💡Okay, so I have this problem where I need to find the maximum number of black squares in an ( n times n ) table. The condition is that for every ( m times m ) sub-table where ( m > 1 ), the number of black squares doesn't exceed the number of white squares. And ( n ) is a natural number divisible by 3. Hmm, interesting.First, let me try to understand the constraints better. Since ( n ) is divisible by 3, I can write ( n = 3k ) for some integer ( k ). That might be useful later on. The main goal is to maximize the number of black squares while ensuring that in every sub-table larger than 1x1, there aren't more black squares than white ones.So, for any ( m times m ) sub-table where ( m > 1 ), the number of black squares should be at most half of the total squares in that sub-table. That means, for example, in a 2x2 sub-table, there can be at most 2 black squares, in a 3x3 sub-table, at most 4 black squares, and so on.I need to figure out a pattern or a way to color the table such that this condition is satisfied everywhere, but also allows for as many black squares as possible. Maybe a periodic pattern that repeats every few rows and columns?Let me think about smaller cases first. Let's say ( n = 3 ). Then, the entire table is 3x3. The condition applies to all 2x2 and 3x3 sub-tables. For the 3x3 table, the maximum number of black squares is 4, as in the example given. So, for ( n = 3 ), the maximum is 4.If I extend this idea, maybe for ( n = 6 ), I can have a similar pattern repeated. So, in each 3x3 block, I can have 4 black squares, and since there are 4 such blocks in a 6x6 table, the total number of black squares would be ( 4 times 4 = 16 ). Wait, but ( 6 times 6 = 36 ), and ( 16 ) is less than half of 36, which is 18. Hmm, maybe I can do better.Wait, actually, the maximum number of black squares in the entire table needs to be as high as possible, but without violating the condition in any sub-table. So, maybe the pattern isn't just repeating the same 3x3 block, but something more intricate.Let me think about the coloring pattern. If I color certain cells black in a way that in every 2x2 and 3x3 sub-table, the number of black squares is limited, perhaps I can maximize the overall number.I recall that in the given example, the cells are colored black if their coordinates satisfy certain modulo conditions. Specifically, cells where ( (a, b) ) are ( (1,2), (2,1), (3,2), (2,3) ) modulo 3 are colored black. That gives 4 black squares in each 3x3 block.Let me visualize this. In a 3x3 grid, the black squares would be at positions:- (1,2)- (2,1)- (3,2)- (2,3)So, it's like a diamond shape in the center of the 3x3 grid. This pattern ensures that in any 2x2 sub-table, there are at most 2 black squares, and in the 3x3, there are 4, which is less than half of 9, so 4.5, which rounds down to 4.If I extend this pattern to a larger grid, say 6x6, by repeating the 3x3 blocks, each 3x3 block contributes 4 black squares. So, in a 6x6 grid, there are 4 such blocks, each contributing 4 black squares, totaling 16. But wait, 16 is less than half of 36, which is 18. So, maybe I can actually have more black squares.Wait, no, because the condition applies to every sub-table, not just the 3x3 ones. So, even in the 6x6 grid, if I have more than 16 black squares, I might end up violating the condition in some 2x2 or 3x3 sub-tables.Let me check. If I try to color more cells black, say 17, would that cause any 2x2 or 3x3 sub-table to have more than half black squares? It's possible, because adding more black squares might cluster them in some sub-tables.Alternatively, maybe the maximum is indeed ( frac{4}{9}n^2 ), as in the example. For ( n = 3 ), that gives 4, which is correct. For ( n = 6 ), that gives ( frac{4}{9} times 36 = 16 ), which seems to fit.But I need to verify if this pattern actually satisfies the condition for all sub-tables. Let's consider a 2x2 sub-table. In the 3x3 block, the 2x2 sub-tables would have at most 2 black squares, which is exactly half. So, that's okay.What about a 4x4 sub-table? Wait, 4 isn't a multiple of 3, so the pattern might not align perfectly. Hmm, this could be a problem. Maybe the pattern needs to be adjusted for larger sub-tables.Alternatively, perhaps the maximum is indeed ( frac{4}{9}n^2 ), and the pattern works because it's periodic with period 3, so any sub-table will have a similar distribution of black squares, ensuring that no sub-table exceeds half black squares.Wait, but if I take a 3x4 sub-table, which is 12 cells, how many black squares would it have? Since the pattern repeats every 3 columns, in 4 columns, we have one full 3-column block and one extra column. So, in the 3 columns, we have 4 black squares, and in the extra column, depending on the alignment, maybe 1 or 2 black squares. So, total black squares could be 5 or 6. Half of 12 is 6, so 6 is okay, but 5 is also fine. Wait, but 6 is exactly half, so that's acceptable.Similarly, for a 4x4 sub-table, which has 16 cells, the number of black squares would be roughly ( frac{4}{9} times 16 approx 7.11 ), so 7 black squares, which is less than half (8). So, that's okay.Wait, but actually, the exact number depends on the alignment. If the 4x4 sub-table is aligned with the 3x3 blocks, it might have a slightly different count. But overall, the density remains ( frac{4}{9} ), so the number of black squares in any sub-table should be at most half.Hmm, I think this pattern works. So, the maximum number of black squares is ( frac{4}{9}n^2 ).But let me try to think if there's a way to have more black squares without violating the condition. Maybe a different coloring pattern?Suppose I try to color every other cell black in a checkerboard pattern. Then, in any 2x2 sub-table, there are exactly 2 black squares, which is half. In a 3x3 sub-table, there would be 5 black squares, which is more than half (4.5). So, that violates the condition.So, the checkerboard pattern doesn't work because it allows some sub-tables to have more than half black squares.What if I color two out of every three cells in each row? For example, in each row, color the first two cells black and the third white, repeating this pattern. Then, in any 3x3 sub-table, there would be 6 black squares, which is more than half (4.5). So, that's no good.Alternatively, maybe a diagonal pattern? If I color cells along certain diagonals, but I need to ensure that in every sub-table, the number of black squares doesn't exceed half.This seems tricky. The initial pattern with 4 black squares in each 3x3 block seems to be the safest bet, as it ensures that in every sub-table, the number of black squares is controlled.Therefore, I think the maximum number of black squares is indeed ( frac{4}{9}n^2 ).**Final Answer**The maximum number of black squares is boxed{dfrac{4n^2}{9}}.

💡Okay, so I need to figure out how many distinct positive factors the number 1320 has. Hmm, I remember that to find the number of factors, I first need to break down the number into its prime factors. Once I have the prime factorization, there's some formula or method to calculate the number of factors. Let me try to recall how that works.Alright, starting with 1320. I think the best way to factorize it is by dividing it by the smallest prime numbers first. Let's see, 2 is the smallest prime. Does 1320 divide by 2? Yes, because it's an even number. So, 1320 divided by 2 is 660. Okay, that's one factor of 2. Now, 660 is still even, so I can divide by 2 again. 660 divided by 2 is 330. That's two factors of 2 so far. 330 is still even, so dividing by 2 once more gives me 165. Now, 165 is not even, so I can't divide by 2 anymore. That gives me three factors of 2 in total.Next, I should try the next smallest prime number, which is 3. Let's see if 165 is divisible by 3. Adding the digits: 1 + 6 + 5 equals 12, and since 12 is divisible by 3, 165 is also divisible by 3. Dividing 165 by 3 gives me 55. So, that's one factor of 3.Now, moving on to the next prime number, which is 5. Checking if 55 is divisible by 5: yes, because it ends with a 5. Dividing 55 by 5 gives me 11. That's one factor of 5.Finally, 11 is a prime number itself, so I can't factorize it any further. So, putting it all together, the prime factors of 1320 are 2, 3, 5, and 11. Specifically, 2 appears three times, 3 appears once, 5 appears once, and 11 appears once.Wait, let me write that down to keep track:- 2^3- 3^1- 5^1- 11^1Okay, so the prime factorization of 1320 is 2^3 * 3^1 * 5^1 * 11^1. Now, how does this help me find the number of distinct positive factors?I think there's a formula where you take the exponents of each prime factor, add one to each of them, and then multiply those together. Let me see if I remember correctly. For example, if a number N has a prime factorization of p1^a * p2^b * p3^c, then the number of distinct positive factors is (a+1)*(b+1)*(c+1). Yeah, that sounds right.So applying that to 1320, which is 2^3 * 3^1 * 5^1 * 11^1, the exponents are 3, 1, 1, and 1. Adding one to each of these exponents gives me 4, 2, 2, and 2. Now, I need to multiply these together: 4 * 2 * 2 * 2.Let me calculate that step by step:- 4 * 2 = 8- 8 * 2 = 16- 16 * 2 = 32So, according to this, 1320 should have 32 distinct positive factors. Hmm, that seems like a lot, but let me double-check my steps to make sure I didn't make a mistake.First, the prime factorization: 1320 divided by 2 three times gives me 165, which is correct. Then, 165 divided by 3 gives me 55, which is correct. 55 divided by 5 gives me 11, which is prime. So, the prime factors are indeed 2^3 * 3^1 * 5^1 * 11^1.Next, applying the formula: (3+1)*(1+1)*(1+1)*(1+1) = 4*2*2*2 = 32. That seems consistent. I don't see any mistakes in my calculations.Just to be thorough, maybe I can list out some of the factors to see if 32 makes sense. Starting with 1 and 1320, obviously. Then, 2 and 660, 3 and 440, 4 and 330, 5 and 264, 6 and 220, 8 and 165, 10 and 132, 11 and 120, 12 and 110, 15 and 88, 20 and 66, 22 and 60, 24 and 55, 30 and 44, 33 and 40. Wait, that's 18 factors already, but I know there are more. Maybe I'm missing some.Alternatively, maybe listing all the factors is too time-consuming, but I can trust the formula since it's a standard method. The formula accounts for all combinations of the prime factors, so it should give the correct number.Therefore, I think my answer is correct. 1320 has 32 distinct positive factors.

💡Alright, so I've got this math problem here: I need to fill in the four spaces between five 10s with the operators "+", "-", "×", and "÷" each exactly once. The goal is to find the maximum possible value of the resulting expression. The options given are (A) 104, (B) 109, (C) 114, and (D) 110.Okay, let me break this down. I have the expression: 10 _ 10 _ 10 _ 10 _ 10. I need to replace the underscores with "+", "-", "×", and "÷" without repeating any of them. Since I have four operators and five numbers, each operator will be used once.First, I should remember the order of operations: multiplication and division come before addition and subtraction. So, wherever I place "×" or "÷", those operations will be performed first.I want to maximize the value of this expression. To do that, I should try to make the larger numbers come from multiplication and addition, and minimize the impact of subtraction and division.Let me think about where to place the multiplication operator "×". Multiplying two 10s will give me 100, which is a big number. So, if I can place "×" between two 10s, that could significantly increase the value.Similarly, division "÷" will reduce the value, so I should place that in a way that it affects the smallest possible part of the expression. Maybe dividing a smaller number by 10, which would give me 1, and then that 1 can be added or subtracted.Subtraction "-" will also reduce the value, so I should place it in a position where it subtracts the smallest possible number. Maybe after a division.Addition "+" will increase the value, so I should place that between two larger numbers.Let me try to construct the expression step by step.First, I'll place "×" between the first two 10s: 10 × 10. That gives me 100.Now, I have three operators left: "+", "-", and "÷". I need to place them between the remaining three 10s.I want to maximize the value, so I should try to add as much as possible. Let's see:If I place "+" next, it would be 10 × 10 + 10. That gives me 110.Then, I have two operators left: "-" and "÷". I need to place them between the last two 10s. If I place "÷" before "-", it would be 10 × 10 + 10 ÷ 10 - 10. But wait, that would be 100 + 1 - 10, which is 91. That's not good.Alternatively, if I place "-" before "÷", it would be 10 × 10 + 10 - 10 ÷ 10. That would be 100 + 10 - 1, which is 109. That's better.But let me check if there's a better arrangement.What if I place "÷" earlier? Let's try:10 ÷ 10 × 10 + 10 - 10. That would be 1 × 10 + 10 - 10, which is 10 + 10 - 10 = 10. That's way too low.What if I place "÷" after "×"? Let's see:10 × 10 ÷ 10 + 10 - 10. That would be 100 ÷ 10 + 10 - 10, which is 10 + 10 - 10 = 10. Again, too low.What if I place "+" after "×"? Let's try:10 × 10 + 10 ÷ 10 - 10. That's 100 + 1 - 10 = 91. Not good.What if I place "-" after "×"? Let's see:10 × 10 - 10 ÷ 10 + 10. That would be 100 - 1 + 10 = 109. Same as before.Hmm, so whether I place "-" before or after "÷", I get 109.Is there a way to get a higher value?What if I place "×" somewhere else? Let's try placing "×" in the middle.10 + 10 × 10 - 10 ÷ 10. That would be 10 + 100 - 1 = 109. Same result.What if I place "×" at the end?10 ÷ 10 + 10 - 10 × 10. That would be 1 + 10 - 100 = -89. That's bad.What if I place "×" between the third and fourth 10?10 + 10 ÷ 10 × 10 - 10. That would be 10 + 10 - 10 = 10. Not good.What if I place "×" between the second and third 10?10 × 10 ÷ 10 + 10 - 10. That's 10 + 10 - 10 = 10. Again, too low.So, it seems that placing "×" between the first two 10s and then arranging the remaining operators to minimize the reduction gives me 109.Is there a way to get higher than 109?Let me think differently. Maybe using division to create a larger number.Wait, if I do 10 ÷ 10, that's 1. If I do 10 × 10, that's 100. If I do 10 + 10, that's 20. If I do 10 - 10, that's 0.So, the largest number I can get from two 10s is 100 by multiplying. So, I should definitely use "×" somewhere to get 100.Then, I need to use the other operators to add as much as possible and subtract as little as possible.So, if I have 10 × 10, that's 100. Then, I have three more 10s and three operators: "+", "-", and "÷".I need to arrange these to maximize the total.If I do 100 + 10 - 10 ÷ 10, that's 100 + 10 - 1 = 109.Alternatively, 100 + 10 ÷ 10 - 10 = 100 + 1 - 10 = 91.Or, 100 - 10 ÷ 10 + 10 = 100 - 1 + 10 = 109.So, the maximum I can get is 109.Wait, what if I arrange the operators differently? Let's try:10 × 10 + 10 ÷ 10 - 10 = 100 + 1 - 10 = 91.No, that's worse.What if I do 10 × (10 + 10) ÷ 10 - 10? But I can't use parentheses, so order of operations applies.Wait, without parentheses, it's just left to right with multiplication and division first.So, 10 × 10 + 10 ÷ 10 - 10 = 100 + 1 - 10 = 91.Same as before.Alternatively, 10 + 10 × 10 ÷ 10 - 10 = 10 + 100 ÷ 10 - 10 = 10 + 10 - 10 = 10.No good.What if I do 10 ÷ 10 + 10 × 10 - 10 = 1 + 100 - 10 = 91.Still not better.So, it seems that 109 is the maximum I can get.Wait, let me try another arrangement:10 × 10 - 10 + 10 ÷ 10 = 100 - 10 + 1 = 91.Nope.What if I do 10 × 10 ÷ 10 + 10 - 10 = 10 + 10 - 10 = 10.Still low.Hmm, seems like 109 is the highest I can get.Wait, what if I place "÷" before "×"? Let's see:10 ÷ 10 × 10 + 10 - 10 = 1 × 10 + 10 - 10 = 10 + 10 - 10 = 10.No good.Alternatively, 10 + 10 ÷ 10 × 10 - 10 = 10 + 1 × 10 - 10 = 10 + 10 - 10 = 10.Still low.Okay, I think I've tried all possible arrangements, and the maximum I can get is 109.So, the answer should be (B) 109.

💡First, I need to understand the relationship between the sides of Rectangle A and Rectangle B. The problem states that the ratios of corresponding sides are equal, specifically a/c = b/d = 3/4.Next, I'll identify the known values. Rectangle B has sides of length 4 and 8. I'll assign c = 4 and d = 8.Using the ratio a/c = 3/4, I can set up the equation a/4 = 3/4. To solve for a, I'll multiply both sides of the equation by 4, which gives me a = 3.Therefore, the length of side a in Rectangle A is 3.

💡Alright, so I have this vector problem here, and I need to find the magnitude of vector b. Let me try to break it down step by step. First, I know that vector a is given as (2,1). That means its components are 2 in the x-direction and 1 in the y-direction. I also know that the magnitude of the sum of vectors a and b is 4. So, |a + b| = 4. Additionally, the dot product of a and b is 1, which is written as a · b = 1.Okay, so I need to find |b|. Hmm, how do I approach this? I remember that when you have the magnitude of the sum of two vectors, you can use the formula that involves the dot product. Let me recall that formula.I think it's something like |a + b|² = |a|² + |b|² + 2(a · b). Yeah, that sounds right. So, if I square both sides of the equation |a + b| = 4, I should get |a + b|² = 16. That makes sense.Now, let's compute |a|². Since a is (2,1), its magnitude squared is 2² + 1², which is 4 + 1 = 5. Got that. So, |a|² is 5.We also know that a · b is 1, so 2(a · b) would be 2*1 = 2. Putting it all together, the equation becomes:16 = 5 + |b|² + 2Wait, that simplifies to 16 = 7 + |b|². So, subtracting 7 from both sides gives |b|² = 9. Taking the square root of both sides, |b| = 3. Hmm, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. Start with |a + b| = 4, so squaring both sides gives |a + b|² = 16.2. Expand |a + b|² using the formula: |a|² + |b|² + 2(a · b) = 16.3. Calculate |a|²: (2)² + (1)² = 4 + 1 = 5.4. Substitute a · b = 1 into the equation: 5 + |b|² + 2*1 = 16.5. Simplify: 5 + |b|² + 2 = 7 + |b|² = 16.6. Subtract 7 from both sides: |b|² = 9.7. Take the square root: |b| = 3.Everything seems to add up. I don't see any errors in my calculations. So, the magnitude of vector b is 3.Just to be thorough, let me think if there's another way to approach this problem. Maybe by expressing vector b in terms of its components and solving the equations. Let's try that.Let’s assume vector b has components (x, y). Then, vector a + vector b would be (2 + x, 1 + y). The magnitude of this vector is given as 4, so:√[(2 + x)² + (1 + y)²] = 4Squaring both sides:(2 + x)² + (1 + y)² = 16Expanding this:(4 + 4x + x²) + (1 + 2y + y²) = 16Combine like terms:4 + 4x + x² + 1 + 2y + y² = 16Simplify:x² + y² + 4x + 2y + 5 = 16Subtract 16 from both sides:x² + y² + 4x + 2y - 11 = 0Now, we also know that the dot product of a and b is 1. The dot product is calculated as:a · b = (2)(x) + (1)(y) = 2x + y = 1So, we have another equation:2x + y = 1Now, we have a system of two equations:1. x² + y² + 4x + 2y - 11 = 02. 2x + y = 1Let me solve equation 2 for y:y = 1 - 2xNow, substitute this into equation 1:x² + (1 - 2x)² + 4x + 2(1 - 2x) - 11 = 0First, expand (1 - 2x)²:1 - 4x + 4x²Now, substitute back:x² + (1 - 4x + 4x²) + 4x + 2(1 - 2x) - 11 = 0Simplify term by term:x² + 1 - 4x + 4x² + 4x + 2 - 4x - 11 = 0Combine like terms:x² + 4x² = 5x²-4x + 4x - 4x = -4x1 + 2 - 11 = -8So, the equation becomes:5x² - 4x - 8 = 0Now, we have a quadratic equation in terms of x:5x² - 4x - 8 = 0Let me try to solve this using the quadratic formula. The quadratic formula is:x = [-b ± √(b² - 4ac)] / (2a)Where a = 5, b = -4, c = -8.So, discriminant D = (-4)² - 4*5*(-8) = 16 + 160 = 176So, x = [4 ± √176] / 10Simplify √176:√176 = √(16*11) = 4√11So, x = [4 ± 4√11] / 10 = [4(1 ± √11)] / 10 = [2(1 ± √11)] / 5So, x = (2 + 2√11)/5 or x = (2 - 2√11)/5Now, let's find y using y = 1 - 2x.First, for x = (2 + 2√11)/5:y = 1 - 2*(2 + 2√11)/5 = 1 - (4 + 4√11)/5 = (5 - 4 - 4√11)/5 = (1 - 4√11)/5Similarly, for x = (2 - 2√11)/5:y = 1 - 2*(2 - 2√11)/5 = 1 - (4 - 4√11)/5 = (5 - 4 + 4√11)/5 = (1 + 4√11)/5So, we have two possible solutions for vector b:b = [(2 + 2√11)/5, (1 - 4√11)/5] or b = [(2 - 2√11)/5, (1 + 4√11)/5]Now, let's compute the magnitude of b in both cases.First, for b = [(2 + 2√11)/5, (1 - 4√11)/5]:|b|² = [(2 + 2√11)/5]^2 + [(1 - 4√11)/5]^2Let's compute each term:[(2 + 2√11)/5]^2 = [4 + 8√11 + 4*11]/25 = [4 + 8√11 + 44]/25 = [48 + 8√11]/25[(1 - 4√11)/5]^2 = [1 - 8√11 + 16*11]/25 = [1 - 8√11 + 176]/25 = [177 - 8√11]/25Adding these together:[48 + 8√11 + 177 - 8√11]/25 = (48 + 177)/25 = 225/25 = 9So, |b|² = 9, which means |b| = 3.Similarly, for the other solution b = [(2 - 2√11)/5, (1 + 4√11)/5], the calculation would be the same because squaring eliminates the sign differences:[(2 - 2√11)/5]^2 = [4 - 8√11 + 44]/25 = [48 - 8√11]/25[(1 + 4√11)/5]^2 = [1 + 8√11 + 176]/25 = [177 + 8√11]/25Adding these together:[48 - 8√11 + 177 + 8√11]/25 = (48 + 177)/25 = 225/25 = 9Again, |b|² = 9, so |b| = 3.Okay, so both solutions give the same magnitude for vector b, which is 3. That confirms my earlier result. I think this thorough approach solidifies that the magnitude of vector b is indeed 3. I don't see any mistakes in my calculations, and both methods—using the dot product formula and solving for components—lead to the same answer. So, I'm confident that the correct answer is 3, which corresponds to option B.

💡I need to calculate the cosine of 60 degrees. I remember that the cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse. For a 60-degree angle, if I consider an equilateral triangle where all sides are equal, splitting it down the middle creates a 30-60-90 triangle. In such a triangle, the sides are in the ratio 1 : √3 : 2, where 1 is the side opposite the 30-degree angle, √3 is opposite the 60-degree angle, and 2 is the hypotenuse. Therefore, the adjacent side to the 60-degree angle is 1, and the hypotenuse is 2. So, the cosine of 60 degrees should be 1 divided by 2, which is 1/2.