💡First, I need to determine the number of participants in the chess tournament. Since each participant plays exactly one game with every other participant, the total number of games can be calculated using the combination formula ( binom{n}{2} ), which simplifies to ( frac{n(n - 1)}{2} ).Given that the total number of games is 630, I set up the equation:[frac{n(n - 1)}{2} = 630]Multiplying both sides by 2 to eliminate the fraction:[n(n - 1) = 1260]Next, I need to find an odd integer ( n ) such that ( n(n - 1) = 1260 ). Factoring 1260, I find that ( 35 times 36 = 1260 ). Since ( n ) must be odd, I choose ( n = 35 ).Now, to determine the number of days the tournament lasts, I consider that each participant plays two games per day. Each participant needs to play 34 games (one against each of the other participants). Therefore, the number of days ( d ) is:[d = frac{34}{2} = 17]Thus, the tournament lasts for 17 days.

💡First, I need to generate all possible three-digit numbers using the digits 2, 0, 7, 5, and 9 without repeating any digit. Since a three-digit number cannot start with 0, I will exclude any numbers that begin with 0.Next, I will list out all the valid three-digit numbers:- Starting with 2: - 207, 205, 209, 270, 275, 279, 250, 257, 259, 290, 297, 295- Starting with 7: - 720, 725, 729, 702, 705, 709, 750, 752, 759, 790, 792, 795- Starting with 5: - 520, 527, 529, 502, 507, 509, 570, 572, 579, 590, 592, 597- Starting with 9: - 920, 927, 925, 902, 907, 905, 950, 952, 957, 970, 972, 975After listing all the numbers, I will arrange them in ascending order:- 205, 207, 209, 250, 257, 259, 270, 275, 279, 290, 295, 297- 502, 507, 509, 520, 527, 529, 570, 572, 579, 590, 592, 597- 702, 705, 709, 720, 725, 729, 750, 752, 759, 790, 792, 795- 902, 905, 907, 920, 927, 925, 950, 952, 957, 970, 972, 975The smallest number (leftmost) is 205, and the fourth number from the left is 250. Adding these two numbers together:205 + 250 = 455Therefore, the sum of the leftmost number and the fourth number from the left is 455.

💡Okay, so I'm trying to figure out this game where two players take turns writing numbers in place of the stars in these six equations. The first player wants all the equations to be true, and the second player wants to stop that from happening. I need to figure out who will win if both play optimally.First, let me write down the equations to get a clear picture:1. *=*2. *=*+*3. *=*+*+*4. *=*+*+*+*5. *=*+*+*+*+*6. *=*+*+*+*+*+*So, each equation has a different number of stars. The first one has two stars, the second one has three, and so on up to seven stars in the last equation.Now, the players take turns replacing the stars with numbers. The first player wants all equations to be true, which means that after all the stars are replaced, each equation should hold mathematically. The second player wants to prevent that.I think the key here is to analyze the number of stars in each equation and how they can be replaced to make the equations true or false. Let's consider each equation one by one.Starting with the first equation: *=*. This is the simplest one. If both stars are replaced by the same number, the equation will hold true. For example, if both are replaced by 1, it becomes 1=1, which is true. So, the first player can easily make this equation true by choosing the same number for both stars.Moving on to the second equation: *=*+*. This equation requires that the number on the left equals the sum of the two numbers on the right. For example, if the left star is 2 and the right stars are 1 and 1, it becomes 2=1+1, which is true. So, the first player can also make this equation true by choosing appropriate numbers.The third equation: *=*+*+*. Here, the left star needs to equal the sum of three stars on the right. For example, if the left star is 3 and the right stars are 1, 1, and 1, it becomes 3=1+1+1, which is true. Again, the first player can choose numbers to make this equation true.Continuing this pattern, the fourth equation: *=*+*+*+*. The left star should equal the sum of four stars on the right. Similarly, the fifth equation requires the left star to equal the sum of five stars, and the sixth equation requires the left star to equal the sum of six stars on the right.Now, considering that the first player wants all equations to be true, they need to ensure that for each equation, the number on the left is the sum of the numbers on the right. The second player, on the other hand, wants to disrupt this by choosing numbers that prevent the equations from being true.I think the number of stars in each equation might play a role here. For example, equations with an even number of stars on the right might be easier to balance, while those with an odd number might be trickier. But I'm not sure yet.Let me think about the total number of stars. The first equation has two stars, the second has three, the third has four, the fourth has five, the fifth has six, and the sixth has seven stars. So, in total, there are 2+3+4+5+6+7 = 27 stars. Since two players are taking turns, the first player will have the first, third, fifth, etc., moves, and the second player will have the second, fourth, sixth, etc.Since there are 27 stars, which is an odd number, the first player will make the last move. That might give the first player an advantage because they can potentially set the final number to ensure an equation is true.But I need to think more carefully. Each equation has a different number of stars, so the first player might focus on certain equations to make them true, while the second player tries to mess up others.Maybe the first player can prioritize the equations with fewer stars first because they are simpler to control. For example, the first equation with two stars can be easily made true by choosing the same number for both. The second player might try to interfere with the equations that have more stars because they are more complex and harder to control.Alternatively, the first player might try to balance the equations in a way that the second player cannot disrupt all of them. Since the first player has the last move, they can potentially fix any equation that the second player tried to mess up.Wait, but the second player might have strategies to make multiple equations invalid, not just one. So, even if the first player can fix one equation, the second player might have already made another equation invalid.I need to think about the parity of the number of stars in each equation. For the first equation, two stars: if both are set by the first player, they can make it true. If the second player sets one of them, the first player can set the other to match.For the second equation, three stars: the first player might set one star, the second player sets another, and the first player sets the third. The first player can ensure that the sum equals the left side.Similarly, for the third equation with four stars, the first player can set two stars, and the second player sets two. The first player can choose numbers such that the sum equals the left side.Wait, but the second player might not cooperate. They might choose numbers that make it impossible for the first player to balance the equation.For example, in the second equation: *=*+*. If the first player sets the left star to 2, and the second player sets one of the right stars to 3, then the first player has to set the last star to -1 to make 2=3+(-1). But if negative numbers are allowed, that's possible. If not, then the first player cannot make it true.But the problem doesn't specify whether numbers have to be positive or can be any integers. If negative numbers are allowed, the first player can always adjust to make the equation true. If only positive numbers are allowed, it might be more challenging.Assuming that numbers can be any integers, including negatives, the first player can always balance the equation by choosing the appropriate number for the last star.But if only positive integers are allowed, the second player can set a star to a very high number, making it impossible for the first player to balance.But since the problem doesn't specify, I think we have to assume that numbers can be any integers, including negatives.Given that, the first player can always balance the equation by choosing the appropriate number for the last star, regardless of what the second player does.Therefore, the first player can ensure that all equations are true by appropriately choosing numbers on their turns, especially since they have the last move overall.But wait, the second player might try to mess up multiple equations at once, making it impossible for the first player to fix all of them.However, since the first player can prioritize fixing the equations one by one, and with the last move advantage, they can ensure that all equations are true.Alternatively, maybe the second player can focus on a single equation and make it impossible for the first player to fix, even with the last move.But if the first player can always adjust the last star in any equation to make it true, regardless of what the second player does, then the first player can win.Wait, but in some equations, the number of stars is odd, so the first player might not have control over all of them.For example, in the second equation with three stars: the first player sets one, the second player sets another, and the first player sets the last one. So, the first player can control the last star to make the equation true.Similarly, in the fourth equation with five stars: the first player sets two, the second player sets two, and the first player sets the last one, allowing them to balance it.In the sixth equation with seven stars: the first player sets three, the second player sets three, and the first player sets the last one, again allowing them to balance it.For the equations with even numbers of stars: the first player sets half, the second player sets half, but since the first player goes first, they can set the first star, and then respond to the second player's moves.Wait, but in equations with even numbers of stars, the second player might have the last move if the number is even. For example, the first equation with two stars: first player sets one, second player sets the other. So, the second player could set it to a different number, making the equation false.But the first player can choose to set both stars on their turn if possible. Wait, no, they take turns setting one star at a time.So, in the first equation, the first player sets one star, say to 1, then the second player sets the other star. If the second player sets it to 2, then the equation becomes 1=2, which is false.Therefore, the second player can make the first equation false by choosing a different number.Similarly, in the third equation with four stars: first player sets two stars, second player sets two stars. The second player can choose numbers such that the sum doesn't match the left side.Wait, but the first player can choose the left side and some right side stars in a way that forces the second player into a corner.This is getting complicated. Maybe I need to think in terms of control over the equations.Each equation has a certain number of stars, and the players alternate setting them. The first player can choose which equation to set a star in, or does the problem specify that they have to set a star in a specific equation?Wait, the problem says they take turns writing numbers in place of the stars, but it doesn't specify that they have to choose a specific equation each time. So, they can choose any equation and any star within that equation to set.If that's the case, the first player can choose to set stars in equations in a way that ensures all equations can be balanced, while the second player can try to set stars in a way that makes some equations impossible to balance.But since the first player has the last move overall, they can potentially fix any equation that the second player tried to mess up.Alternatively, the second player might focus on a single equation and set stars in such a way that the first player cannot fix it, regardless of the last move.But if the first player can always adjust the last star in any equation to make it true, then the first player can ensure all equations are true.However, if the second player can set multiple stars in a single equation in such a way that the first player cannot fix it, even with the last move, then the second player can win.But given that the first player can always set the last star in any equation to balance it, regardless of what the second player does, the first player can win.Wait, but in equations with an even number of stars, the second player might have the last move if the first player doesn't prioritize that equation.For example, in the first equation with two stars: first player sets one, second player sets the other. If the second player chooses a different number, the equation is false.But the first player could choose to set both stars in the first equation on their first two moves, making it true before the second player can interfere.But the problem is that the players take turns setting one star at a time, and they can choose any star in any equation.So, the first player might choose to set a star in the first equation, then the second player can choose to set a star in the same equation, making it potentially false.Alternatively, the first player could spread out their moves across different equations, ensuring that they can control the final value in each.But it's tricky because the second player can focus on a single equation and mess it up.I think the key is that the first player can always ensure that for each equation, they have the last move in setting a star, allowing them to balance it.Since there are six equations, and the first player has the first and last moves overall, they can distribute their moves to control the final value in each equation.Therefore, with optimal play, the first player can ensure that all equations are true, making them the winner.

💡Okay, so I have this function f(x) = |x - 1| + |x + 3|. I need to find its minimum value and then prove something about another expression involving a and b. Let me start with part (1).First, I remember that the sum of absolute value functions can often be minimized by considering the points where the expressions inside the absolute values change their behavior, which are called critical points. For f(x), the critical points are at x = 1 and x = -3 because that's where |x - 1| and |x + 3| change from decreasing to increasing or vice versa.So, I think the minimum of f(x) might occur somewhere between these two points or at one of them. Maybe I should consider different intervals based on these critical points and see how f(x) behaves in each interval.Let me break it down:1. When x < -3: Both |x - 1| and |x + 3| will be negative inside the absolute value, so they become -(x - 1) and -(x + 3). So, f(x) = -(x - 1) - (x + 3) = -x + 1 - x - 3 = -2x - 2. This is a linear function with a slope of -2, which means it's decreasing as x increases. So, as x approaches -3 from the left, f(x) will approach f(-3) = |-3 - 1| + |-3 + 3| = | -4 | + |0| = 4 + 0 = 4.2. When -3 ≤ x ≤ 1: Here, |x + 3| will be non-negative, so it's just x + 3. But |x - 1| will still be negative because x is less than 1, so it becomes -(x - 1). Therefore, f(x) = -(x - 1) + (x + 3) = -x + 1 + x + 3 = 4. So, in this entire interval, f(x) is constant at 4.3. When x > 1: Both |x - 1| and |x + 3| are non-negative, so f(x) = (x - 1) + (x + 3) = 2x + 2. This is a linear function with a slope of 2, which means it's increasing as x increases. So, as x moves away from 1, f(x) increases beyond 4.Putting it all together, f(x) decreases until x = -3, stays constant at 4 between x = -3 and x = 1, and then increases beyond x = 1. Therefore, the minimum value of f(x) is 4.Wait, but I also remember that sometimes the minimum can be found using the triangle inequality. Let me try that approach too to verify.The function f(x) = |x - 1| + |x + 3| can be thought of as the sum of distances from x to 1 and from x to -3 on the real line. The minimum sum of distances occurs when x is between 1 and -3. In fact, the minimal total distance is the distance between 1 and -3, which is |1 - (-3)| = 4. So, that confirms the minimum value is indeed 4.Okay, so part (1) seems solid. The minimum value is 4.Now, moving on to part (2). I need to prove that if a and b are real numbers with |a| < 2 and |b| < 2, then |a + b| + |a - b| < f(x). Since we already found that f(x) is always at least 4, and the minimum is 4, I think I need to show that |a + b| + |a - b| is less than 4.Let me think about |a + b| + |a - b|. I remember that this expression relates to the concept of the norm in linear algebra, specifically the L1 norm and the Linfinity norm. But maybe I can approach it more simply.First, let's compute |a + b| + |a - b|. I recall that this is equal to 2 times the maximum of |a| and |b|. Let me verify that.Case 1: Suppose |a| ≥ |b|. Then, |a + b| + |a - b| = |a + b| + |a - b|. If |a| ≥ |b|, then a + b and a - b will have the same sign as a, so |a + b| + |a - b| = |a + b + a - b| = |2a| = 2|a|.Case 2: Suppose |b| ≥ |a|. Similarly, |a + b| + |a - b| = |b + a| + |b - a|. Since |b| is larger, this becomes |2b| = 2|b|.So, in general, |a + b| + |a - b| = 2 * max(|a|, |b|).Given that |a| < 2 and |b| < 2, the maximum of |a| and |b| is less than 2. Therefore, 2 * max(|a|, |b|) < 2 * 2 = 4.Thus, |a + b| + |a - b| < 4.But wait, in part (1), we found that f(x) is always greater than or equal to 4. So, since |a + b| + |a - b| is less than 4 and f(x) is at least 4, it follows that |a + b| + |a - b| < f(x) for all x.Let me check if there are any edge cases or if I missed something. For example, if a and b are both approaching 2 or -2, does |a + b| + |a - b| approach 4? Let's see.If a approaches 2 and b approaches 2, then |a + b| approaches 4 and |a - b| approaches 0, so the sum approaches 4. Similarly, if a approaches 2 and b approaches -2, |a + b| approaches 0 and |a - b| approaches 4, so the sum again approaches 4. But since |a| and |b| are strictly less than 2, the sum |a + b| + |a - b| is strictly less than 4.Therefore, the inequality |a + b| + |a - b| < f(x) holds because the left side is less than 4 and the right side is at least 4.I think that covers all the cases. So, the proof seems solid.**Final Answer**(1) The minimum value of the function is boxed{4}.(2) The inequality is proven as required.

💡First, I need to calculate the total splash width for each type of rock by multiplying the number of rocks by their respective splash widths.For pebbles:10 pebbles multiplied by 1/5 meter wide equals 2 meters.For rocks:5 rocks multiplied by 2/5 meter wide equals 2 meters.For boulders:4 boulders multiplied by 7/5 meters wide equals 5.6 meters.For mini-boulders:3 mini-boulders multiplied by 4/5 meter wide equals 2.4 meters.For large pebbles:7 large pebbles multiplied by 3/5 meter wide equals 4.2 meters.Finally, I add all these individual splash widths together to find the total splash width:2 meters + 2 meters + 5.6 meters + 2.4 meters + 4.2 meters equals 16.2 meters.

💡First, I need to determine how many bottle caps Marilyn had initially. She shared 36 bottle caps with Nancy and was left with 15 bottle caps.To find the initial number of bottle caps, I can add the number of bottle caps she shared with the number she has left.So, the calculation would be 36 (shared) + 15 (remaining) = 51 bottle caps.Therefore, Marilyn started with 51 bottle caps.