💡First, I need to determine the number of women, men, and children at the beginning of the celebration. The total number of guests is 60, with half being women. This means there are 30 women. There are 15 men, so the remaining guests must be children. Subtracting the number of women and men from the total gives 15 children.Next, I know that 50 people stayed after some men and 5 children left. This means that 10 people left the celebration in total. Since 5 of those who left are children, the remaining 5 must be men.Finally, to find the ratio of the men who left to the total number of men, I divide the number of men who left (5) by the total number of men (15). This simplifies to a ratio of 1:3.

💡Okay, so I have this problem here: when m is divided by 288, the remainder is 47. And when the same m is divided by another number, the remainder is 23. I need to find what that second divisor is. Hmm, let me think about how to approach this.First, I remember that when a number m is divided by another number, say d, and leaves a remainder r, it can be expressed as m = d * q + r, where q is the quotient. So, applying that to the first part, m divided by 288 leaves a remainder of 47. That means m can be written as 288 * k + 47, where k is some integer.Now, the same m when divided by another number, let's call it n, leaves a remainder of 23. So, m can also be written as n * l + 23, where l is another integer. Since both expressions equal m, I can set them equal to each other: 288k + 47 = n * l + 23.If I rearrange this equation, I get n * l = 288k + 47 - 23, which simplifies to n * l = 288k + 24. So, n times some integer l equals 288k plus 24. This suggests that n must be a divisor of 288k + 24.But wait, I don't know what k is. It could be any integer, so maybe I should look for a common factor between 288 and 24. Let me think about the factors of 24 because 24 is a part of the equation. The factors of 24 are 1, 2, 3, 4, 6, 8, 12, and 24.Since the remainder when m is divided by n is 23, n must be greater than 23. Otherwise, the remainder wouldn't make sense because the remainder has to be less than the divisor. So, looking at the factors of 24, the only one greater than 23 is 24 itself. That makes me think that n could be 24.Let me check if that makes sense. If m = 288k + 47, and n is 24, then when I divide m by 24, what's the remainder? Well, 288 is divisible by 24 because 24 * 12 = 288. So, 288k divided by 24 would leave no remainder. Then, the remainder would come from the 47. So, 47 divided by 24 is 1 with a remainder of 23. That matches the given remainder of 23.Wait, that seems to fit perfectly. So, when m is divided by 24, the remainder is indeed 23. Therefore, the second divisor must be 24.I think that makes sense. I started by expressing m in terms of both divisors, set them equal, and then looked for a common factor that would satisfy the remainder condition. Since 24 is the only factor of 24 greater than 23, it must be the answer.

💡Alright, so I've got this problem about a game played on an infinite checkerboard. The board starts with counters filling exactly a rectangle of size 3k × n. The rules are that any counter can jump over an adjacent counter (either vertically or horizontally) into an empty square, and when it does, the counter that was jumped over gets removed. The goal is to prove that it's impossible for exactly one counter to remain on the board.Hmm, okay. Let me try to visualize this. Imagine a big rectangle of counters on an infinite grid. The counters can jump over each other, removing the jumped-over counter. So, it's like a game of checkers but with specific rules about how you can jump and remove counters.First, I need to understand the mechanics of the game better. If a counter jumps over another, it lands in an empty square, and the jumped counter is removed. So, each move reduces the total number of counters by one. That means the number of counters decreases by one each time a move is made.Starting with a rectangle of size 3k × n, the total number of counters initially is 3kn. Since each move removes one counter, the number of counters will always be 3kn minus the number of moves made. So, if we want to end up with exactly one counter, that would mean we've made 3kn - 1 moves. But is that possible?Wait, maybe I should think about this differently. Maybe instead of counting the number of moves, I should look for some invariant or property that remains unchanged throughout the game. If I can find such a property, I can use it to show that having exactly one counter left is impossible.I remember that in some similar problems, people use coloring arguments or parity arguments. Maybe I can color the checkerboard in a certain way and see how the moves affect the counts of counters on different colors.Let me try coloring the checkerboard in three colors, say red, green, and blue, repeating every three squares. So, each row would be red, green, blue, red, green, blue, and so on. Then, the next row would start with green, blue, red, and so on, creating a repeating pattern.If I do this, then in the initial rectangle of size 3k × n, each color class will have exactly kn counters. That's because the rectangle is 3k wide, so each color repeats k times in each row, and there are n rows. So, each color has kn counters initially.Now, let's see how a move affects the counts of each color. Suppose a counter jumps over another counter. The jumping counter moves from its current position to an empty square, and the jumped-over counter is removed. So, the jumping counter moves from one color to another, and the jumped-over counter is removed from its color.Let me think about the colors involved in a jump. If a counter jumps over another counter, it moves two squares in some direction. So, if it starts on a red square, it jumps over a green square and lands on a blue square, or something like that. Depending on the direction of the jump, the colors would change accordingly.Wait, actually, if the board is colored with three colors repeating every three squares, then moving two squares in any direction would cycle through the colors. So, jumping over a counter would mean moving from one color to another, specifically skipping one color.So, if a counter is on a red square and jumps over a green square, it lands on a blue square. Similarly, a counter on a green square jumping over a blue square would land on a red square, and so on.Therefore, each move affects the counts of two color classes: the jumping counter leaves its original color and lands on another color, and the jumped-over counter is removed from its color.So, if we denote the number of counters on red, green, and blue squares as R, G, and B respectively, then each move would decrease one of these by one (the jumped-over counter) and increase another by one (the jumping counter moving to a new color). The third color remains unchanged.Wait, but actually, the jumping counter moves from one color to another, so R decreases by one and B increases by one, for example, while G decreases by one because the jumped-over counter is removed. So, overall, R decreases by one, G decreases by one, and B increases by one. Or depending on the direction of the jump, it could be a different combination.But in any case, each move affects all three color counts: one color loses a counter (the jumped-over one), and another color gains a counter (the jumping one). So, the differences between the color counts change in a specific way.Now, let's think about the initial counts. Since the rectangle is 3k × n, each color has exactly kn counters. So, R = G = B = kn initially.If we make a move, say, R decreases by one, G decreases by one, and B increases by one. So, R becomes kn - 1, G becomes kn - 1, and B becomes kn + 1.Wait, but that's not possible because the total number of counters is 3kn, and after one move, it's 3kn - 1. So, the total counts should still add up to 3kn - 1.But in terms of the color counts, R + G + B = 3kn - 1.But initially, R = G = B = kn, so R + G + B = 3kn.After one move, R = kn - 1, G = kn - 1, B = kn + 1, so R + G + B = (kn - 1) + (kn - 1) + (kn + 1) = 3kn - 1, which matches.Okay, so each move affects the color counts in a way that two colors decrease by one and one color increases by one.Now, let's think about the parity of the color counts. Initially, R, G, and B are all equal to kn, which is an integer. So, depending on whether kn is even or odd, the parities of R, G, and B are the same.If kn is even, then R, G, and B are all even. If kn is odd, then R, G, and B are all odd.Now, let's see how a move affects the parities. Suppose kn is even, so R, G, B are all even.After a move, R decreases by one (becomes odd), G decreases by one (becomes odd), and B increases by one (becomes odd). So, all three color counts become odd.Wait, that's interesting. So, if we start with all even, after one move, all become odd.Similarly, if we start with all odd, after one move, all become even.So, the parities of the color counts toggle with each move.Now, if we want to end up with exactly one counter left, that counter must be on one of the color classes, say red. So, R = 1, and G = B = 0.But initially, R, G, B are all equal to kn, which is either all even or all odd.If kn is even, then R, G, B are all even. After each move, their parities toggle. So, after an odd number of moves, they would all be odd, and after an even number of moves, they would all be even.To end up with R = 1, G = B = 0, we need R to be odd (since 1 is odd) and G and B to be even (since 0 is even). But initially, if kn is even, R, G, B are all even. After an odd number of moves, they would all be odd, which doesn't match the desired final state. After an even number of moves, they would all be even, which also doesn't match.Similarly, if kn is odd, initially R, G, B are all odd. After one move, they all become even. After two moves, they all become odd again, and so on. So, to end up with R = 1 (odd), G = B = 0 (even), we would need R to be odd and G, B to be even. But starting from all odd, after an even number of moves, they would all be even, which doesn't help. After an odd number of moves, they would all be odd, which also doesn't help.Therefore, it's impossible to reach a state where one color has 1 counter and the others have 0, because the parities of the color counts cannot align in that way.Wait, but let me double-check this reasoning. Suppose kn is even, so R, G, B start as even. After one move, they all become odd. After two moves, they all become even again. So, to get to R = 1, which is odd, we need an odd number of moves. But then G and B would also be odd, which contradicts the desired state where G and B are 0 (even). Similarly, if kn is odd, starting with all odd, after one move, all become even, which doesn't help because we need R to be odd.Therefore, regardless of whether kn is even or odd, it's impossible to end up with exactly one counter on the board because the parities of the color counts cannot be satisfied for such a state.So, this seems like a solid argument. By using a three-coloring of the checkerboard and analyzing the parities of the counts on each color, we can show that it's impossible to end up with exactly one counter.I think this makes sense. The key idea is that the moves preserve certain parity conditions among the color counts, and the desired final state violates those conditions. Therefore, it's impossible to achieve.**Final Answer**boxed{text{It is impossible for exactly one counter to remain on the board.}}

💡First, I need to simplify the expression step by step according to the order of operations.Starting with 15 minus 2, which equals 13.Next, adding 4 to 13 gives me 17.Then, dividing 17 by 1 results in 17.After that, dividing 17 by 2 gives me 8.5.Now, the equation is 8.5 multiplied by an unknown number equals 77.To find the unknown number, I divide 77 by 8.5.Calculating this gives me approximately 9.06.So, the number being multiplied after the divisions is approximately 9.06.

💡Alright, so I have this problem where I need to show that it's possible to color some squares of an ( n times n ) board black such that each row and column contains exactly ( k ) black squares, but there's no rectangle with sides parallel to the board that has its four corner squares black. The given ( n ) is ( k^2 - k + 1 ), where ( k ) is a prime number plus one. Hmm, okay, let me try to unpack this step by step.First, let's understand the parameters. If ( k ) is a prime number plus one, that means ( k = p + 1 ) where ( p ) is a prime. So, substituting that into the expression for ( n ), we get:[n = (p + 1)^2 - (p + 1) + 1 = p^2 + 2p + 1 - p - 1 + 1 = p^2 + p + 1]So, ( n = p^2 + p + 1 ). That's interesting because it reminds me of the number of points in a finite projective plane of order ( p ). In finite projective planes, the number of points is ( p^2 + p + 1 ), and each line contains ( p + 1 ) points. Maybe there's a connection here.Let me recall that in finite projective planes, every pair of lines intersects in exactly one point, and every pair of points lies on exactly one line. Also, each line has the same number of points, and each point is on the same number of lines. This seems similar to what we need for our coloring problem.In our case, we need each row (which can be thought of as a line) to have exactly ( k = p + 1 ) black squares (which can be thought of as points). Similarly, each column (another set of lines) should also have exactly ( k ) black squares. Additionally, we don't want any rectangles, which would correspond to two lines intersecting at two different points, forming a rectangle.Wait, in finite projective planes, any two lines intersect at exactly one point. So, if we model our rows and columns as lines in a projective plane, then any two rows would intersect at exactly one column, and any two columns would intersect at exactly one row. This would prevent the formation of rectangles because a rectangle would require two rows intersecting at two different columns, which isn't possible in a projective plane.So, maybe we can use the incidence structure of a finite projective plane to define our coloring. Each row corresponds to a line, and each column corresponds to a point. A black square at position (i, j) would mean that line i passes through point j. Since each line has exactly ( k = p + 1 ) points, each row would have exactly ( k ) black squares. Similarly, since each point is on exactly ( k ) lines, each column would have exactly ( k ) black squares.But does this construction avoid rectangles? Well, as I thought earlier, in a projective plane, any two lines intersect at exactly one point, so there can't be two different columns where two rows both have black squares. That means there can't be a rectangle because a rectangle would require two rows and two columns where all four intersections are black, which would imply two lines intersecting at two different points, which isn't possible in a projective plane.Okay, that seems promising. But I should make sure that this actually works. Let me think about how to explicitly construct such a coloring.First, let's consider the finite projective plane of order ( p ). It has ( n = p^2 + p + 1 ) points and the same number of lines. Each line contains ( k = p + 1 ) points, and each point is on ( k ) lines. The incidence relation between points and lines is such that any two lines meet at exactly one point, and any two points lie on exactly one line.So, if we represent each row as a line and each column as a point, then the incidence matrix of the projective plane would be a ( n times n ) matrix where a 1 indicates that the line (row) passes through the point (column). This matrix would have exactly ( k ) ones in each row and each column, satisfying the first condition.Now, to ensure there are no rectangles, we need to ensure that there are no two rows and two columns such that all four intersections are 1s. In terms of the projective plane, this would mean that there are no two lines that intersect at two different points, which is impossible because any two lines intersect at exactly one point. Therefore, the incidence matrix of the projective plane satisfies both conditions: each row and column has exactly ( k ) ones, and there are no rectangles formed by four ones.But wait, is this always possible? I mean, does a finite projective plane of order ( p ) exist for every prime ( p )? I recall that finite projective planes are known to exist for all orders that are prime powers. Since ( p ) is a prime, ( p ) is certainly a prime power, so a finite projective plane of order ( p ) does exist. Therefore, this construction is valid.Let me try to summarize the steps:1. Recognize that ( n = p^2 + p + 1 ) is the number of points (and lines) in a finite projective plane of order ( p ).2. Note that each line contains ( k = p + 1 ) points, and each point is on ( k ) lines.3. Use the incidence matrix of the projective plane, where rows represent lines and columns represent points.4. In this matrix, each row and column has exactly ( k ) ones, satisfying the first condition.5. Since any two lines intersect at exactly one point, there are no two rows (lines) that share two columns (points), thus avoiding rectangles.6. Conclude that this incidence matrix provides the required coloring.I think that covers it. But just to be thorough, let me consider a small example to see if this works.Let's take ( p = 2 ), so ( k = 3 ) and ( n = 2^2 + 2 + 1 = 7 ). The finite projective plane of order 2 is the Fano plane, which has 7 points and 7 lines, each line containing 3 points, and each point on 3 lines.The incidence matrix of the Fano plane is a 7x7 matrix with exactly 3 ones in each row and column, and no two rows share more than one common column with a one. This means there are no rectangles, as required.For example, one possible incidence matrix for the Fano plane is:[begin{bmatrix}1 & 1 & 1 & 0 & 0 & 0 & 0 1 & 0 & 0 & 1 & 1 & 0 & 0 0 & 1 & 0 & 1 & 0 & 1 & 0 0 & 0 & 1 & 0 & 1 & 1 & 0 0 & 0 & 0 & 1 & 0 & 0 & 1 0 & 0 & 0 & 0 & 1 & 0 & 1 0 & 0 & 0 & 0 & 0 & 1 & 1 end{bmatrix}]Each row has exactly 3 ones, each column has exactly 3 ones, and if you look for any two rows, they share exactly one column with a one. For instance, rows 1 and 2 share column 1, rows 1 and 3 share column 2, and so on. There's no pair of rows that share two columns, which would form a rectangle.This example confirms that the construction works for ( p = 2 ). I can assume it works similarly for higher primes, as the properties of finite projective planes hold generally.One thing I'm a bit fuzzy on is why exactly the finite projective plane avoids rectangles. It's because of the property that any two lines intersect in exactly one point, right? So, in terms of the matrix, any two rows (lines) have exactly one column (point) where they both have a one. Therefore, you can't have two rows sharing two columns, which would be necessary for a rectangle.Yes, that makes sense. So, the key property here is the uniqueness of intersections in the projective plane, which translates to the uniqueness of shared columns between any two rows in the incidence matrix.I also wonder if there are other ways to construct such a coloring without using projective planes. Maybe combinatorial designs or something else? But I think the projective plane gives a straightforward and elegant solution because it inherently satisfies both the regularity condition (each row and column having exactly ( k ) ones) and the no-rectangle condition due to its intersection properties.Another thought: what if ( k ) isn't one more than a prime? Would this construction still work? Well, finite projective planes are known to exist for orders that are prime powers, but it's not known if they exist for all positive integers. In fact, it's a famous open problem in mathematics whether projective planes exist for all orders that are prime powers. But in our case, since ( k ) is one more than a prime, ( p ), and ( p ) is a prime, we know that a projective plane of order ( p ) exists, so our construction is valid.I also recall that the problem statement specifies that ( k ) is a prime plus one, which ensures that ( n = k^2 - k + 1 ) is of the form ( p^2 + p + 1 ), fitting the projective plane construction. If ( k ) weren't of that form, this particular method might not apply, but since it is, we're good.Let me check if I've covered all the bases:1. Understanding the relationship between ( n ), ( k ), and ( p ).2. Recognizing the connection to finite projective planes.3. Verifying that the incidence matrix satisfies the required conditions.4. Using a small example to confirm the reasoning.5. Ensuring that the properties of the projective plane prevent rectangles.I think I've covered all these points. I'm feeling pretty confident about this approach. It might be a bit abstract, but the connection to finite projective planes seems solid, and the example with the Fano plane helps concretize the idea.One last thing: the problem says "show that it is possible to color some squares...", so I don't need to provide an explicit coloring for any ( n ), just to argue that such a coloring exists based on the properties of finite projective planes. Since finite projective planes exist for prime orders, and our ( n ) is constructed from such a prime, the argument holds.Yeah, I think that's a thorough walk-through. I've considered the definitions, connected them to known structures, verified with an example, and ensured that all conditions are satisfied. I don't see any gaps in the logic, so I'm ready to present the solution.

💡Okay, so I have this algebra problem here, and I need to find the value of an expression involving some variables. Let me read it again carefully."Real numbers (a) and (b) are opposite to each other, and (c) and (d) are reciprocals of each other. The absolute value of (x) is (sqrt{3}). Find the value of the algebraic expression (x^{2}+sqrt{a+b+4}-sqrt[3]{{27cd}})."Alright, let's break this down step by step. First, I need to understand what it means for two numbers to be opposites and reciprocals.So, (a) and (b) are opposites. That means if (a) is some number, then (b) is its negative. In other words, (b = -a). So, if I add them together, (a + b), that should be zero because (a + (-a) = 0). Got that.Next, (c) and (d) are reciprocals of each other. That means (d = frac{1}{c}) or (c = frac{1}{d}). So, when you multiply them together, (c times d = c times frac{1}{c} = 1). So, their product is 1. That's useful.Now, the absolute value of (x) is (sqrt{3}). Absolute value means the distance from zero on the number line, so (x) can be either (sqrt{3}) or (-sqrt{3}). But since we're going to square (x) in the expression, the square of either (sqrt{3}) or (-sqrt{3}) will be the same, right? Because squaring a negative number gives a positive result. So, (x^2) will be ((sqrt{3})^2 = 3) regardless of whether (x) is positive or negative.Alright, so let's rewrite the expression with what we know:(x^{2} + sqrt{a + b + 4} - sqrt[3]{27cd})We know (a + b = 0), so the expression inside the square root becomes (0 + 4 = 4). So, (sqrt{4}) is 2.Next, (cd = 1), so the expression inside the cube root becomes (27 times 1 = 27). The cube root of 27 is 3 because (3^3 = 27).So now, substituting all these back into the expression:(x^2 + sqrt{4} - sqrt[3]{27})Which simplifies to:(3 + 2 - 3)Wait, that seems straightforward. Let me double-check each step to make sure I didn't make a mistake.1. (a) and (b) are opposites: (a + b = 0). Correct.2. (c) and (d) are reciprocals: (cd = 1). Correct.3. (|x| = sqrt{3}) implies (x^2 = 3). Correct.4. Substitute into the expression: - (x^2 = 3) - (sqrt{a + b + 4} = sqrt{0 + 4} = 2) - (sqrt[3]{27cd} = sqrt[3]{27 times 1} = sqrt[3]{27} = 3)5. So, (3 + 2 - 3 = 2). That seems right.Hmm, is there any chance I could have messed up the order of operations or misunderstood the problem? Let me see.The expression is (x^{2} + sqrt{a + b + 4} - sqrt[3]{27cd}). I substituted each part correctly. The square root and cube root operations are straightforward. I don't think there's any ambiguity in the expression.Wait, could the cube root of 27 be something else? No, because cube root of 27 is definitely 3 since (3 times 3 times 3 = 27). Similarly, the square root of 4 is 2, not -2, because square roots are defined as the principal (non-negative) root.Also, (x^2) is definitely 3 because whether (x) is positive or negative, squaring it gives a positive result. So, that part is solid.So, putting it all together: 3 (from (x^2)) plus 2 (from the square root) minus 3 (from the cube root) equals 2. Yeah, that seems correct.I think I'm confident that the value of the expression is 2. I don't see any errors in my reasoning.**Final Answer**boxed{2}