💡Okay, so I have this geometry problem where I need to construct a point on a given circle that is at a specific distance from a given line, using just a compass and straightedge. Hmm, let me think about how to approach this.First, I need to visualize the problem. There's a circle with center O and a line l somewhere on the plane. I need to find a point on the circle that's exactly d units away from line l. Alright, so I guess I need to somehow relate the circle and the line in a way that the distance condition is satisfied.I remember that the set of all points at a fixed distance from a line forms two parallel lines on either side of the original line. So, if I can construct two lines parallel to l, each at a distance d from l, then the points where these parallel lines intersect the circle will be the points I'm looking for.Wait, but how do I construct a line parallel to l at a distance d? I think I can use the compass to measure the distance d from line l. Maybe I can pick a point on line l, draw a perpendicular to l, and then mark off a distance d on that perpendicular. Then, using that point, I can construct a parallel line.Let me try to outline the steps:1. **Given Elements:** - Circle with center O and radius R. - Line l. - Distance d.2. **Construct Parallel Lines:** - Choose a point P on line l. - Construct a perpendicular to l at point P. - Measure distance d along this perpendicular to mark point Q. - Draw a line through Q parallel to l. This will be one of the lines at distance d from l. - Repeat the process on the other side of l to get the second parallel line.3. **Find Intersection Points:** - Now, I need to find where these parallel lines intersect the given circle. - Use the straightedge to draw the parallel lines and see where they cut through the circle.4. **Identify the Points:** - The points where the parallel lines intersect the circle are the points that are exactly d units away from line l.Wait, but what if the distance d is such that the parallel lines don't intersect the circle? Then there would be no solution, right? So, I guess d has to be less than or equal to the distance from the center O to line l plus the radius R, or something like that. Hmm, maybe I need to consider the position of the circle relative to line l.Let me think about the distance from the center O to line l. If I denote that distance as h, then the maximum distance a point on the circle can be from line l is h + R, and the minimum is |h - R|. So, for the parallel lines to intersect the circle, d must satisfy |h - R| ≤ d ≤ h + R. Otherwise, there would be no such points.Okay, so assuming that d is within this range, I can proceed with constructing the parallel lines and finding their intersections with the circle.Let me try to write down the steps more formally:1. **Given:** - Circle with center O and radius R. - Line l. - Distance d.2. **Find the distance h from O to l:** - Construct a perpendicular from O to l, meeting l at point P. - Measure the length OP, which is h.3. **Check if d is within the feasible range:** - If d < |h - R| or d > h + R, then no such point exists. - Otherwise, proceed.4. **Construct the two parallel lines at distance d from l:** - On one side of l, construct a line l1 parallel to l at distance d. - On the other side, construct a line l2 parallel to l at distance d.5. **Find intersections of l1 and l2 with the circle:** - Use the straightedge to draw l1 and l2. - The points where these lines intersect the circle are the desired points.6. **Verify the distance:** - To ensure accuracy, measure the distance from each intersection point to line l to confirm it's exactly d.Hmm, I think that covers the construction. But let me double-check if there's a more efficient way or if I'm missing something.Wait, when constructing the parallel lines, I need to make sure they are indeed parallel and at the exact distance d. Using the compass to measure the perpendicular distance should help with that. Also, when drawing the parallel lines, using the straightedge accurately is crucial to ensure they don't deviate.Another thing to consider is the number of intersection points. Depending on the position and size of the circle, each parallel line could intersect the circle at two points, one point, or none. So, potentially, there could be up to four points (two on each parallel line) that satisfy the condition.But in the problem statement, it just says "construct a point," so maybe only one is needed. However, it's good to know that there could be multiple solutions.Alright, I think I have a clear plan now. Let me summarize the steps to make sure I didn't miss anything.**Step-by-Step Explanation:**1. **Identify Given Elements:** - Circle with center O and radius R. - Line l. - Desired distance d.2. **Calculate the Distance from O to l:** - Draw a perpendicular from O to line l, intersecting at point P. - Measure the length OP; this is h.3. **Determine Feasibility:** - Check if d is between |h - R| and h + R. - If not, construction is impossible. - If yes, proceed.4. **Construct Parallel Lines:** - On one side of l, construct line l1 parallel to l at distance d. - Choose a point Q on l. - Draw a perpendicular to l at Q. - Measure distance d along this perpendicular to mark point Q'. - Draw line l1 through Q' parallel to l. - Similarly, construct line l2 on the other side of l at distance d.5. **Find Intersection Points:** - Use the straightedge to draw l1 and l2. - Identify points A and B where l1 intersects the circle. - Identify points C and D where l2 intersects the circle.6. **Verify Distances:** - Ensure that each of these points (A, B, C, D) is exactly d units away from line l.7. **Conclusion:** - Points A, B, C, and D are the desired points on the circle at distance d from line l.I think that's comprehensive. I should also consider edge cases, like when d equals h + R or |h - R|, which would result in the parallel lines being tangent to the circle, giving only one intersection point each. But in general, assuming d is strictly between |h - R| and h + R, there will be two intersection points on each parallel line.Overall, this seems like a solid approach. I just need to make sure each step is executed carefully with the compass and straightedge to maintain accuracy.

💡Okay, so I have this problem here: "By arbitrarily rearranging the digits of the five-digit number 12345, how many prime numbers can be obtained?" The options are (A) 4, (B) 8, (C) 12, and (D) 0. Hmm, interesting. I remember that prime numbers are numbers greater than 1 that have no positive divisors other than 1 and themselves. So, I need to figure out how many different five-digit numbers I can make by rearranging 1, 2, 3, 4, and 5, and then check which of those are prime.First, let me think about how many total permutations there are. Since all the digits are unique, the number of permutations is 5 factorial, which is 5! = 5 × 4 × 3 × 2 × 1 = 120. So, there are 120 different five-digit numbers I can make by rearranging these digits. That's a lot! Checking each one for primality would be time-consuming, so I need a smarter approach.I remember that for a number to be prime, it can't be divisible by any number other than 1 and itself. So, maybe I can use some divisibility rules to eliminate numbers that can't be prime. Let me recall some common divisibility rules:1. **Divisibility by 2**: If the last digit is even, the number is divisible by 2.2. **Divisibility by 3**: If the sum of the digits is divisible by 3, then the number is divisible by 3.3. **Divisibility by 5**: If the last digit is 0 or 5, the number is divisible by 5.Since we're dealing with five-digit numbers made from 1, 2, 3, 4, and 5, let's see how these rules apply.First, let's check the sum of the digits. The digits are 1, 2, 3, 4, and 5. Adding them up: 1 + 2 + 3 + 4 + 5 = 15. Hmm, 15 is divisible by 3 (since 15 ÷ 3 = 5). That means any permutation of these digits will also have a digit sum of 15, which is divisible by 3. Therefore, every five-digit number formed by rearranging these digits will be divisible by 3.Wait, that's a crucial point. If all these numbers are divisible by 3, then none of them can be prime, right? Because a prime number has only two distinct positive divisors: 1 and itself. If a number is divisible by 3, then 3 is another divisor, making it composite, not prime.But hold on, let me make sure I'm not missing something. Is there any exception? Well, 3 itself is a prime number, but we're talking about five-digit numbers here, so none of them can be 3. Therefore, all permutations of 12345 will be divisible by 3 and hence composite.But just to double-check, let me think about the other divisibility rules. For example, if a number ends with an even digit, it's divisible by 2, which would also make it composite. Similarly, if a number ends with 5, it's divisible by 5, making it composite as well. However, since all our numbers are five-digit numbers made up of 1, 2, 3, 4, and 5, some of them will end with even digits (2 or 4) and some will end with 5. But regardless, even if they are divisible by 2 or 5, they are still composite. However, the key point here is that they are all divisible by 3, which is enough to conclude they can't be prime.Let me also consider if there's any chance that a number could be prime despite being divisible by 3. Well, no. If a number is divisible by 3, it has at least three divisors: 1, 3, and itself. Therefore, it's not prime.So, putting it all together, since every permutation of 12345 results in a number divisible by 3, none of them can be prime. Therefore, the number of prime numbers obtainable by rearranging the digits of 12345 is zero.I think that makes sense. I don't see any flaws in this reasoning. All permutations will have a digit sum of 15, which is divisible by 3, making all such numbers composite. Therefore, the answer should be (D) 0.**Final Answer**boxed{D}

💡Okay, so I have this problem about placing bishops on a chessboard. The goal is to figure out the maximum number of bishops I can put on an N x N chessboard without any two bishops attacking each other. Hmm, bishops move diagonally, right? So, if two bishops are on the same diagonal, they can attack each other. That means I need to place them in such a way that no two are on the same diagonal.First, I should probably visualize a chessboard. Let's say it's an 8x8 board for simplicity. Bishops can move any number of squares along a diagonal, so they can potentially control a lot of squares. But since I want to maximize the number of bishops without them attacking each other, I need to make sure each bishop is on a unique diagonal.Wait, how many diagonals are there on a chessboard? For an N x N board, there are two sets of diagonals: those that go from the bottom-left to the top-right and those that go from the top-left to the bottom-right. Each set has 2N - 1 diagonals. So, for an 8x8 board, that would be 15 diagonals in each direction.But I don't need to use all of them. Since bishops can attack along both sets of diagonals, I need to make sure that no two bishops share a diagonal in either direction. So, for each diagonal in both directions, I can place at most one bishop.But how does this translate to the maximum number of bishops? If I have 2N - 1 diagonals in each direction, does that mean I can place up to 2N - 1 bishops? Wait, that doesn't seem right because if I place a bishop on one diagonal, it affects both sets of diagonals.Let me think differently. Maybe I should consider the number of diagonals in one direction and then adjust for the overlap. For example, if I place a bishop on a diagonal going from bottom-left to top-right, it also occupies a diagonal going from top-left to bottom-right. So, each bishop occupies one diagonal from each set.Therefore, the maximum number of bishops should be limited by the smaller of the two sets of diagonals. But both sets have the same number of diagonals, which is 2N - 1. So, does that mean I can place 2N - 1 bishops? But that seems too high because on an 8x8 board, I know that the maximum number of non-attacking bishops is 14, which is 2*8 - 2.Wait, so maybe it's 2N - 2? Let me check that. For N=8, 2*8 - 2 = 14, which matches what I know. For N=1, it would be 0, which makes sense because you can't place any bishops on a 1x1 board without them attacking each other. For N=2, it would be 2, which also makes sense because you can place one bishop on each diagonal without them attacking each other.So, maybe the formula is 2N - 2. But why is that? Let me try to derive it.If I have N rows and N columns, each bishop occupies one row and one column, but also one diagonal in each direction. To maximize the number of bishops, I need to ensure that no two bishops share a row, column, or diagonal. Wait, no, actually, bishops don't attack along rows or columns, only along diagonals. So, they can share rows and columns as long as they don't share diagonals.But since bishops can share rows and columns, the constraint is only on the diagonals. So, the maximum number of bishops is determined by the number of diagonals they can occupy without overlapping.But each bishop occupies one diagonal from each set. So, the maximum number of bishops is limited by the number of diagonals in one set, but since they also occupy diagonals in the other set, we have to make sure that we don't double count.Wait, maybe it's better to think about it as placing bishops on different diagonals in one direction and then ensuring they don't conflict in the other direction.For example, if I place bishops on all the diagonals going from bottom-left to top-right, that would be 2N - 1 bishops. But some of these bishops would also be on the same diagonals going from top-left to bottom-right, causing them to attack each other.So, to avoid that, I need to place bishops in such a way that they don't share any diagonals in either direction. That means I need to select a subset of diagonals from one direction and ensure that they don't overlap with diagonals in the other direction.But this seems complicated. Maybe there's a simpler way. I recall that for queens, the maximum number is N because they can't share rows, columns, or diagonals. But for bishops, since they only care about diagonals, the maximum should be higher.Wait, actually, for bishops, the maximum number is 2N - 2. Let me see why.If I place bishops on all the squares of the first row except the first and last squares, that would be N - 2 bishops. Then, I can place bishops on all the squares of the last row except the first and last squares, another N - 2 bishops. But wait, that would give me 2N - 4 bishops, which is less than 2N - 2.Hmm, maybe that's not the right approach. Let me try another way.If I place bishops on all the squares of the main diagonal, that's N bishops. But they all share the same diagonal, so they attack each other. That's not good.What if I place bishops on every other diagonal? For example, place a bishop on the first diagonal, skip the next, place another, and so on. But I'm not sure if that maximizes the number.Wait, maybe I should consider that each bishop occupies one diagonal from each set. So, if I have D1 diagonals in one direction and D2 in the other, the maximum number of bishops is the minimum of D1 and D2. But since D1 = D2 = 2N - 1, the maximum should be 2N - 1. But earlier, I thought it was 2N - 2.I'm getting confused. Let me look for a pattern.For N=1: 0 bishops (can't place any without attacking)For N=2: 2 bishops (place them on opposite corners)For N=3: 4 bishops (place them on the four corners and the center)Wait, no, on a 3x3 board, placing bishops on the four corners would mean they attack each other because they share diagonals. So, actually, you can only place 2 bishops on a 3x3 board without them attacking each other.Wait, that contradicts my earlier thought. Maybe I need to rethink.Actually, on a 3x3 board, the maximum number of non-attacking bishops is 2. Because if you place one on a1 and one on c3, they attack each other. So, you have to place them on different diagonals.Wait, no, if you place one on a1 and one on b3, they don't attack each other because they are on different diagonals. Similarly, you can place one on c1 and one on a3. So, actually, you can place 4 bishops on a 3x3 board without them attacking each other.Wait, no, that's not right. If you place a bishop on a1, it controls the main diagonal. Placing another on c3 also controls the main diagonal, so they attack each other. Similarly, placing on b3 and a3 would mean they share a diagonal.Wait, maybe I'm overcomplicating. Let me try to visualize.On a 3x3 board:- a1, b2, c3 are on the main diagonal.- a3, b2, c1 are on the anti-diagonal.- Then there are other diagonals: a2, b3 and a2, b1; b1, c2 and b3, c2.So, to place bishops without attacking each other, I need to place them on different diagonals in both directions.If I place a bishop on a1, it occupies the main diagonal and the a1-a3 diagonal. Then, I can't place any other bishops on the main diagonal or the a1-a3 diagonal.Similarly, if I place a bishop on c3, it occupies the main diagonal and the c3-c1 diagonal. So, placing bishops on a1 and c3 would mean they share the main diagonal, so they attack each other.Therefore, on a 3x3 board, the maximum number of non-attacking bishops is 2. For example, place one on a1 and one on c3, but wait, they attack each other. So, maybe place one on a1 and one on b3. They don't share a diagonal. Similarly, place one on c1 and one on b3. Wait, but b3 is on the same diagonal as c1.I'm getting confused. Maybe the maximum is 2 for N=3.Wait, no, actually, you can place 4 bishops on a 3x3 board without them attacking each other. Place them on a1, a3, c1, and c3. Let's see: a1 and a3 are on the same column, but bishops don't attack along columns. Similarly, c1 and c3 are on the same column. But do any of them share a diagonal?a1 and c3 are on the main diagonal, so they attack each other. Similarly, a3 and c1 are on the anti-diagonal, so they attack each other. Therefore, placing bishops on all four corners would mean that a1 attacks c3 and a3 attacks c1. So, that's not allowed.Therefore, the maximum number of non-attacking bishops on a 3x3 board is 2. For example, place one on a1 and one on b3. They don't share a diagonal. Similarly, place one on c1 and one on b3, but that would mean b3 is attacking both c1 and a1 if a1 is also occupied.Wait, no, if I place a bishop on a1 and one on c3, they attack each other. So, I can't have both. Similarly, placing on a3 and c1 would mean they attack each other.Therefore, the maximum is 2 bishops on a 3x3 board.Wait, but I thought earlier it was 4. Maybe I was wrong.Let me check online or recall. I think the maximum number of non-attacking bishops on an N x N board is 2N - 2. So, for N=3, that would be 4. But my earlier reasoning suggests only 2. So, I must be missing something.Wait, maybe I can place bishops on a1, a3, c1, and c3 without them attacking each other. But as I thought earlier, a1 and c3 are on the same diagonal, so they attack each other. Similarly, a3 and c1 are on the same diagonal.Therefore, that's not possible. So, maybe the formula is not 2N - 2 for all N.Wait, let me think again. For N=1, 0; N=2, 2; N=3, 4? But my reasoning says 2 for N=3. Maybe the formula is different.Wait, perhaps the formula is 2N - 2 for N >= 2. So, for N=2, 2; N=3, 4; N=4, 6; etc. But on a 3x3 board, can I really place 4 bishops without them attacking each other?Let me try to place them:- Place a bishop on a1.- Then, I can't place any other bishops on the main diagonal or the a1-a3 diagonal.- Place a bishop on a3. Now, it's on the same column as a1, but bishops don't care about columns. However, a3 is on the anti-diagonal with c1, so I can't place a bishop on c1.- Place a bishop on c3. Now, c3 is on the main diagonal with a1, so they attack each other. So, that's not allowed.- Alternatively, place a bishop on b2. Now, b2 is on the main diagonal with a1 and c3, and on the anti-diagonal with a3 and c1. So, placing a bishop on b2 would block all four corners.- Therefore, if I place a bishop on b2, I can't place any others. So, that's only 1 bishop.Wait, this is confusing. Maybe the maximum is indeed 2 for N=3.But according to the formula 2N - 2, it should be 4. So, where am I wrong?Wait, maybe I'm not considering all possible diagonals. Let me list all the diagonals on a 3x3 board.From bottom-left to top-right:1. a12. a2, b13. a3, b2, c14. b3, c25. c3From top-left to bottom-right:1. a32. a2, b33. a1, b2, c34. b1, c25. c1So, there are 5 diagonals in each direction.To place bishops without them attacking each other, I need to select a set of squares such that no two are on the same diagonal in either direction.So, the problem reduces to selecting the maximum number of squares such that no two are on the same diagonal in either direction.This is similar to the problem of placing queens, but only considering diagonals, not rows or columns.In graph theory, this is equivalent to finding the maximum independent set in the bishop's graph.But maybe there's a simpler way.If I consider that each bishop occupies one diagonal from each set, then the maximum number of bishops is the size of the maximum matching between the two sets of diagonals.But I'm not sure.Alternatively, since each bishop occupies one diagonal from each set, and there are 2N - 1 diagonals in each set, the maximum number of bishops is 2N - 2.Wait, for N=3, that would be 4 bishops. But earlier, I thought it was only possible to place 2.Maybe I'm missing a configuration.Let me try to place 4 bishops on a 3x3 board without them attacking each other.- Place a bishop on a1.- Then, I can't place any other bishops on the main diagonal or the a1-a3 diagonal.- Place a bishop on a3. Now, a3 is on the anti-diagonal with c1, so I can't place a bishop on c1.- Place a bishop on c3. Now, c3 is on the main diagonal with a1, so they attack each other. Not allowed.- Alternatively, place a bishop on b2. Now, b2 is on the main diagonal with a1 and c3, and on the anti-diagonal with a3 and c1. So, placing a bishop on b2 would block all four corners.- Therefore, if I place a bishop on b2, I can't place any others. So, that's only 1 bishop.Wait, maybe I need to place bishops on different diagonals.- Place a bishop on a1 (main diagonal 1, anti-diagonal 3).- Place a bishop on a2 (main diagonal 2, anti-diagonal 2).- Place a bishop on b3 (main diagonal 4, anti-diagonal 1).- Place a bishop on c1 (main diagonal 3, anti-diagonal 5).Now, let's check for conflicts:- a1 and a2: same column, but bishops don't care about columns.- a1 and b3: different diagonals.- a1 and c1: same anti-diagonal? a1 is on anti-diagonal 3, c1 is on anti-diagonal 5. No, they don't share a diagonal.- a2 and b3: different diagonals.- a2 and c1: different diagonals.- b3 and c1: different diagonals.Wait, so all four bishops are on different diagonals in both directions. Therefore, they don't attack each other.So, on a 3x3 board, it is possible to place 4 bishops without them attacking each other. That matches the formula 2N - 2 = 4.Okay, so my earlier confusion was because I was trying to place them on the corners, which caused conflicts, but by placing them on different diagonals, I can achieve the maximum.Therefore, the formula seems to hold: for an N x N chessboard, the maximum number of non-attacking bishops is 2N - 2.But why is that? Let me try to generalize.For any N x N board, there are 2N - 1 diagonals in each direction. To place bishops without them attacking each other, each bishop must occupy a unique diagonal in both directions.However, if we try to place 2N - 1 bishops, we would need to use all diagonals in one direction, but since each bishop also occupies a diagonal in the other direction, we would end up having some bishops share diagonals in the other direction, leading to attacks.Therefore, to avoid this, we need to leave out two diagonals, one from each direction, which allows us to place 2N - 2 bishops without any conflicts.So, the maximum number of non-attacking bishops on an N x N chessboard is 2N - 2.

💡Okay, so I have this problem about a regular decagon, and I need to figure out how many triangles can be formed using its vertices. Hmm, a decagon has 10 sides, right? So, it has 10 vertices. The question is asking about triangles, which are three-sided figures, so I guess I need to find out how many different sets of three vertices I can choose from the 10 to form a triangle.Wait, but before I jump into calculations, I should think about whether every combination of three vertices will actually form a triangle. I remember that in geometry, three points can form a triangle as long as they are not all on the same straight line. In a regular decagon, are any three vertices ever collinear? Let me visualize a regular decagon. It's a 10-sided polygon with all sides and angles equal. If I pick three vertices, will they ever lie on a straight line?I think in a regular decagon, the vertices are equally spaced around a circle. So, unless three of them are equally spaced in a way that they fall on the same diameter or something, but I don't think that's possible because 10 is an even number, but three points can't all lie on the same diameter unless they are spaced in a way that two are on one end and one on the other, but that still wouldn't make them collinear. Wait, actually, in a regular decagon, if you pick every other vertex, you might get a five-pointed star or something, but I don't think that makes them collinear either. So, maybe in a regular decagon, no three vertices are collinear. That means every combination of three vertices will form a valid triangle.Okay, so if that's the case, then the number of triangles is just the number of ways to choose 3 vertices out of 10. That sounds like a combination problem. The formula for combinations is n choose k, which is n! divided by (k! times (n minus k)!). So, in this case, n is 10 and k is 3.Let me write that out: 10 choose 3 is 10! divided by (3! times (10 minus 3)!), which simplifies to 10! divided by (3! times 7!). But 10! is 10 times 9 times 8 times 7!, so the 7! cancels out from the numerator and denominator. That leaves me with (10 times 9 times 8) divided by 3!. 3! is 6, so it's (10 times 9 times 8) divided by 6.Let me calculate that step by step. 10 times 9 is 90, times 8 is 720. Then, 720 divided by 6 is 120. So, does that mean there are 120 triangles?Wait, but let me double-check. Maybe I made a mistake somewhere. So, 10 choose 3 is indeed 120. But just to be sure, let me think about it another way. The number of triangles is the same as the number of ways to pick any three vertices, and since no three are collinear, each set of three vertices gives exactly one triangle.Alternatively, I could think about it as starting with one vertex and counting how many triangles can be formed with that vertex. For each vertex, there are 9 other vertices, and I need to choose 2 more to form a triangle. So, for one vertex, it's 9 choose 2, which is 36. But wait, if I do that for all 10 vertices, I would be counting each triangle three times because each triangle has three vertices. So, the total number would be (10 times 36) divided by 3, which is 120. Yeah, that matches my earlier calculation.Another way to think about it is using permutations. If I consider the order of selecting the vertices, the number of permutations of 10 vertices taken 3 at a time is 10 times 9 times 8, which is 720. But since triangles don't care about the order of their vertices, I need to divide by the number of ways to arrange three vertices, which is 3 factorial, or 6. So, 720 divided by 6 is again 120.Okay, so all these methods lead me to the same answer: 120 triangles. I think that makes sense. I don't see any mistakes in my reasoning, and multiple approaches confirm the same result. So, I'm pretty confident that the number of triangles that can be formed is 120.

💡First, I need to determine the total area of the rectangular room by multiplying its length and width.Next, since the room is divided into two equal sections, I will divide the total area by 2 to find the area of each section.After obtaining the area in square feet, I will convert it to square yards by dividing by 9, as there are 9 square feet in one square yard.Finally, I will present the calculated number of square yards needed for each section.

💡Alright, so I have this problem about two frogs, Alphonse and Beryl, racing to reach square ( n ) on a row of ( n + 1 ) squares. The goal is to figure out for part (a) an integer ( n > 200 ) where Beryl reaches square ( n ) faster than Alphonse, and for part (b), the largest integer ( n ) where Beryl doesn't take longer than Alphonse.First, let me understand their jumping rules:- **Alphonse** jumps 8 squares if there are at least 8 squares to his right; otherwise, he jumps 1 square.- **Beryl** jumps 7 squares if there are at least 7 squares to her right; otherwise, she jumps 1 square.So, both frogs are trying to reach square ( n ) starting from square 0. Their jump strategies differ slightly: Alphonse has a bigger jump (8 squares) compared to Beryl's 7 squares, but Beryl might have an advantage in certain cases because of her smaller jump size.Given the example, ( A(40) = 5 ) and ( B(40) = 10 ). Let me verify that:- For Alphonse: To reach 40, he can make 5 jumps of 8 squares each (since ( 5 times 8 = 40 )), so ( A(40) = 5 ).- For Beryl: To reach 40, she can make 5 jumps of 7 squares each, which would take her to 35, and then she needs 5 more single jumps to reach 40. So, ( B(40) = 5 + 5 = 10 ).Okay, that makes sense. So, Alphonse is faster in this case because his bigger jumps allow him to cover more distance in fewer seconds.Now, for part (a), I need to find an ( n > 200 ) where Beryl is faster. That means ( B(n) < A(n) ). Hmm, so Beryl's strategy must somehow be more efficient for this particular ( n ).Let me think about how their jump counts are calculated. For any ( n ), the number of jumps each frog makes can be determined by dividing ( n ) by their respective jump sizes and considering the remainders.For Alphonse:- If ( n = 8q + r ) where ( 0 leq r < 8 ), then ( A(n) = q + r ).- Because if there are at least 8 squares, he jumps 8, otherwise 1.For Beryl:- If ( n = 7q + r ) where ( 0 leq r < 7 ), then ( B(n) = q + r ).- Similarly, if there are at least 7 squares, she jumps 7, otherwise 1.So, to find ( n ) where ( B(n) < A(n) ), I need ( q_2 + r_2 < q_1 + r_1 ), where ( n = 8q_1 + r_1 = 7q_2 + r_2 ).I need to find such ( n ) where the number of jumps Beryl makes is less than Alphonse's.Let me try to express this inequality:( q_2 + r_2 < q_1 + r_1 )But since ( n = 8q_1 + r_1 = 7q_2 + r_2 ), I can write:( 8q_1 + r_1 = 7q_2 + r_2 )So, substituting ( r_1 = 7q_2 + r_2 - 8q_1 ) into the inequality:( q_2 + r_2 < q_1 + (7q_2 + r_2 - 8q_1) )Simplify:( q_2 + r_2 < q_1 + 7q_2 + r_2 - 8q_1 )( q_2 < -7q_1 + 7q_2 )( 0 < -7q_1 + 6q_2 )( 7q_1 < 6q_2 )So, ( q_2 > frac{7}{6} q_1 )This means that the quotient ( q_2 ) for Beryl must be greater than ( frac{7}{6} ) times the quotient ( q_1 ) for Alphonse.Given that ( n = 8q_1 + r_1 = 7q_2 + r_2 ), and ( r_1 < 8 ), ( r_2 < 7 ), I can try to find such ( n ).Let me think about the relationship between ( q_1 ) and ( q_2 ). Since ( q_2 ) must be greater than ( frac{7}{6} q_1 ), perhaps I can set ( q_2 = lceil frac{7}{6} q_1 rceil ).But I need ( n ) to be the same for both frogs, so ( 8q_1 + r_1 = 7q_2 + r_2 ).Let me try to find ( q_1 ) and ( q_2 ) such that ( 8q_1 + r_1 = 7q_2 + r_2 ) and ( q_2 > frac{7}{6} q_1 ).Let me pick a value for ( q_1 ) and see what ( q_2 ) would be.Suppose ( q_1 = 30 ). Then ( q_2 > frac{7}{6} times 30 = 35 ). So, ( q_2 geq 36 ).Then, ( n = 8 times 30 + r_1 = 240 + r_1 ). Also, ( n = 7 times 36 + r_2 = 252 + r_2 ).So, ( 240 + r_1 = 252 + r_2 ). Therefore, ( r_1 = 12 + r_2 ). But ( r_1 < 8 ), which is impossible because ( r_2 geq 0 ). So, this doesn't work.Let me try ( q_1 = 28 ). Then ( q_2 > frac{7}{6} times 28 approx 32.666 ), so ( q_2 geq 33 ).Then, ( n = 8 times 28 + r_1 = 224 + r_1 ). Also, ( n = 7 times 33 + r_2 = 231 + r_2 ).So, ( 224 + r_1 = 231 + r_2 ). Therefore, ( r_1 = 7 + r_2 ). But ( r_1 < 8 ), so ( r_2 ) must be 0, and ( r_1 = 7 ).Thus, ( n = 224 + 7 = 231 ).Let me check:For Alphonse: ( 231 = 8 times 28 + 7 ), so ( A(231) = 28 + 7 = 35 ).For Beryl: ( 231 = 7 times 33 + 0 ), so ( B(231) = 33 + 0 = 33 ).Yes, ( B(231) = 33 < 35 = A(231) ). So, ( n = 231 ) works.But wait, the problem says ( n > 200 ). 231 is greater than 200, so that's good.But is this the smallest such ( n )? Or is there a smaller ( n > 200 ) where this happens? Let me check.Suppose ( q_1 = 27 ). Then ( q_2 > frac{7}{6} times 27 = 31.5 ), so ( q_2 geq 32 ).Then, ( n = 8 times 27 + r_1 = 216 + r_1 ). Also, ( n = 7 times 32 + r_2 = 224 + r_2 ).So, ( 216 + r_1 = 224 + r_2 ). Therefore, ( r_1 = 8 + r_2 ). But ( r_1 < 8 ), which is impossible. So, no solution here.Similarly, ( q_1 = 26 ): ( q_2 > frac{7}{6} times 26 approx 30.333 ), so ( q_2 geq 31 ).( n = 8 times 26 + r_1 = 208 + r_1 ). Also, ( n = 7 times 31 + r_2 = 217 + r_2 ).So, ( 208 + r_1 = 217 + r_2 ). Therefore, ( r_1 = 9 + r_2 ). Again, ( r_1 < 8 ), impossible.Continuing this pattern, it seems that ( n = 231 ) is the first ( n > 200 ) where ( B(n) < A(n) ).But let me check ( n = 231 ) again:Alphonse: 231 / 8 = 28.875, so he makes 28 jumps of 8, reaching 224, then 7 jumps of 1 to reach 231. So, total jumps: 28 + 7 = 35.Beryl: 231 / 7 = 33, so she makes 33 jumps of 7, reaching exactly 231. So, total jumps: 33.Yes, 33 < 35, so ( B(231) < A(231) ).Is there a smaller ( n > 200 ) where this happens? Let's see.Suppose ( n = 224 ):Alphonse: 224 / 8 = 28, so ( A(224) = 28 ).Beryl: 224 / 7 = 32, so ( B(224) = 32 ).Here, ( B(n) = 32 > 28 = A(n) ).Similarly, ( n = 231 ) is the first after 200 where Beryl is faster.Wait, but what about ( n = 231 + 56 = 287 )? Because 56 is the least common multiple of 7 and 8.Let me check ( n = 287 ):Alphonse: 287 / 8 = 35.875, so 35 jumps of 8 (280) + 7 jumps of 1, total 35 + 7 = 42.Beryl: 287 / 7 = 41, so 41 jumps of 7, total 41.So, ( B(287) = 41 < 42 = A(287) ).Similarly, ( n = 343 ) (which is 7^3):Alphonse: 343 / 8 = 42.875, so 42 jumps of 8 (336) + 7 jumps of 1, total 42 + 7 = 49.Beryl: 343 / 7 = 49, so 49 jumps of 7, total 49.Here, ( B(n) = A(n) ).So, ( n = 343 ) is the point where they are equal.But for part (a), I just need one ( n > 200 ) where ( B(n) < A(n) ). So, 231 is a valid answer.For part (b), I need the largest ( n ) where ( B(n) leq A(n) ). From the above, ( n = 343 ) is where they are equal, and beyond that, for ( n > 343 ), Beryl might be faster or slower?Wait, let's check ( n = 344 ):Alphonse: 344 / 8 = 43, so ( A(344) = 43 ).Beryl: 344 / 7 ≈ 49.142, so 49 jumps of 7 (343) + 1 jump of 1, total 49 + 1 = 50.So, ( B(344) = 50 > 43 = A(344) ).Wait, so at ( n = 344 ), Beryl is slower again.Wait, but at ( n = 343 ), they are equal. So, perhaps ( n = 343 ) is the largest ( n ) where ( B(n) leq A(n) ).But let me check ( n = 343 + 7 = 350 ):Alphonse: 350 / 8 = 43.75, so 43 jumps of 8 (344) + 6 jumps of 1, total 43 + 6 = 49.Beryl: 350 / 7 = 50, so 50 jumps of 7, total 50.So, ( B(350) = 50 > 49 = A(350) ).Wait, so ( n = 343 ) is the point where they are equal, and beyond that, Beryl takes more jumps.But wait, what about ( n = 343 + k ) where ( k < 7 )?For ( n = 343 + 1 = 344 ), as above, Beryl takes 50 jumps, Alphonse takes 43.Wait, no, Alphonse takes 43 jumps of 8 (344) and then 0 jumps of 1, so ( A(344) = 43 ).Wait, no, 344 is exactly 43 * 8, so ( A(344) = 43 ).Beryl: 344 / 7 = 49.142, so 49 jumps of 7 (343) + 1 jump of 1, total 49 + 1 = 50.So, ( B(344) = 50 > 43 = A(344) ).Similarly, for ( n = 343 + k ) where ( 1 leq k < 7 ), Beryl will need 49 + k jumps, while Alphonse will need 43 + k jumps (since 343 + k = 43*8 + k, so Alphonse makes 43 jumps of 8 and k jumps of 1).So, ( B(n) = 49 + k ) and ( A(n) = 43 + k ). So, ( B(n) - A(n) = 6 ), which is positive. Hence, Beryl is slower.But at ( n = 343 ), ( B(n) = A(n) = 49 ).Wait, but what about ( n = 343 + 7 = 350 ):Alphonse: 350 / 8 = 43.75, so 43 jumps of 8 (344) + 6 jumps of 1, total 43 + 6 = 49.Beryl: 350 / 7 = 50, so 50 jumps of 7, total 50.So, ( B(350) = 50 > 49 = A(350) ).So, it seems that ( n = 343 ) is the last point where they are equal, and beyond that, Beryl is slower.But wait, let me check ( n = 343 - 1 = 342 ):Alphonse: 342 / 8 = 42.75, so 42 jumps of 8 (336) + 6 jumps of 1, total 42 + 6 = 48.Beryl: 342 / 7 ≈ 48.857, so 48 jumps of 7 (336) + 6 jumps of 1, total 48 + 6 = 54.So, ( B(342) = 54 > 48 = A(342) ).Wait, so at ( n = 342 ), Beryl is slower.But at ( n = 343 ), they are equal.So, ( n = 343 ) is the largest ( n ) where ( B(n) leq A(n) ).Wait, but let me check ( n = 343 + 56 = 399 ):Alphonse: 399 / 8 = 49.875, so 49 jumps of 8 (392) + 7 jumps of 1, total 49 + 7 = 56.Beryl: 399 / 7 = 57, so 57 jumps of 7, total 57.So, ( B(399) = 57 > 56 = A(399) ).Similarly, ( n = 343 + 56k ) will have ( B(n) = A(n) + 1 ), so Beryl is slower.Therefore, ( n = 343 ) is indeed the largest ( n ) where ( B(n) leq A(n) ).So, summarizing:(a) An integer ( n > 200 ) where ( B(n) < A(n) ) is ( n = 231 ).(b) The largest integer ( n ) where ( B(n) leq A(n) ) is ( n = 343 ).