💡Okay, so I have this problem here: If the complex number ( z = frac{m + i}{1 - i} ) is real, then what is the real number ( m )? The options are A: 0, B: -1, C: -1 or 1, and D: 1. Hmm, let me try to figure this out step by step.First, I remember that a complex number is real if its imaginary part is zero. So, I need to simplify the expression ( frac{m + i}{1 - i} ) and then set the imaginary part equal to zero to solve for ( m ).To simplify the complex fraction, I think I should multiply both the numerator and the denominator by the conjugate of the denominator. The conjugate of ( 1 - i ) is ( 1 + i ). So, let me do that:( z = frac{m + i}{1 - i} times frac{1 + i}{1 + i} )Multiplying the numerators and denominators:Numerator: ( (m + i)(1 + i) )Denominator: ( (1 - i)(1 + i) )Let me expand both the numerator and the denominator.Starting with the numerator:( (m + i)(1 + i) = m times 1 + m times i + i times 1 + i times i )Simplify each term:- ( m times 1 = m )- ( m times i = mi )- ( i times 1 = i )- ( i times i = i^2 )So, putting it all together:( m + mi + i + i^2 )I know that ( i^2 = -1 ), so:( m + mi + i - 1 )Combine like terms:Real parts: ( m - 1 )Imaginary parts: ( mi + i = (m + 1)i )So, the numerator simplifies to:( (m - 1) + (m + 1)i )Now, the denominator:( (1 - i)(1 + i) )This is a difference of squares, so:( 1^2 - (i)^2 = 1 - (-1) = 1 + 1 = 2 )So, the denominator is 2.Putting it all together, the expression for ( z ) becomes:( z = frac{(m - 1) + (m + 1)i}{2} )Which can be written as:( z = frac{m - 1}{2} + frac{m + 1}{2}i )Since ( z ) is a real number, the imaginary part must be zero. That means:( frac{m + 1}{2} = 0 )Solving for ( m ):Multiply both sides by 2:( m + 1 = 0 )Subtract 1 from both sides:( m = -1 )So, the value of ( m ) is -1, which corresponds to option B.Wait, let me double-check my steps to make sure I didn't make any mistakes.1. I started by multiplying numerator and denominator by the conjugate of the denominator, which is correct.2. Expanded the numerator correctly: ( (m + i)(1 + i) ) resulted in ( m + mi + i + i^2 ), which simplifies to ( m - 1 + (m + 1)i ).3. The denominator was correctly simplified to 2.4. Then, I separated the real and imaginary parts and set the imaginary part equal to zero, leading to ( m = -1 ).Everything seems to check out. I don't see any errors in my calculations. Therefore, I'm confident that the correct answer is B: -1.**Final Answer**boxed{B}

💡Okay, so I have this problem here: I need to determine the sum of the highest powers of 10 and 8 that divide 20!. The options are A) 10, B) 8, C) 12, D) 14. Hmm, let me think about how to approach this.First, I remember that to find the highest power of a prime number that divides a factorial, we can use Legendre's formula. Legendre's formula helps us find the exponent of a prime number in the prime factorization of a factorial. The formula is:For a prime number p and a positive integer n, the exponent of p in n! is given by:[sum_{k=1}^{infty} leftlfloor frac{n}{p^k} rightrfloor]This means we keep dividing n by higher powers of p and take the floor of each division, then sum them up until p^k is greater than n.But in this problem, we're dealing with 10 and 8, which are not prime numbers. So, I need to break them down into their prime factors.Starting with 10: 10 is 2 multiplied by 5. So, the highest power of 10 that divides 20! will be determined by the minimum of the exponents of 2 and 5 in the prime factorization of 20!. That is, the exponent of 10 will be the smaller of the exponents of 2 and 5.Similarly, 8 is 2 cubed, so the highest power of 8 that divides 20! will be determined by the exponent of 2 in the prime factorization of 20! divided by 3, taking the floor of that result.So, my plan is:1. Find the exponent of 2 in 20! using Legendre's formula.2. Find the exponent of 5 in 20! using Legendre's formula.3. The exponent of 10 will be the minimum of the exponents of 2 and 5.4. The exponent of 8 will be the exponent of 2 divided by 3, using floor division.5. Finally, add these two exponents together to get the answer.Let me start with step 1: finding the exponent of 2 in 20!.Using Legendre's formula:[leftlfloor frac{20}{2} rightrfloor + leftlfloor frac{20}{4} rightrfloor + leftlfloor frac{20}{8} rightrfloor + leftlfloor frac{20}{16} rightrfloor + leftlfloor frac{20}{32} rightrfloor + dots]Calculating each term:- 20 divided by 2 is 10, floor is 10.- 20 divided by 4 is 5, floor is 5.- 20 divided by 8 is 2.5, floor is 2.- 20 divided by 16 is 1.25, floor is 1.- 20 divided by 32 is 0.625, floor is 0.- Higher powers will also give 0, so we can stop here.Adding them up: 10 + 5 + 2 + 1 = 18. So, the exponent of 2 in 20! is 18.Moving on to step 2: finding the exponent of 5 in 20!.Again, using Legendre's formula:[leftlfloor frac{20}{5} rightrfloor + leftlfloor frac{20}{25} rightrfloor + leftlfloor frac{20}{125} rightrfloor + dots]Calculating each term:- 20 divided by 5 is 4, floor is 4.- 20 divided by 25 is 0.8, floor is 0.- Higher powers will also give 0, so we can stop here.Adding them up: 4 + 0 = 4. So, the exponent of 5 in 20! is 4.Now, step 3: The exponent of 10 in 20! is the minimum of the exponents of 2 and 5. That is, min(18, 4) = 4. So, the highest power of 10 that divides 20! is 10^4.Step 4: The exponent of 8 in 20! is determined by the exponent of 2 divided by 3, using floor division. So, 18 divided by 3 is 6. Therefore, the highest power of 8 that divides 20! is 8^6.Finally, step 5: Adding these two exponents together. The exponent for 10 is 4, and the exponent for 8 is 6. So, 4 + 6 = 10.Looking back at the options, A) 10 is the correct answer.Wait, let me double-check my calculations to make sure I didn't make a mistake.For the exponent of 2:- 20 / 2 = 10- 20 / 4 = 5- 20 / 8 = 2.5 → 2- 20 / 16 = 1.25 → 1- 20 / 32 = 0.625 → 0Total: 10 + 5 + 2 + 1 = 18. Correct.For the exponent of 5:- 20 / 5 = 4- 20 / 25 = 0.8 → 0Total: 4. Correct.Exponent of 10: min(18, 4) = 4. Correct.Exponent of 8: 18 / 3 = 6. Correct.Sum: 4 + 6 = 10. Correct.So, yes, the final answer is 10, which is option A.**Final Answer**boxed{A}

💡Okay, so I have this problem about a 200x200 grid where each cell is colored either black or white. The difference between the number of black and white squares is 404. I need to prove that there's at least one 2x2 square that has an odd number of white squares. Hmm, interesting.Let me start by understanding the problem better. The grid has 200 rows and 200 columns, so there are a total of 200*200 = 40,000 squares. Each square is either black or white. The difference between the number of black and white squares is 404. Let me denote the number of black squares as B and white squares as W. So, we have |B - W| = 404. Since B + W = 40,000, we can set up equations:If B > W, then B = W + 404. Substituting into B + W = 40,000, we get W + 404 + W = 40,000, so 2W + 404 = 40,000. Then, 2W = 39,596, so W = 19,798. Similarly, B = 19,798 + 404 = 20,202.Alternatively, if W > B, then W = B + 404, and similarly, B = 19,798 and W = 20,202. So, regardless, we have 19,798 of one color and 20,202 of the other.Now, the problem is to show that there exists a 2x2 square with an odd number of white squares. So, in other words, in some 2x2 block, the number of white squares is 1 or 3. Because 0, 2, 4 are even, and 1, 3 are odd.I think I should approach this by contradiction. Let me assume that every 2x2 square has an even number of white squares. Then, in every 2x2 square, the number of white squares is 0, 2, or 4.If every 2x2 square has an even number of white squares, what does that imply about the coloring of the entire grid? Maybe there's some periodicity or pattern that emerges.Let me think about the first row. Suppose the first row has some number of black and white squares. Let me denote the number of white squares in the first row as w1 and black squares as b1. So, w1 + b1 = 200.Now, moving to the second row. If every 2x2 square has an even number of white squares, then the coloring of the second row must be related to the first row in a specific way. For example, if I look at the first two rows, each 2x2 block formed by columns 1 and 2 must have an even number of white squares. So, if the first cell in the first row is white, then the cell below it in the second row must be the same color to maintain an even count in that 2x2 block.Wait, let me clarify that. If the first cell in the first row is white, then the cell below it in the second row must also be white to make the number of white squares in that 2x2 block even (either 0, 2, or 4). Similarly, if the first cell is black, the cell below must also be black.So, this suggests that the second row is identical to the first row. Because if they weren't identical, then there would be a 2x2 block with an odd number of white squares.Wait, is that true? Let me think. Suppose the first row is all white. Then, the second row must also be all white to keep every 2x2 block even. Similarly, if the first row is alternating black and white, then the second row must also be alternating in the same way.But what if the first row has some arbitrary pattern? Then, the second row must mirror that pattern exactly to maintain the even count in every 2x2 block. So, in general, the second row must be identical to the first row.Similarly, the third row must be identical to the second row, and so on. Therefore, all rows must be identical. So, the entire grid is made up of identical rows.If that's the case, then the entire grid is just 200 copies of the first row. So, the number of white squares in the entire grid would be 200 times the number of white squares in the first row.But wait, the number of white squares is 19,798 or 20,202, depending on which is more. Let me compute 19,798 divided by 200. 19,798 / 200 = 98.99. Hmm, that's not an integer. Similarly, 20,202 / 200 = 101.01, which is also not an integer.But if all rows are identical, then the number of white squares must be a multiple of 200. Because each row contributes the same number of white squares. But 19,798 and 20,202 are not multiples of 200. Therefore, this leads to a contradiction.Therefore, our initial assumption that every 2x2 square has an even number of white squares must be false. Hence, there must exist at least one 2x2 square with an odd number of white squares.Wait, let me double-check that reasoning. If all rows are identical, then the total number of white squares would be 200 times the number in one row. But 19,798 and 20,202 are not divisible by 200, which is correct because 200*98 = 19,600 and 200*99 = 19,800, which is more than 19,798. Similarly, 200*101 = 20,200, which is less than 20,202. So, neither 19,798 nor 20,202 is a multiple of 200. Therefore, it's impossible for all rows to be identical, which contradicts the assumption that every 2x2 square has an even number of white squares.Therefore, the conclusion is that there must be at least one 2x2 square with an odd number of white squares.I think that makes sense. By assuming the opposite, we arrived at an impossibility, so the original statement must be true.

💡First, I need to evaluate each of the given expressions to determine if they are positive numbers.1. **Evaluate (-left(-3right)):** - The negative of (-3) is (3), which is positive.2. **Evaluate (0):** - Zero is neither positive nor negative.3. **Evaluate (left(-3right)^{2}):** - Squaring (-3) results in (9), which is positive.4. **Evaluate (|-9|):** - The absolute value of (-9) is (9), which is positive.5. **Evaluate (-1^{4}):** - Raising (1) to the fourth power gives (1), and the negative of (1) is (-1), which is negative.After evaluating all the expressions, I find that three of them ((-left(-3right)), (left(-3right)^{2}), and (|-9|)) are positive numbers.

💡Okay, so I've got this math problem here, and I need to figure out the relationship between three variables: a, b, and c. The problem gives me expressions for each of them, and I need to compare their values to see which one is less than the others. The options are A, B, C, or D, each giving a different order. Let me try to break this down step by step.First, let's look at what each variable equals:- a is equal to negative one-third multiplied by nine. So, a = -1/3 × 9.- b is equal to two minus four. So, b = 2 - 4.- c is equal to two divided by negative one-half. So, c = 2 ÷ (-1/2).Alright, let's calculate each one individually.Starting with a: a = -1/3 × 9. Hmm, multiplying fractions can sometimes be tricky, but I think I can handle this. So, -1/3 times 9. Well, 9 divided by 3 is 3, right? So, 1/3 of 9 is 3, and since it's negative, it should be -3. So, a equals -3.Next, b: b = 2 - 4. That seems straightforward. Subtracting 4 from 2. Well, 2 minus 4 is -2. So, b equals -2.Now, c: c = 2 ÷ (-1/2). Dividing by a fraction can be a bit confusing, but I remember that dividing by a fraction is the same as multiplying by its reciprocal. So, 2 divided by (-1/2) is the same as 2 multiplied by (-2/1), right? Because the reciprocal of -1/2 is -2/1. So, 2 times -2 is -4. Therefore, c equals -4.Okay, so now I have the values:- a = -3- b = -2- c = -4Now, I need to figure out the relationship between these three numbers. The question is asking which is less than the others, so I need to compare them in order from least to greatest.Let me list them out:- c = -4- a = -3- b = -2So, starting from the smallest, c is -4, which is less than a, which is -3, and a is less than b, which is -2. So, the order from least to greatest is c < a < b.Looking back at the options given:A: a < b < cB: b < c < aC: c < a < bD: c < b < aSo, the correct order is c < a < b, which is option C.Wait, let me double-check my calculations to make sure I didn't make any mistakes.For a: -1/3 × 9. Yeah, 9 divided by 3 is 3, and with the negative sign, it's -3. That seems right.For b: 2 - 4 is definitely -2. No issues there.For c: 2 ÷ (-1/2). Dividing by a fraction is multiplying by its reciprocal, so 2 × (-2) is -4. That checks out.So, the values are correct, and the order is c < a < b. So, option C is the right answer.I think that's it. I don't see any mistakes in my reasoning, so I'm confident that option C is correct.

💡Okay, so I've got this problem about designing 10 bus routes in a city. The goal is to set up the stops in such a way that if you pick any 8 routes, there's at least one stop that none of those 8 routes go through. But if you pick any 9 routes, they should cover all the stops. Hmm, that sounds a bit tricky, but let me try to break it down.First, I need to visualize what's being asked. We have 10 bus routes, which I can think of as lines on a map. Each route has several stops, which are points where the buses pick up and drop off passengers. The key here is that for any 8 routes, there's at least one stop that's not on any of those 8 routes. But if you take any 9 routes, they should cover all the stops. So, it's like saying the stops are arranged in such a way that no 8 routes can cover everything, but 9 routes can.Maybe I can think of this in terms of sets and subsets. Each bus route is a set of stops, and the problem is about the intersection and union of these sets. Specifically, the union of any 8 sets should not cover the entire set of stops, but the union of any 9 sets should cover all stops.Let me consider how many stops there are in total. If I have 10 routes, and each route has a certain number of stops, how does that relate to the total number of stops? If I have too few stops, it might be hard to ensure that 8 routes miss at least one stop. On the other hand, too many stops might make it difficult for 9 routes to cover everything.Wait, maybe I can use some combinatorial principles here. If I have 10 routes, and each pair of routes intersects at a unique stop, that could be a way to ensure that each stop is shared by exactly two routes. Let me see: if each pair of routes intersects at one stop, then the total number of stops would be the number of ways to choose 2 routes out of 10, which is C(10,2) = 45 stops. So, there would be 45 unique stops, each shared by exactly two routes.Now, if I pick any 8 routes, how many stops do they cover? Each route has 9 stops (since it intersects with the other 9 routes). So, 8 routes would have 8*9 = 72 stops, but since each stop is shared by two routes, I need to divide by 2 to avoid double-counting. So, 72/2 = 36 stops. But there are 45 stops in total, so 45 - 36 = 9 stops are not covered by these 8 routes. That means there are 9 stops that none of the 8 routes go through. But the problem only requires that there exists at least one stop not covered by any of the 8 routes, so this seems to satisfy that condition.Now, what about 9 routes? If I pick any 9 routes, how many stops do they cover? Each route has 9 stops, so 9*9 = 81 stops. Again, since each stop is shared by two routes, I divide by 2, getting 40.5. Wait, that doesn't make sense because the number of stops should be an integer. Maybe I made a mistake here.Let me think again. If I have 9 routes, each intersecting with the other 8 routes, that gives 9*8 = 72 intersections, but each stop is shared by two routes, so 72/2 = 36 stops. But there are also the intersections between these 9 routes and the 10th route, which are 9 more stops. So, in total, 36 + 9 = 45 stops. That makes sense because all 45 stops are covered by the 9 routes. So, any 9 routes will cover all the stops.Wait, but earlier when I calculated for 8 routes, I got 36 stops covered, leaving 9 stops uncovered. But the problem only requires that at least one stop is uncovered, not necessarily all 9. So, this setup seems to satisfy both conditions: any 8 routes miss at least one stop, and any 9 routes cover all stops.But I'm a bit confused about the calculation for 9 routes. Initially, I thought it was 9*9/2 = 40.5, which is not possible, but then I realized that I need to account for the intersections with the 10th route as well. So, it's actually 36 stops from the intersections among the 9 routes plus 9 stops from their intersections with the 10th route, totaling 45 stops. That makes sense.So, in summary, if I design 10 bus routes such that each pair of routes intersects at a unique stop, resulting in 45 stops, then any 8 routes will cover 36 stops, leaving 9 stops uncovered, and any 9 routes will cover all 45 stops. This satisfies the problem's conditions.But I'm still wondering if there's a more efficient way to do this with fewer stops or if this is the minimal number of stops required. Also, I'm curious if this setup works in a real-world scenario, considering that bus routes might not always intersect at unique points, especially in a city with various obstacles and road layouts.Another thought: what if instead of using intersections, I use a different method to assign stops to routes? Maybe using some combinatorial design where each stop is assigned to a specific number of routes, ensuring the coverage properties. But the intersection idea seems to naturally satisfy the conditions, so maybe that's the way to go.I should also consider if there are any potential issues with this design. For example, if two routes share a stop, does that mean buses from both routes will be at that stop, potentially causing congestion? In a real city, that might be a concern, but since the problem doesn't specify any constraints on the number of routes per stop or the frequency of buses, I think it's acceptable for the sake of solving the problem.Additionally, I wonder if this problem is related to any known mathematical concepts or theorems. It seems similar to covering designs or something in combinatorics where you want certain coverage properties. Maybe it's related to the concept of a hypergraph where each hyperedge connects multiple vertices, and we're looking for a specific covering property.Overall, I think the approach of using intersections to create unique stops for each pair of routes works well for this problem. It ensures that any 8 routes miss at least one stop, and any 9 routes cover all stops. I should probably formalize this a bit more to make sure there are no gaps in the logic, but intuitively, it seems solid.