🧠:[<think>Okay, so I need to figure out the maximum value of T in a 999x999 grid where some cells are white and others are red. T is the number of cell groups (C1, C2, C3) where C1 and C2 are in the same row, C2 and C3 are in the same column, C1 and C3 are white, and C2 is red. First, let me parse the problem again to make sure I understand it correctly. We have a grid with 999 rows and 999 columns. Each cell is either white or red. We need to count all possible triplets (C1, C2, C3) such that:1. C1 and C2 are in the same row.2. C2 and C3 are in the same column.3. C1 and C3 are white.4. C2 is red.And our goal is to find the maximum possible value of T given these conditions. So, we need to arrange the white and red cells in such a way that the number of such triplets is maximized.Let me try to visualize this. For each such triplet, C2 is a red cell that connects a white cell C1 in its row to a white cell C3 in its column. So, the triplet forms a right angle, with C2 at the corner, C1 to the left or right in the same row, and C3 above or below in the same column. Therefore, each red cell C2 can potentially be part of multiple triplets. For each red cell at position (i,j), the number of triplets it contributes is equal to the number of white cells in its row (row i) multiplied by the number of white cells in its column (column j). Because for each white cell in the same row as C2 (which is C1), and each white cell in the same column as C2 (which is C3), we have a valid triplet (C1, C2, C3). So, the total T is the sum over all red cells C2 of (number of white cells in C2's row) multiplied by (number of white cells in C2's column).Therefore, T = Σ_{red cells (i,j)} (W_i * W_j'), where W_i is the number of white cells in row i, and W_j' is the number of white cells in column j. Wait, actually, no. Let me clarify. For a red cell at (i,j), the number of white cells in its row i is W_i, and the number of white cells in its column j is W_j. But actually, the cell (i,j) itself is red, so when we count W_i, we are already excluding the cell (i,j) if it's red. Wait, no. If W_i is the number of white cells in row i, then regardless of the color of cell (i,j), W_i is the total white cells in that row. Similarly, W_j is the number of white cells in column j. But the cell (i,j) is red, so in row i, the number of white cells is W_i, and in column j, the number of white cells is W_j.Therefore, for each red cell (i,j), the number of triplets involving it is W_i * W_j. Therefore, the total T is the sum over all red cells (i,j) of W_i * W_j. So, to maximize T, we need to maximize this sum.Now, let's note that W_i is the number of white cells in row i, so R_i = 999 - W_i is the number of red cells in row i. Similarly, for column j, W_j is the number of white cells, and R_j = 999 - W_j is the number of red cells in column j.But the problem is that W_i and W_j are related. Changing the color of a cell affects both its row and column counts. So, we need to set each cell to white or red in such a way that the sum over all red cells of W_i * W_j is maximized.This seems like a problem that might have a known solution or might be related to some combinatorial optimization. Let me think.Alternatively, perhaps we can model this as a matrix. Let me consider the grid as a matrix where each cell is 1 (white) or 0 (red). Then, for each 0 (red cell) at position (i,j), we need to compute the number of 1s in row i and the number of 1s in column j, then multiply them and sum over all 0s.So, T = Σ_{(i,j): cell (i,j)=0} (row_i_ones * column_j_ones)We need to maximize T, given that we can set each cell to 0 or 1, with the constraint that row_i_ones is the number of 1s in row i, column_j_ones is the number of 1s in column j.Alternatively, T can be expressed as:T = Σ_{i=1 to 999} Σ_{j=1 to 999} (1 - cell(i,j)) * (row_i_ones * column_j_ones)Since (1 - cell(i,j)) is 1 if the cell is red (0) and 0 if it's white (1). So, expanding this, we have:T = Σ_{i,j} row_i_ones * column_j_ones - Σ_{i,j} cell(i,j) * row_i_ones * column_j_onesBut Σ_{i,j} row_i_ones * column_j_ones = (Σ_i row_i_ones) * (Σ_j column_j_ones) = (total_white_cells)^2, since Σ_i row_i_ones is the total number of white cells, and similarly for columns. Wait, but rows and columns both count the same total number of white cells. So, total_white_cells = Σ_i row_i_ones = Σ_j column_j_ones. Therefore, Σ_{i,j} row_i_ones * column_j_ones = (total_white_cells)^2.Therefore, T = (total_white_cells)^2 - Σ_{i,j} cell(i,j) * row_i_ones * column_j_onesBut cell(i,j) is 1 if the cell is white and 0 otherwise. Therefore, Σ_{i,j} cell(i,j) * row_i_ones * column_j_ones is the sum over all white cells (i,j) of row_i_ones * column_j_ones.Therefore, T = (total_white_cells)^2 - Σ_{white cells (i,j)} row_i_ones * column_j_onesSo, our problem reduces to maximizing T, which is equivalent to minimizing the sum over white cells of row_i_ones * column_j_ones, given that total_white_cells is fixed? Wait, but total_white_cells isn't fixed. The problem allows us to choose which cells are white or red. So, actually, we can choose how many white cells there are, but with the structure that maximizes T. However, T is given by that expression above. So, perhaps the maximum T is achieved when we arrange the white cells in such a way that the sum over white cells of (row_i_ones * column_j_ones) is minimized, while (total_white_cells)^2 is maximized. But this is a trade-off, because increasing total_white_cells might increase the sum over white cells. So, we need to find a balance.Alternatively, perhaps we can model this problem using linear algebra or matrix operations. Let me consider variables for each row and column. Let me denote x_i as the number of white cells in row i, and y_j as the number of white cells in column j. Then, we have the following:Each cell (i,j) is white or red. If it's white, then it contributes to x_i and y_j. If it's red, then it contributes to the term x_i * y_j in T. Therefore, T is the sum over all red cells (i,j) of x_i * y_j. However, x_i and y_j are dependent on the white cells. The problem is to assign each cell to be white or red such that T is maximized.But since the grid is square (999x999), maybe there's a symmetric solution. Let me think about this.Suppose we arrange the grid such that each row has the same number of white cells, say w, and each column also has the same number of white cells, w. Then, the total number of white cells would be 999 * w. However, for such a configuration to be possible, w must be an integer, and the necessary condition is that 999 * w must be divisible by 999 (since each column also has w white cells). Which it is, because 999 * w / 999 = w. So, such a configuration is possible if we can arrange each row and column to have exactly w white cells. This is similar to a regular graph or a Latin square concept, but here it's a binary matrix with row and column sums equal to w. Such matrices exist due to the Gale-Ryser theorem, I think, as long as the row and column sums are compatible.But maybe assuming such a regular structure might help in maximizing or simplifying T. Let's try that.If every row has w white cells and every column has w white cells, then for each red cell (i,j), which is in a row with w white cells and a column with w white cells, so each red cell contributes w * w to T. The total number of red cells is 999*999 - 999*w = 999*(999 - w). Therefore, T = 999*(999 - w) * w^2.Then, T = 999 * w^2 * (999 - w)To maximize T with respect to w, we can take the derivative. Treating w as a continuous variable, take derivative of T with respect to w:dT/dw = 999 [2w(999 - w) + w^2(-1)] = 999 [2w*999 - 2w^2 - w^2] = 999 [2*999 w - 3w^2]Setting derivative to zero:2*999 w - 3w^2 = 0 => w(2*999 - 3w) = 0 => w=0 or 2*999/3 = 666.Therefore, maximum at w=666. So, if we set each row and column to have 666 white cells, then T = 999 * (666)^2 * (999 - 666) = 999 * 666^2 * 333.But is this achievable? Since 666*999 must be equal to the total number of white cells, which would be 999*666, and each column also needs to have 666 white cells, so the total is also 999*666. Therefore, it's consistent. So, such a matrix exists. Therefore, if we can arrange the white cells such that each row and each column has exactly 666 white cells, then T would be 999 * 666^2 * 333.But wait, let's compute that value. Let's see:First, 666 is 2/3 of 999 (since 999 = 3*333, 666 = 2*333). So, 666 = (2/3)*999. Therefore, the total number of white cells is 999*666 = 999*(2/3)*999 = (2/3)*999^2.But in this case, each red cell (i,j) contributes w * w = 666*666. The number of red cells is 999*999 - 999*666 = 999*(999 - 666) = 999*333. Therefore, T = 999*333 * 666^2 = 999*333*(666)^2.But is this the maximum?Wait, but this assumes that all rows and columns have exactly 666 white cells. However, perhaps arranging the grid in a different way, not regular, might yield a higher T. For example, maybe some rows have more white cells and some have fewer, in a way that the product terms for red cells are larger. But since T is the sum over red cells of (white in row) * (white in column), perhaps concentrating white cells in certain rows and columns could create regions where red cells have higher products.Alternatively, if some rows have more white cells and their corresponding columns also have more white cells, then the red cells at the intersection of those rows and columns would contribute more. However, if you have some rows with many white cells and some with few, and columns similarly, you might end up with red cells that are in high-white rows and high-white columns, contributing a lot, but also red cells in low-white rows and low-white columns, contributing little. The balance might not necessarily be better than the uniform case.Alternatively, perhaps arranging the grid into blocks. Suppose we divide the grid into four blocks: a block of a x a white cells, and the rest being red. But I need to think carefully.Alternatively, let's use Lagrange multipliers to consider the general case. Let me consider variables x_i (number of white cells in row i) and y_j (number of white cells in column j). But the problem is that these variables are linked by the cells: the total number of white cells is the sum over x_i, which is also the sum over y_j. Also, for each cell (i,j), if it is white, then it contributes to both x_i and y_j. If it's red, then it contributes to T as x_i * y_j. So, the total T is sum_{(i,j) red} x_i y_j.But this is a complex optimization problem because the variables x_i and y_j are linked by the grid's cell assignments.But perhaps if we relax the problem to allow fractional white cells, then we can find an optimal solution, and then check if it corresponds to an integer solution.Alternatively, think of this as a bipartite graph between rows and columns, with edges representing white cells. Then, T is the sum over all non-edges (red cells) of the product of the degrees of the two nodes (rows and columns). But this might not directly help.Wait, but perhaps there is a known formula or inequality that can relate this sum. Let me think.The total T can be written as sum_{i,j} (1 - cell(i,j)) * x_i y_jWhich is equal to sum_{i,j} x_i y_j - sum_{i,j} cell(i,j) x_i y_jWe already saw that sum_{i,j} x_i y_j = (sum x_i)(sum y_j) = W^2, where W is the total number of white cells. So, T = W^2 - sum_{white cells} x_i y_jTherefore, to maximize T, we need to minimize sum_{white cells} x_i y_j, given that sum x_i = sum y_j = W.So, the problem reduces to: given that we have a 999x999 grid, assign white cells such that the sum over all white cells of (x_i y_j) is minimized, where x_i is the number of white cells in row i and y_j is the number in column j. Then, T = W^2 - sum_{white cells} x_i y_j.So, maximizing T is equivalent to minimizing sum_{white cells} x_i y_j. Therefore, we need to arrange the white cells in such a way that the sum of x_i y_j over white cells is as small as possible.Hmm. Let's see. Let's denote S = sum_{white cells} x_i y_j. We need to minimize S.Suppose all x_i and y_j are equal. That is, each row has x white cells and each column has x white cells. Then, S would be sum_{white cells} x^2. Since there are W = 999x white cells, then S = W x^2 = 999x * x^2 = 999x^3. Wait, no. Wait, if each white cell is in a row with x white cells and a column with x white cells, then each white cell contributes x * x = x^2. So, sum over all white cells is W * x^2. But W = 999x. So, S = 999x * x^2 = 999x^3.Alternatively, in the regular case where each row and column has x white cells, then W = 999x, and S = 999x * x^2 = 999x^3. Then, T = W^2 - S = (999x)^2 - 999x^3 = 999^2 x^2 - 999x^3 = 999x^2 (999 - x). So, as before.If we take derivative with respect to x, dT/dx = 999[2x(999 - x) + x^2(-1)] = 999[1998x - 2x^2 - x^2] = 999[1998x - 3x^2]. Setting to zero, x=0 or x=1998/3=666. So, maximum T at x=666, as before. So, T=999*(666)^2*(333). But this is under the assumption that all rows and columns have exactly x white cells.But perhaps, if we allow x_i and y_j to vary, maybe we can get a lower S, hence a higher T. For example, if some rows have more white cells and others have fewer, such that when multiplied by their column counts, the total sum S is smaller. How?Suppose we have two types of rows: some rows with high x_i and some with low x_i. Similarly, columns can be split into high y_j and low y_j. If we arrange the white cells such that high x_i rows are paired with low y_j columns and vice versa, then the products x_i y_j would be smaller, leading to a lower S. Hence, a higher T. This is similar to the rearrangement inequality, where sorting two sequences in opposite orders minimizes their dot product. Therefore, to minimize S = sum x_i y_j over white cells, we should pair the largest x_i with the smallest y_j, and the smallest x_i with the largest y_j.Therefore, if we partition the rows into two groups: a group of k rows with x_i = a and the remaining (999 - k) rows with x_i = b. Similarly, partition the columns into two groups: k columns with y_j = b and (999 - k) columns with y_j = a. Then, place all white cells in the intersections of the high x_i rows with low y_j columns and low x_i rows with high y_j columns. This way, each white cell has x_i y_j = a b, which is the same for all white cells.If we do this, then the total number of white cells is k*(999 - k)*a + (999 - k)*k*b = k*(999 - k)*(a + b). Wait, no. Wait, if we have k rows with x_i = a and (999 - k) rows with x_i = b. Similarly, k columns with y_j = b and (999 - k) columns with y_j = a.Then, the white cells are placed in the intersections of the k rows (x_i = a) with the (999 - k) columns (y_j = a). Wait, no. Wait, the idea is to pair high x_i with low y_j. So, if rows with high x_i are in the first k rows, and columns with low y_j are in the first k columns, then place white cells in the off-diagonal blocks.Wait, perhaps it's getting too vague. Let me formalize it.Suppose we divide the rows into two parts: the first m rows with x_i = a, and the remaining (999 - m) rows with x_i = b. Similarly, divide the columns into two parts: the first n columns with y_j = b, and the remaining (999 - n) columns with y_j = a. Then, place white cells only in the intersections of the first m rows with the last (999 - n) columns and the last (999 - m) rows with the first n columns. Thus, each white cell in the first block has x_i = a and y_j = a, contributing a^2 to S. Each white cell in the second block has x_i = b and y_j = b, contributing b^2 to S. Wait, but if we arrange it so that high x_i are paired with low y_j, then the products x_i y_j would be a*b, which is lower than a^2 or b^2 if a and b are arranged appropriately.Wait, maybe I need to adjust the parameters. Let me think again.To minimize S = sum x_i y_j over white cells, we want the white cells to be in positions where x_i is small and y_j is small. But since x_i and y_j are the numbers of white cells in their respective rows and columns, which are determined by the white cells, there's a circular dependency.Alternatively, if we set some rows to have many white cells (high x_i) and others to have few (low x_i), and do the same for columns, but arrange the white cells so that high x_i rows are paired with low y_j columns, then the products x_i y_j would be lower. But how does this affect the total S?Suppose we set the first m rows to have x_i = H and the remaining (999 - m) rows to have x_i = L. Similarly, set the first n columns to have y_j = L and the remaining (999 - n) columns to have y_j = H. Then, place white cells only in the intersections of the first m rows with the first n columns (H * L) and the remaining (999 - m) rows with the remaining (999 - n) columns (L * H). But wait, if white cells are in those intersections, then the counts x_i and y_j would need to satisfy the totals.Wait, this might be too vague. Let's try a specific case. Let me divide the grid into two equal parts. Suppose m = n = 500. But 999 is odd, so maybe 500 and 499. But for simplicity, let me consider splitting into two equal parts, even if it's approximate.Suppose we split the grid into two blocks: the top-left k x k block and the rest. But perhaps this is not the way. Alternatively, suppose we make a bipartition of rows and columns such that the white cells are only in certain blocks.Wait, perhaps the optimal configuration is to have a certain number of rows fully white and the rest fully red, and similarly for columns. But that would not work because if a row is fully white, then all its cells are white, so there are no red cells in that row to contribute to T.Alternatively, if we set a proportion of the grid to be white, arranged in such a way that the product terms are minimized. But maybe this is too vague.Alternatively, going back to the regular case where all rows and columns have the same number of white cells, which gives a certain value of T. If we can show that this regular configuration minimizes S, then that would be the optimal.But how can we compare the regular case with an irregular case? Suppose in the irregular case, some rows have more white cells and some have fewer. Then, the sum S would be higher because the terms x_i y_j for white cells in high x_i rows and high y_j columns would be larger. Hence, the regular case might actually minimize S.Alternatively, using the Cauchy-Schwarz inequality. Let me consider that S = sum_{(i,j) white} x_i y_j.But if we consider the vectors u and v where u_ij = x_i and v_ij = y_j for white cells (i,j), then by Cauchy-Schwarz, (sum u_ij v_ij)^2 ≤ (sum u_ij^2)(sum v_ij^2). But I don't know if that helps.Alternatively, we can think of S as the sum over white cells of x_i y_j. Let's fix the total number of white cells W. Then, to minimize S, we need to distribute the white cells such that the products x_i y_j are as small as possible. If we spread out the white cells evenly across rows and columns, then x_i and y_j are all roughly equal, so each term is x_i y_j ≈ (W/999)^2. So, S ≈ W * (W/999)^2 = W^3 / 999^2.Alternatively, if we cluster the white cells in certain rows and columns, making some x_i and y_j larger and others smaller, then the products x_i y_j for the clustered cells would be higher, but for the sparse cells, lower. However, the overall sum might be higher or lower depending on the distribution.But according to the rearrangement inequality, the sum S is minimized when the white cells are arranged such that larger x_i are paired with smaller y_j and vice versa. So, to minimize S, we need to anti-correlate the x_i and y_j.Therefore, the minimal S occurs when the rows with the most white cells are paired with the columns with the least white cells, and rows with the least white cells are paired with columns with the most white cells. Therefore, suppose we divide the rows into two groups: the first k rows with x_i = a and the remaining (999 - k) rows with x_i = b. Similarly, divide the columns into two groups: the first k columns with y_j = b and the remaining (999 - k) columns with y_j = a. Then, place all white cells in the intersections of the first k rows with the last (999 - k) columns and the last (999 - k) rows with the first k columns. In this configuration, each white cell in the first block (rows 1 to k, columns k+1 to 999) has x_i = a and y_j = a, contributing a*a. Each white cell in the second block (rows k+1 to 999, columns 1 to k) has x_i = b and y_j = b, contributing b*b. Wait, but if we pair high x_i with low y_j, then the products would be a*b. Wait, maybe I need to adjust the column groups.Wait, if we have the first k rows with x_i = a (high), and the first k columns with y_j = b (low), and the remaining columns with y_j = c (high). Then, place white cells in the first k rows and last (999 - k) columns (x_i = a, y_j = c) and in the last (999 - k) rows and first k columns (x_i = d, y_j = b). Then, the products are a*c and d*b. To minimize S, we want a*c and d*b to be as small as possible. However, we need to maintain consistency in the total number of white cells.Alternatively, this might get too complicated. Let's try specific numbers. Let's take k = 333. Then, the first 333 rows have x_i = 666, and the remaining 666 rows have x_i = 333. Similarly, the first 333 columns have y_j = 333, and the remaining 666 columns have y_j = 666. Then, place all white cells in the intersections of the first 333 rows with the last 666 columns (which have y_j = 666) and the last 666 rows with the first 333 columns (which have y_j = 333). In this case, the number of white cells in the first block (333 rows x 666 columns) is 333*666, each contributing x_i y_j = 666*666. The number of white cells in the second block (666 rows x 333 columns) is 666*333, each contributing x_i y_j = 333*333. Therefore, the total S = 333*666*666^2 + 666*333*333^2 = 333*666*(666^2 + 333^2). But wait, no. Wait, for each white cell in the first block, x_i = 666 (since it's in the first 333 rows) and y_j = 666 (since it's in the last 666 columns). So, each such white cell contributes 666*666. The number of white cells in the first block is 333 rows * 666 columns = 333*666. Similarly, in the second block, each white cell is in rows with x_i = 333 and columns with y_j = 333, so each contributes 333*333, and there are 666*333 such cells. Therefore, S = 333*666*(666^2) + 666*333*(333^2) = 333*666*(666^2 + 333^2).But let's compute this. Let's compute S for this configuration and compare it with the regular case where all rows and columns have 666 white cells. In the regular case, S = 999*666 * (666)^2 = 999*666^3.In the partitioned case:S = 333*666*(666^2 + 333^2) = 333*666*(666^2 + 333^2)Let's compute this:First, 333*666 = 333*(666) = 333*(600 + 66) = 333*600 + 333*66 = 199,800 + 21,978 = 221,778.Then, 666^2 = 443,556; 333^2 = 110,889. So, 443,556 + 110,889 = 554,445.Therefore, S = 221,778 * 554,445. Let me compute that:221,778 * 554,445 ≈ 221,778 * 550,000 = 221,778 * 500,000 + 221,778 * 50,000 = 110,889,000,000 + 11,088,900,000 = 121,977,900,000. Then, adjust for the remaining 4,445: 221,778 * 4,445 ≈ 221,778 * 4,000 = 887,112,000; 221,778 * 445 = 221,778 * 400 + 221,778 * 45 = 88,711,200 + 9,980,010 = 98,691,210. Total ≈ 887,112,000 + 98,691,210 = 985,803,210. So total S ≈ 121,977,900,000 + 985,803,210 ≈ 122,963,703,210.In the regular case, S = 999*666^3. Let's compute that:666^3 = 666*666*666 = 443,556*666. Let's compute 443,556*600 = 266,133,600; 443,556*66 = 29,275, 296 (wait, 443,556 * 60 = 26,613,360; 443,556 *6 = 2,661,336. Total 26,613,360 + 2,661,336 = 29,274,696). Therefore, 443,556*666 = 266,133,600 + 29,274,696 = 295,408,296. Then, 999*295,408,296. Since 1000*295,408,296 = 295,408,296,000, subtract 295,408,296 to get 295,408,296,000 - 295,408,296 = 295,112,887,704. So, S_regular = 295,112,887,704.Compare with S_partitioned ≈ 122,963,703,210. So, S_partitioned is much smaller than S_regular. Therefore, T = W^2 - S. So, in the partitioned case, W is the total number of white cells. In the partitioned case, W = 333*666 + 666*333 = 2*333*666 = 2*221,778 = 443,556. Therefore, W^2 = (443,556)^2. Let me compute that:443,556^2. Let's approximate:(440,000 + 3,556)^2 = 440,000^2 + 2*440,000*3,556 + 3,556^2440,000^2 = 193,600,000,0002*440,000*3,556 = 880,000*3,556 ≈ 880,000*3,500 = 3,080,000,000; 880,000*56 = 49,280,000. Total ≈ 3,080,000,000 + 49,280,000 = 3,129,280,0003,556^2 ≈ (3,500 + 56)^2 = 12,250,000 + 2*3,500*56 + 56^2 = 12,250,000 + 392,000 + 3,136 = 12,645,136Therefore, total W^2 ≈ 193,600,000,000 + 3,129,280,000 + 12,645,136 ≈ 196,741,925,136Then, T_partitioned = W^2 - S ≈ 196,741,925,136 - 122,963,703,210 ≈ 73,778,221,926.In the regular case, W = 999*666 = 665,334. Therefore, W^2 = (665,334)^2. Let's approximate:(600,000 + 65,334)^2 ≈ 600,000^2 + 2*600,000*65,334 + 65,334^2600,000^2 = 360,000,000,0002*600,000*65,334 = 1,200,000*65,334 ≈ 1,200,000*60,000 = 72,000,000,000; 1,200,000*5,334 = 6,400,800,000. Total ≈ 72,000,000,000 + 6,400,800,000 = 78,400,800,00065,334^2 ≈ (60,000 + 5,334)^2 ≈ 60,000^2 + 2*60,000*5,334 + 5,334^2 = 3,600,000,000 + 640,080,000 + 28,451,556 ≈ 4,268,531,556Total W^2 ≈ 360,000,000,000 + 78,400,800,000 + 4,268,531,556 ≈ 442,669,331,556Then, T_regular = W^2 - S_regular ≈ 442,669,331,556 - 295,112,887,704 ≈ 147,556,443,852.Comparing the two T values:T_partitioned ≈ 73.78 billionT_regular ≈ 147.56 billionSo, the regular case gives a higher T. Therefore, partitioning the grid into blocks with different densities actually results in a lower T compared to the regular case where all rows and columns have the same number of white cells.This suggests that the regular configuration is better. Therefore, perhaps the maximum T is achieved when all rows and columns have the same number of white cells, which we calculated earlier as T = 999 * (666)^2 * 333.But wait, in the partitioned case, even though S is smaller, W is also smaller. The regular case has a much larger W, so even though S is larger, the term W^2 - S is still larger. So, the regular case benefits from a higher W, even with a higher S. Therefore, increasing W may be more beneficial than decreasing S.But how do we know if there's a sweet spot where increasing W while keeping S manageable gives a higher T? Well, the earlier calculus suggested that when all rows and columns have the same number of white cells, the maximum T is achieved at w=666.But let's verify that with actual numbers. In the regular case with w=666, T = 999 * 666^2 * 333.Compute that:First, 666^2 = 443,556Then, 443,556 * 333 ≈ 443,556 * 300 = 133,066,800; 443,556 * 33 = 14,637, 348 (≈443,556*30=13,306,680 and 443,556*3=1,330,668; total 14,637,348). So total ≈ 133,066,800 + 14,637,348 = 147,704,148.Then, multiply by 999: 147,704,148 * 1,000 = 147,704,148,000; subtract 147,704,148 to get 147,704,148,000 - 147,704,148 = 147,556,443,852. Which matches the earlier approximation of T_regular ≈147.556 billion.In the partitioned case, T ≈73.778 billion, which is about half of the regular case. Therefore, the regular configuration gives a much higher T.Therefore, this suggests that arranging the grid regularly with each row and column having 666 white cells maximizes T.But the question is, can we actually do this? Is it possible to have a 999x999 grid where each row and each column has exactly 666 white cells?Yes, due to the Gale-Ryser theorem. For a binary matrix with row sums r_1, r_2, ..., r_m and column sums c_1, c_2, ..., c_n, such a matrix exists if and only if the sum of row sums equals the sum of column sums, and for every k, the sum of the k largest row sums is at most the sum of the column sums minus the sum of the (n - k) smallest column sums. In our case, all row sums are 666 and all column sums are 666. The total sum is 999*666 on both sides. Also, for any k, the sum of the k largest row sums is 666k, and the sum of the column sums minus the sum of the (999 - k) smallest column sums is 666*999 - 666*(999 -k) = 666k. Therefore, the condition is satisfied. Therefore, such a matrix exists.Therefore, the maximum T is achieved when every row and every column has exactly 666 white cells, leading to T = 999 * 666^2 * 333.But let's compute this value precisely.First, 666 is 2/3 of 999, since 999 = 3*333, and 666 = 2*333. Therefore, 666 = (2/3)*999.Compute 666^2 = (2/3 * 999)^2 = (4/9) * 999^2.Then, T = 999 * (4/9)*999^2 * 333 = 999 * 4/9 * 999^2 * 333.Simplify:999 * 4/9 = (999/9)*4 = 111*4 = 444.Then, T = 444 * 999^2 * 333.But 999 = 1000 - 1, so 999^2 = (1000 - 1)^2 = 1,000,000 - 2,000 + 1 = 998,001.Therefore, T = 444 * 998,001 * 333.Compute 444 * 333 first:444 * 300 = 133,200444 * 33 = 14,652Total = 133,200 + 14,652 = 147,852.Therefore, T = 147,852 * 998,001.Multiply 147,852 by 1,000,000: 147,852,000,000Subtract 147,852 * 1,999: Wait, no. Wait, 998,001 = 1,000,000 - 1,999. Wait, no: 998,001 = 1,000,000 - 1999? Wait, 1,000,000 - 1999 = 998,001. Yes.Therefore, 147,852 * 998,001 = 147,852*(1,000,000 - 1999) = 147,852,000,000 - 147,852*1999.Compute 147,852 * 2000 = 295,704,000Subtract 147,852 to get 295,704,000 - 147,852 = 295,556,148.Therefore, T = 147,852,000,000 - 295,556,148 = 147,852,000,000 - 295,556,148 = 147,556,443,852.So, T = 147,556,443,852.But let's confirm this calculation:First, 147,852 * 1,000,000 = 147,852,000,000147,852 * 1999 = 147,852 * (2000 - 1) = 147,852*2000 - 147,852 = 295,704,000 - 147,852 = 295,556,148Subtracting: 147,852,000,000 - 295,556,148 = 147,556,443,852.Yes, correct.Therefore, the maximum T is 147,556,443,852. But let's write this in terms of factorials or combinations if possible, to see if there's a simpler expression. However, given the problem's size, it's probably acceptable to leave it in the multiplied form. But since the answer requires a boxed number, perhaps we can write it as 999 * 666^2 * 333. Alternatively, we can factor 999 = 3*333, so 666 = 2*333. Therefore:T = 999 * (2*333)^2 * 333 = 999 * 4*333^2 * 333 = 999 * 4 * 333^3.But 999 = 3*333, so:T = 3*333 * 4 * 333^3 = 12 * 333^4.Compute 333^4:333^2 = 110,889.Then, 110,889^2 = (100,000 + 10,889)^2 = 10,000,000,000 + 2*100,000*10,889 + 10,889^2.Compute each term:10,000,000,0002*100,000*10,889 = 200,000*10,889 = 2,177,800,00010,889^2:Compute 10,000^2 = 100,000,0002*10,000*889 = 17,780,000889^2 = 790,321So, 10,889^2 = 100,000,000 + 17,780,000 + 790,321 = 118,570,321.Therefore, 110,889^2 = 10,000,000,000 + 2,177,800,000 + 118,570,321 = 12,296,370,321.Therefore, T = 12 * 12,296,370,321 = 147,556,443,852. Which matches the previous result.Alternatively, we can note that 333^4 = (3*111)^4 = 81*111^4, but this might not help.In any case, the maximum T is 147,556,443,852. However, the problem might expect the answer in terms of 999 * 666^2 * 333, or factored further. Let's verify the components:999 = 3*333666 = 2*333Therefore, 999 * 666^2 * 333 = 3*333 * (2*333)^2 * 333 = 3*333 *4*333^2 *333 = 3*4*333^4 = 12*333^4.But 333 is 3*111, so 333^4 = (3^4)*(111^4) = 81*111^4. Therefore, T = 12*81*111^4 = 972*111^4. However, this might not be necessary. Alternatively, since the problem is about a 999x999 grid, the answer is likely to be presented in terms of 999 and 333, given that 666 = 999 - 333.Alternatively, the answer can be written as 333^3 * 666^2 * 3, since 999 = 3*333, so:T = 999 * 666^2 * 333 = 3*333 * 666^2 * 333 = 3*333^2 * 666^2.Since 666 = 2*333, substitute:T = 3*333^2*(2*333)^2 = 3*333^2*4*333^2 = 12*333^4.Either way, the numerical value is 147,556,443,852. However, the problem might expect the answer in a simplified exponential form with the given numbers. Let's check if 333^4 *12 is equal to 147,556,443,852.Compute 333^4:333^2 = 110,889110,889^2 = 12,296,370,32112,296,370,321 *12 = 147,556,443,852. Yes, correct.Therefore, the maximum T is 12*333^4, which equals 147,556,443,852. Alternatively, since 333 is 999/3, we can write it as 12*(999/3)^4 = 12*(999^4)/(3^4) = 12*999^4/81 = (12/81)*999^4 = (4/27)*999^4. But this seems more complicated.Alternatively, using the original expression:T = 999 * 666^2 * 333. Let's compute each term:999 * 333 = 999 * 300 + 999 * 33 = 299,700 + 32,967 = 332,667666^2 = 443,556Therefore, T = 332,667 * 443,556Compute 332,667 * 400,000 = 133,066,800,000332,667 * 43,556 = ?First, compute 332,667 * 40,000 = 13,306,680,000332,667 * 3,556 = ?332,667 * 3,000 = 998,001,000332,667 * 500 = 166,333,500332,667 * 56 = 18,629, 352 (332,667 * 50 = 16,633,350; 332,667 *6 = 1,996,002; total 16,633,350 + 1,996,002 = 18,629,352)So, total 998,001,000 + 166,333,500 + 18,629,352 = 1,182,963,852.Therefore, 332,667 * 43,556 = 13,306,680,000 + 1,182,963,852 = 14,489,643,852.Then, total T = 133,066,800,000 + 14,489,643,852 = 147,556,443,852. Which matches.Therefore, the maximum value of T is 147,556,443,852.But let me check if there's a smarter way to compute this without going through all these multiplications. Since 666 is 2/3 of 999, and 333 is 1/3 of 999, then:T = 999 * (2/3 * 999)^2 * (1/3 * 999) = 999 * (4/9 * 999^2) * (1/3 * 999) = 999 * 4/9 * 1/3 * 999^3 = (4/27) * 999^4.But 999^4 is a huge number. Let's compute it:999^2 = 998,001999^4 = (998,001)^2 = 996,006,000,001 (Wait, no. Wait, (a - b)^2 = a^2 - 2ab + b^2. Let me compute 998,001^2.Compute (1,000,000 - 1,999)^2 = 1,000,000^2 - 2*1,000,000*1,999 + 1,999^21,000,000^2 = 1,000,000,000,0002*1,000,000*1,999 = 3,998,000,0001,999^2 = (2000 - 1)^2 = 4,000,000 - 4,000 + 1 = 3,996,001Therefore, 998,001^2 = 1,000,000,000,000 - 3,998,000,000 + 3,996,001 = (1,000,000,000,000 - 3,998,000,000) + 3,996,001 = 996,002,000,000 + 3,996,001 = 996,005,996,001.Therefore, T = (4/27)*996,005,996,001.But 996,005,996,001 divided by 27:First, 996,005,996,001 ÷ 9 = 110,667,332, 889. (Since 996,005,996,001 / 9: 9*110,667,332,889 = 996,005,996,001)Then, divide by 3: 110,667,332,889 ÷ 3 = 36,889,110,963.Therefore, 996,005,996,001 /27 = 36,889,110,963.Therefore, T = 4 * 36,889,110,963 = 147,556,443,852. Which matches the previous result.Therefore, regardless of the method, the result is consistent. Hence, the maximum value of T is 147,556,443,852.Final Answerboxed{147556443852}</think>To find the maximum value of ( T ) in a ( 999 times 999 ) grid where ( T ) is the number of cell groups ((C_1, C_2, C_3)) such that ( C_1 ) and ( C_3 ) are white, ( C_2 ) is red, ( C_1 ) and ( C_2 ) are in the same row, and ( C_2 ) and ( C_3 ) are in the same column, we need to consider the optimal arrangement of white and red cells.Key steps:1. Understanding the Problem: ( T ) is the sum over all red cells of the product of the number of white cells in their row and column.2. Formulating the Problem: ( T = sum_{text{red cells } (i,j)} W_i times W_j ), where ( W_i ) and ( W_j ) are the number of white cells in row ( i ) and column ( j ).3. Optimizing the Configuration: To maximize ( T ), we need to balance the number of white cells in rows and columns. The optimal configuration is found by setting each row and column to have an equal number of white cells.4. Regular Configuration: Each row and column has 666 white cells (2/3 of 999). This configuration is possible due to the Gale-Ryser theorem.5. Calculating ( T ): - Each red cell contributes ( 666 times 666 ). - The number of red cells is ( 999 times 333 ). - Total ( T = 999 times 666^2 times 333 ).Final calculation:[T = 999 times 666^2 times 333 = 999 times (2 times 333)^2 times 333 = 999 times 4 times 333^3 = 12 times 333^4]After detailed computation, the maximum value of ( T ) is:[boxed{147556443852}],<think>Okay, so I need to find the maximum value of T in a 999x999 grid where some cells are white and others are red. T is the number of cell groups (C1, C2, C3) such that C1 and C2 are in the same row, C2 and C3 are in the same column, C1 and C3 are white, and C2 is red. Hmm, let's break this down step by step.First, let's visualize the problem. We have a large grid, almost 1000x1000. Each cell can be white or red. The triplet (C1, C2, C3) has specific relationships: C1 and C2 are in the same row, so they share a row but different columns. C2 and C3 are in the same column, so they share a column but different rows. C1 and C3 are white, while C2 is red. So essentially, we're looking for configurations where there's a red cell C2 that is in the same row as a white cell C1 and in the same column as another white cell C3. The triplet is formed by these three cells connected through C2.To maximize T, which counts all such triplets, we need to arrange the white and red cells in such a way that this configuration occurs as often as possible. So, how can we arrange the grid to maximize these triplets?Let me think. For each red cell C2, how many triplets (C1, C2, C3) can it be part of? Well, in its row, there must be at least one white cell C1 (since C1 is in the same row as C2 and is white). Similarly, in its column, there must be at least one white cell C3 (since C3 is in the same column as C2 and is white). Therefore, for each red cell C2, the number of triplets it contributes is equal to the number of white cells in its row multiplied by the number of white cells in its column. So, T can be calculated as the sum over all red cells of (number of white cells in the row of C2) multiplied by (number of white cells in the column of C2). Therefore, T = Σ_{C2 is red} (W_row(C2) * W_col(C2)), where W_row(C2) is the number of white cells in the row of C2, and W_col(C2) is the number of white cells in the column of C2.Our goal is to maximize this sum. So, we need to choose which cells to color white or red such that this sum is as large as possible. Now, the problem is similar to some kind of optimization problem where we need to arrange the white and red cells in the grid so that for each red cell, the product of white cells in its row and column is maximized, but also considering that if we make a row or column have more white cells, that affects all red cells in that row or column. Alternatively, perhaps there's a way to model this with linear algebra or combinatorics. Let me think. If we denote that in row i, there are w_i white cells, and in column j, there are v_j white cells. Then, for each red cell in position (i,j), its contribution is w_i * v_j. Therefore, the total T would be the sum over all red cells (i,j) of w_i * v_j. But each red cell (i,j) is such that cell (i,j) is red, so the cell itself is not white. Therefore, in row i, the number of white cells is w_i = 999 - r_i, where r_i is the number of red cells in row i. Similarly, v_j = 999 - c_j, where c_j is the number of red cells in column j. But wait, maybe not exactly. If the cell (i,j) is red, then in row i, the number of white cells would be 999 minus the number of red cells in row i. But if cell (i,j) is red, then when we compute w_i, it's 999 - r_i, where r_i includes cell (i,j). Similarly for columns.Therefore, T can be expressed as:T = Σ_{(i,j) is red} ( (999 - r_i) * (999 - c_j) )But we need to maximize this sum. The variables here are the r_i and c_j, but they are related because the total number of red cells in row i, r_i, is the sum over j of red cells in that row, and similarly for columns. However, since each red cell is in both a row and a column, there's a dependency here. So, this seems like a problem where we need to choose the red cells such that the sum over all red cells of (999 - r_i)(999 - c_j) is maximized.This is starting to look like a problem that might have a known maximum under certain constraints. Perhaps arranging the grid in a way that the red cells are arranged such that the products (999 - r_i)(999 - c_j) are maximized for each red cell. But how?Alternatively, perhaps if we consider that for each red cell, (999 - r_i) is the number of white cells in row i, and (999 - c_j) is the number of white cells in column j. So, the problem reduces to maximizing the sum over all red cells of (white cells in their row) * (white cells in their column).To maximize this sum, perhaps we need to have red cells placed in rows and columns that have as many white cells as possible. But since white cells are 999 - red cells in the row, and similarly for columns, this seems a bit circular. Maybe it's better to consider the entire grid's configuration. If we can set certain rows and columns to have a specific number of white cells, and then place red cells in intersections where their row and column have high numbers of white cells.Wait, but the more white cells in a row, the fewer red cells in that row. Similarly, the more white cells in a column, the fewer red cells in that column. So, if we have a row with many white cells (i.e., few red cells), then any red cell in that row would have a high (999 - r_i) term. Similarly, a column with many white cells (few red cells) would have a high (999 - c_j) term. Therefore, placing red cells in rows and columns that have few red cells (i.e., many white cells) would maximize each term in the sum.But we need to maximize the overall sum. So, if we can arrange red cells such that they are in rows and columns with as few other red cells as possible, then each red cell would contribute a large product term. However, since each red cell in a row increases r_i, thereby decreasing (999 - r_i) for all other red cells in that row. Similarly, each red cell in a column increases c_j, decreasing (999 - c_j) for all other red cells in that column.This seems like a trade-off. Adding a red cell in a row with few red cells gives a high contribution for that cell but slightly reduces the contribution of other cells in the same row. Similarly for columns. This problem resembles an optimization problem where each red cell's contribution depends on the number of red cells in its row and column. The challenge is to place red cells such that the sum of these contributions is maximized.Alternatively, maybe there's a way to model this with linear algebra. Let me denote x_i as the number of red cells in row i, and y_j as the number of red cells in column j. Then, the total number of red cells is Σx_i = Σy_j. However, not all combinations of x_i and y_j are possible, since the total number of red cells must be consistent across rows and columns.But maybe we can relax the problem and assume that x_i and y_j can be chosen independently, and then check if such a configuration is possible. This might lead us to an upper bound, and then we can check if the bound is achievable.If we consider that T = Σ_{(i,j) is red} ( (999 - x_i) * (999 - y_j) )But the total number of red cells is Σx_i = Σy_j = R (total red cells). However, we cannot have x_i and y_j completely independent because of overlapping cells. But maybe using some inequality here.Alternatively, perhaps we can use the Cauchy-Schwarz inequality or AM-QM inequality to find the maximum.Alternatively, think of T as the sum over all red cells of (white in row) * (white in column). To maximize this, perhaps arrange the red cells such that they are in rows and columns with as many white cells as possible. But if a row has many white cells, it has few red cells. Similarly, a column with many white cells has few red cells.Therefore, perhaps the optimal configuration is to have all red cells concentrated in a small number of rows and columns, but that seems counterintuitive because if red cells are in few rows and columns, then those rows and columns would have many red cells, leading to (999 - x_i) and (999 - y_j) being small for those rows and columns. Therefore, the contribution of each red cell would be small.Alternatively, if red cells are spread out as much as possible, so that each row and column has approximately the same number of red cells, then (999 - x_i) and (999 - y_j) would be similar across all rows and columns, leading to a more uniform contribution. But is this the maximum?Wait, let's think in terms of variables. Suppose each row has k red cells, and each column has l red cells. Then, for each red cell, the contribution is (999 - k)*(999 - l). If all rows have exactly k red cells and all columns have exactly l red cells, then the total number of red cells would be 999*k = 999*l, so k = l. Therefore, each row and column has the same number of red cells, say k. Then, T would be 999*k*(999 - k)^2.But is this the case? Wait, no. If each row has k red cells, then total red cells is 999*k. Similarly, each column has k red cells, so total red cells is 999*k. Therefore, to have such a regular grid where each row and column has exactly k red cells, we need a kind of Latin square or a balanced incomplete block design. However, such a design is possible only if certain conditions are met. For example, for a grid where each row and column has exactly k red cells, it's similar to a biadjacency matrix of a regular bipartite graph. Such a graph exists if the obvious divisibility conditions hold, i.e., 999 divides 999*k, which is always true. So, for any k from 0 to 999, such a configuration is possible. Therefore, if we arrange the grid such that each row and column has exactly k red cells, then the total T is 999*k*(999 - k)^2. Then, to find the maximum T, we can treat this as a function of k and find its maximum.Wait, but is this the actual maximum? Because if we allow rows and columns to have different numbers of red cells, maybe we can get a higher T. For example, suppose some rows have more red cells and others have fewer. Then, the rows with fewer red cells would contribute more per red cell, but there are fewer red cells in those rows. Similarly, columns with fewer red cells would contribute more per red cell.But perhaps arranging all rows and columns to have the same number of red cells (i.e., a regular configuration) would maximize the product due to the convexity or concavity of the function. Let's explore this.Assume that all rows have x red cells, so each row contributes x red cells. Then, each column must also have x red cells (since total red cells is 999x, and there are 999 columns, so 999x / 999 = x). Therefore, in such a configuration, T = 999x * (999 - x)^2. To maximize this, we can consider it as a function of x:f(x) = 999x(999 - x)^2We can find the maximum of this function by taking the derivative. Let's treat x as a real variable for the moment.f(x) = 999x(999 - x)^2Take the derivative f’(x):f’(x) = 999[(999 - x)^2 + x*2(999 - x)(-1)]= 999[(999 - x)^2 - 2x(999 - x)]= 999(999 - x)[(999 - x) - 2x]= 999(999 - x)(999 - 3x)Set f’(x) = 0:Either 999 - x = 0 => x = 999 (minimum), or 999 - 3x = 0 => x = 333.So, critical points at x = 333 and x = 999. Since x must be an integer between 0 and 999, the maximum occurs at x = 333.Therefore, in the regular configuration where each row and column has 333 red cells, the total T would be:T = 999 * 333 * (999 - 333)^2 = 999 * 333 * 666^2Compute this value:First, 666 = 2 * 3 * 3 * 37, but perhaps we can leave it as 666 for now.But maybe we can factor 999:999 = 27 * 37 = 9 * 111 = 9 * 3 * 37Similarly, 333 = 3 * 111 = 3 * 3 * 37666 = 6 * 111 = 2 * 3 * 3 * 37So, 999 * 333 * 666^2 = (27 * 37) * (3 * 3 * 37) * (2^2 * 3^4 * 37^2) = 2^2 * 3^(1+2+4) * 37^(1+1+2) * 27 ??? Wait, maybe this approach is complicating. Let's compute it step by step.Wait, 999 = 9 * 111, 333 = 3 * 111, 666 = 6 * 111So, 999 * 333 * 666^2 = 9*111 * 3*111 * (6*111)^2= 9*3*6^2 * 111 * 111 * 111^2= 9*3*36 * 111^4= (27*36) * 111^427*36 = 972So, 972 * 111^4But 111^2 = 12321, so 111^4 = 12321^2 = 12321*12321. However, perhaps we don't need to compute the exact number since the problem asks for the maximum value of T, which might be expressible in terms of 999, 333, etc., but likely we need to present it as a product or in a simplified form.However, the problem states that the grid is 999x999. The question is asking for the maximum T. If the regular configuration where each row and column has exactly 333 red cells is possible, then T would be 999 * 333 * (999 - 333)^2 = 999 * 333 * 666^2. But before finalizing this, we need to check whether such a regular configuration is possible.In a 999x999 grid, arranging each row and column to have exactly 333 red cells is equivalent to a 999x999 biadjacency matrix of a 333-regular bipartite graph. Since 999 is divisible by 333 (999 / 333 = 3), such a regular bipartite graph exists. For example, we can divide each set of 999 nodes into 3 groups of 333 and have each node in one group connect to all nodes in another group. However, in grid terms, this might correspond to dividing the grid into blocks. Wait, but in a grid, it's a bit different. Each row is connected to each column via the cell at their intersection. To have each row have exactly 333 red cells and each column also have exactly 333 red cells, we can use the concept of a projective plane or finite geometry, but perhaps more straightforwardly, using a block design.Alternatively, since 999 = 3 * 333, we can divide the grid into 3x3 blocks where each block is 333x333. Then, set the red cells in a diagonal pattern. For example, in the first third of rows, put red cells in the first third of columns; in the second third of rows, put red cells in the second third of columns; and in the last third of rows, put red cells in the last third of columns. However, this would result in each row and column having 333 red cells. But each row would only have red cells in one third of the columns, and each column would only have red cells in one third of the rows. This would satisfy the regular condition.Yes, such a configuration is possible. Therefore, the maximum T is achieved when each row and column has exactly 333 red cells, leading to T = 999 * 333 * 666^2.But wait, let's confirm this. Suppose we arrange the grid so that each row has 333 red cells and each column has 333 red cells. Then, for each red cell, the number of white cells in its row is 999 - 333 = 666, and the number of white cells in its column is also 666. Therefore, each red cell contributes 666 * 666 to T. Since there are 999 * 333 red cells in total (because each row has 333 red cells and there are 999 rows), the total T is 999 * 333 * 666 * 666.Which is 999 * 333 * 666². So, that's the calculation.But is this the maximum? Suppose that instead of having exactly 333 red cells per row and column, we have some variation. Maybe some rows have more red cells and some have fewer. Let's see.Suppose we have two rows: one with 334 red cells and one with 332 red cells. Let's see how this affects the total T.If we take a row with 334 red cells instead of 333, then each red cell in that row would contribute (999 - 334) * (999 - c_j), where c_j is the number of red cells in column j. If the columns still have 333 red cells each, then (999 - c_j) is still 666. However, (999 - 334) = 665. Therefore, each red cell in that row would contribute 665 * 666 instead of 666 * 666. So, the total contribution from that row would be 334 * 665 * 666. Whereas, if the row had 333 red cells, it would contribute 333 * 666 * 666. Let's compute the difference.334 * 665 * 666 vs. 333 * 666 * 666.Compute the ratio: (334 * 665) / (333 * 666) ≈ (334/333) * (665/666) ≈ (1.003) * (0.998) ≈ 1.001. Wait, this is approximately 1.001, which is slightly greater than 1. But that can't be. Wait, let's compute it exactly.334 * 665 = 334 * (666 - 1) = 334*666 - 334 = 222,444 - 334 = 222,110333 * 666 = 333*(600 + 66) = 333*600 + 333*66 = 199,800 + 21,978 = 221,778So, 222,110 vs. 221,778. So, 222,110 - 221,778 = 332. Therefore, 334 * 665 is 332 more than 333 * 666. Therefore, if we replace a row of 333 red cells with a row of 334 red cells, keeping the columns at 333, the contribution from that row increases by 332 * 666 (since each red cell's column contribution is still 666). Wait, no. Wait, in the previous calculation, 334*665 vs 333*666, the difference is 332. But since each red cell in the row contributes (665 * 666) vs (666 * 666). Wait, no. Wait, the total contribution from the row with 334 red cells is 334 * 665 * 666, and from 333 red cells is 333 * 666 * 666.Compute the difference:334*665*666 - 333*666*666 = 666*(334*665 - 333*666)Compute 334*665 - 333*666:334*665 = (333 + 1)*665 = 333*665 + 665333*666 = 333*(665 + 1) = 333*665 + 333Therefore, 334*665 - 333*666 = (333*665 + 665) - (333*665 + 333) = 665 - 333 = 332Therefore, the difference is 666*332 = 221, 112.So, increasing a row from 333 to 334 red cells (and compensating another row to keep column counts at 333) would increase T by 221,112. But wait, how does this affect the columns? If we increase a row's red cells by 1, we have to add a red cell in some column, which would increase that column's red cell count by 1, thereby decreasing (999 - c_j) for that column from 666 to 665. Therefore, all red cells in that column would now have their contribution reduced by 1 in the column term.Therefore, the net effect is not just the gain from the row but also the loss from the column. Let's model this properly.Suppose we take one row and increase its red cells from 333 to 334. To keep the column counts at 333, we have to compensate by reducing another row's red cells by 1. Wait, no. If we add a red cell in a column, that column's count increases by 1, so to keep all columns at 333, we can't just add a red cell anywhere. Therefore, the assumption that columns remain at 333 is invalid unless we adjust another red cell in that column.This complicates things. Therefore, changing the number of red cells in a row affects the column counts, which in turn affect the contributions of other red cells. Therefore, the previous calculation is incomplete because it didn't account for the necessary changes in column counts.Therefore, perhaps the regular configuration where all rows and columns have exactly 333 red cells is indeed optimal because any deviation would require adjusting columns, leading to some red cells having lower contributions which might offset the gains.Alternatively, perhaps the function f(x) = 999x(999 - x)^2 is maximized at x = 333 even when considering integer values, and hence the regular configuration gives the maximum T.Therefore, under the assumption that such a regular configuration is possible, the maximum T is 999 * 333 * 666².But let's confirm this with another approach. Suppose we have a grid where each cell is red with probability p and white with probability 1 - p. Then, the expected value of T would be:E[T] = Number of red cells * E[W_row * W_col | cell is red]But since the grid is large, and we're looking for maximum T, which is a deterministic value, this might not help directly, but perhaps gives insight. If each cell is red independently, then W_row and W_col are binomial variables. However, to maximize T, we need a structured arrangement rather than a random one.Alternatively, use Lagrange multipliers to maximize T with respect to variables x_i and y_j, but this might be complex.Alternatively, consider the following identity:Σ_{(i,j)} (W_row(i) * W_col(j)) = Σ_i W_row(i) * Σ_j W_col(j) = (Σ_i W_row(i)) * (Σ_j W_col(j))But Σ_i W_row(i) = Σ_j W_col(j) = Total white cells = W.Therefore, Σ_{(i,j)} (W_row(i) * W_col(j)) = W^2.However, T is equal to Σ_{(i,j) is red} (W_row(i) * W_col(j)).Therefore, we can write:T = Σ_{(i,j) is red} (W_row(i) * W_col(j)) = Σ_{(i,j)} (W_row(i) * W_col(j)) - Σ_{(i,j) is white} (W_row(i) * W_col(j))= W^2 - Σ_{(i,j) is white} (W_row(i) * W_col(j))Therefore, to maximize T, we need to minimize Σ_{(i,j) is white} (W_row(i) * W_col(j)).Given that W is fixed (since W = total white cells = 999^2 - R), where R is the total red cells. Wait, but no, actually, W is not fixed because the problem allows us to choose any number of white and red cells. Wait, no, in the problem statement, "some cells are white, and the others are red." So, we can choose any configuration, not necessarily a fixed number of white or red cells. Therefore, W can vary from 0 to 999^2. But since we need to maximize T, which depends on both the number of red cells and their placement.However, using the identity above, T = W^2 - Σ_{(i,j) is white} (W_row(i) * W_col(j))Therefore, maximizing T is equivalent to minimizing Σ_{(i,j) is white} (W_row(i) * W_col(j)) given that W is variable. But this seems complicated. Alternatively, perhaps we can bound Σ_{(i,j) is white} (W_row(i) * W_col(j)).By Cauchy-Schwarz inequality, Σ_{(i,j) is white} (W_row(i) * W_col(j)) >= (Σ_{(i,j) is white} W_row(i)) * (Σ_{(i,j) is white} W_col(j)) / (999^2)Wait, not sure. Alternatively, maybe use convexity.Alternatively, since each white cell (i,j) contributes W_row(i) * W_col(j), and W_row(i) is the number of white cells in row i, which includes the cell (i,j) itself. Similarly, W_col(j) includes cell (i,j). Therefore, for a white cell, W_row(i) and W_col(j) are both at least 1.But how does this help? Maybe consider that for each white cell, its contribution is W_row(i) * W_col(j). To minimize the sum over white cells of W_row(i)*W_col(j), perhaps arrange the white cells such that they are in rows and columns with as few other white cells as possible. That is, cluster the white cells together in a block. For example, if all white cells are in a k x k subgrid, then each white cell is in a row with k white cells and a column with k white cells, so each white cell contributes k * k, and the total sum is k^2 * k^2 = k^4. Whereas if the white cells are spread out, each in different rows and columns, then each white cell would be in a row with 1 white cell and column with 1 white cell, contributing 1*1=1, so total sum would be k^2. Therefore, spreading out white cells minimizes the sum Σ_{(i,j) is white} (W_row(i) * W_col(j)), which in turn maximizes T = W^2 - Σ... Therefore, to minimize Σ..., spread out white cells as much as possible. Wait, but this contradicts. If we spread out white cells, then each white cell is in a row and column with many white cells, but no, if they are spread out so that each row and column has exactly one white cell, then W_row(i) = 1 and W_col(j) = 1 for each white cell, leading to Σ... = W, which is minimal. Whereas if they are clustered, Σ... is larger. Therefore, to minimize Σ..., we need to spread out white cells such that each row and column has as few white cells as possible. However, given that W is the total number of white cells, the minimal Σ... is achieved when the white cells are spread out as evenly as possible.Wait, if W is fixed, then according to the Cauchy-Schwarz inequality, Σ W_row(i)*W_col(j) over white cells is minimized when the white cells are spread out such that each row and column has as equal number of white cells as possible. This is similar to the concept in the inequality that the product sum is minimized when the distributions are as uniform as possible.But since we can choose W freely, how does this affect T? Wait, T = W^2 - Σ..., so to maximize T, for a given W, we need to minimize Σ..., but W itself can vary. Therefore, we need to find the value of W (number of white cells) that maximizes T = W^2 - Σ....However, this seems complex. Alternatively, perhaps the maximum T occurs when all cells are red except for a small number, but then W is small, so W^2 is small, but Σ... is also small, but maybe not. Alternatively, if all cells are white, then T = 0 because there are no red cells. If all cells are red, then T = 0 because there are no white cells. So, the maximum T must occur somewhere in between.But this approach might not be the most straightforward. Let's return to the initial idea of the regular configuration where each row and column has 333 red cells, leading to T = 999 * 333 * 666^2. Let's compute this value.First, 999 * 333 = (1000 - 1) * 333 = 333,000 - 333 = 332,667.Then, 666^2 = 443,556.Therefore, T = 332,667 * 443,556.Compute 332,667 * 443,556:This is a large number. However, the problem might expect the answer in terms of these factors rather than the multiplied out form. But let me check if the expression can be simplified.Note that 999 = 3 * 333, so 999 * 333 = 3 * 333 * 333 = 3 * 333^2.Similarly, 666 = 2 * 333.Therefore, T = 3 * 333^2 * (2 * 333)^2 = 3 * 333^2 * 4 * 333^2 = 12 * 333^4.But 12 * 333^4 = 12 * (333^2)^2. However, 333^2 = 110,889. Therefore, 12 * (110,889)^2. But again, maybe the problem expects the answer as 999 * 333 * 666^2 or in terms of 333^4 * 12.Alternatively, notice that 333 = 999 / 3, so substituting back:T = 999 * (999/3) * (2*999/3)^2 = 999 * (999/3) * (4*999²/9) = 999 * 999/3 * 4*999²/9.Simplify:= 999 * 999/3 * 4 * 999² / 9= 999^(1 + 1 + 2) * 4 / (3 * 9)= 999^4 * 4 / 27= 999^4 * (4/27)But 4/27 is approximately 0.148, but in terms of exact fraction, it's 4/27.Alternatively, 4/27 = (4/9) * (1/3) = but not sure. Alternatively, 999^4 * 4 / 27 can be written as (4/27) * 999^4. However, whether this is the simplest form is unclear.Alternatively, 4/27 is equal to (2^2)/(3^3). So, perhaps write it as 2² * 999^4 / 3³. Since 999 = 3³ * 37, substitute:999 = 3^3 * 37Therefore, 999^4 = (3^3 * 37)^4 = 3^12 * 37^4Therefore, T = 2² * 3^12 * 37^4 / 3³ = 2² * 3^(12 - 3) * 37^4 = 4 * 3^9 * 37^4.But this seems more complicated. The problem might not require simplifying beyond the initial expression. However, the answer likely expects the form as 333^4 multiplied by some factor. However, let's check.Wait, 12 * 333^4 = 12 * (3 * 111)^4 = 12 * 3^4 * 111^4 = 12 * 81 * 111^4 = 972 * 111^4, which matches the earlier calculation.Alternatively, the answer can be written as 999 * 333 * 666². Given that 999, 333, and 666 are all multiples of 111, perhaps factor that out:999 = 9 * 111333 = 3 * 111666 = 6 * 111Therefore, T = 9*111 * 3*111 * (6*111)^2 = 9*3*6² * 111^4 = 9*3*36 * 111^4 = 972 * 111^4.Therefore, another way to write T is 972 * (111)^4. However, this might still not be the simplest form. But since the problem asks for the maximum value of T, and given the grid size, the answer is likely expected to be presented in the form of 333^4 multiplied by 12, or 972 * 111^4, but in the context of the problem, maybe it's acceptable to write it as 999 * 333 * 666².But before confirming, let's check if there exists a configuration that could give a higher T than the regular configuration.Suppose we divide the grid into two parts: a block of a x a cells which are all white, and the rest of the grid is red. Then, for the red cells outside the a x a block, their rows and columns have W_row = a (since the a white cells are in a block), and W_col = a. Therefore, each red cell outside the block contributes a * a = a². The number of red cells is 999² - a². Therefore, T = (999² - a²) * a². To maximize this, take derivative with respect to a:d/da [ (999² - a²) * a² ] = d/da [ 999² a² - a^4 ] = 2*999² a - 4a³Set to zero:2*999² a - 4a³ = 0 => a(2*999² - 4a²) = 0Solutions a=0 or 2*999² -4a²=0 => 4a²=2*999² => a² = (999²)/2 => a = 999/√2 ≈ 706. Therefore, maximum at a≈706. Then, T ≈ (999² - (706)^2)*(706)^2 ≈ (998001 - 498,436)*498,436 ≈ (499,565)*498,436 ≈ 249,282, 034, 540.Compare this to the regular configuration T = 972 * 111^4. Let's compute 111^4:111^2 = 12,321111^4 = 12,321^2 = 12,321*12,321.Compute 12,321*12,321:First, 12,000*12,000 = 144,000,00012,000*321 = 3,852,000321*12,000 = 3,852,000321*321 = 103,041Therefore, total = 144,000,000 + 3,852,000 + 3,852,000 + 103,041 = 144,000,000 + 7,704,000 + 103,041 = 151,704,000 + 103,041 = 151,807,041.Therefore, 111^4 = 151,807,041.Then, T = 972 * 151,807,041. Compute 972 * 150,000,000 = 145,800,000,000972 * 1,807,041 = ?Compute 972 * 1,800,000 = 1,749,600,000972 * 7,041 = ?972 * 7,000 = 6,804,000972 * 41 = 39,852Total = 6,804,000 + 39,852 = 6,843,852Therefore, 972 * 1,807,041 = 1,749,600,000 + 6,843,852 = 1,756,443,852Therefore, total T = 145,800,000,000 + 1,756,443,852 = 147,556,443,852 ≈ 1.47556 x 10^11Compare this to the previous block method's T ≈ 2.4928 x 10^11, which is larger. Wait, this suggests that the block configuration might give a higher T. But this contradicts our initial assumption.Wait, but this seems suspicious. If we make a large block of white cells, then all the red cells outside the block have W_row = a and W_col = a, so each red cell contributes a². The number of red cells is 999² - a². Therefore, T = a²*(999² - a²). This function's maximum is at a = 999/√2 ≈706, as we found earlier, leading to T ≈ (999² / 2)^2 = (999^4)/4. Whereas in the regular configuration, T = 12 * 333^4 ≈ 12*( (999/3)^4 ) = 12*(999^4)/81 = (12/81)*999^4 = (4/27)*999^4 ≈0.148*999^4. Comparing this to the block method's T≈0.25*999^4. Therefore, the block method gives a higher T. Therefore, our initial assumption was incorrect.This suggests that the regular configuration is not optimal, and a different configuration where white cells are arranged in a large block gives a higher T. But why is that?Wait, perhaps because in the block configuration, the red cells have a high number of white cells in their rows and columns (since the white cells are concentrated in a block), whereas in the regular configuration, each red cell has a moderate number of white cells in their row and column. Therefore, even though there are fewer red cells in the block method (since a large block is white), each red cell contributes a higher value (a² vs (666)^2). Wait, but in the block method, the number of red cells is 999² - a², which when a is around 700, is about 999² - 700² ≈ (999 -700)(999 +700) ≈ 299 * 1699 ≈ 507,401. So, around 500,000 red cells, each contributing 700² = 490,000. So total T ≈ 500,000 * 490,000 = 245,000,000,000.Whereas in the regular configuration, we have 999 *333 ≈332,667 red cells, each contributing 666²≈443,556. So total T≈332,667 *443,556≈147,556,443,852, which is about half of the block method's T. Therefore, the block method gives a higher T. But this contradicts our initial analysis. Therefore, perhaps the maximum T is indeed achieved by the block configuration. However, this seems to be the case, so why did we initially think the regular configuration is optimal? Because we assumed regularity, but the block configuration, even though irregular, gives a higher T.But wait, in the block configuration, the white cells are concentrated, so the W_row and W_col for the red cells are high, but the number of red cells is lower. However, the product a²*(N² -a²) is maximized when a² = N²/2, so when a = N/√2. For N=999, a≈706. So, T≈(706)^2*(999^2 - (706)^2)≈(706)^2*(999^2 - (706)^2). This is the maximum of the function f(a) = a²*(N² -a²), which is (N^4)/4 when a = N/√2. However, since a must be an integer, it's approximately that value.But in this case, the white cells are all concentrated in a block, so the W_row for red cells outside the block is a (the number of white cells in their row), which is 706 if the white block is in the first 706 rows and columns. However, this is not the case. If the white cells are in an a x a block, then the red cells are divided into those in the same rows as the white block and those in different rows. Wait, no. If the white block is, say, the first a rows and first a columns, then:- Rows 1 to a: each has a white cells (in columns 1 to a) and 999 - a red cells (columns a+1 to 999).- Rows a+1 to 999: each has 0 white cells and 999 red cells.Similarly, columns 1 to a: each has a white cells (rows 1 to a) and 999 - a red cells (rows a+1 to 999).Columns a+1 to 999: each has 0 white cells and 999 red cells.Therefore, the W_row for a red cell in row i:- If i is in 1 to a, then W_row(i) = a.- If i is in a+1 to 999, then W_row(i) = 0.Similarly, W_col(j) for a red cell in column j:- If j is in 1 to a, then W_col(j) = a.- If j is in a+1 to 999, then W_col(j) = 0.Therefore, a red cell in position (i,j) where i <=a and j <=a: but those are white cells, so not red.A red cell in (i,j) where i <=a and j >a: W_row(i) =a, W_col(j)=0.A red cell in (i,j) where i >a and j <=a: W_row(i)=0, W_col(j)=a.A red cell in (i,j) where i >a and j >a: W_row(i)=0, W_col(j)=0.Therefore, the only red cells that contribute to T are those in (i,j) where either i <=a and j >a, or i >a and j <=a. For these cells, the contribution is either a*0=0 or 0*a=0. Therefore, in this block configuration, T=0.Wait, that can't be. If we have a block of white cells, then the red cells outside the block are in rows and columns that have all white cells in the block. Wait, no. If the white block is a x a, then rows 1 to a have a white cells, so W_row(i) = a for i <=a, and W_row(i) = 0 for i >a. Similarly, columns 1 to a have a white cells, so W_col(j) = a for j <=a, and W_col(j)=0 for j >a.Therefore, a red cell in (i,j):- If i <=a and j >a: W_row(i)=a, W_col(j)=0. Contribution = a*0=0.- If i >a and j <=a: W_row(i)=0, W_col(j)=a. Contribution =0*a=0.- If i >a and j >a: W_row(i)=0, W_col(j)=0. Contribution=0*0=0.Therefore, in this block configuration, T=0. Which is terrible. Therefore, my previous analysis was incorrect.This is because the red cells outside the white block are in rows or columns that either have a white cells (if they are in the first a rows or columns) or 0 white cells (if they are outside). But in either case, the product W_row * W_col is zero for all red cells. Therefore, T=0.Therefore, the block configuration is actually the worst possible for T, not the best. This is a critical mistake in the previous analysis. Therefore, the earlier approach of using the regular configuration is correct.Therefore, returning to the regular configuration where each row and column has exactly 333 red cells, leading to T = 999 * 333 * 666^2.But why did the block configuration give T=0? Because the red cells were in rows or columns that had no white cells in the orthogonal direction.Therefore, to maximize T, the red cells must be in rows and columns that have many white cells. The optimal way is to spread out the red cells so that each red cell is in a row and column with as many white cells as possible, but without overcrowding rows or columns with too many red cells which would reduce the white cell counts.This brings us back to the regular configuration. By having each row and column contain the same number of red cells, we ensure that each red cell is in a row and column with a high number of white cells (666), and the total number of red cells is maximized for that regular structure.Therefore, the regular configuration likely provides the maximum T. To confirm, let's consider another configuration where half the grid is red and half is white, arranged in a checkerboard pattern. In this case, each red cell is in a row with 499 or 500 white cells and a column with 499 or 500 white cells. The contribution per red cell would be approximately 500*500 = 250,000. With about half of the grid red (around 500,000 red cells), T would be approximately 500,000 * 250,000 = 125,000,000,000, which is less than the regular configuration's T of approximately 147,556,443,852. Therefore, the regular configuration still gives a higher T.Therefore, it seems that the regular configuration where each row and column has exactly 333 red cells is indeed optimal, leading to T = 999 * 333 * 666².Thus, the maximum value of T is 999 * 333 * 666², which can be simplified to 12 * 333⁴ or 972 * 111⁴, but in the problem's context, the answer is likely expected to be expressed in terms of the given grid size (999) and the optimal number of red cells per row/column (333). Therefore, the answer is:T = 999 * 333 * 666² = 999 * 333 * (666)^2Alternatively, simplifying:Since 666 = 2 * 333,T = 999 * 333 * (2 * 333)^2 = 999 * 333 * 4 * 333² = 4 * 999 * 333³But 999 = 3 * 333,T = 4 * 3 * 333 * 333³ = 12 * 333⁴Which is another way to write it. However, both forms are equivalent. The problem might accept either form, but likely prefers the form with the original numbers given in the problem (999, 666).Therefore, the maximum value of T is boxed{999 times 333 times 666^2}.Final Answerboxed{999 times 333 times 666^2}</think>To find the maximum value of ( T ) in a ( 999 times 999 ) grid where ( T ) is the number of cell groups ((C_1, C_2, C_3)) such that ( C_1 ) and ( C_2 ) are in the same row, ( C_2 ) and ( C_3 ) are in the same column, ( C_1 ) and ( C_3 ) are white, and ( C_2 ) is red, we need to consider the optimal arrangement of red and white cells.1. Understanding the Contribution of Each Red Cell: - For each red cell ( C_2 ), the number of triplets ((C_1, C_2, C_3)) it contributes is the product of the number of white cells in its row and the number of white cells in its column. - If each row and column has a balanced number of red and white cells, the contribution of each red cell is maximized.2. Regular Configuration: - Consider a configuration where each row and each column has exactly 333 red cells. This means each row and column has ( 999 - 333 = 666 ) white cells. - For each red cell, the contribution to ( T ) is ( 666 times 666 ).3. Calculating the Total Contribution: - There are ( 999 times 333 ) red cells in total (since each of the 999 rows has 333 red cells). - Each red cell contributes ( 666^2 ).4. Final Calculation: - The total number of triplets ( T ) is the product of the number of red cells and the contribution per red cell: [ T = 999 times 333 times 666^2 ]Thus, the maximum value of ( T ) is (boxed{999 times 333 times 666^2}).]

🔑:3. First, prove: For an n times n grid, we haveT leqslant frac{4 n^{4}}{27} text {. }Let the i-th row have a_{i} white cells, and the j-th column have b_{j} white cells. Let R be the set of red cells. For each red cell (i, j), there are a_{i} b_{j} valid quadruples left(C_{1}, C_{2}, C_{3}right), where C_{2}=(i, j). Therefore,T=sum_{(i, j) in R} a_{i} b_{j} text {. }By the AM-GM inequality,begin{array}{l}T leqslant frac{1}{2} sum_{(i, j) in R}left(a_{i}^{2}+b_{j}^{2}right) =frac{1}{2} sum_{i=1}^{n}left(n-a_{i}right) a_{i}^{2}+frac{1}{2} sum_{j=1}^{n}left(n-b_{j}right) b_{j}^{2},end{array}This is because the i-th row has n-a_{i} red cells, and the j-th column has n-b_{j} red cells.For 0 leqslant x leqslant n, by the AM-GM inequality,begin{array}{l}(n-x) x^{2}=frac{1}{2}(2 n-2 x) x cdot x leqslant frac{1}{2}left(frac{2 n}{3}right)^{3}=frac{4 n^{3}}{27},end{array}with equality if and only if x=frac{2 n}{3}.Thus, T leqslant frac{n}{2} cdot frac{4 n^{3}}{27}+frac{n}{2} cdot frac{4 n^{3}}{27}=frac{4 n^{4}}{27}.If n=999, then x=frac{2 n}{3}=666.For a coloring where each row and each column has 666 white cells, the equality in the inequality for T holds.The following example ensures that each row and each column has 666 white cells. Ifi-j equiv 1,2, cdots, 666(bmod 999),then color the cell (i, j) white, and color the other cells red. Thus, the maximum value of T isfrac{4 times 999^{4}}{27}=148 times 999^{3} .

🧠:[<think>Okay, so I need to find the distance between two points in 3D space, right? The points are A(1, 0, -2) and B(-2, 4, 3). Hmm, I remember there's a formula for the distance between two points in three dimensions. Let me recall. In 2D, the distance formula is the square root of [(x2 - x1)^2 + (y2 - y1)^2]. So in 3D, I think we just add the z-component as well. So it should be the square root of [(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Let me make sure that's right. Yeah, I think that's correct because in 3D space each coordinate difference contributes to the overall distance. So the formula extends naturally from 2D to 3D by including the third dimension.Alright, let me apply this formula to points A and B. Let's label the coordinates first. Point A has coordinates (1, 0, -2), so x1 = 1, y1 = 0, z1 = -2. Point B has coordinates (-2, 4, 3), so x2 = -2, y2 = 4, z2 = 3. Now I need to compute the differences between each corresponding coordinate.Starting with the x-coordinates: x2 - x1 = -2 - 1 = -3. Then the y-coordinates: y2 - y1 = 4 - 0 = 4. Finally, the z-coordinates: z2 - z1 = 3 - (-2) = 3 + 2 = 5. So the differences are -3, 4, and 5. Now I need to square each of these differences.Squaring each difference: (-3)^2 = 9, (4)^2 = 16, and (5)^2 = 25. Adding those squared differences together: 9 + 16 + 25. Let me compute that step by step. 9 + 16 is 25, and then 25 + 25 is 50. So the sum of the squared differences is 50. Then the distance is the square root of this sum. So sqrt(50). Hmm, sqrt(50) can be simplified. Since 50 is 25*2, sqrt(25*2) is 5*sqrt(2). So the distance should be 5 times the square root of 2.Wait, let me double-check my calculations. Let me go through each step again to make sure I didn't make a mistake. First, the coordinates: A is (1, 0, -2) and B is (-2, 4, 3). Subtracting x-coordinates: -2 - 1 = -3. Squared: (-3)^2 = 9. Subtracting y-coordinates: 4 - 0 = 4. Squared: 16. Subtracting z-coordinates: 3 - (-2) = 5. Squared: 25. Summing them: 9 + 16 is 25, plus 25 is 50. Square root of 50 is indeed 5*sqrt(2). That seems right.Is there another way to verify this? Maybe by visualizing the 3D coordinates? Well, in 3D, the distance formula is a straight line between the two points, taking into account all three dimensions. So if I imagine moving from point A to point B, I'm moving -3 units in the x-direction, 4 units in the y-direction, and 5 units in the z-direction. The distance is like the hypotenuse of a rectangular prism with sides of lengths 3, 4, and 5. The formula accounts for all three directions. So the calculation is correct.Alternatively, maybe plugging into the distance formula again. Let me write it out step by step:Distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]Plugging in the values:sqrt[(-2 - 1)^2 + (4 - 0)^2 + (3 - (-2))^2] = sqrt[(-3)^2 + 4^2 + 5^2] = sqrt[9 + 16 + 25] = sqrt[50] = 5*sqrt(2).Yes, that's exactly what I did before. So unless I made a mistake in the arithmetic, which I checked twice, this should be the correct answer. Let me confirm sqrt(50) is 5*sqrt(2). Since 50 is 25*2, sqrt(25*2) is sqrt(25)*sqrt(2) = 5*sqrt(2). Correct. So the distance is 5√2.Just to make sure, maybe calculate the decimal value. sqrt(2) is approximately 1.4142, so 5*1.4142 ≈ 7.071. Let me see if that makes sense. The differences in coordinates are -3, 4, 5, so each component's square is 9, 16, 25. Adding those gives 50, sqrt(50) ≈ 7.071, which is about 7.07 units. That seems reasonable given the coordinate differences. If I think of a 3-4-5 triangle scaled up, but here it's 3-4-5 in each axis, so the diagonal would indeed be sqrt(50). Yeah, that seems right.I don't think I made any mistakes here. The process is straightforward, and I verified each step. So I'm confident the distance between points A and B is 5 times the square root of 2.Final AnswerThe distance from point A to point B is boxed{5sqrt{2}}.</think>To find the distance from point ( A(1, 0, -2) ) to point ( B(-2, 4, 3) ) in the rectangular coordinate system, we use the 3D distance formula:[text{Distance} = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}]Substituting the coordinates of points ( A ) and ( B ):[begin{align*}x_2 - x_1 &= -2 - 1 = -3, y_2 - y_1 &= 4 - 0 = 4, z_2 - z_1 &= 3 - (-2) = 5.end{align*}]Squaring each difference:[begin{align*}(-3)^2 &= 9, 4^2 &= 16, 5^2 &= 25.end{align*}]Summing these squared differences:[9 + 16 + 25 = 50]Taking the square root of the sum:[sqrt{50} = 5sqrt{2}]Thus, the distance from point ( A ) to point ( B ) is (boxed{5sqrt{2}}).,<think>Okay, so I need to find the distance between point A(1, 0, -2) and point B(-2, 4, 3) in 3D space. Hmm, I remember there's a formula for the distance between two points in three dimensions. Let me think. In 2D, the distance formula is the square root of [(x2 - x1)^2 + (y2 - y1)^2], right? So maybe in 3D, it's similar but with an extra term for the z-coordinate. That makes sense because each coordinate represents a dimension, so adding the z difference squared should account for the third dimension. Alright, so the formula should be distance = sqrt[(x2 - x1)^2 + (y2 - y1)^2 + (z2 - z1)^2]. Let me confirm that. Maybe I can check my notes or textbook, but since I can't do that now, I'll try to visualize it. If you have two points in space, the distance between them is like the hypotenuse of a rectangular prism, where the sides are the differences in each coordinate. So yeah, that formula sounds right.Let me write down the coordinates for points A and B. Point A is (1, 0, -2) and point B is (-2, 4, 3). So labeling the coordinates: for point A, x1 = 1, y1 = 0, z1 = -2. For point B, x2 = -2, y2 = 4, z2 = 3.Now I need to compute the differences in each coordinate. Let's do each one step by step. First, the x-coordinate difference: x2 - x1 = (-2) - 1. Hmm, subtracting 1 from -2. Wait, no, it's x2 minus x1, so that's -2 - 1, which is -3. Okay, so (x difference)^2 will be (-3)^2 = 9.Next, the y-coordinate difference: y2 - y1 = 4 - 0 = 4. So (y difference)^2 = 4^2 = 16.Then the z-coordinate difference: z2 - z1 = 3 - (-2) = 3 + 2 = 5. Therefore, (z difference)^2 = 5^2 = 25.Now, adding up these squared differences: 9 + 16 + 25. Let me compute that. 9 + 16 is 25, and 25 + 25 is 50. So the sum is 50.Then, the distance is the square root of 50. Hmm, sqrt(50). Let me simplify that. 50 can be written as 25*2, so sqrt(25*2) = sqrt(25)*sqrt(2) = 5*sqrt(2). That's a simplified radical form. So the distance is 5 times the square root of 2.Wait, let me verify each step again to make sure I didn't make any mistakes. Starting with the coordinates:Point A: (1, 0, -2). Point B: (-2, 4, 3). x difference: -2 - 1 = -3. Squared: 9. Correct.y difference: 4 - 0 = 4. Squared: 16. Correct.z difference: 3 - (-2) = 5. Squared: 25. Correct.Sum: 9 + 16 + 25 = 50. Square root of 50 is indeed 5*sqrt(2). Okay, that seems right. But just to double-check, maybe I should compute sqrt(50) numerically to see what that is approximately. 5*sqrt(2) is about 5*1.4142 ≈ 7.071. Let me check with a calculator. Wait, sqrt(50) is approximately 7.071, which matches 5*sqrt(2). So that's correct.So I think that's all. The key steps are calculating the differences in each coordinate, squaring them, summing, and taking the square root. It's straightforward once you remember the formula. I just need to make sure I don't mix up the coordinates or make a sign error when subtracting. For example, subtracting x1 from x2, not the other way around, which in this case, since both coordinates are different, the order matters for the sign, but since we square it anyway, the sign doesn't affect the result. However, in general, it's good practice to keep the order consistent with the formula.Wait, let's test that thought. Suppose I did x1 - x2 instead of x2 - x1. Then, x1 - x2 would be 1 - (-2) = 3. Squared, that's still 9. So the square makes it positive regardless. So actually, the order doesn't matter here because we square it. So even if I subtract in the reverse order, the square would negate the negative. So perhaps that's why the formula works regardless of the order. But in the formula, it's written as (x2 - x1)^2, but even if we do (x1 - x2)^2, it's the same value. So that's a good point. So maybe I don't need to worry about the order as long as I subtract consistently? Wait, no, because for the other terms, like if I did (y1 - y2)^2 instead, but in this case, since it's squared, it still becomes positive. So in fact, the distance formula is symmetric. So the distance from A to B is the same as from B to A. So whether I subtract x2 - x1 or x1 - x2, when squared, it's the same. So that's why the formula is written that way. So maybe even if I messed up the order, the result would still be the same. So that's reassuring.So, in any case, my calculation seems correct. The differences were calculated properly, squared, summed, and square-rooted. So the answer should be 5*sqrt(2). Let me just check once again:Compute each difference:x: -2 - 1 = -3, squared is 9.y: 4 - 0 = 4, squared is 16.z: 3 - (-2) = 5, squared is 25.Total sum: 9 + 16 + 25 = 50.sqrt(50) = 5*sqrt(2). Yep, that's correct.So I think that's the right answer. No mistakes in the steps. So the distance between point A and point B is 5 times the square root of 2.Final AnswerThe distance from point A(1, 0, -2) to point B(-2, 4, 3) is boxed{5sqrt{2}}.</think>To find the distance from point ( A(1, 0, -2) ) to point ( B(-2, 4, 3) ) in 3D space, we use the distance formula:[text{distance} = sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2 + (z_2 - z_1)^2}]Substituting the coordinates of points ( A ) and ( B ):- ( x_1 = 1 ), ( y_1 = 0 ), ( z_1 = -2 )- ( x_2 = -2 ), ( y_2 = 4 ), ( z_2 = 3 )Calculate the differences in each coordinate:- ( x )-difference: ( -2 - 1 = -3 )- ( y )-difference: ( 4 - 0 = 4 )- ( z )-difference: ( 3 - (-2) = 5 )Square each difference:- ( (-3)^2 = 9 )- ( 4^2 = 16 )- ( 5^2 = 25 )Sum the squared differences:[9 + 16 + 25 = 50]Take the square root of the sum:[sqrt{50} = 5sqrt{2}]Thus, the distance from point ( A(1, 0, -2) ) to point ( B(-2, 4, 3) ) is (boxed{5sqrt{2}}).]

🔑:Solution: The distance from point A(1, 0, -2) to point B(-2, 4, 3) is:d = sqrt{(1 - (-2))^2 + (0 - 4)^2 + (-2 - 3)^2} = 5sqrt{2}.Hence, the answer is: boxed{5sqrt{2}}.This problem is solved directly using the distance formula between two points. The method of finding the distance between two points in this problem tests basic knowledge of the distance formula and computational problem-solving skills, making it a fundamental question.

🧠:[<think>Okay, let me try to figure out this problem. The question says: "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is (A) 688. (B) 686. (C) 684. (D) 658. (E) 630." Hmm, wait, I need to make sure I understand the problem correctly. First, the problem mentions "32 ⋅ 28 is less than 1000." Let me check that. 32 times 28. Let me compute that: 32*28. 30*28 is 840, and 2*28 is 56, so 840 + 56 = 896. So, 896 is indeed less than 1000. Maybe this is just a setup for the range of numbers we're considering? Wait, but the problem doesn't specify a range. Hmm, maybe the original problem was different? Wait, the problem is from the 17th American High School Mathematics Examination, 1966. Let me check if there's a misinterpretation here.Wait, the original problem might be phrased differently. The user wrote: "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is..." That seems a bit confusing. Maybe the actual problem is: "The number of positive integers less than 1000 that cannot be divided by 5 or 7 is..." and then the options. Because 32*28=896 is less than 1000, but maybe that's part of a different problem. Wait, maybe the problem is in Chinese or another language, and the translation is a bit off. Wait, no, the user mentioned it's from the 17th American High School Mathematics Examination, 1966. Let me check if that's a real contest.Yes, the American High School Mathematics Examination (AHSME) was the predecessor of the AMC 12. The 17th edition would be 1966. Let me try to look up the problem to confirm the exact wording. However, since I can't access external resources, I need to work with what's given.The problem as presented says: "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is (A) 688. (B) 686. (C) 684. (D) 658. (E) 630."Hmm. The first part says "32 ⋅ 28 is less than 1000"—which we verified is 896. Maybe that's a red herring or maybe the original problem had a different structure. Alternatively, perhaps the problem is asking for the number of positive integers less than 32⋅28 (which is 896) that cannot be divided by 5 and 7? That would make sense. Because if it's just "positive integers" in general, the number would be infinite. So, maybe the problem is: "The number of positive integers less than 32⋅28 (which is 896 and less than 1000) that cannot be divided by 5 or 7 is..." So, the correct answer would be 686 or another of the options given. Let me proceed with that assumption.So, restating the problem: Find the number of positive integers less than 896 that are not divisible by 5 or 7. Then the answer choices are given. Let's check the answer choices: 688, 686, 684, 658, 630. So, I need to compute the number of integers from 1 to 895 inclusive that are not divisible by 5 or 7. To solve this, I can use the principle of inclusion-exclusion. First, calculate the total number of integers in the range, which is 895. Then subtract the numbers divisible by 5 and the numbers divisible by 7, and then add back the numbers divisible by both 5 and 7 (since they were subtracted twice).Total numbers: 895Numbers divisible by 5: floor(895/5) = 179 (since 5*179 = 895)Numbers divisible by 7: floor(895/7). Let's compute 7*127 = 889, so 127 numbers. 895 - 889 = 6, so floor(895/7)=127.Numbers divisible by both 5 and 7: These are numbers divisible by 35. So floor(895/35). Let's compute 35*25 = 875, 35*26 = 910, which is over 895. So 25 numbers.Therefore, using inclusion-exclusion:Number of numbers divisible by 5 or 7 = 179 + 127 - 25 = 281.Therefore, numbers not divisible by 5 or 7 = Total - divisible by 5 or 7 = 895 - 281 = 614.Wait, but 895 - 281 = 614. But none of the answer choices are 614. The options are 688, 686, 684, 658, 630. Hmm, so this suggests that my assumption is wrong. So maybe the upper limit is 1000 instead of 896? Let's check.If the problem is asking for numbers less than 1000, then total numbers are 999. Let's compute that.Total numbers: 999Numbers divisible by 5: floor(999/5) = 199 (since 5*199=995)Numbers divisible by 7: floor(999/7). 7*142=994, so 142.Numbers divisible by 35: floor(999/35). 35*28=980, 35*29=1015. So 28.So numbers divisible by 5 or 7: 199 + 142 - 28 = 313.Numbers not divisible by 5 or 7: 999 - 313 = 686. Which is option B. So that's one of the answer choices. But in the problem statement, it says "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is..." So maybe there was a translation error, and the original problem was asking for numbers less than 32*28=896, but in the user's translation, it's presented as "32 ⋅ 28 is less than 1000 and..." which is confusing. But according to the answer choices, if we compute for numbers less than 1000, we get 686, which is option B. Since that's one of the choices, and 686 is a standard answer for such a problem, maybe the correct upper limit is 1000. Let me check again.Alternatively, maybe the problem is as stated: "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is..." But this is confusing because the first part seems like a statement, and the second part is the question. Maybe the problem is in Chinese, and in the original, it's structured as two separate parts? Maybe "Compute 32*28 (which is less than 1000) and find the number of positive integers that cannot be divided by 5 and 7." But that still doesn't make sense. Wait, maybe the problem is actually in two parts: first confirming that 32*28 is less than 1000, and then asking the number of positive integers up to 1000 that cannot be divided by 5 and 7. If that's the case, then the second part is about numbers up to 1000, which gives 686 as the answer, which is option B. Alternatively, perhaps up to 32*28=896, which would give 614, but that's not one of the options. So maybe the problem is in Chinese, and the translation is slightly off. Alternatively, maybe "divided by 5 and 7" is a mistranslation and should be "divided by 5 or 7". Because if it's "cannot be divided by 5 and 7", that would mean numbers that are not divisible by both 5 and 7, which is a different interpretation. Let me check that.Wait, if the problem is asking for numbers that cannot be divided by 5 and 7, i.e., numbers that are not divisible by both 5 and 7, then the count would be total numbers minus numbers divisible by both 5 and 7. That is, numbers not divisible by 35. So, if the range is up to 999, then numbers not divisible by 35 would be 999 - floor(999/35) = 999 - 28 = 971. Which is not among the options. So that's not it.Alternatively, if it's "numbers that cannot be divided by 5 or 7", meaning numbers not divisible by 5 and not divisible by 7, then that's the standard inclusion-exclusion problem. Which for up to 999 gives 686. So that must be it. So perhaps the problem was miswritten here as "cannot be divided by 5 and 7" but should be "cannot be divided by 5 or 7". That would make sense. So with that correction, the answer is 686. So option B.Alternatively, if the problem is indeed as stated, "cannot be divided by 5 and 7", then in logical terms, that would be numbers not (divisible by 5 and divisible by 7), which by De Morgan's laws is equivalent to numbers not divisible by 5 or not divisible by 7. Which is the same as numbers not divisible by 5 or not divisible by 7, which is all numbers except those divisible by both 5 and 7. Which would be total numbers minus numbers divisible by 35, which for 999 numbers is 999 - 28 = 971. But that's not an answer choice. Hence, the problem must have meant "numbers not divisible by 5 or 7", meaning neither divisible by 5 nor by 7. Therefore, the correct approach is inclusion-exclusion as I did earlier for numbers up to 999, giving 686. Therefore, the answer is B.Therefore, even though the problem statement is a bit confusing due to translation or formatting issues, the likely correct interpretation is numbers less than 1000 (as 32*28 is less than 1000 is just a parenthetical statement), and the answer is 686, option B.Final Answerboxed{B}</think>To solve the problem, we need to determine the number of positive integers less than 1000 that cannot be divided by 5 or 7. First, we calculate the total number of positive integers less than 1000, which is 999.Next, we use the principle of inclusion-exclusion to find the number of integers divisible by 5 or 7:1. Numbers divisible by 5: [ leftlfloor frac{999}{5} rightrfloor = 199 ]2. Numbers divisible by 7: [ leftlfloor frac{999}{7} rightrfloor = 142 ]3. Numbers divisible by both 5 and 7 (i.e., 35): [ leftlfloor frac{999}{35} rightrfloor = 28 ]Using inclusion-exclusion, the number of integers divisible by 5 or 7 is:[199 + 142 - 28 = 313]Subtracting this from the total number of integers less than 1000 gives the number of integers not divisible by 5 or 7:[999 - 313 = 686]Thus, the number of positive integers less than 1000 that cannot be divided by 5 or 7 is (boxed{B}).,<think>Okay, let me try to figure out this problem. So, the question says: "32 cdot 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is (A) 688. (B) 686. (C) 684. (D) 658. (E) 630." Hmm, first, I need to understand exactly what the problem is asking.Wait, the first part says "32 cdot 28 is less than 1000". Let me check that. 32 times 28. Let me compute that. 30*28 is 840, and 2*28 is 56, so 840+56=896. So 32*28=896, which is indeed less than 1000. Maybe that's just a setup for the problem? Then the main question is about the number of positive integers that cannot be divided by 5 and 7. Wait, "cannot be divided by 5 and 7"—does that mean numbers not divisible by both 5 and 7, or numbers not divisible by either 5 or 7?Hmm, the wording is a bit ambiguous. Let me think. If it's "cannot be divided by 5 and 7", it might mean numbers that are not divisible by both 5 and 7. But usually, when we talk about numbers not divisible by either 5 or 7, we use "or". Maybe the problem is asking for numbers not divisible by either 5 or 7. Let me verify.Alternatively, maybe it's saying the numbers that are not divisible by 5 and not divisible by 7. That would be numbers not divisible by either. So, in other words, numbers coprime to both 5 and 7, so they are not multiples of 5 or 7. If that's the case, then the number of such integers up to N is N minus the numbers divisible by 5 or 7. Using inclusion-exclusion principle.But wait, the problem says "the number of positive integers that cannot be divided by 5 and 7". The wording is a bit confusing. If it's "cannot be divided by 5 and 7", does that mean numbers that are not divisible by both 5 and 7? So, numbers that are not multiples of 5 and also not multiples of 7. Wait, actually, "cannot be divided by 5 and 7" is ambiguous. It could mean that the numbers are not (divisible by both 5 and 7), or it could mean that the numbers are not divisible by 5 and not divisible by 7. The wording is unclear.But in the context of such problems, when they ask for numbers not divisible by two numbers, it's usually numbers not divisible by either, so that would be the inclusion-exclusion case. So, maybe the problem is asking for numbers from 1 to N (where N is 32*28=896) that are not divisible by 5 or 7. Then, the answer would be total numbers minus numbers divisible by 5 or 7.But let me check the problem again. "The number of positive integers that cannot be divided by 5 and 7 is..." Hmm. So if it's "cannot be divided by 5 and 7", it's a bit ambiguous. If it's "numbers that cannot be divided by 5 and cannot be divided by 7", that would be numbers not divisible by 5 and not divisible by 7, which is the same as numbers not divisible by either. If it's "numbers that cannot be divided by both 5 and 7", that would be numbers not divisible by 5 and 7 simultaneously. But being divisible by both 5 and 7 is the same as being divisible by LCM(5,7)=35. So numbers not divisible by 35. But then the answer would be total numbers minus numbers divisible by 35.But the options are in the 600s, so let's see. If N is 896, then total numbers is 896. If we subtract numbers divisible by 5 or 7, that would be 896 - (numbers divisible by 5 + numbers divisible by 7 - numbers divisible by both 5 and 7). Let me compute that.First, numbers divisible by 5 up to 896: floor(896/5) = 179 (since 5*179=895). Numbers divisible by 7: floor(896/7) = 128 (since 7*128=896). Numbers divisible by both 5 and 7, which is 35: floor(896/35) = 25 (since 35*25=875). So, using inclusion-exclusion: numbers divisible by 5 or 7 is 179 + 128 - 25 = 282. Therefore, numbers not divisible by 5 or 7 would be 896 - 282 = 614. Hmm, but 614 is not one of the options. The options are 688, 686, 684, 658, 630. So that can't be right. So maybe my interpretation is wrong.Alternatively, if the problem is asking for numbers not divisible by both 5 and 7, i.e., numbers not divisible by 35. Then total numbers minus numbers divisible by 35: 896 - 25 = 871, which is also not in the options. So that's not correct either.Wait, maybe the problem is referring to numbers that cannot be divided by 5 or 7. Wait, "cannot be divided by 5 and 7". If "and" is used here, in some contexts, it might mean "neither 5 nor 7". Like "the numbers that cannot be divided by 5 and cannot be divided by 7". So that's equivalent to numbers not divisible by 5 and not divisible by 7, which is the same as numbers not divisible by either 5 or 7. So, that would be the inclusion-exclusion case. Then the calculation I did earlier gives 614, which isn't an option. So maybe my N is wrong.Wait, the problem says "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is...". Maybe the problem is saying that 32*28 is less than 1000, so N is 32*28, which is 896, and then the number of positive integers up to 896 that cannot be divided by 5 and 7. So, as I calculated, 614. But the options don't have 614. So something's wrong here.Wait, perhaps I made a calculation mistake. Let me double-check the numbers.First, N is 32*28. Let me compute that again. 32*28: 30*28=840, 2*28=56, so 840+56=896. So N=896.Numbers divisible by 5: floor(896/5). 5*179=895, so 179 numbers.Numbers divisible by 7: floor(896/7). 7*128=896, so 128 numbers.Numbers divisible by both 5 and 7 (i.e., 35): floor(896/35). 35*25=875, 35*26=910, which is more than 896, so 25 numbers.Therefore, numbers divisible by 5 or 7: 179 + 128 - 25 = 282. So numbers not divisible by 5 or 7: 896 - 282 = 614. But 614 is not among the options. So perhaps the problem is not up to N=896, but up to 1000?Wait, the problem says "32 ⋅ 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is...". Maybe the first part is just a statement, and the actual N is 1000? Wait, but the first part is "32 ⋅ 28 is less than 1000"—so 32*28=896 < 1000. So perhaps the number of positive integers less than 1000 that cannot be divided by 5 and 7? But the options are around 600, so 1000 would be too big.Wait, no. If N=1000, the number of numbers not divisible by 5 or 7 would be 1000 - (200 + 142 - 28) = 1000 - 314 = 686. Wait, 200 is floor(1000/5), 142 is floor(1000/7), and floor(1000/35)=28. So 1000 - (200+142-28)=1000-314=686. Which is option B. But the problem says "32 ⋅ 28 is less than 1000". So maybe N is 32*28, which is 896, but the answer options are designed to trick you into thinking N is 1000? But in the problem statement, does it specify the range of positive integers? Wait, the problem says "the number of positive integers that cannot be divided by 5 and 7 is". It doesn't specify a range. That seems odd. Normally, such problems specify a range, like "up to N". Maybe the problem is in Chinese, and the translation is a bit off? Wait, no, it's the 17th American High School Mathematics Examination, 1966.Wait, let me check the original problem again. The user wrote: "32 cdot 28 is less than 1000 and the number of positive integers that cannot be divided by 5 and 7 is (A) 688. (B) 686. (C) 684. (D) 658. (E) 630." So, perhaps the problem is in two parts? First, verifying that 32*28 is less than 1000, and then the number of positive integers (maybe up to 32*28) that cannot be divided by 5 and 7. So the first part is just a statement to set N=32*28=896, and then compute the number of positive integers up to 896 not divisible by 5 or 7. But as I computed earlier, that gives 614, which is not an option. So either I'm misunderstanding the problem or miscalculating.Wait, perhaps the problem is saying "the number of positive integers less than 32*28 (which is 896) that cannot be divided by 5 or 7". But "cannot be divided by 5 and 7"—if it's using "and", maybe it's numbers not divisible by 5 and not divisible by 7. Then, that would be the same as numbers not divisible by 5 or 7. Wait, in logic, "not A and not B" is equivalent to "neither A nor B". So "numbers that cannot be divided by 5 and cannot be divided by 7" is "numbers not divisible by 5 and not divisible by 7", which is the same as numbers not divisible by 5 or 7. So that's the inclusion-exclusion case. So my calculation should be correct.But if so, 896 - (179 + 128 -25) = 614. But none of the options are 614. So maybe the original problem had a different number? Wait, but the user provided options are A) 688, B) 686, C) 684, D) 658, E) 630. So perhaps the problem is not up to 896, but up to 1000? Let me check. If N=1000, then numbers not divisible by 5 or 7: 1000 - (200 + 142 -28) = 1000 - 314 = 686, which is option B. So that's one of the options. But why does the problem mention 32*28 is less than 1000? Maybe that's a hint or part of the problem's setup. Wait, maybe it's a problem in Chinese where the translation is a bit confusing? Wait, the original problem is from the 17th American High School Mathematics Examination, 1966, so it's in English. Let me check if I can find the original problem.Wait, since the user mentioned it's the 17th AHSME 1966, problem number? Maybe the problem is: "The number of positive integers less than 1000 which are not divisible by 5 or 7 is: (A) 688 (B) 686 (C) 684 (D) 658 (E) 630." But then why mention 32*28 is less than 1000? Maybe the original problem says something like "Since 32*28=896 is less than 1000, the number of positive integers less than 1000 not divisible by 5 or 7 is..." So the 32*28 is just an example showing that 32*28 is less than 1000, but the actual N is 1000. Then the answer would be 686, which is option B. But why mention 32*28? Unless the problem is actually in Chinese and the translation is off. Alternatively, maybe the problem is from a different source.Alternatively, maybe the problem is combining two separate statements. The first is "32*28 is less than 1000" (which is true, 896 < 1000) and the second part is "the number of positive integers that cannot be divided by 5 and 7 is..." So maybe the problem is split into two parts, but the translation or formatting is confusing. If the second part is a separate question, then the number of positive integers (in general?) that cannot be divided by 5 and 7. But that would be infinite, so that doesn't make sense. So maybe the problem is "the number of positive integers less than 32*28 (which is 896) that cannot be divided by 5 or 7 is..." Then the answer would be 614, which is not an option. So maybe there is a miscalculation here.Wait, let me check again. If N=896.Numbers divisible by 5: floor(896/5)=179 (since 5*179=895)Numbers divisible by 7: floor(896/7)=128 (since 7*128=896)Numbers divisible by both 5 and 7 (35): floor(896/35)=25 (since 35*25=875)So numbers divisible by 5 or 7: 179 + 128 -25 = 282Numbers not divisible by 5 or 7: 896 - 282 = 614. So 614. Not an option.If N=1000, numbers divisible by 5: 200, numbers divisible by 7: 142 (since 7*142=994), numbers divisible by 35: 28 (35*28=980). So numbers divisible by 5 or 7: 200 + 142 -28=314. Numbers not divisible by 5 or 7: 1000 -314=686. That's option B. So maybe the actual N is 1000, but the problem mentions 32*28=896 is less than 1000 as part of the problem's introduction. Maybe the problem is: "Given that 32*28=896 is less than 1000, find the number of positive integers less than 1000 not divisible by 5 or 7." Then the answer is 686. So perhaps the problem is presented as two statements: first verifying that 32*28 is less than 1000, then asking the main question. The original problem in the 1966 AHSME might have been structured that way, perhaps as two separate parts, but in the user's transcription, it's presented as a single problem. Alternatively, maybe the problem is in Chinese and the translation is a bit confusing, leading to the mention of 32*28 as a setup.Alternatively, maybe there's a typo in the problem. If the problem is "the number of positive integers less than 32*28 (896) not divisible by 5 and 7", but the answer options are for N=1000, then the answer would be 686. Alternatively, maybe the original problem is in another language, and the translation mixed up the numbers.Alternatively, maybe the problem is referring to numbers that cannot be divided by both 5 and 7, meaning numbers not divisible by 35, and the count up to N=32*28=896. Then numbers not divisible by 35 would be 896 - floor(896/35)=896 -25=871, which is not an option. So that's not helpful.Alternatively, maybe the problem is referring to numbers that are not divisible by either 5 or 7 up to 1000, and the mention of 32*28=896 is just a red herring. Then, as calculated earlier, the answer is 686, which is option B. Since 686 is an option, and 32*28=896 being less than 1000 might be just part of the problem's setup (maybe in the original problem, there were multiple parts, and part a) was verifying that 32*28 is less than 1000, and part b) was the main question). So the answer is B) 686.Therefore, the correct answer is boxed{B}.Final Answerboxed{B}</think>The problem states that (32 cdot 28) is less than 1000 and asks for the number of positive integers that cannot be divided by 5 and 7. First, we verify that (32 cdot 28 = 896) is indeed less than 1000. Next, we need to determine the number of positive integers up to 1000 that are not divisible by either 5 or 7. This is calculated using the inclusion-exclusion principle:1. The number of integers divisible by 5 up to 1000 is (leftlfloor frac{1000}{5} rightrfloor = 200).2. The number of integers divisible by 7 up to 1000 is (leftlfloor frac{1000}{7} rightrfloor = 142).3. The number of integers divisible by both 5 and 7 (i.e., divisible by 35) up to 1000 is (leftlfloor frac{1000}{35} rightrfloor = 28).Using the inclusion-exclusion principle, the number of integers divisible by either 5 or 7 is:[200 + 142 - 28 = 314]Therefore, the number of integers not divisible by either 5 or 7 is:[1000 - 314 = 686]Thus, the correct answer is (boxed{B}).]

🔑:[Solution] If [x] represents the greatest integer not exceeding x, note that:the number of positive integers less than n (a natural number) is n-1;the number of positive integers less than n and divisible by d does not exceed left[frac{n-1}{d}right].Thus, among the 999 positive integers less than 1000, the number of positive integers that are not divisible by 5 and 7 is999-left[frac{999}{5}right]-left[frac{999}{7}right]+left[frac{999}{5 cdot 7}right]=999-199-142+28=686 text {. }Note that the positive integers divisible by 5 cdot 7 are subtracted twice. Therefore, the answer is (B).

🧠:[<think>Okay, so I need to find the smallest period of the function y = cos¹⁰x + sin¹⁰x. Hmm, let's see. First, I remember that both sine and cosine functions have a period of 2π, but when they are raised to powers, sometimes the period can become smaller. For example, sin²x or cos²x have a period of π instead of 2π because squaring them makes the function repeat more frequently. So maybe raising them to higher powers like 10 might also affect the period? But I need to check carefully.Let me start by recalling that the period of a function f(x) is the smallest positive number T such that f(x + T) = f(x) for all x. So, I need to find the smallest T where cos¹⁰(x + T) + sin¹⁰(x + T) = cos¹⁰x + sin¹⁰x for all x.Since the function is a combination of cos¹⁰x and sin¹⁰x, the period of the entire function will be the least common multiple (LCM) of the periods of cos¹⁰x and sin¹⁰x, provided they are different. But if both have the same period, then that will be the period of the sum.Now, let's analyze cos¹⁰x and sin¹⁰x individually. For cosine raised to an even power, like cos²x, the period becomes π because cos²x = (1 + cos2x)/2, which clearly has a period of π. Similarly, sin²x = (1 - cos2x)/2 also has a period of π. So, raising sine or cosine to an even power halves the period. But does this pattern continue for higher even powers?Wait, let's test this. Let's take cos⁴x. Using power-reduction formulas, cos⁴x can be written in terms of cos2x and cos4x. Let's recall the identity: cos²x = (1 + cos2x)/2, so cos⁴x = (cos²x)² = [(1 + cos2x)/2]^2 = (1 + 2cos2x + cos²2x)/4. Then, cos²2x can be written as (1 + cos4x)/2. So, cos⁴x = (1 + 2cos2x + (1 + cos4x)/2)/4 = (2 + 4cos2x + 1 + cos4x)/8 = (3 + 4cos2x + cos4x)/8. So cos⁴x is expressed in terms of cos2x and cos4x, which have periods π and π/2, respectively. The overall period would be the LCM of π and π/2, which is π. So cos⁴x still has a period of π. Hmm, interesting.Similarly, cos⁶x would be (cos²x)^3, and expanding that would involve cos2x, cos4x, cos6x, right? Let me check. If we use the identity again, cos³x can be written as (3cosx + cos3x)/4, but for even powers, maybe it's different. Wait, cos⁶x can be considered as (cos²x)^3 = [(1 + cos2x)/2]^3. Expanding this would give (1 + 3cos2x + 3cos²2x + cos³2x)/8. Then, each term can be expressed further. For example, cos²2x is (1 + cos4x)/2 and cos³2x can be expressed as (3cos2x + cos6x)/4. So substituting back, cos⁶x becomes [1 + 3cos2x + 3(1 + cos4x)/2 + (3cos2x + cos6x)/4]/8. Simplifying all terms, let's compute numerator:1 + 3cos2x + (3/2 + 3/2 cos4x) + (3/4 cos2x + 1/4 cos6x)Combine constants: 1 + 3/2 = 5/2Combine cos2x terms: 3cos2x + 3/4 cos2x = (12/4 + 3/4)cos2x = 15/4 cos2xcos4x term: 3/2 cos4xcos6x term: 1/4 cos6xSo numerator is 5/2 + 15/4 cos2x + 3/2 cos4x + 1/4 cos6xTherefore, cos⁶x = [5/2 + 15/4 cos2x + 3/2 cos4x + 1/4 cos6x]/8 = 5/16 + 15/32 cos2x + 3/16 cos4x + 1/32 cos6x.So the expression for cos⁶x includes cos6x, which has period π/3. However, the other terms have periods π, π/2, etc. The LCM of π, π/2, and π/3 is π. Wait, LCM of π and π/2 is π, and LCM of π and π/3 is π. So the period remains π? Wait, that's confusing. Because if there's a term with period π/3, then the overall period would need to be a multiple of π/3 as well. Hmm.Wait, but LCM of π and π/3 is π. Because π is a multiple of π/3 (3*(π/3) = π). So even if there is a term with period π/3, the overall function will have period π because π is a multiple of π/3. Wait, let's check.Suppose a function has components with periods π, π/2, and π/3. The periods are π, π/2, π/3. To find the LCM, convert them to numbers. Let’s take π as a unit. Then the periods are 1, 1/2, 1/3. The LCM of 1, 1/2, 1/3. LCM of fractions is LCM(numerators)/GCD(denominators). Numerators: 1,1,1. Denominators: 1,2,3. GCD of denominators is 1. So LCM is 1/1 = 1. So LCM of 1, 1/2, 1/3 is 1. Translating back, LCM is π. Therefore, even with the π/3 term, the overall period is π. So cos⁶x still has period π.So even for higher even powers, the period remains π? Wait, but if I take cos^10x, will it also have period π? Let's think. If cos²x has period π, then cos^n x where n is even, maybe also has period π? Because when you raise cosine to an even power, it becomes a combination of multiple cosine terms with frequencies that are multiples of 2x, so periods π, π/2, etc., but LCM remains π. So perhaps even for higher powers, the period is still π? Similarly, sin^n x would also have period π for even n. So then, if both cos¹⁰x and sin¹⁰x have period π, then the function y = cos¹⁰x + sin¹⁰x would also have period π. But wait, maybe there's a smaller period?Wait, let me test for specific cases. Let's take y = cos^2x + sin^2x. That's equal to 1, which has any period, but technically the function is constant, so its period is undefined or any real number. But in this problem, since we're dealing with cos^10x + sin^10x, which is not constant. Wait, cos^2x + sin^2x = 1, but for higher powers, they are not equal to 1. For example, cos^4x + sin^4x = (cos²x + sin²x)^2 - 2cos²x sin²x = 1 - 0.5 sin²2x, which is not constant. Similarly, cos^10x + sin^10x is definitely not constant, so it will have a certain period.But going back, for the function y = cos^10x + sin^10x, if both terms have period π, then the sum should have period π. However, maybe the sum has a smaller period? For example, if there's some symmetry in the function that causes it to repeat more frequently.Wait, let's check for π/2. Let me compute y(x + π/2) and see if it equals y(x). Let's recall that sin(x + π/2) = cosx and cos(x + π/2) = -sinx. So:y(x + π/2) = cos^10(x + π/2) + sin^10(x + π/2) = (-sinx)^10 + (cosx)^10 = sin^10x + cos^10x = y(x). So y(x + π/2) = y(x). Therefore, π/2 is a period. Wait, so π/2 is a period? Then the period is at most π/2. But is π/2 the minimal period?Wait, maybe even smaller? Let's check π/4. Let's compute y(x + π/4). Then:cos(x + π/4) = cosx cosπ/4 - sinx sinπ/4 = (cosx - sinx)/√2sin(x + π/4) = sinx cosπ/4 + cosx sinπ/4 = (sinx + cosx)/√2Therefore, cos(x + π/4) = (cosx - sinx)/√2, sin(x + π/4) = (cosx + sinx)/√2So cos^10(x + π/4) = [(cosx - sinx)/√2]^10sin^10(x + π/4) = [(cosx + sinx)/√2]^10Therefore, y(x + π/4) = [(cosx - sinx)^10 + (cosx + sinx)^10]/(√2)^10Simplify denominator: (√2)^10 = (2^(1/2))^10 = 2^(5) = 32Numerator: (cosx - sinx)^10 + (cosx + sinx)^10Hmm, this looks complicated. Let's see if these terms are equal to the original function. Probably not, because when we shift by π/4, the expression changes. Therefore, y(x + π/4) is not equal to y(x), so π/4 is not a period. Let me verify with x = 0:Compute y(0) = cos^10(0) + sin^10(0) = 1 + 0 = 1y(0 + π/4) = cos^10(π/4) + sin^10(π/4) = ( (√2/2)^10 ) + ( (√2/2)^10 ) = 2*( (2^(1/2)/2)^10 ) = 2*( (2^(5))/2^10 ) ) = 2*( 2^(-5) ) = 2^(-4) = 1/16But y(0) = 1 and y(π/4) = 1/16, which are different, so π/4 is not a period. So shifting by π/4 changes the value, so π/4 can't be a period.How about π/2? Let's check if π/2 is a period. Let's take x = 0 again:y(0) = 1 + 0 = 1y(0 + π/2) = cos^10(π/2) + sin^10(π/2) = 0 + 1 = 1, same as y(0)Check another point, like x = π/4:y(π/4) = ( (√2/2)^10 + (√2/2)^10 ) = 2*(2^(5)/2^10) ) = 2*(1/32) = 1/16y(π/4 + π/2) = y(3π/4) = cos^10(3π/4) + sin^10(3π/4) = ( (-√2/2)^10 + (√2/2)^10 ) = same as before, 1/16. So same value.Another test: x = π/6y(π/6) = cos^10(π/6) + sin^10(π/6) = ( (√3/2)^10 + (1/2)^10 )Compute (√3/2)^10: (3^(1/2)/2)^10 = 3^5 / 2^10 = 243 / 1024(1/2)^10 = 1/1024So y(π/6) = 243/1024 + 1/1024 = 244/1024 = 61/256 ≈ 0.238y(π/6 + π/2) = y(2π/3) = cos^10(2π/3) + sin^10(2π/3)cos(2π/3) = -1/2, sin(2π/3) = √3/2So cos^10(2π/3) = ( -1/2 )^10 = 1/1024sin^10(2π/3) = ( √3/2 )^10 = 243/1024Thus y(2π/3) = 1/1024 + 243/1024 = 244/1024 = 61/256 ≈ 0.238 same as y(π/6). So it seems that shifting by π/2 gives the same value. So π/2 is indeed a period.But is π/2 the minimal period? Let's check if there's a smaller period, say π/4. But we saw that shifting by π/4 changes the value. So π/4 is not a period. How about π/3? Let's check.Take x = 0. y(0) = 1. y(0 + π/3) = cos^10(π/3) + sin^10(π/3) = (1/2)^10 + ( (√3/2)^10 ) = 1/1024 + 243/1024 = 244/1024 = 61/256 ≠ 1. So y(0 + π/3) ≠ y(0), so π/3 is not a period.How about π/1? Wait, π is a period, but we already saw that π/2 is a period. So the minimal period must be a divisor of π/2. Wait, divisors of π/2 would be π/2, π/4, π/6, etc. But we saw that π/4 isn't a period. Let me check another candidate, π/2 divided by an integer. Let's test T = π/2.Suppose T is the minimal period. To confirm, we need to check if T/2 is not a period.Suppose we check T = π/2. Let's see if T/2 = π/4 is a period. As before, when we checked x=0, y(0 + π/4) = 1/16 ≠ 1 = y(0). So π/4 is not a period. How about T = π/3? We saw it's not a period. How about T = π/2? Let's see if there's a smaller period.Alternatively, perhaps the minimal period is π/2. Wait, but maybe there's a symmetry that causes the function to repeat more often.Alternatively, let's try to express the function in terms of multiple angles, which might reveal its period.Given y = cos¹⁰x + sin¹⁰x. Let's try to use power-reduction formulas or multiple-angle identities to simplify this expression.First, note that cos¹⁰x + sin¹⁰x can be written as (cos²x)^5 + (sin²x)^5. Hmm, perhaps we can factor this expression using the formula for a^5 + b^5.Recall that a^5 + b^5 = (a + b)(a^4 - a^3b + a^2b^2 - ab^3 + b^4). But since a = cos²x and b = sin²x, and a + b = cos²x + sin²x = 1, so this simplifies to 1*(a^4 - a^3b + a^2b^2 - ab^3 + b^4) = a^4 + b^4 - a^3b - ab^3 + a^2b^2.But this seems complicated. Maybe another approach.Alternatively, use the binomial theorem to expand cos¹⁰x and sin¹⁰x, but that might not be helpful. Alternatively, use the identity cos²x = 1 - sin²x, so maybe express everything in terms of sinx or cosx, but that might not help.Alternatively, use the exponential form: cosx = (e^(ix) + e^(-ix))/2 and sinx = (e^(ix) - e^(-ix))/(2i). Then, cos¹⁰x + sin¹⁰x can be written as [(e^(ix) + e^(-ix))/2]^10 + [(e^(ix) - e^(-ix))/(2i)]^10. But this might be messy, but perhaps reveals the period.Alternatively, note that both terms cos¹⁰x and sin¹⁰x can be expressed in terms of multiple-angle cosines. For example, cos²x = (1 + cos2x)/2, so cos¹⁰x = ( (1 + cos2x)/2 )^5. Similarly, sin²x = (1 - cos2x)/2, so sin¹⁰x = ( (1 - cos2x)/2 )^5. Then, y = [ (1 + cos2x)^5 + (1 - cos2x)^5 ] / 2^5.Let me compute that. Let's expand (1 + cos2x)^5 and (1 - cos2x)^5 using the binomial theorem:(1 + cos2x)^5 = 1 + 5cos2x + 10cos²2x + 10cos³2x + 5cos⁴2x + cos⁵2x(1 - cos2x)^5 = 1 - 5cos2x + 10cos²2x - 10cos³2x + 5cos⁴2x - cos⁵2xAdding these two expressions:[1 + 1] + [5cos2x - 5cos2x] + [10cos²2x + 10cos²2x] + [10cos³2x - 10cos³2x] + [5cos⁴2x + 5cos⁴2x] + [cos⁵2x - cos⁵2x]Which simplifies to:2 + 0 + 20cos²2x + 0 + 10cos⁴2x + 0 = 2 + 20cos²2x + 10cos⁴2xTherefore, y = [2 + 20cos²2x + 10cos⁴2x] / 32 = (2 + 20cos²2x + 10cos⁴2x)/32 = (1 + 10cos²2x + 5cos⁴2x)/16So now, y = [1 + 10cos²2x + 5cos⁴2x]/16.Now, let's try to simplify this expression further using power-reduction identities.First, cos²2x can be written as (1 + cos4x)/2, and cos⁴2x can be written as ( (1 + cos4x)/2 )² = (1 + 2cos4x + cos²4x)/4.Substituting these into y:y = [1 + 10*(1 + cos4x)/2 + 5*(1 + 2cos4x + cos²4x)/4 ] / 16Let me compute each term:First term: 1Second term: 10*(1 + cos4x)/2 = 5*(1 + cos4x) = 5 + 5cos4xThird term: 5*(1 + 2cos4x + cos²4x)/4 = (5/4)*(1 + 2cos4x + cos²4x) = 5/4 + (5/2)cos4x + (5/4)cos²4xNow, sum all terms:1 + 5 + 5cos4x + 5/4 + (5/2)cos4x + (5/4)cos²4xCombine constants: 1 + 5 + 5/4 = 6 + 5/4 = 29/4Combine cos4x terms: 5cos4x + (5/2)cos4x = (10/2 + 5/2)cos4x = 15/2 cos4xThen, the cos²4x term: (5/4)cos²4xSo, y = [29/4 + 15/2 cos4x + (5/4)cos²4x]/16Multiply numerator and denominator by 4 to eliminate denominators:= [29 + 30cos4x + 5cos²4x]/64Now, apply the power-reduction formula to cos²4x:cos²4x = (1 + cos8x)/2Substitute:y = [29 + 30cos4x + 5*(1 + cos8x)/2 ] / 64Calculate inside the brackets:29 + 30cos4x + 5/2 + (5/2)cos8xConvert 29 to 58/2 and 5/2 remains:58/2 + 5/2 = 63/2So:= [63/2 + 30cos4x + (5/2)cos8x]/64Factor out 1/2:= [ (63 + 60cos4x + 5cos8x)/2 ] /64= (63 + 60cos4x + 5cos8x)/128So finally, y = 63/128 + (60/128)cos4x + (5/128)cos8xSimplify fractions:60/128 = 15/32, 5/128 remains as is.So y = 63/128 + (15/32)cos4x + (5/128)cos8xTherefore, the function is a constant plus cosine terms with frequencies 4x and 8x. So the periods of these cosine terms are:For cos4x: period = 2π /4 = π/2For cos8x: period = 2π /8 = π/4The constant term doesn't affect the period. The overall period of the function is the least common multiple (LCM) of the periods of the individual cosine terms. So LCM of π/2 and π/4. Since π/4 is a divisor of π/2, the LCM is π/2. Therefore, the function y has period π/2.But wait, before we conclude, we need to check if there's a smaller period. For example, even though LCM of π/2 and π/4 is π/2, maybe when combined, the function could have a smaller period? Let's test if π/4 is a period. Let's take x = 0:y(0) = 63/128 + 15/32 *1 + 5/128 *1 = 63/128 + 60/128 + 5/128 = 128/128 = 1y(0 + π/4) = y(π/4) = 63/128 + 15/32 cos(4*(π/4)) + 5/128 cos(8*(π/4)) = 63/128 + 15/32 cosπ + 5/128 cos2πcosπ = -1, cos2π = 1. So:= 63/128 + 15/32*(-1) + 5/128*(1) = 63/128 - 15/32 + 5/128Convert 15/32 to 60/128:= 63/128 - 60/128 + 5/128 = (63 - 60 + 5)/128 = 8/128 = 1/16But y(0) = 1 and y(π/4) = 1/16, which are different. Therefore, π/4 is not a period. So the LCM of π/2 and π/4 is π/2, and since π/4 is not a period, the minimal period is π/2.Therefore, the smallest period of the function y = cos¹⁰x + sin¹⁰x is π/2.But wait, let me confirm once more. Let's check if T = π/2 is indeed a period. We saw before that shifting by π/2 gives the same function. Let's see using the simplified expression:y(x + T) = 63/128 + (15/32)cos4(x + T) + (5/128)cos8(x + T)If T = π/2, then:cos4(x + π/2) = cos(4x + 2π) = cos4xcos8(x + π/2) = cos(8x + 4π) = cos8xTherefore, y(x + π/2) = y(x). Hence, T = π/2 is indeed a period.Is there a smaller period? Suppose we take T = π/2 divided by some integer, say T = π/2 /2 = π/4. But we already saw that π/4 is not a period. How about T = π/3? Let's check.If T = π/3, then cos4(x + π/3) = cos(4x + 4π/3) and cos8(x + π/3) = cos(8x + 8π/3). Unless 4π/3 and 8π/3 are multiples of 2π, which they aren't. So cos(4x + 4π/3) ≠ cos4x, so y(x + π/3) ≠ y(x). Hence, π/3 is not a period.What about T = π/2 divided by another integer, say T = π/2 /3 = π/6. Let's check x = 0:y(0) = 1y(π/6) = 61/256 ≈ 0.238, which is different. So π/6 is not a period.Therefore, the smallest period is π/2.Final AnswerThe smallest period of the function is boxed{dfrac{pi}{2}}.</think>To find the smallest period of the function ( y = cos^{10} x + sin^{10} x ), we start by analyzing the periods of the individual terms (cos^{10} x) and (sin^{10} x). Both (cos x) and (sin x) have a fundamental period of (2pi). However, when raised to an even power, their periods can be reduced. For example, (cos^2 x) and (sin^2 x) have a period of (pi). This pattern continues for higher even powers, where the period remains (pi) due to the use of power-reduction formulas, which introduce higher frequency terms but result in the least common multiple (LCM) of the periods being (pi).However, we need to check if the sum (cos^{10} x + sin^{10} x) has a smaller period. We test (pi/2) by substituting (x + pi/2) into the function:- (cos(x + pi/2) = -sin x)- (sin(x + pi/2) = cos x)Thus, (cos^{10}(x + pi/2) = (-sin x)^{10} = sin^{10} x) and (sin^{10}(x + pi/2) = cos^{10} x). Therefore, (y(x + pi/2) = sin^{10} x + cos^{10} x = y(x)), confirming (pi/2) is a period.Next, we express the function using power-reduction identities:[cos^{10} x + sin^{10} x = frac{1}{16} left( 1 + 10 cos^2 2x + 5 cos^4 2x right)]Further simplification using (cos^2 2x = frac{1 + cos 4x}{2}) and (cos^4 2x = frac{3 + 4 cos 4x + cos 8x}{8}) results in:[y = frac{63}{128} + frac{15}{32} cos 4x + frac{5}{128} cos 8x]The periods of (cos 4x) and (cos 8x) are (pi/2) and (pi/4) respectively. The LCM of (pi/2) and (pi/4) is (pi/2), confirming the smallest period is (pi/2).Thus, the smallest period of the function ( y = cos^{10} x + sin^{10} x ) is (boxed{dfrac{pi}{2}}).,<think>Alright, so I need to find the smallest period of the function y = cos¹⁰x + sin¹⁰x. Hmm, okay. Let's start by recalling what the period of a function means. The period is the smallest positive number T such that f(x + T) = f(x) for all x. So, I need to find the smallest T where adding T to x doesn't change the value of the function.First, I know that both sine and cosine functions individually have a period of 2π. However, when you raise them to a power, their periods can change. For example, sin²x and cos²x have a period of π instead of 2π because squaring them makes the function repeat more often. Similarly, higher even powers might have even smaller periods? Wait, but maybe not. Let me think.For sin²x and cos²x, the period halves because sin²x = (1 - cos2x)/2 and cos²x = (1 + cos2x)/2. So, they have a period of π. But what happens when the exponent is higher? Let's see. For example, sin⁴x and cos⁴x. Let me try to express them using multiple-angle identities.Maybe I can use power-reduction formulas here. For cos¹⁰x and sin¹⁰x, it's going to get complicated, but perhaps I can find a way to express them in terms of multiple angles. That might help in determining the period.Alternatively, since both terms are raised to the 10th power, their individual periods might influence the overall period of the function. Let's consider cos¹⁰x and sin¹⁰x separately.First, let's analyze cos¹⁰x. The standard cosine function has period 2π. When you raise it to an even power, the graph becomes non-negative and more "peaked," but how does that affect the period? Well, similar to squaring, the period could potentially halve because the function becomes symmetric around the peaks. For example, cos²x has period π. Similarly, cos⁴x would also have period π, right? Because whether it's squared or to the fourth power, the even power makes the function symmetric over half the original period. Wait, let's verify this.Take cos²x: using the identity cos²x = (1 + cos2x)/2. Here, the function inside the cosine is 2x, so the period is π. Similarly, if I take cos⁴x, I can write it as (cos²x)² = [(1 + cos2x)/2]^2. Expanding this would involve terms with cos2x and cos²2x. The cos²2x term can be written as (1 + cos4x)/2, so cos⁴x would involve terms with cos4x, which has a period of π/2. However, the entire expression for cos⁴x would be a combination of constants and cos4x. So, the period would be π/2? Wait, but when you have multiple cosine terms with different frequencies, the overall period is the least common multiple (LCM) of their individual periods.For cos²x, we have cos2x, which has period π. But when we go to cos⁴x, we get cos4x, which has period π/2. However, the entire expression for cos⁴x is (3/8) + (1/2)cos2x + (1/8)cos4x. So, there's a cos2x term with period π and a cos4x term with period π/2. The LCM of π and π/2 is π, so the overall period of cos⁴x would be π. Wait, is that correct? Because the presence of both cos2x and cos4x would require the period to be a multiple of both π and π/2. The LCM of π and π/2 is π. So, even though there's a term with period π/2, the overall period is π. But is that true?Wait, let's test this. Suppose f(x) = cos4x. Then f(x + π/2) = cos4(x + π/2) = cos(4x + 2π) = cos4x = f(x). So, π/2 is a period. But if I have a function that's a combination of cos2x and cos4x, would π/2 still be a period? Let's check:Let f(x) = cos2x + cos4x.Then f(x + π/2) = cos2(x + π/2) + cos4(x + π/2) = cos(2x + π) + cos(4x + 2π) = -cos2x + cos4x.Which is not equal to f(x). So, π/2 is not a period here. What about π?f(x + π) = cos2(x + π) + cos4(x + π) = cos(2x + 2π) + cos(4x + 4π) = cos2x + cos4x = f(x). So, π is a period. Therefore, even if you have higher frequency terms, the overall period can be determined by the LCM of the individual periods of each term.Therefore, for cos⁴x, which can be written as (3/8) + (1/2)cos2x + (1/8)cos4x, the periods of the trigonometric terms are π (for cos2x) and π/2 (for cos4x). The LCM of π and π/2 is π, so the period is π.Similarly, for cos¹⁰x, if we expand it using power-reduction formulas, we might get a combination of cosines with different multiples of x. Each term would have a period of 2π divided by the coefficient of x. The key is to find the LCM of all these periods to determine the overall period of cos¹⁰x.But this seems complicated. Maybe there's a pattern here. For even powers of cosine, like cos²ⁿx, is the period π? Wait, for cos²x it's π, for cos⁴x, we saw it's π as well. Let's check cos⁶x. If we expand cos⁶x, using power-reduction formulas, we would get terms like cos2x, cos4x, cos6x. The periods would be π, π/2, π/3. The LCM of π, π/2, π/3 is π. So, the period is π. Wait, is that true?Wait, let's take cos⁶x. Suppose we write it as (cos²x)^3 = [(1 + cos2x)/2]^3. Expanding this, we get (1/8)(1 + 3cos2x + 3cos²2x + cos³2x). Then, cos²2x can be written as (1 + cos4x)/2 and cos³2x can be written using the identity for cos³θ = (3cosθ + cos3θ)/4. So, cos³2x = (3cos2x + cos6x)/4. Putting it all together, cos⁶x would involve constants, cos2x, cos4x, and cos6x. The periods here are π (for cos2x), π/2 (for cos4x), and π/3 (for cos6x). The LCM of π, π/2, and π/3 is π. Therefore, the period would be π.Wait, but cos6x has period π/3. If we take LCM of π, π/2, and π/3, we need to find the smallest number T such that T is a multiple of π, π/2, and π/3. The LCM of these is π. Because π is a multiple of π/2 (since π = 2*(π/2)) and a multiple of π/3 (since π = 3*(π/3))? Wait, no. π is not an integer multiple of π/2 or π/3. Wait, LCM for numbers a and b is the smallest positive number that is a multiple of both a and b. But here, we're dealing with periods, which are real numbers, not integers. So, maybe the concept is slightly different.Alternatively, if we think in terms of functions, the period of a sum of functions with different periods is the LCM of their individual periods. So, if we have cos6x with period π/3, then to satisfy f(x + T) = f(x), T must be a multiple of π/3. Similarly, for cos4x, T must be a multiple of π/2, and for cos2x, a multiple of π. So, the LCM of π, π/2, and π/3 is the smallest T such that T = kπ, T = mπ/2, T = nπ/3 for integers k, m, n. So, solving for T:From T = kπ and T = mπ/2, we get kπ = mπ/2 => 2k = m. So, m must be even.From T = kπ and T = nπ/3, we get kπ = nπ/3 => 3k = n.Therefore, T must be a multiple of π, and k must be chosen such that 2k and 3k are integers. The smallest k for which 2k and 3k are integers is k=1, but then 2k=2 and 3k=3, which are integers. Wait, but T = kπ. If k=1, then T=π. But is π a multiple of π/2 and π/3?No, π is not an integer multiple of π/2 or π/3. Wait, maybe my approach here is flawed. Let me think again.Suppose we have a function f(x) = cos2x + cos4x + cos6x. Then, to find the period of f(x), we need the smallest T > 0 such that f(x + T) = f(x) for all x. That means T must be a period for each term. So, T must be a multiple of π (for cos2x), π/2 (for cos4x), and π/3 (for cos6x). Therefore, T must be the least common multiple of π, π/2, π/3.But how do we compute LCM for real numbers? It's not the same as integers. Alternatively, we can express each period in terms of a base unit. Let’s let π be the base unit. Then:- The period of cos2x is π, which is 1π.- The period of cos4x is π/2, which is (1/2)π.- The period of cos6x is π/3, which is (1/3)π.So, we need to find the smallest T such that T is a multiple of 1π, (1/2)π, and (1/3)π. In terms of the base unit π, we can think of T as kπ, where k is a positive real number. Then, kπ must be a multiple of 1π, (1/2)π, and (1/3)π. Which means k must be a multiple of 1, 1/2, and 1/3. The smallest such k is the least common multiple of 1, 1/2, and 1/3. But LCM for fractions is found by taking LCM of the numerators divided by GCD of the denominators. Wait, maybe not. Let me recall.Alternatively, the LCM of 1, 1/2, and 1/3 is the smallest positive number that is an integer multiple of 1, 1/2, and 1/3. To find this, convert them to fractions: 1 = 1/1, 1/2, 1/3. The LCM of fractions is LCM(numerators)/GCD(denominators). The numerators are 1,1,1, so LCM(1,1,1)=1. The denominators are 1,2,3, so GCD(1,2,3)=1. Therefore, LCM is 1/1 = 1. So, T = 1π. Therefore, the LCM is π. Therefore, the period is π.But wait, earlier when we tested f(x) = cos2x + cos4x, adding π/2 didn't work, but π did. So, even though there are higher frequency terms, the period is dictated by the lowest frequency component? Wait, in the case of cos2x + cos4x, the frequencies are 2 and 4. The periods are π and π/2. So, the LCM of π and π/2 is π. So, the period is π. Similarly, adding cos6x with period π/3, LCM of π, π/2, π/3 is π.So, perhaps, for any even power of cosine or sine, when expressed in terms of multiple angles, the highest period term is π, so the LCM is π. Therefore, maybe cos¹⁰x also has a period of π. Similarly, sin¹⁰x would also have a period of π. Therefore, their sum, y = cos¹⁰x + sin¹⁰x, would have a period of π. But wait, I need to verify this.Wait, but maybe not. Let's think. Suppose cos¹⁰x has period π and sin¹⁰x also has period π. Then their sum would have period π. But perhaps there is a smaller period. For example, if both functions have period π, but they might also have a common smaller period. For instance, if each has period π, but they are also even or odd functions, maybe the combination has a smaller period.Wait, for example, consider cos²x + sin²x. We know that's equal to 1, which has any period, but the functions individually have period π. But their sum is a constant function, which technically has all real numbers as periods, but the fundamental period is undefined. But in this problem, y = cos¹⁰x + sin¹⁰x isn't a constant. Wait, but let's check with lower exponents. For example, cos⁴x + sin⁴x. What's its period?Let me compute cos⁴x + sin⁴x. Using identities: cos⁴x + sin⁴x = (cos²x + sin²x)^2 - 2cos²x sin²x = 1 - (1/2)sin²2x. Since sin²2x has period π/2, so 1 - (1/2)sin²2x would have period π/2. Therefore, the period is π/2. So, in that case, even though cos⁴x and sin⁴x each have period π, their sum has period π/2. Interesting.Therefore, maybe when we combine cos¹⁰x and sin¹⁰x, their sum could have a smaller period than π. So, perhaps the individual periods are π, but the sum has a smaller period. So, I need to check if the sum is π/2 periodic or something else.Therefore, perhaps the same approach applies here. Let's try to express y = cos¹⁰x + sin¹⁰x in terms of multiple angles to see what the period is.First, let's recall that cos¹⁰x and sin¹⁰x can be expressed using the binomial theorem and power-reduction formulas. However, expanding these terms might get very cumbersome. Alternatively, we can use the identity that for even exponents, cos²ⁿx and sin²ⁿx can be written as a sum of cosines of multiple angles.Alternatively, we can use the formula for cos^{2n}x and sin^{2n}x. For example, the general formula for cos^{2n}x is a combination of cos2kx terms, where k ranges from 0 to n. Similarly for sin^{2n}x, which can be expressed as a combination involving cos2kx terms with alternating signs.Alternatively, since both cos¹⁰x and sin¹⁰x are even functions, their sum will also be even. Therefore, the function y is even, meaning it's symmetric around the y-axis. However, being even doesn't necessarily affect the period, but it might help in simplifying the expression.Alternatively, let's note that cos¹⁰x + sin¹⁰x can be rewritten using the identity sin²x = 1 - cos²x. So, substituting sin²x = 1 - cos²x into sin¹⁰x, we can express everything in terms of cos²x. Let's try that.Let’s set t = cos²x. Then sin²x = 1 - t. Therefore, cos¹⁰x = (cos²x)^5 = t^5 and sin¹⁰x = (sin²x)^5 = (1 - t)^5. So, y = t^5 + (1 - t)^5. Let's expand this:t^5 + (1 - t)^5 = t^5 + 1 - 5t + 10t² - 10t³ + 5t⁴ - t⁵ = 1 - 5t + 10t² - 10t³ + 5t⁴.The t^5 and -t^5 terms cancel out. So, y = 1 - 5t + 10t² - 10t³ + 5t⁴, where t = cos²x.But since t = cos²x, we can substitute back:y = 1 - 5cos²x + 10cos⁴x - 10cos⁶x + 5cos⁸x.Hmm, so now we have y expressed in terms of cos²x, cos⁴x, cos⁶x, and cos⁸x. Each of these can be expressed using multiple-angle formulas. Let's try to do that step by step.First, let's recall that cos²x = (1 + cos2x)/2.Then, cos⁴x = (cos²x)^2 = [(1 + cos2x)/2]^2 = (1 + 2cos2x + cos²2x)/4 = (1/4) + (1/2)cos2x + (1/4)cos²2x.Similarly, cos⁶x = (cos²x)^3 = [(1 + cos2x)/2]^3. Expanding this:= (1/8)(1 + 3cos2x + 3cos²2x + cos³2x).Again, cos²2x can be written as (1 + cos4x)/2, and cos³2x can be written as (3cos2x + cos6x)/4.Similarly, cos⁸x = (cos²x)^4 = [(1 + cos2x)/2]^4. Expanding this would involve more terms, but each term can be expressed in terms of multiple angles.But this seems like a lot of work, but maybe manageable. Let's proceed step by step.First, let's substitute cos²x = (1 + cos2x)/2 into the expression for y.y = 1 - 5cos²x + 10cos⁴x - 10cos⁶x + 5cos⁸x.Let me compute each term:1. -5cos²x = -5*(1 + cos2x)/2 = -5/2 - (5/2)cos2x.2. 10cos⁴x: As above, cos⁴x = (1/4) + (1/2)cos2x + (1/4)cos²2x. Therefore,10cos⁴x = 10*(1/4 + 1/2cos2x + 1/4cos²2x) = (10/4) + 5cos2x + (10/4)cos²2x = 2.5 + 5cos2x + 2.5cos²2x.3. -10cos⁶x: Let's expand cos⁶x first.cos⁶x = (cos²x)^3 = [(1 + cos2x)/2]^3 = (1/8)(1 + 3cos2x + 3cos²2x + cos³2x).Then, cos³2x can be written using the identity cos³θ = (3cosθ + cos3θ)/4. Wait, but θ here is 2x, so cos³2x = (3cos2x + cos6x)/4.Therefore, cos⁶x = (1/8)[1 + 3cos2x + 3cos²2x + (3cos2x + cos6x)/4] = (1/8)[1 + 3cos2x + 3cos²2x + (3/4)cos2x + (1/4)cos6x].Combine like terms:= (1/8)[1 + (3 + 3/4)cos2x + 3cos²2x + (1/4)cos6x] = (1/8)[1 + (15/4)cos2x + 3cos²2x + (1/4)cos6x].Multiply through:= 1/8 + (15/32)cos2x + (3/8)cos²2x + (1/32)cos6x.Therefore, -10cos⁶x = -10*(1/8 + 15/32cos2x + 3/8cos²2x + 1/32cos6x) = -10/8 - (150/32)cos2x - 30/8cos²2x - 10/32cos6x = -5/4 - (75/16)cos2x - 15/4cos²2x - 5/16cos6x.4. 5cos⁸x: This will be even more complicated, but let's try.cos⁸x = (cos²x)^4 = [(1 + cos2x)/2]^4. Expanding this:= (1/16)(1 + 4cos2x + 6cos²2x + 4cos³2x + cos⁴2x).Now, cos³2x can be written as (3cos2x + cos6x)/4, as before.cos⁴2x = (cos²2x)^2 = [(1 + cos4x)/2]^2 = (1 + 2cos4x + cos²4x)/4.So, substituting back into cos⁸x:= (1/16)[1 + 4cos2x + 6cos²2x + 4*(3cos2x + cos6x)/4 + (1 + 2cos4x + cos²4x)/4]Simplify term by term:- The term 4cos³2x becomes 4*(3cos2x + cos6x)/4 = 3cos2x + cos6x.- The term cos⁴2x becomes (1 + 2cos4x + cos²4x)/4.Therefore,cos⁸x = (1/16)[1 + 4cos2x + 6cos²2x + 3cos2x + cos6x + (1/4) + (1/2)cos4x + (1/4)cos²4x]Combine like terms:First, constants: 1 + 1/4 = 5/4.cos2x terms: 4cos2x + 3cos2x = 7cos2x.cos²2x terms: 6cos²2x.cos6x term: cos6x.cos4x term: (1/2)cos4x.cos²4x term: (1/4)cos²4x.So,cos⁸x = (1/16)[5/4 + 7cos2x + 6cos²2x + cos6x + (1/2)cos4x + (1/4)cos²4x]Multiply through:= 5/(16*4) + 7/(16)cos2x + 6/(16)cos²2x + 1/(16)cos6x + (1)/(32)cos4x + (1)/(64)cos²4xSimplify fractions:= 5/64 + 7/16cos2x + 3/8cos²2x + 1/16cos6x + 1/32cos4x + 1/64cos²4x.Therefore, 5cos⁸x = 5*(5/64 + 7/16cos2x + 3/8cos²2x + 1/16cos6x + 1/32cos4x + 1/64cos²4x) = 25/64 + 35/16cos2x + 15/8cos²2x + 5/16cos6x + 5/32cos4x + 5/64cos²4x.Now, let's combine all the terms for y:1. The constant terms:1 (from the original y) + (-5/2) + 2.5 + (-5/4) + 25/64.Wait, let's convert all to fractions with denominator 64 for ease:1 = 64/64-5/2 = -160/642.5 = 160/64-5/4 = -80/6425/64 = 25/64Adding these up:64/64 - 160/64 + 160/64 - 80/64 + 25/64 = (64 - 160 + 160 - 80 + 25)/64 = (64 - 80 + 25)/64 = ( -16 + 25)/64 = 9/64.2. The cos2x terms:-5/2cos2x (from -5cos²x) + 5cos2x (from 10cos⁴x) -75/16cos2x (from -10cos⁶x) + 35/16cos2x (from 5cos⁸x).Convert all coefficients to sixteenths:-5/2 = -40/165 = 80/16-75/16 = -75/1635/16 = 35/16Adding them up:-40/16 + 80/16 -75/16 +35/16 = (-40 + 80 -75 +35)/16 = (80 -40) + (-75 +35)/16 = 40 -40/16 = 0/16 = 0.So, the cos2x terms cancel out.3. The cos²2x terms:2.5cos²2x (from 10cos⁴x) -15/4cos²2x (from -10cos⁶x) +15/8cos²2x (from 5cos⁸x).Convert to eighths:2.5 = 20/8-15/4 = -30/815/8 = 15/8Adding them up:20/8 -30/8 +15/8 = (20 -30 +15)/8 = 5/8.So, 5/8cos²2x.But cos²2x can be written as (1 + cos4x)/2. Therefore:5/8cos²2x = 5/8*(1 + cos4x)/2 = 5/16 + 5/16cos4x.4. The cos4x terms:From 5cos⁸x: 5/32cos4x.From cos²2x term above: 5/16cos4x.So total cos4x terms:5/32cos4x + 5/16cos4x = 5/32 + 10/32 = 15/32cos4x.5. The cos6x terms:-5/16cos6x (from -10cos⁶x) +5/16cos6x (from 5cos⁸x).These cancel out: -5/16 +5/16 = 0.6. The cos²4x term:From 5cos⁸x: 5/64cos²4x.cos²4x = (1 + cos8x)/2, so 5/64cos²4x = 5/128 +5/128cos8x.7. Any remaining terms?From the original expansion, the constants, cos2x, cos4x, cos6x, and cos8x. Let's make sure we didn't miss anything.So putting it all together:y = 9/64 (constant) + 0 (cos2x) + 5/16 (from cos²2x) + 15/32cos4x + 0 (cos6x) + 5/128 +5/128cos8x.Wait, no. Wait:Wait, the cos²2x term became 5/16 +5/16cos4x. Then the cos4x terms are 15/32cos4x from the original cos4x term plus the 5/16cos4x from the cos²2x. Wait, hold on. Let me check again.Wait, the cos²2x term was converted to 5/16 +5/16cos4x, and then there was a separate cos4x term from 5cos⁸x, which was 5/32cos4x. So total cos4x terms:5/16cos4x +5/32cos4x = (10/32 +5/32)cos4x =15/32cos4x.Additionally, the cos²4x term from 5cos⁸x became 5/128 +5/128cos8x.So, compiling all terms:Constants: 9/64 +5/16 +5/128.cos4x: 15/32cos4x.cos8x:5/128cos8x.Let's compute the constants:Convert all to 128 denominator:9/64 = 18/1285/16 = 40/1285/128 =5/128Total constants: 18 +40 +5 =63/128.Therefore, y = 63/128 +15/32cos4x +5/128cos8x.Simplify the coefficients:15/32 = 60/128, so:y = 63/128 +60/128cos4x +5/128cos8x.Factor out 1/128:y = (63 +60cos4x +5cos8x)/128.Therefore, the function simplifies to:y = [63 +60cos4x +5cos8x]/128.So, now we have y expressed as a sum of cos4x and cos8x terms plus a constant. Now, let's analyze the periods of each term:- The constant term 63/128 has no period (or all periods).- The term cos4x has period 2π/4 = π/2.- The term cos8x has period 2π/8 = π/4.Therefore, the function y is a combination of terms with periods π/2 and π/4. The overall period of y is the least common multiple (LCM) of π/2 and π/4.To find the LCM of π/2 and π/4, note that π/4 is a divisor of π/2 because π/2 = 2*(π/4). Therefore, the LCM of π/2 and π/4 is π/2. Hence, the period of y is π/2.But wait, hold on. Let me check. If we have cos4x with period π/2 and cos8x with period π/4, then the LCM of π/2 and π/4 is π/2 because π/2 is a multiple of π/4 (specifically, π/2 = 2*(π/4)). Therefore, the LCM is π/2. Therefore, the function y repeats every π/2.But let's verify this by checking if y(x + π/2) = y(x).Compute y(x + π/2):cos4(x + π/2) = cos(4x + 2π) = cos4x.Similarly, cos8(x + π/2) = cos(8x + 4π) = cos8x.Therefore, both cos4x and cos8x are periodic with period π/2, so y(x + π/2) = y(x). Therefore, π/2 is indeed a period.But is π/2 the minimal period? Let's check if a smaller period, say π/4, works.Compute y(x + π/4):cos4(x + π/4) = cos(4x + π) = -cos4x.cos8(x + π/4) = cos(8x + 2π) = cos8x.Therefore, y(x + π/4) = [63 +60*(-cos4x) +5cos8x]/128 = [63 -60cos4x +5cos8x]/128.Compare this to y(x) = [63 +60cos4x +5cos8x]/128. These are not equal unless cos4x = 0 for all x, which it isn't. Therefore, π/4 is not a period.Similarly, check T = π/3. Wait, but the periods here are π/2 and π/4, so LCM is π/2. There's no point in checking π/3 since it's not a multiple of the individual periods. Similarly, trying T = π/2/2 = π/4, which we saw doesn't work.Therefore, the minimal period is π/2.Wait a second, but earlier, when we considered cos⁴x + sin⁴x, which simplified to 1 - (1/2)sin²2x, which has period π/2. So, similar logic here. After simplifying y = cos¹⁰x + sin¹⁰x, we get a function with period π/2. Therefore, the minimal period is π/2.But let me confirm once more. Let's take specific values. Let's compute y(0) and y(π/2):y(0) = cos¹⁰0 + sin¹⁰0 = 1 + 0 = 1.y(π/2) = cos¹⁰(π/2) + sin¹⁰(π/2) = 0 + 1 = 1.Now, check y(π/4):y(π/4) = cos¹⁰(π/4) + sin¹⁰(π/4). Since cos(π/4) = sin(π/4) = √2/2, so each term is (√2/2)^10 = (2^(1/2)/2)^10 = (2^(-1/2))^10 = 2^(-5) = 1/32. Therefore, y(π/4) = 1/32 +1/32 = 1/16.Now, if the period were π/2, then y(x + π/2) = y(x). Let's check a point shifted by π/2. For example, y(π/2 + 0) = y(0) =1, which matches. Similarly, y(π/2 + π/4) = y(3π/4). Compute y(3π/4):cos(3π/4) = -√2/2, sin(3π/4)=√2/2. So cos¹⁰(3π/4) = (√2/2)^10 =1/32, sin¹⁰(3π/4)=1/32. So y(3π/4)=1/32 +1/32=1/16, same as y(π/4). So y(3π/4) = y(π/4). Similarly, y(π/2 + π/2) = y(π) = cos¹⁰π + sin¹⁰π = (-1)^10 +0=1, which is equal to y(0). So that's consistent.But to check if π/2 is indeed the minimal period, let's see if there's a smaller period. Let's suppose T is the minimal period, T < π/2. Let's assume T = π/4. Let's check y(0) =1, y(π/4)=1/16, y(π/2)=1. If T=π/4, then y(0 + π/4) should equal y(0). But y(π/4)=1/16 ≠ y(0)=1. Therefore, π/4 is not a period.Similarly, check T=π/3. y(π/3) = cos¹⁰(π/3) + sin¹⁰(π/3). cos(π/3)=1/2, sin(π/3)=√3/2. So cos¹⁰(π/3) = (1/2)^10 =1/1024, sin¹⁰(π/3)= (√3/2)^10 = (3/4)^5 = 243/1024. So y(π/3)= (1 +243)/1024 =244/1024=61/256≈0.238.If T=π/3 were a period, then y(π/3 + π/3)=y(2π/3) should equal y(π/3). Compute y(2π/3): cos(2π/3)= -1/2, sin(2π/3)=√3/2. So cos¹⁰(2π/3)= ( -1/2)^10=1/1024, sin¹⁰(2π/3)= (√3/2)^10=243/1024. So y(2π/3)=1/1024 +243/1024=244/1024=61/256≈0.238, which is equal to y(π/3). Hmm, interesting. But does T=π/3 work for all x? Let's check another point. For example, x=0. y(0)=1. If T=π/3 is a period, then y(0 + π/3)=y(π/3)=61/256, which should equal y(0)=1. But 61/256 ≠1. Therefore, T=π/3 is not a period. So that test fails.Therefore, the minimal period is indeed π/2.Alternatively, going back to the expression y = [63 +60cos4x +5cos8x]/128. The function cos4x has period π/2, cos8x has period π/4. The presence of cos8x with period π/4 might suggest that the period is π/4. However, when we tested π/4, it didn't work because shifting by π/4 introduced a negative sign in the cos4x term. However, in the expression [63 +60cos4x +5cos8x]/128, when we shift x by π/4, cos4(x + π/4) = cos(4x + π) = -cos4x, and cos8(x + π/4)=cos(8x + 2π)=cos8x. Therefore, y(x + π/4) = [63 -60cos4x +5cos8x]/128, which is not equal to y(x). Therefore, π/4 is not a period.However, when we shift by π/2:cos4(x + π/2) = cos(4x + 2π) = cos4x,cos8(x + π/2) = cos(8x +4π) = cos8x.Therefore, y(x + π/2) = y(x). Therefore, π/2 is indeed a period.Is there a smaller period than π/2? Let's check T=π/4: already checked, doesn't work. T=π/3: already checked, doesn't work. T=π/6: let's check.Compute y(π/6):cos(π/6)=√3/2, sin(π/6)=1/2. So cos¹⁰(π/6)= (√3/2)^10 = (3/4)^5 =243/1024, sin¹⁰(π/6)=(1/2)^10=1/1024. So y(π/6)=244/1024=61/256≈0.238.If T=π/6 were a period, then y(π/6 +π/6)=y(π/3)=61/256, which is equal. But y(0 +π/6)=y(π/6)=61/256≠y(0)=1. Therefore, not a period.Alternatively, is there a period between π/4 and π/2? For example, π/3, which we checked. But since periods have to divide evenly into the function's period, and π/2 is LCM of π/2 and π/4, which is π/2. Therefore, the minimal period is π/2.Therefore, the smallest period of the function y = cos¹⁰x + sin¹⁰x is π/2.Final AnswerThe smallest period of the function is boxed{dfrac{pi}{2}}.</think>To find the smallest period of the function ( y = cos^{10} x + sin^{10} x ), we start by recognizing that both (cos^{10} x) and (sin^{10} x) are even-powered trigonometric functions. We use power-reduction formulas and multiple-angle identities to simplify the expression.First, we express (cos^{10} x) and (sin^{10} x) in terms of multiple angles. By using the identity (cos^2 x = frac{1 + cos 2x}{2}) and expanding (cos^{10} x) and (sin^{10} x) through binomial theorem and power-reduction formulas, we eventually simplify the function ( y ).After simplification, we find:[y = frac{63 + 60 cos 4x + 5 cos 8x}{128}]Next, we determine the periods of the individual cosine terms:- The term (cos 4x) has a period of (frac{pi}{2}).- The term (cos 8x) has a period of (frac{pi}{4}).The least common multiple (LCM) of (frac{pi}{2}) and (frac{pi}{4}) is (frac{pi}{2}). This is because (frac{pi}{2}) is a multiple of (frac{pi}{4}) (specifically, (frac{pi}{2} = 2 times frac{pi}{4})).We verify this by checking that ( y(x + frac{pi}{2}) = y(x) ) for all ( x ), confirming that (frac{pi}{2}) is indeed a period. Testing smaller periods like (frac{pi}{4}) shows they do not satisfy the periodicity condition.Thus, the smallest period of the function ( y = cos^{10} x + sin^{10} x ) is (boxed{dfrac{pi}{2}}).]

🔑:Answer: frac{pi}{2}. Solution. First, we check that frac{pi}{2} is a period of the function. Indeed,yleft(x+frac{pi}{2}right)=left(cos ^{2}left(x+frac{pi}{2}right)right)^{5}+left(sin ^{2}left(x+frac{pi}{2}right)right)^{5}=sin ^{10} x+cos ^{10} xTo prove that this is the smallest period, we make the substitution u=sin ^{2} x. Then y=(1-u)^{5}+u^{5}. Considering the function y(u) on the interval [0,1] (this corresponds to the change of x over the interval from 0 to frac{pi}{2}), we get that y(u) decreases on the interval left[0, frac{1}{2}right] and increases on the interval left[frac{1}{2}, 1right] (this can be verified using the derivative: it is equal to left.-5(1-u)^{4}+5 u^{4}right). From this, it follows that the function y(x)=cos ^{10} x+sin ^{10} x cannot have a period smaller than frac{pi}{2}.

🧠:[<think>Alright, so I need to figure out the last digits of these three expressions. Let me tackle them one by one.Starting with the first one: 135^x + 31^y + 56^(x+y), where x and y are natural numbers. Hmm, okay. The question is about the last digit of each of these terms when added together. Since we're dealing with exponents, the key here is probably to look at the patterns of the last digits when these bases are raised to different powers.Let's break it down term by term.First term: 135^x. The base here is 135, and the last digit of 135 is 5. I remember that any number ending with 5 raised to any positive integer power will always end with 5. For example, 5^1=5, 5^2=25, 5^3=125, etc. So 135^x will end with 5 regardless of x.Second term: 31^y. The last digit of 31 is 1. Similarly, any number ending in 1 raised to any power will end with 1. Like 1^1=1, 1^2=1, and so on. So 31^y will end with 1 no matter the value of y.Third term: 56^(x+y). The last digit of 56 is 6. Now, numbers ending with 6 also have a consistent pattern. 6^1=6, 6^2=36, 6^3=216, so it always ends with 6. Therefore, 56^(x+y) will end with 6 irrespective of the exponent.So adding them together: 5 (from 135^x) + 1 (from 31^y) + 6 (from 56^(x+y)) = 12. But we only care about the last digit of the sum, which is 2. Wait, 5 + 1 is 6, plus 6 is 12. So the last digit is 2. Therefore, regardless of the values of x and y (as long as they are natural numbers), the sum will always end with 2. So the answer for part 1) should be 2.Moving on to the second problem: 142 + 142^2 + 142^3 + ... + 142^20. Again, we need the last digit of this entire sum. Since each term is a power of 142, we can look at the last digit of each power and sum those up, then take the last digit of the total sum.The last digit of 142 is 2. So let's consider the last digit of 2^n for n from 1 to 20. The pattern for powers of 2 cycles every 4: 2^1=2, 2^2=4, 2^3=8, 2^4=6, then 2^5=2, and so on. So the cycle is 2, 4, 8, 6, repeating every 4 exponents.Therefore, for each term 142^k, the last digit is the same as 2^k. So we can model each term's last digit as per the cycle. Let's figure out how many full cycles there are in 20 terms and the remainder.Since the cycle length is 4, 20 divided by 4 is 5 with no remainder. So there are 5 full cycles. Each cycle has the digits 2, 4, 8, 6. The sum of one cycle is 2 + 4 + 8 + 6 = 20. Therefore, 5 cycles would sum up to 5 * 20 = 100. The last digit of 100 is 0. However, wait, we need to confirm if this approach is correct.Wait, but actually, each term from k=1 to 20 corresponds to exponents 1 through 20 of 2. So each term's last digit is as follows:Term 1: 2^1 = 2Term 2: 2^2 = 4Term 3: 2^3 = 8Term 4: 2^4 = 6Term 5: 2^5 = 2Term 6: 2^6 = 4... and so on up to Term 20: 2^20 = 6 (since 20 divided by 4 is 5, so 2^4 ends with 6, 2^8 ends with 6, ..., 2^20 ends with 6).So each group of 4 terms (from k=1 to 4, 5 to 8, etc.) sums to 20. Since there are 5 groups (because 20 terms divided by 4 is 5), the total sum of all last digits is 5*20=100. The last digit of 100 is 0. Therefore, the entire sum 142 + 142^2 + ... + 142^20 ends with 0. So the answer for part 2) is 0.Wait, hold on a second. Let me verify this. If each cycle's sum is 20, which ends with 0, then 5 cycles would be 100, ending with 0. But 142^1 is 142, which is 2, 142^2 is 4, etc. So adding up the units digits as 2,4,8,6 repeating five times gives sum 100. So yes, the last digit is 0. Seems right.Third problem: 34^x + 34^(x+1) + 34^(2x), where x is a natural number. Need the last digit of this sum. Again, let's analyze each term's last digit.First term: 34^x. The last digit of 34 is 4. The pattern for powers of 4 is cyclical: 4^1=4, 4^2=16 (ends with 6), 4^3=64 (ends with 4), 4^4=256 (ends with 6), so the cycle is 4,6 repeating every 2 exponents. So if x is odd, 34^x ends with 4; if x is even, ends with 6.Second term: 34^(x+1). Similarly, the exponent here is x+1. So if x is odd, x+1 is even, so the last digit is 6; if x is even, x+1 is odd, last digit is 4.Third term: 34^(2x). The exponent here is 2x. Let's see, 2x is always even, regardless of x. So 34^(2x) will end with 6, because as per the cycle, even exponents of 4 end with 6. So regardless of x, 34^(2x) ends with 6.So let's consider the cases where x is odd and x is even.Case 1: x is odd.First term: 4Second term: x+1 is even, so 6Third term: 6Sum: 4 + 6 + 6 = 16. Last digit is 6.Case 2: x is even.First term: 6Second term: x+1 is odd, so 4Third term: 6Sum: 6 + 4 + 6 = 16. Last digit is 6.Wait, in both cases, the sum is 16, so the last digit is 6. Therefore, regardless of whether x is odd or even, the last digit is 6. So the answer for part 3) is 6.Let me check once again to make sure.If x is odd:34^x ends with 434^(x+1) ends with 6 (since exponent becomes even)34^(2x) is 34^(even) ends with 64 + 6 + 6 = 16 → 6If x is even:34^x ends with 634^(x+1) ends with 4 (exponent odd)34^(2x) ends with 66 + 4 + 6 = 16 → 6Yep, same result. So regardless of x, the last digit is 6. Therefore, the answers are 2, 0, 6.Final Answer1) boxed{2}2) boxed{0}3) boxed{6}</think>To determine the digits at the end of the decimal representations of the given numbers, we analyze each term's last digit and sum them up.1) For (135^x + 31^y + 56^{x+y}): - The last digit of (135^x) is 5, as any number ending in 5 raised to any power ends in 5. - The last digit of (31^y) is 1, as any number ending in 1 raised to any power ends in 1. - The last digit of (56^{x+y}) is 6, as any number ending in 6 raised to any power ends in 6. - Summing these last digits: (5 + 1 + 6 = 12). The last digit of 12 is 2. Therefore, the answer is (boxed{2}).2) For (142 + 142^2 + 142^3 + ldots + 142^{20}): - The last digit of 142 is 2. The last digits of powers of 2 cycle every 4: 2, 4, 8, 6. - Each cycle (2, 4, 8, 6) sums to 20. There are 5 full cycles in 20 terms. - Summing these cycles: (5 times 20 = 100). The last digit of 100 is 0. Therefore, the answer is (boxed{0}).3) For (34^x + 34^{x+1} + 34^{2x}): - The last digit of (34^x) cycles between 4 (for odd (x)) and 6 (for even (x)). - The last digit of (34^{x+1}) cycles between 6 (for even (x+1)) and 4 (for odd (x+1)). - The last digit of (34^{2x}) is always 6, as (2x) is even. - For odd (x): (4 + 6 + 6 = 16), last digit 6. - For even (x): (6 + 4 + 6 = 16), last digit 6. Therefore, the answer is (boxed{6}).,<think>Okay, so I need to figure out the last digits of these three different expressions. Let me take them one by one.Starting with the first one: 135^x + 31^y + 56^(x+y), where x and y are natural numbers. Hmm, the question is asking for the digits that the decimal representations of these numbers end with. So, essentially, I need to find the last digit of each term and then add them up, right? Because the last digit of a sum depends only on the last digits of the addends.Alright, let's break it down. Let's consider each term separately.First term: 135 raised to the power of x. The base here is 135, and since the last digit of 135 is 5, any power of a number ending with 5 will also end with 5. For example, 5^1 = 5, 5^2 = 25, 5^3 = 125, and so on. So regardless of the exponent, as long as it's a natural number, 135^x should end with 5.Second term: 31^y. The last digit here is 1. Similar logic applies: any power of a number ending with 1 will end with 1. Like 1^1=1, 1^2=1, etc. So 31^y will always end with 1, no matter what y is.Third term: 56^(x+y). The last digit of 56 is 6. Now, powers of numbers ending with 6 also have a pattern. 6^1=6, 6^2=36, 6^3=216, etc. So any power of 6 will always end with 6. Therefore, 56^(x+y) will end with 6 regardless of the exponent.So now, putting it all together: the last digits are 5 (from 135^x), 1 (from 31^y), and 6 (from 56^(x+y)). Adding those together: 5 + 1 + 6 = 12. But wait, the last digit is the units digit of this sum. So 12 has a units digit of 2. Therefore, the entire expression should end with 2. So the answer to the first part is 2.Wait, but let me double-check. Let me test with some small exponents. Let's pick x=1 and y=1. Then:135^1 = 135, last digit 5.31^1 = 31, last digit 1.56^(1+1) = 56^2 = 3136, last digit 6.Adding the last digits: 5 + 1 + 6 = 12, so last digit 2. Correct.Another test: x=2, y=3.135^2 = 18225, last digit 5.31^3 = 29791, last digit 1.56^(2+3) = 56^5. Let's see, 56^1=56, 56^2=3136, 56^3=175616, 56^4=9834496, 56^5=550731776. Last digit 6. So again, 5 + 1 + 6 = 12, last digit 2. Yep, still works.Okay, so first problem's answer is 2.Moving on to the second problem: 142 + 142^2 + 142^3 + … + 142^20. Need to find the last digit of this sum.Alright, each term here is a power of 142. The last digit of 142 is 2. So, each term's last digit is the last digit of 2 raised to some power. Let's recall the pattern for powers of 2:2^1 = 2, last digit 22^2 = 4, last digit 42^3 = 8, last digit 82^4 = 16, last digit 62^5 = 32, last digit 2Then the cycle repeats every 4 exponents: 2, 4, 8, 6, 2, 4, 8, 6, etc.So, the last digit of 142^k is the same as the last digit of 2^k. Therefore, each term in the sum 142 + 142^2 + ... + 142^20 will have last digits following the cycle [2,4,8,6] for exponents 1 through 20.So, we can compute the last digit of each term 142^k by looking at k modulo 4:If k ≡1 mod 4, last digit is 2k ≡2 mod 4, last digit is 4k ≡3 mod 4, last digit is 8k ≡0 mod 4, last digit is 6Therefore, for exponents from 1 to 20, each cycle of 4 exponents will contribute 2,4,8,6. Since 20 terms, let's see how many complete cycles there are and any remaining terms.20 divided by 4 is 5, so there are 5 complete cycles. Each cycle (4 terms) has last digits 2,4,8,6. The sum of one cycle's last digits is 2 + 4 + 8 + 6 = 20. Therefore, 5 cycles would contribute 5 * 20 = 100. But wait, the total sum's last digit is the last digit of 100, which is 0. But wait, hold on.Wait, but actually, each term in the sum is 142^k, so k goes from 1 to 20. So each term's last digit is 2^1, 2^2, ..., 2^20. So the last digits are [2,4,8,6,2,4,8,6,2,4,8,6,2,4,8,6,2,4,8,6]. So 20 terms, 5 cycles of 4 terms each. Each cycle's last digits sum to 2 + 4 + 8 + 6 = 20. So total sum is 5 * 20 = 100. But since we're only concerned with the last digit of the entire sum, 100's last digit is 0. Therefore, the answer is 0.Wait, but let me check with a small example. Let's take k from 1 to 4:142 + 142^2 + 142^3 + 142^4. The last digits are 2 + 4 + 8 + 6 = 20, last digit 0. Then the next four terms would be same, sum 20 again, last digit 0. So every four terms contribute 0 to the last digit. Hence, 5 cycles (20 terms) would be 5*0=0. So the total last digit is 0. That seems correct.But wait, let's compute manually for a smaller exponent. Let's take k=1 to 2:142 + 142^2. Last digits 2 + 4 = 6. If we do 142 + 142^2 = 142 + 20164 = 20306. Last digit 6. Correct. Then k=1 to 3: 2 + 4 + 8 = 14, last digit 4. 142 + 142^2 + 142^3 = 142 + 20164 + 2863288 = 2883594. Last digit 4. Correct. Then adding 142^4: last digit 6. 2883594 + 142^4. 142^4 is 142*142*142*142. We know the last digit is 6. So 2883594 + ...6 = ...0. So last digit 4 + 6 = 10, last digit 0. Correct. So each cycle of 4 terms brings the last digit back to 0. Therefore, with 20 terms, which is 5 cycles, the last digit is 0. So answer is 0.But wait, let me check for k=1 to 4:Sum of last digits: 2 + 4 + 8 + 6 = 20, last digit 0. Then, sum of next four: 2 + 4 + 8 + 6 = 20, total sum so far 40, last digit 0. Continuing, each cycle adds another 20, so after 5 cycles: 20*5=100, last digit 0. Correct. So yes, the answer is 0.Third problem: 34^x + 34^(x+1) + 34^(2x), where x is a natural number. Find the last digit.So again, need to compute last digits of each term and sum them.First, the base is 34, which ends with 4. Let's recall the pattern for powers of 4:4^1 = 4, last digit 44^2 = 16, last digit 64^3 = 64, last digit 44^4 = 256, last digit 6So the cycle is 4,6,4,6,... repeating every 2 exponents.Therefore, the last digit of 34^k depends on k:If k is odd, last digit is 4.If k is even, last digit is 6.So let's analyze each term:First term: 34^x. If x is odd, last digit 4; if even, 6.Second term: 34^(x+1). The exponent is x+1. So if x is odd, x+1 is even, so last digit 6. If x is even, x+1 is odd, so last digit 4.Third term: 34^(2x). The exponent here is 2x. Since 2x is always even, regardless of x. So 34^(2x) will have a last digit of 6.So putting this together, let's consider two cases: x is odd or x is even.Case 1: x is odd.First term: 4Second term: x+1 is even, so 6Third term: 6Sum: 4 + 6 + 6 = 16. Last digit 6.Case 2: x is even.First term: 6Second term: x+1 is odd, so 4Third term: 6Sum: 6 + 4 + 6 = 16. Last digit 6.Wait, in both cases, the sum is 16, so last digit is 6. Therefore, regardless of whether x is odd or even, the last digit is 6.Wait, let me check with some examples.Take x=1 (odd):34^1 + 34^(1+1) + 34^(2*1) = 34 + 34^2 + 34^2.Compute each term:34: last digit 4.34^2: 1156, last digit 6.34^2 again: 6.So sum: 4 + 6 + 6 = 16. Last digit 6. Correct.Take x=2 (even):34^2 + 34^3 + 34^4.34^2: last digit 6.34^3: last digit 4 (since 3 is odd).34^4: last digit 6.Sum: 6 + 4 + 6 = 16. Last digit 6. Correct.Another test: x=3.34^3 (4) + 34^4 (6) + 34^6 (6). Sum: 4 + 6 + 6 = 16. Last digit 6. Correct.So yeah, regardless of x, the last digit is 6. So answer is 6.Therefore, compiling all the answers:1) 22) 03) 6Final Answer1) boxed{2}2) boxed{0}3) boxed{6}</think>To determine the last digits of the given expressions, we analyze each term's contribution based on the properties of powers and their cycles:1) For (135^x + 31^y + 56^{x+y}): - The last digit of (135^x) is always 5 (since any power of a number ending in 5 ends in 5). - The last digit of (31^y) is always 1 (since any power of a number ending in 1 ends in 1). - The last digit of (56^{x+y}) is always 6 (since any power of a number ending in 6 ends in 6). - Summing these last digits: (5 + 1 + 6 = 12), so the last digit is 2.2) For (142 + 142^2 + 142^3 + ldots + 142^{20}): - The last digit of (142^k) follows the cycle of powers of 2: 2, 4, 8, 6 (repeating every 4 terms). - There are 20 terms, which is 5 complete cycles of 4 terms each. - Each cycle sums to (2 + 4 + 8 + 6 = 20), contributing a last digit of 0. - Five cycles contribute (5 times 20 = 100), so the last digit is 0.3) For (34^x + 34^{x+1} + 34^{2x}): - The last digit of (34^k) follows the cycle 4, 6 (repeating every 2 terms). - For (34^x): last digit is 4 if (x) is odd, 6 if (x) is even. - For (34^{x+1}): last digit is 6 if (x) is odd, 4 if (x) is even. - For (34^{2x}): last digit is always 6 (since (2x) is even). - Summing these last digits in both cases (odd or even (x)): (4 + 6 + 6 = 16) or (6 + 4 + 6 = 16), so the last digit is 6.Final answers:1) boxed{2}2) boxed{0}3) boxed{6}]

🔑:12. 3) mathrm{Solution.} mathrm{If} mathrm{a} mathrm{number} mathrm{ends} mathrm{in} mathrm{four,} mathrm{then} even powers of it end in 6, and odd powers end in 4. Therefore, one of the first two terms ends in four, and the other ends in six. The third term ends in six, so the decimal representation of the sum ends in six.

🧠:[<think>Okay, so I need to solve this problem about two squares. The problem says that the difference in their perimeters is 8 cm, and the difference in their areas is 16 cm². I have to find the sum of their areas. Hmm, let's see. Let me start by recalling some properties of squares. A square has all sides equal, so the perimeter of a square is 4 times the side length, and the area is the side length squared. Let me denote the side lengths of the two squares as, maybe, ( a ) and ( b ), where ( a > b ). Then the perimeter of the first square would be ( 4a ), and the second one ( 4b ). The difference in perimeters is 8 cm, so that would be ( 4a - 4b = 8 ). Similarly, the areas are ( a^2 ) and ( b^2 ), and their difference is 16 cm², so ( a^2 - b^2 = 16 ).Okay, so I have two equations here:1. ( 4a - 4b = 8 )2. ( a^2 - b^2 = 16 )Hmm, maybe I can simplify these equations. Let's start with the first equation. If I factor out the 4, it becomes ( 4(a - b) = 8 ). Dividing both sides by 4 gives ( a - b = 2 ). So that's a simpler equation: the difference between the sides is 2 cm. Now, looking at the second equation: ( a^2 - b^2 = 16 ). I remember that ( a^2 - b^2 ) factors into ( (a - b)(a + b) ). That's a difference of squares. So substituting that in, we get ( (a - b)(a + b) = 16 ). But from the first equation, we already know that ( a - b = 2 ). So we can substitute that value into this equation. So substituting ( a - b = 2 ), the equation becomes ( 2(a + b) = 16 ). Then, dividing both sides by 2, we get ( a + b = 8 ). Alright, so now we have two equations:1. ( a - b = 2 )2. ( a + b = 8 )These are two equations with two variables, so we can solve them simultaneously. Let me add the two equations together. Adding the left sides: ( (a - b) + (a + b) = 2a ). Adding the right sides: ( 2 + 8 = 10 ). So, ( 2a = 10 ), which means ( a = 5 ). Once we know ( a = 5 ), we can substitute back into one of the equations to find ( b ). Let's use ( a + b = 8 ). Substituting ( a = 5 ) gives ( 5 + b = 8 ), so ( b = 3 ). Let me check if these values satisfy both original equations. First, the perimeters: ( 4a - 4b = 4*5 - 4*3 = 20 - 12 = 8 ). That's correct. Second, the areas: ( a^2 - b^2 = 25 - 9 = 16 ). That's also correct. So the side lengths are 5 cm and 3 cm. The problem asks for the sum of their areas. So, ( a^2 + b^2 = 25 + 9 = 34 ). Wait, but let me double-check. Sometimes when dealing with differences, you have to be careful which one is larger. But in this case, we assumed ( a > b ), so the difference in areas is ( a^2 - b^2 ), which is 16, so yes, that gives the correct result. Alternatively, if I had considered ( b > a ), the difference in perimeters would be negative, but since the problem states "difference" without specifying order, perhaps absolute difference. But since they mentioned the difference in areas is 16 cm², similarly, they might mean absolute difference, but in the equations, I considered ( a > b ), so the difference is positive. But since both differences came out positive, that's consistent. So, the sum of the areas is 25 + 9 = 34 cm². Let me verify again:Perimeter difference: 4*5 = 20, 4*3 = 12, 20 - 12 = 8 cm. Correct.Area difference: 25 - 9 = 16 cm². Correct.Therefore, sum of areas: 34 cm². I think that's solid. Maybe I can think of another way to approach the problem, just to confirm. Let's see. Suppose we let the sides be x and y, with x > y. Then 4x - 4y = 8 => x - y = 2. And x² - y² = 16. Then x² - y² factors as (x - y)(x + y) = 16. Since x - y = 2, then 2*(x + y) = 16 => x + y = 8. Then solving x - y = 2 and x + y = 8, adding gives 2x = 10 => x = 5, and then y = 3. So same result. Alternatively, maybe using substitution. From the first equation, x = y + 2. Substitute into the second equation: (y + 2)^2 - y^2 = 16. Expand that: y² + 4y + 4 - y² = 16 => 4y + 4 = 16 => 4y = 12 => y = 3. Then x = 5. So again, same answer. Therefore, the sum of their areas is 25 + 9 = 34. So the final answer is 34 cm². Final AnswerThe sum of their areas is boxed{34}.</think>Let the side lengths of the two squares be ( a ) and ( b ) where ( a > b ). The given conditions are:1. The difference in their perimeters is ( 8 text{cm} ): [ 4a - 4b = 8 implies a - b = 2 ]2. The difference in their areas is ( 16 text{cm}^2 ): [ a^2 - b^2 = 16 ]Using the difference of squares, we factor the second equation: [ (a - b)(a + b) = 16 ]Substituting ( a - b = 2 ) into the equation: [ 2(a + b) = 16 implies a + b = 8 ]We now have the system of equations:1. ( a - b = 2 )2. ( a + b = 8 )Adding these equations: [ (a - b) + (a + b) = 2 + 8 implies 2a = 10 implies a = 5 ]Substituting ( a = 5 ) back into ( a + b = 8 ): [ 5 + b = 8 implies b = 3 ]The areas of the squares are ( a^2 = 25 text{cm}^2 ) and ( b^2 = 9 text{cm}^2 ). The sum of their areas is: [ 25 + 9 = 34 ]Thus, the sum of their areas is (boxed{34}).,<think>Okay, let's see. I need to find the sum of the areas of two squares when given the difference in their perimeters and the difference in their areas. Hmm, let's start by breaking down the problem.First, let me recall that the perimeter of a square is 4 times the length of one side, right? So if I call the side length of the first square "a" and the side length of the second square "b", then their perimeters would be 4a and 4b respectively. The problem says the difference in their perimeters is 8 cm. So, I can write that as:4a - 4b = 8 cmOr maybe |4a - 4b| = 8 cm, but since it just says the difference, maybe it doesn't matter which one is bigger. Wait, but probably the first square has a larger perimeter since the difference is positive. Hmm, but maybe I should keep it as 4a - 4b = 8 for now. Let's go with that. If that's not correct, I can adjust later.Similarly, the difference in their areas is 16 cm². The area of a square is side squared, so that would be a² - b² = 16 cm². Again, assuming a is bigger than b. But maybe the problem just states the difference, so it's |a² - b²| = 16. But the way the problem is phrased, since it says "difference in their areas is 16 cm²," maybe they subtract the smaller from the larger, so if we take a as larger, then a² - b² = 16.So, I have two equations:1) 4a - 4b = 82) a² - b² = 16I need to solve these two equations to find a and b, then compute a² + b².Alright, let's simplify the first equation. 4(a - b) = 8, so dividing both sides by 4 gives:a - b = 2So, a = b + 2. That's straightforward.Now, the second equation is a² - b² = 16. But I remember that a² - b² can be factored into (a - b)(a + b). So:(a - b)(a + b) = 16We already know that a - b = 2 from the first equation. So substituting that in:2*(a + b) = 16Divide both sides by 2:a + b = 8So now we have a system of equations:a - b = 2a + b = 8If we add these two equations together, the b terms will cancel:(a - b) + (a + b) = 2 + 82a = 10Therefore, a = 5Then, substituting back into a + b = 8:5 + b = 8b = 3So the sides are 5 cm and 3 cm. Let's check if this works.First, perimeters: 4*5 = 20 cm, 4*3 = 12 cm. The difference is 20 - 12 = 8 cm. That's correct.Areas: 5² = 25 cm², 3² = 9 cm². The difference is 25 - 9 = 16 cm². That's also correct.So the sum of their areas is 25 + 9 = 34 cm². So the answer should be 34.Wait, but let me make sure I didn't make any assumptions here. I assumed that a > b, which seems valid because the differences came out positive. If it were the other way around, the equations would be 4b - 4a = 8 and b² - a² = 16, which would lead to negative differences, but since difference is a positive quantity, probably the way I did it is correct.Alternatively, if the problem had said the perimeter of the first square is 8 cm less than the second, then we would set up the equations differently, but since it just says the difference is 8 cm, and areas difference is 16 cm², we can assume that the larger square has the larger perimeter and area. So I think my solution is solid.Therefore, the sum of their areas is 34 cm².Final AnswerThe sum of their areas is boxed{34} (mathrm{cm}^2).</think>Given the problem, we need to find the sum of the areas of two squares when the difference in their perimeters is (8 mathrm{~cm}) and the difference in their areas is (16 mathrm{~cm}^2).1. Let the side lengths of the two squares be (a) and (b) with (a > b).2. The difference in their perimeters is given by: [ 4a - 4b = 8 implies a - b = 2 ]3. The difference in their areas is given by: [ a^2 - b^2 = 16 ]4. Using the difference of squares, we factorize the area difference: [ (a - b)(a + b) = 16 ]5. Substituting (a - b = 2) into the equation: [ 2(a + b) = 16 implies a + b = 8 ]6. Solving the system of equations (a - b = 2) and (a + b = 8): - Adding the equations: (2a = 10 implies a = 5) - Substituting (a = 5) into (a + b = 8): (5 + b = 8 implies b = 3)7. Calculating the areas of the squares: - Area of the first square: (a^2 = 5^2 = 25 mathrm{~cm}^2) - Area of the second square: (b^2 = 3^2 = 9 mathrm{~cm}^2)8. Sum of the areas: [ 25 + 9 = 34 ]Thus, the sum of their areas is (boxed{34}) (mathrm{cm}^2).]

🔑:B2. The difference in perimeters should be 4 a-4 b=8 mathrm{~cm}, and the difference in areas a^{2}-b^{2}=16 mathrm{~cm}^{2}. From the first equation, we get a-b=2, and from the second, due to (a-b)(a+b)=16, we also get a+b=8. From this, we arrive at a=5 mathrm{~cm} and b=3 mathrm{~cm}. The sum of both areas is 5^{2}+3^{2}=34 mathrm{~cm}^{2}.Writing the difference in perimeters 4 a-4 b=8. 1 pointWriting the difference in areas a^{2}-b^{2}=16 1 pointCorrect solving of the system 1 pointSolutions a=5 mathrm{~cm} and b=3 mathrm{~cm}. 1+1 pointsCalculated sum of areas: S_{1}+S_{2}=34 mathrm{~cm}^{2} 1 point