💡Okay, so I'm trying to figure out this question about Deng Xiaoping's thoughts in 1979. The question gives a quote from Deng where he talks about wanting a prosperous socialism and mentions that socialism can engage in a market economy. Then there are four options about what these ideas facilitated.First, I need to understand the context. Deng Xiaoping was a key leader in China during the late 20th century, and he's known for initiating economic reforms and opening up the country to the outside world. The year 1979 is significant because it's just after the Cultural Revolution, and China was starting to shift its economic policies.Looking at the options:A: Facilitated the establishment and development of China's Special Economic ZonesB: Set the goal of establishing a market economy system as the reform objectiveC: Marked the final formation of New China's decision to open up to the outside worldD: Promoted the implementation of the agricultural production responsibility system in Anhui and SichuanI know that the Special Economic Zones (SEZs) were established in 1980, so Deng's 1979 statements probably influenced their creation. That makes A a strong candidate.Option B talks about setting the goal of a market economy. I remember that the goal of a socialist market economy was formally set later, maybe in the 1990s, so B might not be correct.Option C mentions the final decision to open up. I think the decision to open up was made earlier, around the late 1970s, so C might not be the best fit.Option D refers to the agricultural responsibility system in Anhui and Sichuan. I believe that reform started earlier, around 1978, so D might not be directly related to Deng's 1979 statements.So, putting it all together, A seems to be the correct answer because Deng's ideas in 1979 likely led to the creation of the SEZs in 1980.

💡Okay, so I have this problem where I need to figure out how many of the sets ( T_0, T_1, T_2, ldots, T_{499} ) do not contain a perfect square. Each set ( T_i ) consists of integers ( n ) such that ( 200i leq n < 200(i + 1) ). For example, ( T_4 ) is the set ( {800, 801, 802, ldots, 999} ). First, I need to understand what each set ( T_i ) represents. Each ( T_i ) is a block of 200 consecutive integers starting from ( 200i ) up to ( 200(i + 1) - 1 ). So, ( T_0 ) is ( {0, 1, 2, ldots, 199} ), ( T_1 ) is ( {200, 201, ldots, 399} ), and so on, up to ( T_{499} ), which is ( {99800, 99801, ldots, 99999} ).The question is asking how many of these 500 sets do not contain any perfect squares. So, my goal is to count how many of these blocks ( T_i ) have no perfect squares within them.To approach this, I think I should first figure out the range of perfect squares that lie within the entire range of ( T_0 ) to ( T_{499} ). The smallest number in ( T_0 ) is 0, and the largest number in ( T_{499} ) is 99999. So, I need to find all perfect squares between 0 and 99999.The smallest perfect square is ( 0^2 = 0 ), and the largest perfect square less than 100000 is ( 316^2 = 99856 ) because ( 317^2 = 100489 ), which is greater than 100000. So, the perfect squares we are considering are ( 0^2, 1^2, 2^2, ldots, 316^2 ).Now, each perfect square ( k^2 ) will lie in some set ( T_i ). Specifically, ( k^2 ) will be in ( T_i ) where ( i = lfloor frac{k^2}{200} rfloor ). Therefore, each perfect square corresponds to a specific ( T_i ). If a ( T_i ) contains at least one perfect square, it is "hit" by that square. We need to find how many ( T_i ) are not hit by any perfect square.So, the total number of sets is 500 (from ( T_0 ) to ( T_{499} )). If I can find how many of these sets are hit by at least one perfect square, then subtracting that number from 500 will give me the number of sets that do not contain any perfect squares.To find how many sets are hit, I can consider the distribution of perfect squares across the sets ( T_i ). Since each ( T_i ) is 200 numbers long, the difference between consecutive squares is important here. The difference between ( (k+1)^2 ) and ( k^2 ) is ( 2k + 1 ). For smaller ( k ), this difference is small. For example, ( 1^2 = 1 ) and ( 2^2 = 4 ), so the difference is 3. As ( k ) increases, the difference between consecutive squares increases. When ( k ) is 100, the difference is ( 2*100 + 1 = 201 ). So, starting from ( k = 100 ), the difference between consecutive squares exceeds 200. This means that beyond ( k = 100 ), each perfect square is more than 200 apart from the next one.This is significant because if the difference between consecutive squares is greater than 200, it means that each square will fall into a different ( T_i ) set. For example, ( 100^2 = 10000 ) is in ( T_{50} ) because ( 10000 / 200 = 50 ). The next square is ( 101^2 = 10201 ), which is in ( T_{51} ) because ( 10201 / 200 = 51.005 ), so ( lfloor 51.005 rfloor = 51 ). Similarly, ( 102^2 = 10404 ) is in ( T_{52} ), and so on.Therefore, starting from ( k = 100 ), each perfect square falls into a unique ( T_i ) set. This means that from ( k = 100 ) to ( k = 316 ), each ( T_i ) will contain exactly one perfect square, provided that the squares don't overlap into the same ( T_i ). Since the difference between consecutive squares is greater than 200, they won't overlap.Now, let's calculate how many perfect squares there are from ( k = 100 ) to ( k = 316 ). That would be ( 316 - 100 + 1 = 217 ) perfect squares. Each of these corresponds to a unique ( T_i ) set from ( T_{50} ) to ( T_{158} ) because ( 316^2 = 99856 ), and ( 99856 / 200 = 499.28 ), so ( lfloor 499.28 rfloor = 499 ). Wait, that doesn't seem right because ( 316^2 ) is 99856, which is in ( T_{499} ) since ( 499*200 = 99800 ), so 99856 is between 99800 and 99999.Wait, so starting from ( k = 100 ), the squares go from ( T_{50} ) up to ( T_{499} ). So, how many sets is that? From ( T_{50} ) to ( T_{499} ), that's ( 499 - 50 + 1 = 450 ) sets. But we only have 217 perfect squares from ( k = 100 ) to ( k = 316 ). So, each of these 217 squares falls into a unique ( T_i ) set, but since 217 is less than 450, there are many ( T_i ) sets beyond ( T_{158} ) that don't contain any perfect squares.Wait, I think I made a mistake here. Let me clarify. The number of perfect squares from ( k = 100 ) to ( k = 316 ) is 217, each corresponding to a unique ( T_i ) set. So, these 217 sets ( T_i ) each contain exactly one perfect square. Therefore, these 217 sets are "hit" by a perfect square.But before ( k = 100 ), i.e., from ( k = 0 ) to ( k = 99 ), the difference between consecutive squares is less than 200. So, in these cases, it's possible that multiple perfect squares fall into the same ( T_i ) set. Therefore, we need to check how many ( T_i ) sets are hit by these smaller squares.So, let's break this down into two parts:1. Perfect squares from ( k = 0 ) to ( k = 99 ): These are squares from 0 to ( 99^2 = 9801 ). Each of these squares falls into some ( T_i ) set. Since the difference between consecutive squares is less than 200, it's possible that multiple squares fall into the same ( T_i ).2. Perfect squares from ( k = 100 ) to ( k = 316 ): These are squares from ( 10000 ) to ( 99856 ). Each of these squares falls into a unique ( T_i ) set because the difference between consecutive squares is greater than 200.So, first, let's handle the squares from ( k = 0 ) to ( k = 99 ). Each square ( k^2 ) falls into ( T_i ) where ( i = lfloor frac{k^2}{200} rfloor ). We need to find how many unique ( T_i ) sets are hit by these squares.To do this, we can note that for ( k ) from 0 to 99, ( k^2 ) ranges from 0 to 9801. Each ( T_i ) set is 200 numbers long, so the number of ( T_i ) sets hit by these squares is roughly the number of squares divided by the average number of squares per set. However, since the squares are increasing quadratically, the number of squares per set decreases as ( k ) increases.Alternatively, we can calculate the number of unique ( T_i ) sets hit by squares from ( k = 0 ) to ( k = 99 ). Let's consider that each square ( k^2 ) is in ( T_i ) where ( i = lfloor frac{k^2}{200} rfloor ). So, for each ( k ), compute ( i ) and count the unique ( i ) values.But doing this for each ( k ) from 0 to 99 would be tedious. Instead, we can find the range of ( k ) such that ( k^2 ) falls into the same ( T_i ). For example, ( T_0 ) contains squares from ( k = 0 ) to ( k = 14 ) because ( 14^2 = 196 ) and ( 15^2 = 225 ), which is in ( T_1 ). So, ( T_0 ) contains squares from ( k = 0 ) to ( k = 14 ).Similarly, ( T_1 ) contains squares from ( k = 15 ) to ( k = 22 ) because ( 22^2 = 484 ) and ( 23^2 = 529 ), which is in ( T_2 ). Wait, let me check: ( 200*1 = 200 ), so ( T_1 ) is 200 to 399. The squares in this range are from ( k = 15 ) (225) up to ( k = 19 ) (361), because ( 20^2 = 400 ) is in ( T_2 ). So, ( T_1 ) contains squares from ( k = 15 ) to ( k = 19 ).Wait, this is getting complicated. Maybe a better approach is to find the maximum ( k ) such that ( k^2 < 200(i + 1) ) for each ( i ). But this might not be straightforward.Alternatively, we can note that for ( k ) from 0 to 99, the squares ( k^2 ) are spread across ( T_0 ) to ( T_{49} ) because ( 99^2 = 9801 ), and ( 9801 / 200 = 49.005 ), so ( lfloor 49.005 rfloor = 49 ). Therefore, the squares from ( k = 0 ) to ( k = 99 ) fall into ( T_0 ) to ( T_{49} ).Now, how many unique ( T_i ) sets are hit by these squares? Each ( T_i ) from ( T_0 ) to ( T_{49} ) may contain multiple squares, but we need to count how many unique ( T_i ) sets are hit. Since each square falls into some ( T_i ), and the squares are increasing, each ( T_i ) can be hit by multiple squares, but we only need to count each ( T_i ) once if it contains at least one square.Therefore, the number of unique ( T_i ) sets hit by squares from ( k = 0 ) to ( k = 99 ) is 50, because ( T_0 ) to ( T_{49} ) are all hit by at least one square. Wait, is that true? Let me check.For example, ( T_0 ) contains squares from ( k = 0 ) to ( k = 14 ). ( T_1 ) contains squares from ( k = 15 ) to ( k = 19 ). ( T_2 ) contains squares from ( k = 20 ) to ( k = 22 ) because ( 22^2 = 484 ) is in ( T_2 ) (200*2=400 to 600). Wait, 22^2 is 484, which is in ( T_2 ) (400-599). Then ( 23^2 = 529 ) is also in ( T_2 ). Wait, no, 529 is in ( T_2 ) because 529 is between 400 and 599. Then ( 24^2 = 576 ) is also in ( T_2 ). ( 25^2 = 625 ) is in ( T_3 ) because 625 is between 600 and 799.Wait, so ( T_2 ) contains squares from ( k = 20 ) to ( k = 24 ). Similarly, ( T_3 ) contains squares from ( k = 25 ) to ( k = 27 ) because ( 27^2 = 729 ) is in ( T_3 ) (600-799), and ( 28^2 = 784 ) is also in ( T_3 ). ( 29^2 = 841 ) is in ( T_4 ) because 841 is between 800 and 999.Wait, so each ( T_i ) from ( T_0 ) to ( T_{49} ) is hit by at least one square. Therefore, all 50 sets ( T_0 ) to ( T_{49} ) contain at least one perfect square.Now, moving on to the squares from ( k = 100 ) to ( k = 316 ). As established earlier, each of these squares falls into a unique ( T_i ) set because the difference between consecutive squares is greater than 200. Therefore, each square from ( k = 100 ) to ( k = 316 ) corresponds to a unique ( T_i ) set.The number of such squares is ( 316 - 100 + 1 = 217 ). Therefore, these 217 squares hit 217 unique ( T_i ) sets. Since the squares from ( k = 100 ) to ( k = 316 ) fall into ( T_{50} ) to ( T_{499} ), these 217 sets are within ( T_{50} ) to ( T_{499} ).So, in total, the number of ( T_i ) sets hit by perfect squares is the sum of the sets hit by squares from ( k = 0 ) to ( k = 99 ) and from ( k = 100 ) to ( k = 316 ). That is, 50 sets (from ( T_0 ) to ( T_{49} )) plus 217 sets (from ( T_{50} ) to ( T_{499} )).Wait, but hold on. If the squares from ( k = 100 ) to ( k = 316 ) hit 217 sets starting from ( T_{50} ), then the total number of sets hit is 50 + 217 = 267 sets.But wait, the total number of sets is 500 (from ( T_0 ) to ( T_{499} )). So, the number of sets not containing any perfect squares would be 500 - 267 = 233 sets.However, the initial thought process mentioned 234, so I might have made a miscalculation here.Let me double-check the numbers:- Squares from ( k = 0 ) to ( k = 99 ): 100 squares, but they hit 50 unique ( T_i ) sets (from ( T_0 ) to ( T_{49} )).- Squares from ( k = 100 ) to ( k = 316 ): 217 squares, each hitting a unique ( T_i ) set from ( T_{50} ) to ( T_{499} ).So, total sets hit: 50 + 217 = 267.Total sets: 500.Sets without any perfect squares: 500 - 267 = 233.But wait, the initial thought process said 234. So, where is the discrepancy?Let me check the calculation of the number of squares from ( k = 100 ) to ( k = 316 ). It's ( 316 - 100 + 1 = 217 ). That seems correct.And the number of sets hit by squares from ( k = 0 ) to ( k = 99 ) is 50 sets (from ( T_0 ) to ( T_{49} )). That also seems correct.So, 50 + 217 = 267 sets hit. Therefore, 500 - 267 = 233 sets not hit.But the initial thought process concluded 234. So, perhaps there was an off-by-one error somewhere.Wait, let's check the calculation of the number of squares from ( k = 100 ) to ( k = 316 ). If ( k = 100 ) is included, and ( k = 316 ) is included, then the count is 316 - 100 + 1 = 217. That's correct.But perhaps the initial thought process considered that ( T_{50} ) is the first set without a perfect square, but actually, ( T_{50} ) does contain a perfect square because ( 100^2 = 10000 ) is in ( T_{50} ) (since ( 50*200 = 10000 )). Wait, no, ( T_{50} ) is ( 10000 leq n < 12000 ). So, ( 100^2 = 10000 ) is the first number in ( T_{50} ). Therefore, ( T_{50} ) does contain a perfect square.Wait, but in the initial thought process, it was said that ( T_{50} ) is the first set potentially without a perfect square, but that's incorrect because ( T_{50} ) does contain ( 100^2 ).Therefore, the initial thought process might have made a mistake there. So, perhaps the first set without a perfect square is ( T_{51} ), but no, because ( 101^2 = 10201 ) is in ( T_{51} ). Similarly, ( 102^2 = 10404 ) is in ( T_{52} ), and so on.Wait, so actually, all sets from ( T_{50} ) to ( T_{499} ) contain exactly one perfect square each, because the squares from ( k = 100 ) to ( k = 316 ) fall into these sets, each in their own unique ( T_i ).But that can't be because the number of squares from ( k = 100 ) to ( k = 316 ) is 217, and the number of sets from ( T_{50} ) to ( T_{499} ) is 450 sets. So, 217 sets are hit, and the remaining 450 - 217 = 233 sets are not hit.Wait, but 500 total sets minus 267 hit sets equals 233 sets not hit. So, the answer should be 233.But the initial thought process said 234. So, perhaps there was a miscalculation there.Alternatively, maybe the initial thought process considered that ( T_{50} ) does not contain a perfect square, but that's incorrect because ( 100^2 = 10000 ) is in ( T_{50} ).Wait, let me recast the problem.Total sets: 500 (from ( T_0 ) to ( T_{499} )).Sets hit by squares from ( k = 0 ) to ( k = 99 ): 50 sets (from ( T_0 ) to ( T_{49} )).Sets hit by squares from ( k = 100 ) to ( k = 316 ): 217 sets (from ( T_{50} ) to ( T_{499} )).Total hit sets: 50 + 217 = 267.Therefore, sets not hit: 500 - 267 = 233.So, the answer should be 233.But the initial thought process concluded 234. So, perhaps the initial thought process had a miscalculation.Alternatively, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case because ( T_{50} ) contains ( 100^2 ).Wait, let me check the squares around ( T_{50} ).( T_{50} ) is ( 10000 leq n < 12000 ).( 100^2 = 10000 ) is in ( T_{50} ).( 101^2 = 10201 ) is in ( T_{51} ) because ( 10201 / 200 = 51.005 ), so ( lfloor 51.005 rfloor = 51 ).Similarly, ( 102^2 = 10404 ) is in ( T_{52} ), and so on.Therefore, ( T_{50} ) does contain a perfect square, so the first set without a perfect square would be after ( T_{50} ), but since each subsequent square falls into the next ( T_i ), there are no gaps until ( k = 316 ).Wait, but ( 316^2 = 99856 ) is in ( T_{499} ) because ( 499*200 = 99800 ), so 99856 is in ( T_{499} ).Therefore, from ( T_{50} ) to ( T_{499} ), each ( T_i ) from ( T_{50} ) to ( T_{499} ) is hit by exactly one square from ( k = 100 ) to ( k = 316 ). But wait, that would mean 450 sets (from ( T_{50} ) to ( T_{499} )) are hit by 217 squares, which is not possible because 217 < 450. Therefore, only 217 sets are hit, and the remaining 450 - 217 = 233 sets are not hit.Therefore, the total number of sets not hit is 233.But wait, the initial thought process said 234. So, perhaps the initial thought process considered that ( T_{50} ) is not hit, but that's incorrect because ( T_{50} ) is hit by ( 100^2 ).Alternatively, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case because ( T_{51} ) is hit by ( 101^2 ).Wait, perhaps the initial thought process made a miscalculation in the number of squares from ( k = 100 ) to ( k = 316 ). Let me recalculate that.( k ) ranges from 100 to 316 inclusive. So, the number of squares is ( 316 - 100 + 1 = 217 ). That's correct.Therefore, the number of sets hit by these squares is 217.Adding the 50 sets hit by squares from ( k = 0 ) to ( k = 99 ), we get 267 sets hit.Total sets: 500.Sets not hit: 500 - 267 = 233.Therefore, the correct answer should be 233.But the initial thought process concluded 234. So, perhaps there was an off-by-one error in the initial thought process.Alternatively, perhaps the initial thought process considered that ( T_{50} ) is not hit, but that's incorrect because ( T_{50} ) is hit by ( 100^2 ).Wait, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case because ( T_{51} ) is hit by ( 101^2 ).Alternatively, perhaps the initial thought process considered that the squares from ( k = 100 ) to ( k = 316 ) hit 216 sets instead of 217, leading to 50 + 216 = 266 sets hit, and 500 - 266 = 234 sets not hit.But why would that be? Let me check the number of squares from ( k = 100 ) to ( k = 316 ).( k = 100 ): 100^2 = 10000.( k = 316 ): 316^2 = 99856.So, the number of squares is 316 - 100 + 1 = 217.Therefore, 217 sets are hit by these squares.Therefore, the initial thought process might have made a miscalculation, perhaps subtracting 216 instead of 217.Alternatively, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case because ( T_{51} ) is hit by ( 101^2 ).Therefore, the correct number of sets without any perfect squares is 233.But to be thorough, let me consider the exact distribution.From ( k = 0 ) to ( k = 99 ): 100 squares, hitting 50 sets (T0-T49).From ( k = 100 ) to ( k = 316 ): 217 squares, hitting 217 sets (T50-T499).Total hit sets: 50 + 217 = 267.Total sets: 500.Sets without any perfect squares: 500 - 267 = 233.Therefore, the answer should be 233.But the initial thought process said 234. So, perhaps the initial thought process had a miscalculation.Alternatively, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case because ( T_{51} ) is hit by ( 101^2 ).Wait, perhaps the initial thought process considered that the first set without a perfect square is ( T_{50} ), but that's incorrect because ( T_{50} ) is hit by ( 100^2 ).Therefore, the correct answer is 233.But to be absolutely sure, let me consider the exact number of sets hit by squares from ( k = 100 ) to ( k = 316 ).Each square from ( k = 100 ) to ( k = 316 ) falls into a unique ( T_i ) set because the difference between consecutive squares is greater than 200. Therefore, each square corresponds to a unique ( T_i ).The number of such squares is 217, so 217 sets are hit.Therefore, the total number of sets hit is 50 (from k=0 to 99) + 217 (from k=100 to 316) = 267.Total sets: 500.Sets without any perfect squares: 500 - 267 = 233.Therefore, the answer is 233.But the initial thought process concluded 234, so perhaps there was a miscalculation there. Alternatively, perhaps the initial thought process considered that the first set without a perfect square is ( T_{51} ), but that's not the case.Therefore, after careful consideration, the correct number of sets without any perfect squares is 233.

💡Alright, so I've got this functional equation problem from the USAMO 2010. It says:Find all functions ( f: mathbb{Z} rightarrow mathbb{Z} ) such that for all integers ( x, y ) (with ( x neq 0 )) we have:[ x f(2 f(y) - x) + y^2 f(2 x - f(y)) = frac{f(x)^2}{x} + f(y f(y)) ]Okay, functional equations can be tricky, but let's try to break this down step by step.First, I remember that for functional equations, substituting specific values for variables can often simplify things and give us clues about the form of ( f ). Let me try substituting some simple values for ( x ) and ( y ).Let's start by setting ( y = 0 ). That might simplify the equation because ( y ) is multiplied by something. So, substituting ( y = 0 ):[ x f(2 f(0) - x) + 0^2 f(2x - f(0)) = frac{f(x)^2}{x} + f(0 cdot f(0)) ]Simplifying this, the ( y^2 ) term disappears, and ( f(0 cdot f(0)) ) becomes ( f(0) ). So we have:[ x f(2 f(0) - x) = frac{f(x)^2}{x} + f(0) ]Hmm, that's still a bit complicated, but maybe we can get more information by choosing a specific value for ( x ). Let's pick ( x = 2 f(0) ). Wait, but ( x ) can't be zero, so as long as ( f(0) ) isn't zero, this should be fine. Let's try that.Substituting ( x = 2 f(0) ):[ 2 f(0) f(2 f(0) - 2 f(0)) + 0^2 f(4 f(0) - f(0)) = frac{f(2 f(0))^2}{2 f(0)} + f(0) ]Simplifying inside the functions:[ 2 f(0) f(0) = frac{f(2 f(0))^2}{2 f(0)} + f(0) ]So:[ 2 f(0)^2 = frac{f(2 f(0))^2}{2 f(0)} + f(0) ]Multiply both sides by ( 2 f(0) ) to eliminate the denominator:[ 4 f(0)^3 = f(2 f(0))^2 + 2 f(0)^2 ]Hmm, that gives us a relationship between ( f(0) ) and ( f(2 f(0)) ). Let me rearrange this:[ f(2 f(0))^2 = 4 f(0)^3 - 2 f(0)^2 ]Factor out ( f(0)^2 ):[ f(2 f(0))^2 = f(0)^2 (4 f(0) - 2) ]So, ( f(2 f(0)) ) must be equal to ( f(0) sqrt{4 f(0) - 2} ). But since ( f ) maps integers to integers, ( sqrt{4 f(0) - 2} ) must also be an integer. Let me denote ( k = f(0) ), so:[ sqrt{4k - 2} ]must be an integer. Let's set ( m = sqrt{4k - 2} ), so:[ m^2 = 4k - 2 ][ 4k = m^2 + 2 ][ k = frac{m^2 + 2}{4} ]Since ( k ) must be an integer, ( m^2 + 2 ) must be divisible by 4. Let's see what possible ( m ) can satisfy this.Looking at squares modulo 4:- If ( m ) is even, say ( m = 2n ), then ( m^2 = 4n^2 equiv 0 mod 4 ).- If ( m ) is odd, say ( m = 2n + 1 ), then ( m^2 = 4n^2 + 4n + 1 equiv 1 mod 4 ).So, ( m^2 ) is either 0 or 1 modulo 4. Therefore, ( m^2 + 2 ) is either 2 or 3 modulo 4. But 2 and 3 modulo 4 are not divisible by 4, which means ( k ) cannot be an integer unless ( m^2 + 2 ) is divisible by 4. But from above, it's not possible. Therefore, the only way this can hold is if ( f(0) = 0 ).So, ( f(0) = 0 ). That's a good start.Now, let's substitute ( y = 0 ) back into the original equation with ( f(0) = 0 ):[ x f(2 f(0) - x) + 0^2 f(2x - f(0)) = frac{f(x)^2}{x} + f(0) ]Simplifying:[ x f(-x) = frac{f(x)^2}{x} ]Multiply both sides by ( x ):[ x^2 f(-x) = f(x)^2 ]Interesting. So, ( f(-x) ) is related to ( f(x) ). Let's keep this in mind.Now, let's try substituting ( x = -x ) into the equation ( x^2 f(-x) = f(x)^2 ). So, replacing ( x ) with ( -x ):[ (-x)^2 f(x) = f(-x)^2 ][ x^2 f(x) = f(-x)^2 ]But from earlier, we have ( x^2 f(-x) = f(x)^2 ). So, combining these two equations:From ( x^2 f(-x) = f(x)^2 ), we can write ( f(-x) = frac{f(x)^2}{x^2} ).Substituting this into the second equation:[ x^2 f(x) = left( frac{f(x)^2}{x^2} right)^2 ][ x^2 f(x) = frac{f(x)^4}{x^4} ]Multiply both sides by ( x^4 ):[ x^6 f(x) = f(x)^4 ]Bring all terms to one side:[ x^6 f(x) - f(x)^4 = 0 ][ f(x) (x^6 - f(x)^3) = 0 ]So, for each ( x ), either ( f(x) = 0 ) or ( f(x)^3 = x^6 ). Therefore, ( f(x) = 0 ) or ( f(x) = x^2 ).Hmm, so for each ( x ), ( f(x) ) is either 0 or ( x^2 ). But we need to check if this holds globally for all ( x ) and ( y ). Maybe the function is identically zero, or identically ( x^2 ), or a mix? Let's see.First, let's check if ( f(x) = x^2 ) works. Let's substitute ( f(x) = x^2 ) into the original equation:Left-hand side (LHS):[ x f(2 f(y) - x) + y^2 f(2x - f(y)) ][ = x f(2 y^2 - x) + y^2 f(2x - y^2) ][ = x (2 y^2 - x)^2 + y^2 (2x - y^2)^2 ]Right-hand side (RHS):[ frac{f(x)^2}{x} + f(y f(y)) ][ = frac{(x^2)^2}{x} + f(y cdot y^2) ][ = frac{x^4}{x} + f(y^3) ][ = x^3 + (y^3)^2 ][ = x^3 + y^6 ]So, let's compute LHS:First term: ( x (2 y^2 - x)^2 )[ = x (4 y^4 - 4 x y^2 + x^2) ][ = 4 x y^4 - 4 x^2 y^2 + x^3 ]Second term: ( y^2 (2x - y^2)^2 )[ = y^2 (4 x^2 - 4 x y^2 + y^4) ][ = 4 x^2 y^2 - 4 x y^4 + y^6 ]Adding both terms:[ 4 x y^4 - 4 x^2 y^2 + x^3 + 4 x^2 y^2 - 4 x y^4 + y^6 ]Simplify:- ( 4 x y^4 - 4 x y^4 = 0 )- ( -4 x^2 y^2 + 4 x^2 y^2 = 0 )- Remaining: ( x^3 + y^6 )So, LHS simplifies to ( x^3 + y^6 ), which matches RHS. Therefore, ( f(x) = x^2 ) is indeed a solution.Now, let's check if ( f(x) = 0 ) for all ( x ) is a solution. Substitute ( f(x) = 0 ):LHS:[ x f(2 f(y) - x) + y^2 f(2x - f(y)) ][ = x f(-x) + y^2 f(2x) ]Since ( f ) is zero everywhere:[ = x cdot 0 + y^2 cdot 0 = 0 ]RHS:[ frac{f(x)^2}{x} + f(y f(y)) ][ = frac{0^2}{x} + f(0) ][ = 0 + 0 = 0 ]So, LHS = RHS = 0. Therefore, ( f(x) = 0 ) is also a solution.But wait, could there be a mix where ( f(x) ) is sometimes 0 and sometimes ( x^2 )? Let's see.Suppose there exists some ( a neq 0 ) such that ( f(a) = 0 ). Let's substitute ( y = a ) into the original equation:[ x f(2 f(a) - x) + a^2 f(2x - f(a)) = frac{f(x)^2}{x} + f(a f(a)) ]Since ( f(a) = 0 ), this simplifies to:[ x f(-x) + a^2 f(2x) = frac{f(x)^2}{x} + f(0) ]But ( f(0) = 0 ), so:[ x f(-x) + a^2 f(2x) = frac{f(x)^2}{x} ]From earlier, we have ( x^2 f(-x) = f(x)^2 ), so ( f(-x) = frac{f(x)^2}{x^2} ). Therefore, substituting into the equation:[ x cdot frac{f(x)^2}{x^2} + a^2 f(2x) = frac{f(x)^2}{x} ]Simplify:[ frac{f(x)^2}{x} + a^2 f(2x) = frac{f(x)^2}{x} ]Subtract ( frac{f(x)^2}{x} ) from both sides:[ a^2 f(2x) = 0 ]Since ( a neq 0 ), ( a^2 neq 0 ), so ( f(2x) = 0 ) for all ( x ).Therefore, ( f ) is zero for all even integers. Now, what about odd integers?Suppose ( x ) is odd. Let's see if we can express ( x ) in terms of some other variable to relate it to even integers. For example, let ( x = 2k + 1 ) for some integer ( k ). Maybe we can find a relationship.Alternatively, let's consider substituting ( x = 1 ) into the original equation. Let's set ( x = 1 ):[ 1 cdot f(2 f(y) - 1) + y^2 f(2 cdot 1 - f(y)) = frac{f(1)^2}{1} + f(y f(y)) ]Simplify:[ f(2 f(y) - 1) + y^2 f(2 - f(y)) = f(1)^2 + f(y f(y)) ]If ( f(y) ) is zero for even ( y ), but for odd ( y ), ( f(y) ) could be ( y^2 ). Wait, but if ( f ) is zero for all even integers, let's see what happens when ( y ) is even or odd.Case 1: ( y ) is even. Then, ( f(y) = 0 ). So:[ f(2 cdot 0 - 1) + y^2 f(2 - 0) = f(1)^2 + f(y cdot 0) ]Simplify:[ f(-1) + y^2 f(2) = f(1)^2 + f(0) ]But ( f(2) = 0 ) and ( f(0) = 0 ), so:[ f(-1) = f(1)^2 ]But from earlier, ( f(-1) = frac{f(1)^2}{1^2} = f(1)^2 ). So, this holds.Case 2: ( y ) is odd. Let ( y = 2k + 1 ). Then, ( f(y) ) could be ( y^2 ) or 0. But if ( f(y) = y^2 ), then:[ f(2 y^2 - 1) + y^2 f(2 - y^2) = f(1)^2 + f(y^3) ]But ( 2 y^2 - 1 ) is odd, and ( 2 - y^2 ) is even (since ( y ) is odd, ( y^2 ) is odd, so ( 2 - y^2 ) is odd minus odd, which is even). Therefore, ( f(2 - y^2) = 0 ) because it's even. Similarly, ( f(y^3) ) depends on whether ( y^3 ) is even or odd. Since ( y ) is odd, ( y^3 ) is odd, so ( f(y^3) ) could be ( (y^3)^2 ) or 0.But let's see:Left-hand side (LHS):[ f(2 y^2 - 1) + y^2 cdot 0 = f(2 y^2 - 1) ]Right-hand side (RHS):[ f(1)^2 + f(y^3) ]So:[ f(2 y^2 - 1) = f(1)^2 + f(y^3) ]But ( 2 y^2 - 1 ) is odd, so ( f(2 y^2 - 1) ) could be ( (2 y^2 - 1)^2 ) or 0. Similarly, ( f(y^3) ) could be ( (y^3)^2 ) or 0.If ( f(2 y^2 - 1) = 0 ), then:[ 0 = f(1)^2 + f(y^3) ]But ( f(1)^2 ) is non-negative, and ( f(y^3) ) is either 0 or a square, so the only way this holds is if both ( f(1)^2 = 0 ) and ( f(y^3) = 0 ). Therefore, ( f(1) = 0 ) and ( f(y^3) = 0 ).But if ( f(1) = 0 ), then from ( f(-1) = f(1)^2 ), we have ( f(-1) = 0 ).Similarly, ( f(y^3) = 0 ) for all odd ( y ). Therefore, ( f ) is zero for all integers, because:- For even ( x ), ( f(x) = 0 ).- For odd ( x ), ( f(x) = 0 ) because ( x ) can be written as ( y^3 ) for some integer ( y ) (since cube is bijective over integers), so ( f(x) = 0 ).Therefore, the only possible solutions are ( f(x) = 0 ) for all ( x ) or ( f(x) = x^2 ) for all ( x ).Wait, but earlier we considered the possibility of ( f(x) ) being zero for some ( x ) and ( x^2 ) for others. But the above reasoning shows that if ( f ) is zero at any non-zero ( x ), then ( f ) must be zero everywhere. Therefore, the only possible solutions are the constant zero function or the function ( f(x) = x^2 ).Let me double-check if there are any other possibilities. Suppose ( f ) is not identically zero, then ( f(x) = x^2 ) for all ( x ). We already verified that this works. If ( f ) is identically zero, that also works. So, no other functions seem to satisfy the equation.Therefore, the solutions are ( f(x) = 0 ) and ( f(x) = x^2 ).

💡First, identify the number of loads washed each day based on the given information.On Wednesday, Vincent washed 6 loads.On Thursday, he washed double the number of loads from Wednesday:6 loads × 2 = 12 loads.On Friday, he washed half the number of loads from Thursday:12 loads ÷ 2 = 6 loads.On Saturday, he washed a third of the number of loads from Wednesday:6 loads ÷ 3 = 2 loads.Finally, sum up the loads washed each day to find the total for the week:6 + 12 + 6 + 2 = 26 loads.

💡Okay, so Kolya has seven ancient coins: four identical doubloons and three identical crowns. He doesn't remember the exact weight of each type of coin, but he knows that a doubloon weighs either 5 or 6 grams, and a crown weighs either 7 or 8 grams. The challenge is to figure out the exact weight of each type of coin using only two weighings on a balance scale without any additional weights. Alright, let's break this down. First, I need to understand what a balance scale can tell us. A balance scale can compare the weight of two groups of items and tell us which group is heavier, or if they're equal. Since we don't have standard weights, we have to use the coins themselves to compare against each other.Kolya has four doubloons and three crowns. Each doubloon is either 5 or 6 grams, and each crown is either 7 or 8 grams. So, the possible total weights for the doubloons are 4×5=20 grams or 4×6=24 grams. For the crowns, the possible total weights are 3×7=21 grams or 3×8=24 grams.Hmm, interesting. So, the total weight of the doubloons can be either 20 or 24 grams, and the total weight of the crowns can be either 21 or 24 grams. Let me think about how to use this information. Maybe I can weigh all the doubloons against all the crowns in the first weighing. So, first weighing: four doubloons vs. three crowns.If the doubloons and crowns balance, that means their total weights are equal. Looking at the possible totals, the only overlapping total weight is 24 grams. So, if they balance, both the doubloons and crowns weigh 24 grams. Therefore, each doubloon is 6 grams, and each crown is 8 grams. That would solve the problem in just one weighing.But what if they don't balance? If the doubloons are heavier, that means the total weight of the doubloons is 24 grams, and the crowns are lighter, so their total weight must be 21 grams. Therefore, each doubloon is 6 grams, and each crown is 7 grams. On the other hand, if the crowns are heavier, that means the total weight of the crowns is 24 grams, and the doubloons are lighter, so their total weight must be 20 grams. Therefore, each doubloon is 5 grams, and each crown is 8 grams.Wait a minute, so in the first weighing, depending on which side is heavier or if they balance, we can already determine the exact weights of the coins. But the problem says Kolya can only use two weighings. So, if the first weighing doesn't balance, do we need a second weighing to confirm?Let me think. If the first weighing shows that the doubloons are heavier, we know the doubloons are 6 grams and crowns are 7 grams. If the crowns are heavier, we know the doubloons are 5 grams and crowns are 8 grams. If they balance, both are 24 grams, so doubloons are 6 grams and crowns are 8 grams.So, actually, with just one weighing, we can determine the exact weights. But the problem says two weighings. Maybe I'm missing something here.Wait, perhaps I need to consider that Kolya doesn't know which side is which. So, he might not know whether the doubloons are heavier or lighter. But no, he can see which side is heavier. So, he can interpret the result accordingly.Alternatively, maybe I need to think differently. Perhaps the first weighing isn't all doubloons vs. all crowns. Maybe a different combination would be better.Let me consider another approach. Suppose in the first weighing, Kolya weighs a certain number of doubloons against a certain number of crowns. For example, he could weigh two doubloons against two crowns.So, first weighing: two doubloons vs. two crowns.Possible total weights for two doubloons: 2×5=10 grams or 2×6=12 grams.Possible total weights for two crowns: 2×7=14 grams or 2×8=16 grams.So, if he weighs two doubloons against two crowns, the possible outcomes are:1. Left side (doubloons) lighter: 10 vs. 14 or 10 vs. 16 or 12 vs. 14 or 12 vs. 16.2. Right side (crowns) lighter: Not possible because the minimum crowns can weigh is 14 grams, which is more than the maximum doubloons can weigh, which is 12 grams.Wait, that's not right. If two doubloons can be 10 or 12 grams, and two crowns can be 14 or 16 grams, then the crowns will always be heavier. So, this weighing won't give us any useful information because the crowns will always outweigh the doubloons.That's not helpful. So, maybe that's not a good first weighing.Alternatively, what if Kolya weighs three doubloons against two crowns?First weighing: three doubloons vs. two crowns.Possible total weights for three doubloons: 3×5=15 grams or 3×6=18 grams.Possible total weights for two crowns: 2×7=14 grams or 2×8=16 grams.So, possible outcomes:1. Left side (doubloons) lighter: 15 vs. 16.2. Left side (doubloons) heavier: 18 vs. 14 or 18 vs. 16.3. Balanced: Not possible because 15 vs. 14 or 15 vs. 16 or 18 vs. 14 or 18 vs. 16.Wait, so if the left side is lighter, that means three doubloons weigh 15 grams and two crowns weigh 16 grams. Therefore, each doubloon is 5 grams and each crown is 8 grams.If the left side is heavier, there are two possibilities: either three doubloons are 18 grams and two crowns are 14 grams, meaning each doubloon is 6 grams and each crown is 7 grams, or three doubloons are 18 grams and two crowns are 16 grams, meaning each doubloon is 6 grams and each crown is 8 grams.So, if the first weighing shows the left side is lighter, we know the exact weights: doubloons are 5 grams and crowns are 8 grams.If the left side is heavier, we have two possibilities: either crowns are 7 grams or 8 grams, with doubloons being 6 grams in both cases.So, in this case, we would need a second weighing to distinguish between these two possibilities.What could be the second weighing? Maybe weigh a different combination.Suppose in the second weighing, Kolya weighs one doubloon against one crown.So, second weighing: one doubloon vs. one crown.Possible outcomes:1. Left side (doubloon) lighter: 5 vs. 7 or 5 vs. 8.2. Right side (crown) lighter: 6 vs. 7 or 6 vs. 8.3. Balanced: Not possible because doubloons are either 5 or 6, and crowns are either 7 or 8.Wait, so if in the second weighing, the doubloon is lighter, that means the doubloon is 5 grams and the crown is either 7 or 8 grams. But from the first weighing, we already know that if the left side was heavier, the doubloons are 6 grams. So, this seems contradictory.Wait, no. If in the first weighing, the left side was heavier, meaning three doubloons vs. two crowns, and that could be either 18 vs. 14 or 18 vs. 16. So, if in the second weighing, one doubloon vs. one crown, if the doubloon is lighter, that would mean the crown is heavier, so the crown must be 8 grams, making the doubloons 6 grams. If the crown is lighter, that would mean the crown is 7 grams, making the doubloons 6 grams.Wait, but if the doubloon is lighter, that would mean the crown is heavier, so crown is 8 grams. If the crown is lighter, that would mean the crown is 7 grams. So, in either case, we can determine the exact weight of the crowns, and thus the doubloons.So, in this case, the second weighing would help us determine whether the crowns are 7 or 8 grams.Therefore, this approach would work: first weighing three doubloons vs. two crowns, and second weighing one doubloon vs. one crown.But let's see if this is necessary. Because if in the first weighing, the left side is lighter, we already know the exact weights. If the left side is heavier, we have two possibilities, and the second weighing can distinguish between them.Alternatively, maybe there's a more efficient way.Wait, let's go back to the first idea: weighing all four doubloons against all three crowns.If they balance, then both are 24 grams, so doubloons are 6 grams and crowns are 8 grams.If the doubloons are heavier, then doubloons are 24 grams (6 grams each) and crowns are 21 grams (7 grams each).If the crowns are heavier, then crowns are 24 grams (8 grams each) and doubloons are 20 grams (5 grams each).So, in this case, just one weighing would suffice because each outcome corresponds to a unique combination of weights.But the problem says two weighings. Maybe the problem assumes that Kolya doesn't know which side is which, but that doesn't make sense because he can see which side is heavier.Alternatively, maybe the problem is designed to require two weighings regardless, perhaps to confirm the result.But in reality, with the first weighing, Kolya can already determine the exact weights based on the result. So, maybe the answer is yes, he can determine the exact weights with two weighings, but actually, he can do it with just one.But since the problem specifies two weighings, perhaps the intended solution is to use two weighings, even though one might be sufficient.Alternatively, maybe the first weighing isn't all four doubloons vs. all three crowns, but a different combination.Wait, let's think about another approach. Suppose Kolya weighs one doubloon against one crown in the first weighing.First weighing: one doubloon vs. one crown.Possible outcomes:1. Left side (doubloon) lighter: 5 vs. 7 or 5 vs. 8.2. Right side (crown) lighter: 6 vs. 7 or 6 vs. 8.3. Balanced: Not possible because doubloons are either 5 or 6, and crowns are either 7 or 8.So, if the doubloon is lighter, it's either 5 vs. 7 or 5 vs. 8.If the crown is lighter, it's either 6 vs. 7 or 6 vs. 8.So, from the first weighing, we can narrow it down to two possibilities for the weights.Then, for the second weighing, Kolya can weigh a different combination to distinguish between the possibilities.For example, if the first weighing showed the doubloon is lighter, meaning it's either 5 vs. 7 or 5 vs. 8, then in the second weighing, he can weigh three doubloons against two crowns.So, second weighing: three doubloons vs. two crowns.If the doubloons are 5 grams and crowns are 7 grams, then three doubloons would be 15 grams and two crowns would be 14 grams, so the left side would be heavier.If the doubloons are 5 grams and crowns are 8 grams, then three doubloons would be 15 grams and two crowns would be 16 grams, so the right side would be heavier.Therefore, based on the second weighing, Kolya can determine whether the crowns are 7 or 8 grams.Similarly, if the first weighing showed the crown is lighter, meaning it's either 6 vs. 7 or 6 vs. 8, then in the second weighing, he can weigh three doubloons against two crowns.If the doubloons are 6 grams and crowns are 7 grams, then three doubloons would be 18 grams and two crowns would be 14 grams, so the left side would be heavier.If the doubloons are 6 grams and crowns are 8 grams, then three doubloons would be 18 grams and two crowns would be 16 grams, so the left side would still be heavier, but by a different amount.Wait, but in this case, both possibilities would result in the left side being heavier, just by different margins. So, Kolya wouldn't be able to distinguish between 7 and 8 grams for the crowns based on whether the left side is heavier by a certain amount because he doesn't have standard weights to measure the difference.Therefore, this approach might not work because he can't quantify the difference in weight, only compare which side is heavier.So, maybe a different second weighing is needed.Alternatively, if the first weighing showed the crown is lighter, meaning it's either 6 vs. 7 or 6 vs. 8, then in the second weighing, Kolya could weigh two doubloons against one crown.So, second weighing: two doubloons vs. one crown.If the doubloons are 6 grams and crowns are 7 grams, then two doubloons would be 12 grams and one crown would be 7 grams, so the left side would be heavier.If the doubloons are 6 grams and crowns are 8 grams, then two doubloons would be 12 grams and one crown would be 8 grams, so the left side would still be heavier, but by a different margin.Again, Kolya can't quantify the difference, so he can't distinguish between 7 and 8 grams.Hmm, this is tricky.Maybe another approach is needed. Let's go back to the first idea of weighing all four doubloons against all three crowns.If they balance, we're done: doubloons are 6 grams, crowns are 8 grams.If the doubloons are heavier, then doubloons are 6 grams, crowns are 7 grams.If the crowns are heavier, then doubloons are 5 grams, crowns are 8 grams.So, in this case, one weighing suffices. But since the problem allows two weighings, maybe Kolya can use the second weighing to confirm.But actually, with one weighing, he can already determine the exact weights, so perhaps the answer is yes, he can determine the exact weights with two weighings, but in reality, he only needs one.But the problem says "using two weighings," so maybe the intended answer is yes, and the method is as follows:First weighing: all four doubloons vs. all three crowns.If they balance, done.If not, then based on which side is heavier, we know the exact weights.But to be thorough, maybe a second weighing is used to confirm.Alternatively, perhaps the problem assumes that Kolya doesn't know which side is which, but that doesn't make sense because he can see the result.Alternatively, maybe the problem is designed to require two weighings, even though one is sufficient, perhaps to ensure accuracy.In any case, based on the first weighing, Kolya can determine the exact weights, so the answer is yes, he can determine the exact weights with two weighings.But to be precise, he can do it with one weighing, but since the problem allows two, it's still possible.So, the answer is yes, Kolya can determine the exact weights of the coins with two weighings.

💡First, I need to identify the correct operation the student was supposed to perform, which is finding (frac{4}{5}) of the number (N). This can be expressed as (frac{4}{5} times N).However, the student incorrectly divided the number by (frac{4}{5}), which is equivalent to multiplying by the reciprocal, (frac{5}{4}). This means the student calculated (frac{5}{4} times N).According to the problem, the student's incorrect result is 36 more than the correct answer. This leads to the equation:[frac{5}{4} times N = frac{4}{5} times N + 36]To solve for (N), I'll first subtract (frac{4}{5} times N) from both sides:[frac{5}{4} times N - frac{4}{5} times N = 36]Next, I'll find a common denominator to combine the fractions. The common denominator for 4 and 5 is 20:[frac{25}{20} times N - frac{16}{20} times N = 36]Simplifying the left side:[frac{9}{20} times N = 36]Finally, I'll solve for (N) by dividing both sides by (frac{9}{20}):[N = 36 div frac{9}{20} = 36 times frac{20}{9} = 80]