the pressed switches in such a way that the overlap is even or odd appropriately. Hmm.Alternatively, think about the following: the total number of possible subsets of switches is 2^70. Each subset corresponds to a vector in GF(2)^15 indicating which lamps are on. We need to show that there exists a subset of size 19 whose corresponding vector has weight at least 8.But perhaps we can use the probabilistic method. Let me try that.Suppose we randomly select 19 switches. For each lamp, the probability that it is toggled an odd number of times is equal to the probability that an odd number of the 19 switches are connected to it. Since each lamp is connected to 35 switches out of 70. So the probability that a particular lamp is toggled an odd number of times is equal to the sum over k odd of C(35, k)*C(35, 19 -k)/C(70, 19). Hmm, this is the hypergeometric distribution.Alternatively, since each switch is chosen independently, but we are choosing exactly 19 switches. Wait, no, it's without replacement. So hypergeometric.But hypergeometric calculations can be messy. Maybe approximate it as a binomial distribution. If n is large and p is the probability that a randomly selected switch is connected to the lamp. Here, for each lamp, the probability that a randomly selected switch is connected to it is 35/70 = 0.5. So, if we pick 19 switches, the number of switches connected to a particular lamp follows a hypergeometric distribution with parameters N=70, K=35, n=19. The probability that this number is odd is approximately 0.5, especially since N is large and n is not too small. But exact value?Actually, for hypergeometric distribution, the probability that the number of successes is odd can be calculated using generating functions or inclusion-exclusion. However, there's a trick for symmetric cases. Since the total number of switches connected to the lamp is 35, exactly half of the 70 switches. So, when selecting 19 switches, the probability that an even number are connected to the lamp is equal to the probability that an odd number are connected. Wait, is that true?Wait, in general, for a hypergeometric distribution with N=2K, the probability of even or odd number of successes in a sample of size n is equal? Let me test with small numbers. Suppose N=2, K=1. If we sample n=1, then probability of odd is 1/2, even is 1/2. If n=2, then probability of even is 1 (since you have to pick both), odd is 0. Hmm, not equal. So in general, it's not necessarily equal.But in our case, N=70, K=35, n=19. Let me see. The probability mass function for hypergeometric is C(35, k)*C(35, 19 -k)/C(70, 19). The probability that k is odd is sum_{k odd} C(35, k)C(35, 19 -k)/C(70,19). Similarly for even.Is there a symmetry here? Let's see. For each lamp, the switches are split into 35 connected and 35 not connected. The number of ways to choose 19 switches with k connected and 19 -k not connected. If we replace k with 19 -k, but since connected and not connected are symmetric (both 35), then sum_{k} C(35, k)C(35,19 -k) = sum_{k} C(35,19 -k)C(35,k) = same. So the distribution is symmetric around k=9.5? Not exactly, but perhaps for the parity.Wait, maybe considering generating functions. The generating function for the hypergeometric distribution is [C(35,0) + C(35,1)x + ... + C(35,35)x^35] * [C(35,0) + C(35,1)x + ... + C(35,35)x^35] divided by C(70,19). Wait, no. Actually, the generating function for the number of connected switches in the sample is [ (1 + x)^35 ] * [ (1 + x)^35 ] / (1 + x)^{70} } but I think I need to think differently.Alternatively, consider that the total number of subsets of 19 switches is C(70,19). The number of subsets with an odd number of connected switches for a particular lamp is sum_{k odd} C(35,k)C(35,19 -k). Let me compute this sum.This is equivalent to 1/2 [ (C(35,0)C(35,19) + C(35,1)C(35,18) + ... + C(35,19)C(35,0)) - (C(35,0)C(35,19) - C(35,1)C(35,18) + ... + (-1)^19 C(35,19)C(35,0)) ].But this might be complicated. Alternatively, use the identity that sum_{k=0}^n (-1)^k C(m, k)C(m, n -k) = 0 when n is odd? Wait, not sure.Alternatively, use the principle of inclusion-exclusion with generating functions.Wait, let me recall that in hypergeometric distribution, when the population is split evenly (N=2K), the probability of choosing an odd number of successes can be calculated as follows.Let’s use the generating function approach. The generating function for the number of connected switches is (1 + x)^{35} * (1 + x)^{35} / (1 + x)^{70} }? Wait, no. Wait, the generating function for the hypergeometric distribution isn't straightforward. Alternatively, think of it as a convolution.Alternatively, consider that each switch has a 50% chance of being connected to the lamp. If the selection is without replacement, but since 35 out of 70 are connected, the probability that any particular switch is connected is 0.5. However, the selections are dependent.But for large N, the hypergeometric distribution approximates the binomial distribution. So, if we approximate the number of connected switches in the sample as a binomial(n=19, p=0.5), then the probability that it's odd is 0.5. Because for a binomial distribution with p=0.5, the probability of odd number of successes is 0.5. So, maybe approximately 0.5 per lamp.If we assume independence between lamps (which they are not, since the switches are connected to multiple lamps), the expected number of lamps turned on would be approximately 15 * 0.5 = 7.5. So, on average, we expect 7.5 lamps to be on. But we need at least 8. By the probabilistic method, if the expectation is 7.5, there must exist some selection where the number is at least 8? Wait, no. The expectation is the average, so there could be some below and some above. But since it's not an integer, maybe there's at least one with 8 or more. However, the expectation being 7.5 doesn't strictly guarantee that. For example, maybe half the time it's 7 and half the time 8, giving expectation 7.5. But in reality, the actual distribution might be different.However, if we can show that the variance is low, then maybe we can use Chebyshev's inequality or something else. But this might get complicated. Alternatively, use linearity of expectation and show that the probability of having less than 8 is less than 1, so there must exist a selection with at least 8.Wait, let's formalize this. Let X be the number of lamps turned on by a random selection of 19 switches. Then E[X] = sum_{i=1}^{15} Pr(lamp i is on). Each Pr(lamp i is on) = Pr(odd number of connected switches are pressed). Let's denote this probability as p. Then E[X] = 15p.If we can compute p, then we can find E[X]. If E[X] is close to 8, then maybe we can use the probabilistic method. But first, let's compute p.For a single lamp, we have 35 connected switches and 35 disconnected. We choose 19 switches. The probability that an odd number of connected switches are chosen is p = [sum_{k odd} C(35, k)C(35, 19 -k)] / C(70, 19).This seems difficult to compute exactly, but perhaps we can find a closed-form expression or use a combinatorial identity.Recall that in general, sum_{k=0}^n C(a, k)C(b, n -k) = C(a + b, n). This is the Vandermonde's identity. But here, we have sum_{k odd} C(35, k)C(35, 19 -k). Let's denote this sum as S.Then, S = [C(70, 19) - sum_{k even} C(35, k)C(35, 19 -k)] / 2.Alternatively, use generating functions. Consider (1 + x)^35 * (1 + x)^35 = (1 + x)^70. The coefficient of x^19 is C(70, 19). But we want the sum of coefficients where k is odd in the product C(35, k)C(35, 19 -k). This can be found by evaluating ( (1 + x)^35 * (1 - x)^35 ) at x = 1 and x = -1.Wait, let's see. Let f(x) = (1 + x)^{35} * (1 + x)^{35} = (1 + x)^{70}. But if we consider f(1) = 2^{70}, f(-1) = 0. But if we want to compute the sum over k odd C(35, k)C(35, 19 -k), we can use the following trick:Consider ( (1 + x)^{35} ) * ( (1 - x)^{35} ) = (1 - x^2)^{35}. Then, expanding this, the coefficient of x^{19} is sum_{k=0}^{19} C(35, k)C(35, 19 -k)(-1)^{19 -k}. But this seems complicated.Alternatively, note that sum_{k even} C(35, k)C(35, 19 -k) - sum_{k odd} C(35, k)C(35, 19 -k) = [sum_{k} (-1)^k C(35, k)C(35, 19 -k)].But this is equal to the coefficient of x^{19} in (1 - x)^{35}(1 + x)^{35} = (1 - x^2)^{35}. Therefore, sum_{k} (-1)^k C(35, k)C(35, 19 -k) = coefficient of x^{19} in (1 - x^2)^{35}.But (1 - x^2)^{35} = sum_{m=0}^{35} C(35, m)(-1)^m x^{2m}. Therefore, the coefficient of x^{19} is zero, since 19 is odd. Therefore, sum_{k} (-1)^k C(35, k)C(35, 19 -k) = 0.Therefore, sum_{k even} C(35, k)C(35, 19 -k) = sum_{k odd} C(35, k)C(35, 19 -k). But since sum_{k even} + sum_{k odd} = C(70, 19), then each sum is equal to C(70, 19)/2. Therefore, the probability p = [C(70, 19)/2] / C(70, 19) = 1/2.Wait, that's brilliant! So, for any particular lamp, the probability that an odd number of connected switches are pressed is 1/2. Therefore, E[X] = 15 * 1/2 = 7.5.So, the expected number of lamps turned on is 7.5. Therefore, by the probabilistic method, there exists at least one selection of 19 switches that results in at least 8 lamps being on. Because if all selections resulted in at most 7 lamps, then the expectation would be at most 7, but it's 7.5. Therefore, there must exist some selections with 8 or more.Therefore, this proves the existence. Hence, it's possible to press some 19 switches to turn on at least eight lamps.But wait, let me check this logic again. The key step is that for each lamp, the probability it's on is 1/2, hence expectation 7.5. Then, since the expectation is 7.5, there must be at least one selection where the number of lamps on is at least 8. Because if all selections had ≤7 lamps on, the expectation would be ≤7, contradicting 7.5. Therefore, such a selection exists. Therefore, the proof is complete.But the problem is stated as "prove that it is possible to press some 19 switches..." So this probabilistic method argument suffices. However, in an Olympiad problem, would this be sufficient? Or do they expect a constructive proof? But given the problem's nature, existence is enough, so the probabilistic method is a valid approach.Alternatively, maybe use linear algebra. Since the number of switches (70) is more than twice the number of lamps (15). The dimension is 15, so by the pigeonhole principle, there must be dependencies. But how to connect this to pressing 19 switches.Alternatively, consider that each switch corresponds to a vector in GF(2)^15. We need a linear combination of 19 vectors (coefficients 1 or 0, since pressing a switch is adding the vector once) such that the sum has weight at least 8. The total number of vectors is 70. The dimension is 15, so the number of possible distinct vectors is 2^15. However, since we have 70 vectors, which is much larger than 2^15, there must be a lot of dependencies.But how to ensure that a combination of 19 vectors gives a high weight. Maybe by the pigeonhole principle, considering that the number of low-weight vectors is limited.Wait, the number of vectors with weight less than 8 is sum_{k=0}^7 C(15, k). For example, sum_{k=0}^7 C(15, k) = C(15,0) + C(15,1) + ... + C(15,7). This is equal to 2^14 = 16384. Since 2^15 = 32768, so half of the vectors have weight ≤7. But if we can generate enough vectors through combinations of 19 switches, but since the total number of possible combinations is C(70,19), which is astronomically larger than 2^15, then by the pigeonhole principle, there must be overlaps. However, we need a specific combination of 19 switches that results in a high-weight vector.But this line of reasoning isn't straightforward. The probabilistic method seems more direct.Therefore, the answer is proven by showing that the expected number of lamps turned on is 7.5, hence there must exist at least one subset of 19 switches that achieves at least 8.But to make this rigorous, let's write it formally.Let S be a randomly chosen subset of 19 switches. For each lamp L_i, let X_i be an indicator variable which is 1 if L_i is on after pressing the switches in S, and 0 otherwise. Then, the total number of lamps on is X = sum_{i=1}^{15} X_i.As calculated, for each X_i, Pr(X_i = 1) = 1/2, so E[X] = 15*(1/2) = 7.5.Since the expectation is 7.5, there must exist at least one subset S for which X ≥ 7.5. But since X must be an integer, there exists a subset S where X ≥ 8. Hence, proved.Yes, this seems correct. Therefore, the proof is complete.

✅:To prove that pressing some 19 switches can turn on at least eight lamps, we use the probabilistic method as follows:1. Setup and Expectation Calculation: - Each lamp is connected to 35 switches. When selecting 19 switches randomly, the probability that a specific lamp is toggled an odd number of times (thus turned on) is ( frac{1}{2} ). This is due to symmetry in the hypergeometric distribution when the population (70 switches) is evenly split (35 connected and 35 not connected to the lamp). - Let ( X_i ) be an indicator variable for lamp ( i ) being on. The expected number of lamps on is: [ mathbb{E}[X] = sum_{i=1}^{15} mathbb{E}[X_i] = 15 times frac{1}{2} = 7.5. ]2. Existence Argument: - Since the expected number of lamps on is ( 7.5 ), there must exist at least one selection of 19 switches where the actual number of lamps on is at least ( 8 ). If all selections resulted in ( leq 7 ) lamps, the expectation would be ( leq 7 ), contradicting ( 7.5 ).Conclusion: By the probabilistic method, there exists a subset of 19 switches that turns on at least eight lamps.boxed{text{Such a subset of 19 switches exists as proven by expectation and the probabilistic method.}}

🔑:1. Understanding the Structure: - We have 70 switches and 15 lamps. - Each lamp is connected to 35 switches. - No two switches are connected to the same set of lamps. - Pressing a switch toggles (turns on or off) all the lamps it is connected to. - Initially, all lamps are off.2. Constructing Sets and Pairs: - Consider any set of 19 switches. - For any single lamp, consider the 35 switches it is connected to. - Each of these 35 switches can be paired uniquely with another switch among these 35, hence forming 17 pairs and one remaining switch.3. Examining the Impact: - Each of the 19 switches, because 19 is an odd number, will uniquely contribute to toggling its corresponding lamp an odd number of times whenever that particular lamp's corresponding switches within the 19 chosen are toggled. - This implies there is at least one switch in our chosen 19 that will ensure its connected lamp is toggled. 4. Quantifying the Effect: - Let's look at our chosen sets of 19 switches: - Since each switch uniquely impacts certain lamps, and the distribution of switches prevents them from all affecting the same sets of lamps identically, there must be some switches in our set of 19 that cause corresponding lamps to toggle. - Structuring these into a table with lamps as rows and sets of 19 switches as columns, we find the overlapping toggles of lamps per set.5. Ensuring Minimum Number of Lamps Lit: - Since each of our sets works to toggle an odd number of specific lamps' states and no set perfectly mirrors another, it's guaranteed by construction we will affect the lamps. It's shown half the toggles would naturally light up switching elements based on these structural constraints.6. Conclusion: - Therefore, it is proved by combining the combinatorial logic and unpaired toggle distributions that there exists at least one configuration or set of 19 switches that will ensure 8 or more lamps are on. This is well-backed by counting the simple parity and mutual exclusiveness in toggles.[boxed{text{The statement is proved true under given conditions.}}]

🧠:Okay, let's see. Bogdan wants to measure the area of a rectangular carpet, but he doesn't have a ruler. Instead, he uses his shoe. The problem says that the shoe fits exactly 15 times along one edge and 10 times along another. Later, he measures the shoe to be 28 cm long. We need to find the area of the carpet in square centimeters. The answer choices are given, so that might help if I get stuck, but I should try to solve it first without just guessing.Alright, so the carpet is rectangular, so the area should be length multiplied by width. But Bogdan didn't measure the length and width directly; he used his shoe. If the shoe fits 15 times along one edge, that means the length of that edge is 15 shoe lengths. Similarly, the other edge is 10 shoe lengths. Since each shoe length is 28 cm, I can convert the number of shoes into centimeters by multiplying.Let me write this down step by step. Let's call the length of the shoe 's', which is 28 cm. The length of the carpet along one edge is 15 times the shoe length, so that's 15s. The other edge is 10 times the shoe length, so 10s. Then the area would be 15s multiplied by 10s, which is 15 * 10 * s * s. That simplifies to 150s².Wait, s is 28 cm, so substituting that in, the area is 150 * (28 cm)². Let me compute that. First, calculate 28 squared. 28 times 28. Hmm, 28 squared is 784, right? Because 20 squared is 400, and 8 squared is 64, and then the cross term is 2*20*8=320, so 400 + 320 + 64 = 784. Yeah, that's correct.So then, 150 multiplied by 784. Let me do that multiplication step by step. 150 times 700 is 105,000. Then 150 times 84. Let's see, 150*80=12,000 and 150*4=600. So 12,000 + 600 = 12,600. Adding that to 105,000 gives 105,000 + 12,600 = 117,600. So the area is 117,600 cm².Looking at the answer choices, option E is 117600 cm². So that must be the answer. Let me just double-check my calculations to make sure I didn't make a mistake.First, the shoe is 28 cm. The length is 15 shoes, so 15*28 = 420 cm. The width is 10 shoes, so 10*28 = 280 cm. Then the area is 420 cm * 280 cm. Let me compute that. 420 times 280. Well, 420 times 200 is 84,000. 420 times 80 is 33,600. Adding those together: 84,000 + 33,600 = 117,600 cm². Yep, same result. So that's correct. The area of the carpet is 117,600 cm², which is option E.Wait, but hold on. Let me check the answer choices again. The options given are:(A) 150 cm²(D) 630000 cm²(B) 4200 cm²(E) 117600 cm²(C) 22500 cm²Yes, option E is there. So that's definitely the right answer. But just to make sure, let me think if there's any chance that the problem might have said "fits exactly 15 times along one edge and 10 times along another" but perhaps in some different way. For example, maybe he walked the shoe along the edges and counted the number of steps? But the problem says "fits exactly 15 times along one edge", which sounds like laying the shoe end to end 15 times. So each edge is 15 shoe lengths and 10 shoe lengths, respectively. So converting both to centimeters by multiplying by 28, then multiplying together gives the area. That seems right.Alternatively, if someone thought that the total number of shoes is 15 + 10 = 25, but that would be perimeter, but the question is about area, which is multiplication. But the problem clearly states "along one edge" and "another", so it's two different edges. So length and width. So 15 shoes for length, 10 shoes for width, so area is 15*10*(shoe length)^2. Which is 150*(28)^2 = 150*784=117,600. So yes, that's correct.Therefore, the answer is E.

✅:To determine the area of the rectangular carpet using Bogdan's shoe measurements, follow these steps:1. Convert shoe lengths to centimeters: - The shoe length is 28 cm. - One edge of the carpet is 15 shoe lengths: ( 15 times 28 , text{cm} = 420 , text{cm} ). - The other edge is 10 shoe lengths: ( 10 times 28 , text{cm} = 280 , text{cm} ).2. Calculate the area: - Area = Length × Width ( = 420 , text{cm} times 280 , text{cm} ). - Compute ( 420 times 280 ): ( 420 times 200 = 84,000 ) and ( 420 times 80 = 33,600 ). Summing these: ( 84,000 + 33,600 = 117,600 , text{cm}^2 ).Final Answer: (boxed{E}) (117600 , text{cm}^2)

🔑:1. Given the problem, we are tasked with finding the area of a rectangular carpet. The dimensions of the carpet in terms of a shoe are 15 shoes and 10 shoes respectively. The length of the shoe is also given as 28 cm.2. First, calculate the actual length of each dimension of the carpet using the shoe length: - One dimension (length) of the carpet: [ text{Length} = 15 times 28 , text{cm} = 420 , text{cm} ] - The other dimension (width) of the carpet: [ text{Width} = 10 times 28 , text{cm} = 280 , text{cm} ]3. Next, compute the area of the carpet using the calculated dimensions: [ text{Area} = text{Length} times text{Width} = 420 , text{cm} times 280 , text{cm} = 117600 , text{cm}^2 ]Conclusion:[boxed{117600 , text{cm}^2}] 2 (Alternative Approach):1. Using another method, we can initially calculate the area of the carpet in terms of the number of shoes. Since the carpet is 15 shoes long and 10 shoes wide, the number of square shoes is: [ text{Area in square shoes} = 15 times 10 = 150 , text{square shoes} ]2. Each shoe is a square with an area equivalent to its length squared. Given the shoe length is 28 cm: [ text{1 square shoe} = 28 , text{cm} times 28 , text{cm} = 784 , text{cm}^2 ]3. Convert the area from square shoes to square centimeters by multiplying by the area of one square shoe: [ text{Area in square centimeters} = 150 , text{square shoes} times 784 , text{cm}^2/text{square shoe} = 117600 , text{cm}^2 ]Conclusion:[boxed{117600 , text{cm}^2}]

🧠:Alright, so I need to find the sum of all positive integers that can't be written in the form 50m + 3n, where m and n are non-negative integers. Hmm, this reminds me of the coin problem, where you find the largest amount that can't be formed with coins of certain denominations. I think it's called the Frobenius number. But here, instead of two coins, it's like two numbers, 50 and 3. Wait, but the Frobenius number formula only applies when the two numbers are coprime, right? Let me check if 50 and 3 are coprime. The GCD of 50 and 3 is 1, since 50 is 2*5^2 and 3 is prime, so they share no common factors besides 1. That means the formula should work here. The formula for the Frobenius number when you have two coprime integers a and b is ab - a - b. So if I plug in 50 and 3, that would be 50*3 - 50 - 3 = 150 - 53 = 97. So 97 is the largest number that can't be expressed as 50m + 3n. But wait, the problem isn't asking for the largest such number, but the sum of all positive integers not in S. So I need to find all numbers from 1 up to 97 that can't be written in this form and then sum them up. But how do I approach this? The Frobenius number tells me the largest number that can't be expressed, but there could be other numbers below 97 that also can't be expressed. I need a systematic way to list all such numbers and add them. First, let's recall that once we have two coprime numbers a and b, every number greater than or equal to (a-1)(b-1) can be expressed as a combination of a and b. Wait, is that correct? Or is it a different formula? Let me think. The formula for the Frobenius number is ab - a - b, which is 97 in this case. So numbers greater than 97 can all be expressed. Therefore, the numbers that cannot be expressed are from 1 up to 97. But actually, even some numbers below 97 can be expressed. So I need to find all numbers from 1 to 97 that can't be written as 50m + 3n and sum them. Alternatively, maybe there's a formula for the sum of the non-representable numbers when you have two coprime integers. I remember that there is a formula: if a and b are coprime, then the sum of the numbers that cannot be expressed is (ab - a - b + 1)(ab - a - b)/2 divided by something? Wait, maybe not. Let me check.Actually, the formula for the sum of the non-representable numbers is ( (a-1)(b-1)(2ab - a - b - 1) ) / 12. Wait, I'm not sure. Maybe I should look this up, but since I can't, I need to derive it. Alternatively, perhaps it's easier to compute the numbers that can be expressed and subtract from the total sum up to 97. Wait, so the total sum of numbers from 1 to 97 is 97*98/2 = 4753. Then if I can compute the sum of numbers that can be expressed as 50m + 3n from 1 to 97, subtract that from 4753 to get the desired sum. That might be a feasible approach. But how do I compute the sum of all numbers that can be expressed as 50m + 3n where m, n are non-negative integers, and the numbers are less than or equal to 97? Let's think. First, note that 50 and 3 are the coefficients. Since 50 is much larger than 3, for each m, n can range such that 50m + 3n <= 97. Let's try to iterate over possible m values and find the possible n.For m=0: Then n can range such that 3n <=97 => n <=32. So all multiples of 3 up to 96 are included. 96 is 3*32. But 97 is not a multiple of 3. So the numbers here are 0,3,6,...,96. But we need positive integers, so starting from 3.For m=1: 50*1 + 3n <=97 => 3n <=47 => n <=15.666, so n=0 to 15. Thus the numbers here are 50, 53, 56, ..., 50+45=95. Because 3*15=45, so 50+45=95.For m=2: 50*2 +3n <=97 => 100 +3n <=97, which is impossible. So m can only be 0 or 1. So the numbers in S up to 97 are the union of the numbers from m=0 (multiples of 3 starting at 3) and m=1 (50 + multiples of 3 up to 95). Therefore, the numbers not in S are the numbers from 1 to 97 that are neither multiples of 3 nor 50 plus multiples of 3. But we need to be careful because some numbers might be in both sets. For example, 50 + 3n might overlap with the multiples of 3. Let's check if 50 is a multiple of 3? 50 divided by 3 is 16.666, so no. So 50 is not a multiple of 3. Then, adding multiples of 3 to 50 will produce numbers congruent to 50 mod 3. 50 mod 3 is 2, so all numbers in the m=1 case are congruent to 2 mod 3. Whereas the multiples of 3 (m=0 case) are congruent to 0 mod 3. Therefore, these two sets are disjoint. Therefore, the total numbers in S up to 97 are the multiples of 3 from 3 to 96 (inclusive) and the numbers 50,53,...,95. So first, let's compute the sum of the multiples of 3 from 3 to 96. The number of terms is (96 -3)/3 +1 = (93)/3 +1=31 +1=32 terms. The sum is 3*(1 + 2 + ... +32) = 3*(32*33)/2 = 3*528=1584.Next, compute the sum of numbers from 50 to 95 with step 3. Let's see: starting at 50, each term increases by 3. The last term is 95. How many terms? The formula for the nth term is 50 +3(n-1)=95 => 3(n-1)=45 =>n-1=15 =>n=16 terms. So the sum is (number of terms)/2*(first term + last term) =16/2*(50 +95)=8*145=1160. But wait, we need to check if these numbers are all less than or equal to 97. The largest term in the second set is 95, which is less than 97. So that's okay. Therefore, the total sum of numbers in S up to 97 is 1584 +1160=2744. But wait, hold on, is 0 included? The problem states that m and n are non-negative integers, so 0 is allowed. But in the problem statement, they mention positive integers not in S. So S includes numbers like 0 (when m=0,n=0), but since we are considering positive integers, we need to subtract 0 from the set. Wait, but when we calculated the sum of multiples of 3 from 3 to 96, that's correct because we are excluding 0. Similarly, the numbers 50,53,...,95 are all positive. So the total sum of elements in S up to 97 is indeed 1584 +1160=2744. Therefore, the sum of all positive integers not in S up to 97 is total sum from 1 to97 minus 2744. The total sum from 1 to97 is 97*98/2=4753. So 4753 -2744=2009. Wait, but is that correct? Let me double-check my calculations.First, sum from 1 to97: 97*98/2. 97*98=9506, divided by 2 is 4753. That's correct.Sum of multiples of 3 from 3 to96: first term 3, last term96, number of terms: (96-3)/3 +1= (93)/3 +1=31+1=32. Sum is (3 +96)*32/2=99*16=1584. Correct.Sum of numbers starting at50, step3, 16 terms: first term50, last term95. Sum=(50+95)*16/2=145*8=1160. Correct.Total sum in S:1584 +1160=2744. Then 4753-2744=2009. Hmm. But wait, 4753 -2744 is indeed 2009. But let me verify with actual numbers. For example, 1,2,4,5,7, etc., are not in S. Let's check some numbers. Is 1 in S? 1=50m +3n. Since 50m is 0 or 50, 3n would have to be 1, but 1 isn't a multiple of3. So 1 is not in S. Similarly, 2 is not. 4: 4=3n +50m. 3n=4, no. 50m=4, no. So 4 is not. 5: same. 7: same. So yes, these numbers are not in S. But wait, when m=1 and n=0, we have 50. So 50 is in S. Then 53=50+3. 56=50+6, etc. So numbers like 50,53,56,...95. Then the multiples of3:3,6,9,...96. But what about numbers like 51? 51=3*17, which is a multiple of3. So 51 is in S. Similarly, 54 is 3*18. So yes, covered. So numbers not in S would be numbers not multiples of3 and not 50+3k. So numbers congruent to1 or2 mod3, except those that are 50+3k. But since 50 is congruent to2 mod3, adding 3k preserves that. So numbers congruent to2 mod3 from50 upwards (50,53,56,...) are included, but numbers congruent to2 mod3 below50 are not. Similarly, numbers congruent to1 mod3 are never in S, since S only includes numbers congruent to0 or2 mod3. Therefore, all numbers congruent to1 mod3 from1 to97 cannot be expressed, and numbers congruent to2 mod3 from1 to49 cannot be expressed. So let's calculate how many numbers are not in S. Numbers congruent to1 mod3 between1 and97: starting at1,4,7,...,97. Wait, 97 divided by3 is32.333, so the last term is97. 1 +3k <=97 =>k <=32. So numbers:1 +3*0 up to1 +3*32=97. So 33 numbers. Their sum is sum_{k=0}^{32} (1 +3k)=33*1 +3*sum_{k=0}^{32}k=33 +3*(32*33)/2=33 +3*528=33 +1584=1617.Then numbers congruent to2 mod3 between1 and49. Since 50 is the first number in the 50 +3k sequence. So numbers congruent to2 mod3 from1 to49: starting at2,5,8,...,49- let's see. The numbers are2 +3k where2 +3k <=49. So 3k <=47 =>k <=15.666, so k=0 to15. Thus, numbers:2,5,8,...,47 (since2 +3*15=47). Number of terms:16 terms. Sum: sum_{k=0}^{15} (2 +3k)=16*2 +3*sum_{k=0}^{15}k=32 +3*(15*16)/2=32 +3*120=32 +360=392.Therefore, total sum of numbers not in S is1617 +392=2009. Which matches the previous result. So that's good, cross-verified.Therefore, the answer is2009. But wait, let me make sure there are no overlaps or mistakes. For example, the numbers congruent to1 mod3 from1 to97 sum to1617, and numbers congruent to2 mod3 from1 to49 sum to392, so total2009. Correct. Alternatively, using the inclusion-exclusion principle: the total numbers not in S are those congruent to1 mod3 (all of them) and those congruent to2 mod3 below50. So numbers in S are multiples of3 and numbers congruent to2 mod3 starting from50. So yes, the non-representable numbers are the ones mentioned. Therefore, the sum is1617 +392=2009. So the answer isboxed{2009}.Final Answerboxed{2009}

✅:To find the sum of all positive integers not in ( S ), where ( S ) is the set of integers that can be written in the form ( 50m + 3n ) with ( m ) and ( n ) as non-negative integers, we start by noting that 50 and 3 are coprime. The Frobenius number for 50 and 3 is ( 50 times 3 - 50 - 3 = 97 ), meaning 97 is the largest number that cannot be expressed as ( 50m + 3n ).We need to sum all positive integers from 1 to 97 that cannot be written in the form ( 50m + 3n ). These numbers are either congruent to 1 modulo 3 or congruent to 2 modulo 3 but less than 50.1. Numbers congruent to 1 modulo 3 from 1 to 97: - These numbers are ( 1, 4, 7, ldots, 97 ). - There are 33 terms (since ( 97 = 1 + 3 times 32 )). - The sum is ( sum_{k=0}^{32} (1 + 3k) = 33 times 1 + 3 times frac{32 times 33}{2} = 33 + 1584 = 1617 ).2. Numbers congruent to 2 modulo 3 from 1 to 49: - These numbers are ( 2, 5, 8, ldots, 47 ). - There are 16 terms (since ( 47 = 2 + 3 times 15 )). - The sum is ( sum_{k=0}^{15} (2 + 3k) = 16 times 2 + 3 times frac{15 times 16}{2} = 32 + 360 = 392 ).Adding these sums gives the total sum of all positive integers not in ( S ):[ 1617 + 392 = 2009 ]Thus, the sum of all positive integers not in ( S ) is (boxed{2009}).

🔑:1. Understanding the Set (S): (S) is the set of integers that can be expressed as (50m + 3n) where (m, n) are non-negative integers. We need to determine the smallest integers modulo 3 that belong to (S).2. Smallest Multiples and Their Properties: - For (i equiv 0 pmod{3}): The smallest integer is (3). - For (i equiv 1 pmod{3}): The smallest integer is (50). - For (i equiv 2 pmod{3}): The smallest integer is (50 + 50 = 100).3. Generating Subsequent Numbers in (S): Since (3) is the smallest multiple of 3 in (S), any number (x) in (S) can be written as (x + 3k). Similarly, numbers smaller than (x) but having the same residue modulo 3 are not in (S).4. Identifying Positive Integers Not in (S): - Numbers that are (1 pmod{3}): (1, 4, 7, ldots, 97) - Numbers that are (2 pmod{3}): (2, 5, 8, ldots, 47)5. Summing Up the Series of Numbers Not in (S): - Sum of the sequence (1, 4, 7, ldots, 97): [ a = 1, , l = 97, , d = 3 ] [ n = frac{97 - 1}{3} + 1 = 33 ] Using the sum formula for an arithmetic series (S_n = frac{n}{2} (a + l)): [ sum_1 = frac{33}{2} (1 + 97) = frac{33 times 98}{2} = 33 times 49 = 1617 ] - Sum of the sequence (2, 5, 8, ldots, 47): [ a = 2, , l = 47, , d = 3 ] [ n = frac{47 - 2}{3} + 1 = 16 ] Using the sum formula for an arithmetic series (S_n = frac{n}{2} (a + l)): [ sum_2 = frac{16}{2} (2 + 47) = 8 times 49 = 392 ]6. Summing the Series to Get the Final Answer: [ text{Total Sum} = sum_1 + sum_2 = 1617 + 392 = 2009 ] Conclusion:[boxed{2009}]

🧠:Okay, so I need to prove that for every integer n, the fraction (21n + 4)/(14n + 3) is irreducible. Hmm, irreducible means that the numerator and the denominator have no common divisors other than 1, right? So, I need to show that the greatest common divisor (GCD) of 21n + 4 and 14n + 3 is 1 for all integers n. Let me recall that the Euclidean algorithm is a good method for finding the GCD of two numbers. Maybe I can apply that here. The Euclidean algorithm works by repeatedly applying the division algorithm to reduce the problem into smaller numbers. So, if I take the two numbers 21n + 4 and 14n + 3, I can start by dividing the larger one by the smaller one and then take the remainder. Let me try that step by step.First, let me denote a = 21n + 4 and b = 14n + 3. Then, GCD(a, b) should be the same as GCD(b, a mod b). Let's compute a mod b. To find a mod b, I need to find the remainder when a is divided by b. So, let's divide 21n + 4 by 14n + 3. Let's perform the division:21n + 4 divided by 14n + 3. The coefficient of n in the numerator is 21, and in the denominator, it's 14. So, 21 divided by 14 is 1.5, but since we are dealing with integers, the quotient would be 1. So, multiplying the denominator by 1 gives 14n + 3, and subtracting this from the numerator:(21n + 4) - 1*(14n + 3) = 21n + 4 - 14n - 3 = 7n + 1.So, the remainder is 7n + 1. Therefore, GCD(21n + 4, 14n + 3) = GCD(14n + 3, 7n + 1).Now, we repeat the process with b = 14n + 3 and the new remainder r = 7n + 1. So, compute GCD(14n + 3, 7n + 1). Let's divide 14n + 3 by 7n + 1. The coefficient here is 14 divided by 7, which is exactly 2, so the quotient is 2. Multiply 7n + 1 by 2: 14n + 2. Subtract that from 14n + 3:(14n + 3) - 2*(7n + 1) = 14n + 3 - 14n - 2 = 1.So, the remainder is 1. Hence, GCD(14n + 3, 7n + 1) = GCD(7n + 1, 1). But the GCD of any number and 1 is 1, because 1 divides every integer, and the only positive divisor of 1 is itself. Therefore, GCD(7n + 1, 1) = 1.Putting it all together, GCD(21n + 4, 14n + 3) = 1. Therefore, the fraction cannot be reduced further for any integer n. That seems straightforward, but let me verify if I made any mistakes in the steps.First, I applied the Euclidean algorithm correctly. Divided 21n + 4 by 14n + 3, got a remainder of 7n + 1. Then, divided 14n + 3 by 7n + 1, got a remainder of 1. Then, GCD reduces to 1. So, yes, the conclusion is that the GCD is 1. Therefore, the fraction is irreducible.Wait, but maybe I should check with specific values of n to ensure that this works. Let's test n = 0. Then, numerator is 4, denominator is 3. GCD(4, 3) is 1. So, irreducible. Good.n = 1: numerator is 25, denominator is 17. GCD(25, 17) = 1. Correct.n = -1: numerator is -21 + 4 = -17, denominator is -14 + 3 = -11. GCD(17, 11) = 1. Still works.n = 2: numerator 42 + 4 = 46, denominator 28 + 3 = 31. GCD(46, 31) = 1. Yep.n = 10: 21*10 + 4 = 214, 14*10 + 3 = 143. GCD(214, 143). Let's compute that. 214 divided by 143 is 1 with remainder 71. Then, 143 divided by 71 is 2 with remainder 1. Then GCD is 1. So, works here too.Hmm, all these examples check out. So, my reasoning seems correct.But let me think if there's another way to approach this. Maybe by assuming that there's a common divisor d > 1 and reaching a contradiction.Suppose d divides both 21n + 4 and 14n + 3. Then, d divides their linear combinations. For example, d divides (21n + 4) - k*(14n + 3) for any integer k. Let me choose k such that the coefficients of n cancel out or reduce. Let's try to eliminate n. For instance, if we take 21n + 4 - 1*(14n + 3) = 7n + 1, as before. So, d divides 7n + 1. Then, since d divides 14n + 3 and 7n + 1, let's take another combination: 14n + 3 - 2*(7n + 1) = 14n + 3 - 14n - 2 = 1. Therefore, d divides 1. Hence, d = 1. Therefore, the GCD must be 1. That's another way to see it, same result.This method also leads to the conclusion that the GCD is 1, which confirms the previous result.Alternatively, using the property that GCD(a, b) = GCD(a - b, b). So, GCD(21n + 4, 14n + 3) = GCD((21n + 4) - (14n + 3), 14n + 3) = GCD(7n + 1, 14n + 3). Then, GCD(7n + 1, 14n + 3) = GCD(7n + 1, (14n + 3) - 2*(7n + 1)) = GCD(7n + 1, 1). Which again gives GCD 1.So, all these methods converge to the same conclusion. Therefore, it's solidifying my confidence that the fraction is indeed irreducible for all integers n.Wait a second, but maybe I should also consider that n can be any integer, positive, negative, or zero. Does the proof hold for all integers? Let me check with a negative n. Let's say n = -2. Then numerator is 21*(-2) + 4 = -42 + 4 = -38. Denominator is 14*(-2) + 3 = -28 + 3 = -25. GCD of -38 and -25. GCD is the same as GCD of 38 and 25. 38 divided by 25 is 1 with remainder 13. Then, GCD(25,13). 25 divided by 13 is 1 with remainder 12. GCD(13,12). Then GCD(12,1) which is 1. So, yes, still 1. Another negative example: n = -3. Numerator: 21*(-3) +4 = -63 +4 = -59. Denominator: 14*(-3)+3 = -42 +3 = -39. GCD(59,39). 59 divided by 39 is 1 with remainder 20. GCD(39,20). 39 divided by 20 is 1 with remainder 19. GCD(20,19). Then GCD(19,1)=1. So, works here too.So, regardless of n being positive or negative, the GCD is 1. What if n is zero? As checked before, GCD(4,3)=1. Therefore, the proof seems robust across all integers. Is there another angle to approach this? Maybe using Bézout's identity, which states that if there exist integers x and y such that ax + by = 1, then GCD(a, b) = 1. So, if we can find integers x and y such that (21n +4)x + (14n +3)y =1, then their GCD is 1. Let's see if we can find such x and y. Let me try to find x and y. Let's express 1 as a combination:(21n + 4)x + (14n + 3)y = 1.We can rearrange terms:n*(21x + 14y) + (4x + 3y) = 1.For this equation to hold for all integers n, the coefficients of n and the constant term must satisfy:21x + 14y = 0,4x + 3y = 1.So, we have a system of equations:1. 21x + 14y = 02. 4x + 3y = 1Let me solve this system. From the first equation, divide both sides by 7: 3x + 2y = 0. So, 3x = -2y. Let me express x in terms of y: x = (-2/3)y. Substitute into the second equation:4*(-2/3)y + 3y = 1 => (-8/3)y + 3y = 1 => (-8/3 + 9/3)y = 1 => (1/3)y = 1 => y = 3.Then, x = (-2/3)*3 = -2.So, x = -2 and y = 3. Let me check these values in the original equation:(21n +4)(-2) + (14n +3)(3) = -42n -8 + 42n +9 = ( -42n +42n ) + ( -8 +9 ) = 0 +1 =1.Yes, it works. Therefore, there exist integers x = -2 and y = 3 such that the linear combination equals 1. Therefore, by Bézout's identity, GCD(21n +4,14n +3)=1. Hence, the fraction is irreducible.This method also confirms the result. So, using Bézout's identity is another valid approach here.Alternatively, maybe using prime factors. Suppose that there's a prime p that divides both numerator and denominator. Then, p divides (21n +4) and p divides (14n +3). Then, p divides their difference: (21n +4) - (14n +3) =7n +1. So, p divides 7n +1. Then, since p divides 14n +3 and 7n +1, p divides 2*(7n +1) - (14n +3) =14n +2 -14n -3= -1. So, p divides -1, which means p divides 1. But the only prime divisor of 1 is none, so there's no such prime p. Therefore, GCD is 1.This is another way to see it. So, that's a concise proof.So, multiple methods lead to the same conclusion. Therefore, the fraction cannot be simplified further for any integer n. All the approaches—using the Euclidean algorithm, linear combinations, Bézout's identity, prime factors—confirm that the GCD is 1. Therefore, I can confidently say that the fraction (frac{21n + 4}{14n + 3}) is irreducible for all integers n.

✅:To prove that the fraction (frac{21n + 4}{14n + 3}) cannot be reduced further for any integer (n), we need to show that the greatest common divisor (GCD) of the numerator (21n + 4) and the denominator (14n + 3) is always 1. Here's a step-by-step explanation:1. Apply the Euclidean Algorithm: - Let (a = 21n + 4) and (b = 14n + 3). - Compute (a mod b): [ (21n + 4) - 1 cdot (14n + 3) = 7n + 1 ] Thus, (gcd(21n + 4, 14n + 3) = gcd(14n + 3, 7n + 1)).2. Repeat the Euclidean Algorithm: - Now, let (a' = 14n + 3) and (b' = 7n + 1). - Compute (a' mod b'): [ (14n + 3) - 2 cdot (7n + 1) = 1 ] Thus, (gcd(14n + 3, 7n + 1) = gcd(7n + 1, 1)).3. Final GCD: - Since (gcd(k, 1) = 1) for any integer (k), we conclude: [ gcd(21n + 4, 14n + 3) = 1 ]4. Verification via Bézout's Identity: - Find integers (x) and (y) such that ((21n + 4)x + (14n + 3)y = 1). - Solving the system: [ 21x + 14y = 0 quad text{and} quad 4x + 3y = 1 ] yields (x = -2) and (y = 3), confirming that the GCD is 1.Conclusion: The GCD of the numerator and denominator is always 1, so the fraction (frac{21n + 4}{14n + 3}) is irreducible for all integers (n).(boxed{text{The fraction } frac{21n + 4}{14n + 3} text{ is irreducible for all integers } n.})

🔑:1. Goal: Prove that the fraction frac{21n+4}{14n+3} is in its simplest form for any integer n. This means we need to show that the greatest common divisor (gcd) of the numerator and denominator is 1, i.e., gcd(21n+4, 14n+3) = 1.2. Applying Bézout's identity: We will use Bézout's identity which states that if a and b are integers, then gcd(a, b) = d if and only if there exist integers x and y such that ax + by = d. In our case, we need to demonstrate there exist integers x and y such that: [ (21n + 4)x + (14n + 3)y = 1 ]3. Find such integers: Consider the specific integers x = -2 and y = 3. Calculate the expression: [ 3(14n + 3) - 2(21n + 4) ] Perform the operations within the expressions: [ 3(14n + 3) = 42n + 9 ] [ -2(21n + 4) = -42n - 8 ] Adding these two results together gives: [ 42n + 9 - 42n - 8 = 9 - 8 = 1 ]4. Conclusion: Since we have found integers x = -2 and y = 3 such that (3(14n + 3) - 2(21n + 4) = 1), it follows from Bézout’s identity that the gcd, gcd(21n + 4, 14n + 3), is indeed 1. This completes the proof that the fraction frac{21n+4}{14n+3} cannot be reduced any further.[boxed{text{The fraction is in its simplest form.}}]

🧠:Okay, so I need to find the second derivative of the function y = sin²x. Hmm, let me start by recalling what the second derivative means. The second derivative is just the derivative of the first derivative, right? So first, I have to find the first derivative of y with respect to x, and then differentiate that result again to get the second derivative.Alright, let's start with the first derivative. The function is y = sin²x. I remember that this is the same as (sinx)². To differentiate this, I should use the chain rule. The chain rule states that if you have a composite function, like f(g(x)), then the derivative is f'(g(x)) * g'(x). So here, the outer function is something squared, and the inner function is sinx. Applying the chain rule, the derivative of the outer function (something squared) is 2*sinx, and then multiply by the derivative of the inner function, which is cosx. So putting that together, the first derivative dy/dx should be 2*sinx*cosx. Let me write that down:dy/dx = 2 sinx cosxWait, but maybe there's another way to write 2 sinx cosx? Oh right, that's the double-angle identity for sine. 2 sinx cosx is equal to sin(2x). So alternatively, dy/dx = sin(2x). That might make taking the second derivative easier. Hmm, but I can proceed either way. Let me think. If I keep it as 2 sinx cosx, then I need to use the product rule for the second derivative. If I rewrite it as sin(2x), then I can use the chain rule again. Which is simpler?The product rule states that the derivative of u*v is u'v + uv'. So if I have 2 sinx cosx, then the derivative would be 2*(d/dx sinx * cosx + sinx * d/dx cosx). Let's compute that. The derivative of sinx is cosx, so the first term is cosx * cosx = cos²x. The derivative of cosx is -sinx, so the second term is sinx * (-sinx) = -sin²x. Therefore, the derivative would be 2*(cos²x - sin²x). Then multiply by the 2 in front, so 2*(cos²x - sin²x). Alternatively, cos²x - sin²x is another double-angle identity, equal to cos(2x). Therefore, this simplifies to 2*cos(2x). So the second derivative is 2 cos(2x).Alternatively, if I use the expression dy/dx = sin(2x), then taking the derivative of that would require the chain rule. The derivative of sin(u) is cos(u)*u', where u = 2x, so the derivative is cos(2x)*2, so that's 2 cos(2x). So either way, I get the same result. That seems consistent.Therefore, the second derivative d²y/dx² is 2 cos(2x). Let me double-check my steps to make sure I didn't make a mistake. First derivative: starting with sin²x, apply chain rule: 2 sinx cosx. Correct. Then, second derivative: if I use the product rule on 2 sinx cosx, factor out the 2, then differentiate sinx cosx. The product rule gives cosx*cosx + sinx*(-sinx) = cos²x - sin²x. Multiply by 2, gives 2(cos²x - sin²x). Which is equal to 2 cos(2x), since cos(2x) = cos²x - sin²x. Alternatively, if I first rewrite the first derivative as sin(2x), then derivative is 2 cos(2x). So same result. Therefore, the second derivative is 2 cos(2x). Wait, but just to make sure, let me verify with an example. Suppose x = 0. Then y = sin²0 = 0. First derivative dy/dx at x=0 is 2 sin0 cos0 = 0. The second derivative would be 2 cos(0) = 2*1 = 2. Alternatively, if I compute the second derivative as 2 cos(2x), at x=0, that's 2 cos0 = 2*1 = 2. Let me check another point. Let’s take x = π/4. Then y = sin²(π/4) = (√2/2)² = 1/2. First derivative dy/dx is 2*(√2/2)*(√2/2) = 2*(1/2) = 1. But using the sin(2x) version, at x=π/4, sin(2*(π/4)) = sin(π/2) = 1. Then the second derivative should be 2 cos(2*(π/4)) = 2 cos(π/2) = 0. Let me compute via the other expression: 2(cos²x - sin²x) at x=π/4. cos²(π/4) = (√2/2)² = 1/2, sin²(π/4) = 1/2, so 2*(1/2 - 1/2) = 0. Correct. So both methods give the same result. So seems like the second derivative is indeed 2 cos(2x). Therefore, after careful consideration and verification, the second derivative of y = sin²x is 2 cos(2x).

✅:To find the second derivative of ( y = sin^2 x ), follow these steps:1. First Derivative: - Apply the chain rule to ( y = sin^2 x ): [ frac{dy}{dx} = 2 sin x cdot cos x = sin 2x quad (text{using the identity } 2 sin x cos x = sin 2x). ]2. Second Derivative: - Differentiate ( frac{dy}{dx} = sin 2x ) using the chain rule: [ frac{d^2y}{dx^2} = frac{d}{dx} (sin 2x) = 2 cos 2x. ]Verification:- Using the product rule on ( 2 sin x cos x ): [ frac{d^2y}{dx^2} = 2 left( cos^2 x - sin^2 x right) = 2 cos 2x quad (text{since } cos 2x = cos^2 x - sin^2 x). ]Both methods confirm the result.Final Answer:[boxed{2 cos 2x}]

🔑:1. To find the second derivative of the function ( y = sin^2(x) ), we begin by finding the first derivative. 2. We use the chain rule for differentiation: [ y = (sin(x))^2 ] Let ( u = sin(x) ), then ( y = u^2 ). The derivative of ( y ) with respect to ( x ) is: [ frac{dy}{dx} = frac{d}{dx} (u^2) = 2u frac{du}{dx} ] Substituting back ( u = sin(x) ) and ( frac{du}{dx} = cos(x) ): [ y' = 2 sin(x) cos(x) ] Recognizing the double-angle identity for sine, (sin(2x) = 2 sin(x) cos(x)), we get: [ y' = sin(2x) ]3. Next, to find the second derivative, we differentiate ( y' = sin(2x) ) with respect to ( x ): [ y'' = frac{d}{dx} (sin(2x)) ] Again, using the chain rule, let ( v = 2x ). Then: [ y'' = frac{d}{dx} (sin(v)) = cos(v) frac{dv}{dx} ] Substituting back ( v = 2x ) and ( frac{dv}{dx} = 2 ): [ y'' = cos(2x) cdot 2 = 2 cos(2x) ]Conclusion:[boxed{2 cos(2x)}]

🧠:Okay, so I have this problem here: Given integers a, b, c, d, and m, such that the polynomial a m³ + b m² + c m + d is divisible by 5, and d is not divisible by 5. I need to prove that there's always an integer n such that d n³ + c n² + b n + a is also divisible by 5. Hmm, interesting.First, let's parse the problem. We have two polynomials. The first one is a cubic polynomial evaluated at m, and it's divisible by 5. The second one is another cubic polynomial with coefficients reversed (d, c, b, a) evaluated at some n, and we need to show such an n exists that this is also divisible by 5. The key point here is that d is not divisible by 5, so d is invertible modulo 5. That might be important.Since we're dealing with divisibility by 5, it might be helpful to work modulo 5. Let me recall that if something is divisible by 5, it's congruent to 0 mod 5. So, the first condition is:a m³ + b m² + c m + d ≡ 0 mod 5.And we need to find an n such that:d n³ + c n² + b n + a ≡ 0 mod 5.Also, since d is not divisible by 5, d mod 5 is 1, 2, 3, or 4. So, d is invertible modulo 5. Maybe we can manipulate these equations somehow.Let me think. If I have the first equation, maybe I can solve for a in terms of the other variables modulo 5. Let's try that.From the first equation:a m³ ≡ - (b m² + c m + d) mod 5.So, a ≡ - (b m² + c m + d) / m³ mod 5. Wait, but division is multiplication by the inverse. So, assuming that m is invertible mod 5, which would be the case if m is not divisible by 5. But m could be divisible by 5, right? Wait, but if m is divisible by 5, then the original equation a m³ + b m² + c m + d ≡ 0 mod 5 reduces to d ≡ 0 mod 5, which contradicts the given condition that d is not divisible by 5. Therefore, m cannot be divisible by 5. So, m is invertible modulo 5.Therefore, m ≡ 1, 2, 3, or 4 mod 5. So, m is invertible, which is good. Therefore, we can write:a ≡ - (b m² + c m + d) / m³ mod 5.Alternatively, multiplying both sides by m³:a m³ ≡ - (b m² + c m + d) mod 5.But maybe there's a better way to approach this. Let's consider the original polynomial P(x) = a x³ + b x² + c x + d. Given that P(m) ≡ 0 mod 5. We need to show that the reversed polynomial Q(x) = d x³ + c x² + b x + a has a root mod 5. Since we need to find n such that Q(n) ≡ 0 mod 5.Since we're working modulo 5, which is a prime, we can use properties of finite fields. But maybe even simpler, since there are only 5 possible residues mod 5 (0,1,2,3,4). We can check each residue to see if Q(n) ≡ 0 mod 5 for some n. But since the problem states that such an n must exist, perhaps we can use some symmetry or relation between P(m) and Q(n).Alternatively, maybe relate n to the inverse of m modulo 5. Let's see. Let me think: If I set n = m^{-1} mod 5, since m is invertible. Then, n is the multiplicative inverse of m mod 5. Let's see what happens if we substitute n = m^{-1} into Q(n):Q(n) = d n³ + c n² + b n + a.If n = m^{-1}, then n = 1/m mod 5. Let's compute Q(n):= d (1/m)^3 + c (1/m)^2 + b (1/m) + a.Multiply numerator and denominator by m³ to eliminate denominators:= (d + c m + b m² + a m³) / m³.But the numerator is exactly P(m) = a m³ + b m² + c m + d, which is congruent to 0 mod 5. Therefore, Q(n) = P(m)/m³ ≡ 0 / m³ ≡ 0 mod 5. Wait, that seems promising.But hold on, division by m³ is allowed because m is invertible modulo 5, so m³ is also invertible. Therefore, if we set n = m^{-1} mod 5, then Q(n) = P(m)/m³ ≡ 0 mod 5. Therefore, n = m^{-1} mod 5 would work. Thus, such an n exists.But let me verify this step by step to ensure I haven't made a mistake.Starting with Q(n) where n = m^{-1} mod 5:Q(n) = d n³ + c n² + b n + a.Substitute n = 1/m:= d (1/m)^3 + c (1/m)^2 + b (1/m) + a.Multiply numerator and denominator by m³:= (d + c m + b m² + a m³) / m³.But the numerator is exactly P(m) = a m³ + b m² + c m + d, which is congruent to 0 mod 5. Therefore, Q(n) = 0 / m³ ≡ 0 mod 5, since m³ is invertible (as m is invertible), so dividing zero by an invertible element still gives zero.Therefore, n = m^{-1} mod 5 is the desired integer. Since m is invertible mod 5, n exists (specifically, n is the multiplicative inverse of m modulo 5, which exists because 5 is prime and m is not divisible by 5). Therefore, such an n always exists.But let me check with an example to make sure this works.Take m = 2 mod 5. Then n = 3 mod 5 because 2*3=6 ≡1 mod5.Suppose P(m) = a*8 + b*4 + c*2 + d ≡0 mod5.Then Q(n) = d*27 + c*9 + b*3 + a.But 27 mod5 is 2, 9 mod5 is 4, 3 mod5 is 3. So Q(n) = 2d +4c +3b +a.Now, from P(m) ≡0 mod5: 8a +4b +2c +d ≡0 mod5. Let's compute 8a ≡3a, 4b, 2c, d. So 3a +4b +2c +d ≡0 mod5.Now, if we take Q(n)=2d +4c +3b +a. Let's compare with P(m). Let me see:If we multiply the equation 3a +4b +2c +d ≡0 mod5 by 2 (the inverse of 3 mod5 is 2, since 3*2=6≡1 mod5). Then:2*(3a +4b +2c +d) ≡0 mod5 ⇒ 6a +8b +4c +2d ≡ a + 3b +4c +2d ≡0 mod5.But Q(n)=2d +4c +3b +a, which is a +3b +4c +2d, which is the same as above. Therefore, Q(n)=0 mod5. So in this case, it works.Another example: let m=1 mod5. Then n=1^{-1}=1 mod5. Then Q(1)=d +c +b +a. But P(1)=a +b +c +d≡0 mod5. Therefore, Q(1)=P(1)=0 mod5. So n=1 works here.Similarly, if m=3 mod5, then n=2 mod5 (since 3*2=6≡1 mod5). Let's compute Q(2).Q(2)=d*8 +c*4 +b*2 +a.8 mod5=3, 4 mod5=4, 2 mod5=2. So Q(2)=3d +4c +2b +a.Now, P(m)=a*27 +b*9 +c*3 +d. 27 mod5=2, 9 mod5=4, 3 mod5=3. So P(3)=2a +4b +3c +d≡0 mod5.If we multiply this equation by 3^{-1} mod5, which is 2, since 3*2=6≡1 mod5:2*(2a +4b +3c +d)=4a +8b +6c +2d≡4a +3b +c +2d≡0 mod5.But Q(2)=3d +4c +2b +a. Let's see if these are related.Wait, not sure. Wait, maybe there's a better way. Let's compute Q(2):3d +4c +2b +a.From P(3)=2a +4b +3c +d≡0 mod5.Let me solve for a from P(3):2a ≡ - (4b +3c +d) mod5 ⇒ a ≡ - (4b +3c +d)*3 mod5 (since 2^{-1}=3 mod5).Therefore, a ≡ -12b -9c -3d ≡ (-12b mod5)= (-12 mod5)= -2, so -2b; (-9c mod5)= -4c; (-3d mod5)= -3d. So a ≡ -2b -4c -3d mod5.Now plug into Q(2)=3d +4c +2b +a:=3d +4c +2b + (-2b -4c -3d) mod5.Simplify:3d -3d +4c -4c +2b -2b ≡0 mod5.So yes, Q(2)≡0 mod5. So that works.Another example: m=4 mod5, inverse of 4 is 4, since 4*4=16≡1 mod5. So n=4.Compute Q(4)=d*64 +c*16 +b*4 +a. 64 mod5=4, 16 mod5=1, 4 mod5=4. So Q(4)=4d +c +4b +a.P(4)=a*64 +b*16 +c*4 +d. 64 mod5=4, 16=1, 4=4. So P(4)=4a +b +4c +d≡0 mod5.Let me solve for a from P(4):4a ≡ - (b +4c +d) mod5 ⇒ a ≡ - (b +4c +d)*4^{-1} mod5. Since 4^{-1}=4 mod5 (because 4*4=16≡1), so:a ≡ - (b +4c +d)*4 ≡ -4b -16c -4d ≡ -4b -1c -4d mod5.Now plug into Q(4)=4d +c +4b +a:=4d +c +4b + (-4b -c -4d) mod5.Simplify:4d -4d +c -c +4b -4b ≡0 mod5.Again, Q(4)≡0 mod5. So that works too.Therefore, in all cases, taking n = m^{-1} mod5 works, which makes sense from the algebraic manipulation earlier. Therefore, such an n always exists, specifically the inverse of m modulo 5.Therefore, the proof is as follows: Since am³ + bm² + cm + d ≡0 mod5 and d≢0 mod5, m cannot be 0 mod5 (as that would force d≡0). Thus, m is invertible mod5. Let n = m⁻¹ mod5. Then, evaluating Q(n) = dn³ + cn² + bn + a, substitute n = m⁻¹:Q(n) = d(m⁻¹)³ + c(m⁻¹)² + b(m⁻¹) + a.Multiply numerator and denominator by m³:Q(n) = (d + cm + bm² + am³)/m³.But the numerator is P(m) ≡0 mod5, so Q(n) ≡0/m³ ≡0 mod5. Hence, such an n exists.Therefore, the existence is guaranteed by taking the inverse of m modulo 5.Final Answerboxed{There always exists such an integer ( n ).}

✅:Given integers (a, b, c, d,) and (m) such that (a m^3 + b m^2 + c m + d) is divisible by 5, and (d) is not divisible by 5, we need to prove there exists an integer (n) such that (d n^3 + c n^2 + b n + a) is also divisible by 5.1. Initial Condition: The polynomial (P(m) = a m^3 + b m^2 + c m + d equiv 0 pmod{5}). Since (d notequiv 0 pmod{5}), (m) cannot be 0 modulo 5 (otherwise, (d) would be 0 modulo 5). Thus, (m) is invertible modulo 5.2. Inverse of (m): Let (n) be the multiplicative inverse of (m) modulo 5, i.e., (n equiv m^{-1} pmod{5}). This inverse exists because 5 is prime and (m) is not divisible by 5.3. Evaluating the Reversed Polynomial: Consider the polynomial (Q(n) = d n^3 + c n^2 + b n + a). Substitute (n = m^{-1}): [ Q(n) = d (m^{-1})^3 + c (m^{-1})^2 + b (m^{-1}) + a ]4. Multiplying by (m^3): Multiply numerator and denominator by (m^3) to clear the denominators: [ Q(n) = frac{d + c m + b m^2 + a m^3}{m^3} ]5. Numerator Equivalence: The numerator (d + c m + b m^2 + a m^3) is exactly (P(m)), which is congruent to 0 modulo 5: [ P(m) = a m^3 + b m^2 + c m + d equiv 0 pmod{5} ]6. Result: Since the numerator is 0 modulo 5 and (m^3) is invertible modulo 5, the entire expression is 0 modulo 5: [ Q(n) equiv frac{0}{m^3} equiv 0 pmod{5} ]Thus, there always exists an integer (n) (specifically (n = m^{-1} mod 5)) such that (d n^3 + c n^2 + b n + a) is divisible by 5.[boxed{There always exists such an integer ( n ).}]

🔑:1. Establishing the problem constraints: We are given that (a, b, c, d, text{ and } m) are integers such that the polynomial (a m^3 + b m^2 + c m + d) is divisible by 5, but (d) is not divisible by 5. We need to prove that there exists an integer (n) such that (d n^3 + c n^2 + b n + a) is also divisible by 5.2. Analyzing the divisibility condition: Let's start with the given polynomial (a m^3 + b m^2 + c m + d): [ 5 mid a m^3 + b m^2 + c m + d ] This implies that when (m) is such that (d) is not divisible by 5, the entirety of (a m^3 + b m^2 + c m + d) is divisible by 5.3. Considering the values of (m mod 5): Given that (m) cannot be divisible by 5 (as (d) is not divisible by 5), we represent (m) in the form: [ m = 5k + r, quad text{where } r in {1, 2, 3, 4} ]4. Finding (n) such that (mn equiv 1 mod 5): Notice the following congruences modulo 5: [ 1 cdot 1 equiv 1 pmod{5} ] [ 2 cdot 3 equiv 1 pmod{5} ] [ 4 cdot 4 equiv 1 pmod{5} ] Given these possibilities, if (m = 5k + 1), then (n) can be (5t + 1). Similarly: [ begin{aligned} &text{If } m = 5k + 2, text{ choose } n = 5t + 3 text{ which gives } mn equiv 1 mod 5 &text{If } m = 5k + 3, text{ choose } n = 5t + 2 text{ which gives } mn equiv 1 mod 5 &text{If } m = 5k + 4, text{ choose } n = 5t + 4 text{ which gives } mn equiv 1 mod 5 end{aligned} ]5. Setting up the expressions for (A) and (B): Let: [ A = a m^3 + b m^2 + c m + d ] [ B = d n^3 + c n^2 + b n + a ] Since (5 mid A), we will transform the expressions of (B) in a way to show that it is also divisible by 5.6. Transform given expression: To prove (B) is divisible by 5, consider eliminating (d): [ A n^3 - B equiv (mn - 1) left[a(m^2 n^2 + mn + 1) + b n(mn + 1) + c n^2right] pmod{5} ]7. Final argument of divisibility: As (m n equiv 1 pmod{5}) and given that (5 mid A): [ 5 mid (A n^3 - B) ] Given (5 mid A), it follows that: [ 5 mid B = d n^3 + c n^2 + b n + a ] Conclusion:[boxed{text{There exists an integer } n text{ such that } d n^3 + c n^2 + b n + a text{ is divisible by 5.}}]