💡First, I recognize that the total work can be represented as ( W ).Peter can complete the work in 35 days, so his work rate is ( frac{W}{35} ) per day.Matt and Peter together can complete the work in 20 days, so their combined work rate is ( frac{W}{20} ) per day.Let ( x ) be the number of days Matt and Peter worked together. During these ( x ) days, they completed ( x times frac{W}{20} ) of the work.After Matt stops, Peter works alone for 14 days, completing ( 14 times frac{W}{35} ) of the work.The sum of the work done by Matt and Peter together and by Peter alone should equal the total work ( W ):[x times frac{W}{20} + 14 times frac{W}{35} = W]To simplify, divide both sides by ( W ):[frac{x}{20} + frac{14}{35} = 1]Next, find a common denominator for the fractions. The least common multiple (LCM) of 20 and 35 is 140.Convert the fractions:[frac{7x}{140} + frac{4 times 14}{140} = 1]Simplify:[7x + 56 = 140]Solve for ( x ):[7x = 84][x = 12]Therefore, Matt and Peter worked together for 12 days.

💡First, I need to determine how many cards Aaron had initially. I know that after finding 62 cards, Aaron ended up with 67 cards.To find the initial number of cards Aaron had, I can subtract the number of cards he found from the total number of cards he ended up with.So, I'll subtract 62 from 67 to find the initial number of cards Aaron had.

💡First, I note that the train's speed without stoppages is 60 km/h, and with stoppages, it is 40 km/h. This means the train's effective speed is reduced by 20 km/h due to stoppages.Next, I calculate the distance the train would cover in one hour without any stoppages:[text{Distance without stoppages} = 60 text{ km/h} times 1 text{ hour} = 60 text{ km}]Then, I determine the distance the train actually covers in one hour with stoppages:[text{Distance with stoppages} = 40 text{ km/h} times 1 text{ hour} = 40 text{ km}]The difference in distance covered is due to stoppages:[text{Distance difference} = 60 text{ km} - 40 text{ km} = 20 text{ km}]To find out how long it takes the train to cover this 20 km distance at its original speed of 60 km/h:[text{Time} = frac{20 text{ km}}{60 text{ km/h}} = frac{1}{3} text{ hour}]Finally, I convert this time from hours to minutes:[text{Time in minutes} = frac{1}{3} text{ hour} times 60 text{ minutes/hour} = 20 text{ minutes}]Therefore, the train stops for 20 minutes per hour.

💡First, identify the initial number of cells, which is 5.Next, determine how many times the cells split over the 12-day period. Since the cells split every 3 days, there are 12 / 3 = 4 splitting events.Calculate the number of cells after each splitting event:- After the first split (3 days): 5 * 2 = 10 cells- After the second split (6 days): 10 * 2 = 20 cells- After the third split (9 days): 20 * 2 = 40 cells- After the fourth split (12 days): 40 * 2 = 80 cellsAt the end of the ninth day, before the final split, 3 cells die. Subtract these from the total:80 - 3 = 77 cellsTherefore, the total number of cells at the end of the 12th day is 77.

💡Alright, so I'm trying to figure out this matchstick game problem. There are three boxes, each with 2016 matches. Two players take turns taking any number of matches from any one box, and the person who takes the last match wins. I need to prove that the first player can always win, no matter how the second player plays.Okay, let's start by understanding the game mechanics. Each box has the same number of matches, which is 2016. Players alternate turns, and on each turn, a player can take any number of matches from one box. The goal is to be the one who takes the last match.Hmm, so if I'm the first player, what should I do to ensure I win? Maybe I need to find a strategy that forces the second player into a losing position. I remember something about symmetry in games like this. Maybe if I can mirror the second player's moves somehow, I can control the game.Let me think. If I take all the matches from one box on my first move, that leaves two boxes with 2016 matches each. Then, whatever the second player does, I can respond in a way that maintains some kind of balance. For example, if the second player takes some matches from one of the remaining boxes, I can take the same number from the other box. This way, I'm always keeping the two boxes equal in terms of the number of matches.Wait, but if I take all the matches from one box initially, that might leave the second player with two boxes, and they can choose to take from either one. If I mirror their moves, I can ensure that I'm always the one to take the last match. Let me test this idea with a smaller number of matches to see if it makes sense.Suppose each box has 2 matches instead of 2016. I take both matches from one box, leaving two boxes with 2 matches each. Now, if the second player takes 1 match from one box, I take 1 match from the other box, leaving both boxes with 1 match each. Then, if the second player takes the last match from one box, I take the last match from the other box and win. That seems to work.What if the second player takes all the matches from one box on their first turn? If I have two boxes with 2 matches each, and the second player takes both from one box, then I have one box with 2 matches left. I can take both and win. So that still works.Okay, maybe this strategy of taking all matches from one box first and then mirroring the opponent's moves on the other boxes ensures that I can always win. But I need to make sure this works for any number of matches, not just 2.Let me think about it more generally. If I have three boxes with equal numbers of matches, and I take all matches from one box, leaving two boxes with the same number. Then, whatever the second player does to one box, I do the same to the other. This way, I'm always maintaining the balance between the two boxes, and eventually, I'll be the one to take the last match.But wait, what if the second player decides to take matches from the box that I left untouched? For example, if I take all matches from box A, leaving boxes B and C with 2016 each. If the second player takes some matches from box B, I take the same number from box C. If they take from box C, I take from box B. If they take from box A, but box A is already empty, so they can't take from it. So, the second player is forced to take from either B or C, and I can always mirror their move on the other box.This seems solid. By reducing the problem to two boxes and maintaining symmetry, I can control the game's outcome. Essentially, I'm turning the game into a two-box game where I can always respond to the second player's moves, ensuring that I'm the one to take the last match.But let me consider another angle. What if instead of taking all matches from one box, I take a different number? Would that still work? For example, if I take 1008 matches from one box, leaving 1008 in that box and 2016 in the others. Then, the second player could disrupt the symmetry by taking from the box with 1008, and I might not be able to mirror effectively. So, taking all matches from one box seems like the safest strategy to start with.Another thought: what if the second player tries to take all matches from one of the remaining boxes? If I have two boxes with 2016 matches each, and the second player takes all from one, then I have one box left with 2016 matches. I can then take all of them and win. So, that still works in my favor.Wait, but what if the second player takes a different number each time? Like, they take 1 match from one box, then 2 from the other, and so on. But since I'm mirroring their moves, I can always respond by taking the same number from the other box, maintaining the balance. This ensures that I'm always one step ahead, and eventually, I'll be the one to take the last match.I think this strategy works because it forces the second player into a position where they can't avoid me taking the last match. By controlling the balance between the two remaining boxes, I can always respond in a way that leads me to victory.But let me try to formalize this a bit more. Let's say after my first move, there are two boxes with 2016 matches each. Let's denote these as Box B and Box C. The second player can choose to take any number of matches from either Box B or Box C. Suppose they take 'k' matches from Box B. Then, I take 'k' matches from Box C. This keeps the number of matches in Box B and Box C equal after my turn.If the second player takes all matches from Box B, then Box C still has 2016 matches, and I can take all of them and win. If they take some matches from Box B, I take the same number from Box C, maintaining the equality. This ensures that I can always respond to their moves in a way that leads me to take the last match.This seems like a solid strategy. By reducing the problem to two boxes and maintaining symmetry, I can control the game's flow and ensure victory. I don't see any flaws in this approach, so I think this is the way to go.In summary, the strategy is:1. On the first move, take all matches from one box, leaving two boxes with equal numbers of matches.2. On each subsequent move, mirror the second player's move by taking the same number of matches from the other box.3. This ensures that I always have a move that maintains the balance, leading me to take the last match and win.I think this proves that the first player can always win, regardless of the second player's strategy.

💡Alright, so I have this problem here: A rectangle has a perimeter of 40 units, and its length and width are whole numbers. I need to find the maximum possible area of the rectangle in square units. Hmm, okay, let's break this down step by step.First, I remember that the perimeter of a rectangle is calculated by the formula P = 2*(length + width). In this case, the perimeter is given as 40 units. So, I can set up the equation like this:2*(l + w) = 40Where l is the length and w is the width. If I simplify this equation, I can divide both sides by 2 to get:l + w = 20Okay, so the sum of the length and width is 20. That means if I know one of them, I can find the other by subtracting from 20. For example, if the length is 10, then the width would also be 10, right? Because 10 + 10 = 20.Now, the problem asks for the maximum possible area. I recall that the area of a rectangle is given by A = l * w. So, I need to find the values of l and w that will make this product as large as possible, given that l + w = 20 and both l and w are whole numbers.Let me think about how to approach this. One way is to express one variable in terms of the other using the perimeter equation. So, from l + w = 20, I can express w as:w = 20 - lThen, substitute this into the area formula:A = l * (20 - l)Simplifying this, I get:A = 20l - l^2Hmm, this is a quadratic equation in terms of l. It looks like a downward-opening parabola because the coefficient of l^2 is negative. I remember that the maximum value of a quadratic function ax^2 + bx + c occurs at x = -b/(2a). In this case, a = -1 and b = 20, so:l = -20/(2*(-1)) = -20/(-2) = 10So, the length that maximizes the area is 10 units. Then, the width would be:w = 20 - l = 20 - 10 = 10Wait a minute, so both the length and width are 10 units? That means the rectangle is actually a square. I remember that for a given perimeter, the shape that maximizes the area is a square. So, this makes sense.Let me double-check the area. If l = 10 and w = 10, then:A = 10 * 10 = 100So, the maximum area is 100 square units.But just to be thorough, let me try some other whole number values for l and w to see if I can get a larger area.For example, if l = 11, then w = 20 - 11 = 9. The area would be 11 * 9 = 99, which is less than 100.If l = 12, then w = 8, and the area is 12 * 8 = 96, still less than 100.What about l = 9, w = 11? That's the same as l = 11, w = 9, so area is still 99.If l = 8, w = 12, area is 96.Going further, l = 7, w = 13, area = 91.l = 6, w = 14, area = 84.l = 5, w = 15, area = 75.l = 4, w = 16, area = 64.l = 3, w = 17, area = 51.l = 2, w = 18, area = 36.l = 1, w = 19, area = 19.And if l = 0, w = 20, area = 0.So, as I increase the length beyond 10, the area decreases, and the same happens if I decrease the length below 10. Therefore, the maximum area occurs when l = w = 10, making it a square with an area of 100 square units.I think that's pretty much it. I've considered different possibilities and confirmed that the maximum area is indeed achieved when the rectangle is a square with sides of 10 units each.