answer：To solve the problem, I start with the given equation ( a - b + 3 = 0 ). First, I'll rearrange this equation to express ( a ) in terms of ( b ). Subtracting ( b ) and 3 from both sides gives ( a = b - 3 ).Next, I'll substitute this expression for ( a ) into the algebraic expression ( 2 - 3a + 3b ). Replacing ( a ) with ( b - 3 ) results in ( 2 - 3(b - 3) + 3b ).Now, I'll simplify the expression by distributing the (-3) across the parentheses: ( 2 - 3b + 9 + 3b ).Combining like terms, the (-3b) and (3b) cancel each other out, leaving ( 2 + 9 ).Finally, adding these constants together gives the value of the expression as 11.

answer：First, let's define the variables. Let ( x ) be the number of marbles Tyrone gave to Eric.Initially, Tyrone has 97 marbles and Eric has 11 marbles. After Tyrone gives ( x ) marbles to Eric:- Tyrone's marbles become ( 97 - x ).- Eric's marbles become ( 11 + x ).According to the problem, after the transfer, Tyrone has twice as many marbles as Eric. This gives us the equation:[97 - x = 2(11 + x)]Expanding the right side:[97 - x = 22 + 2x]Adding ( x ) to both sides to gather like terms:[97 = 22 + 3x]Subtracting 22 from both sides:[75 = 3x]Dividing both sides by 3 to solve for ( x ):[x = 25]Therefore, Tyrone gave Eric 25 marbles.

answer：Alright, so I have this problem where an ant starts at vertex A of a cube and wants to get to vertex B, moving only along the edges. The catch is that the ant has to take the shortest possible route. I need to figure out how many different paths the ant can take. Hmm, okay, let me think through this step by step.First, I should visualize the cube. A cube has 8 vertices and 12 edges. Each vertex is connected to three other vertices. So, from any given vertex, the ant has three possible edges to choose from. But since the ant is looking for the shortest path, it can't just wander around; it needs to move towards B in the most efficient way.Wait, but what exactly are vertices A and B? Are they opposite corners of the cube? I think so, because if they were adjacent, the shortest path would just be one edge, and there wouldn't be much to count. So, assuming A and B are diagonally opposite on the cube, meaning they are not connected directly by an edge, nor are they on the same face. Instead, they are as far apart as possible on the cube.In a cube, the shortest path between two opposite vertices (also called the space diagonal) isn't along the edges, but if we have to move along edges, then the shortest path would consist of moving along three edges. Because each edge is a unit length, so the shortest path would be 3 units long. So, the ant needs to traverse three edges to get from A to B.Now, how do we count the number of such paths? Let me think. Each path consists of three moves, each time choosing an edge to go along. But we have to make sure that the ant doesn't revisit any vertex, because that would make the path longer than necessary.So, starting from A, the ant has three choices for the first move. Let me denote these as edges leading to three different vertices, say, A1, A2, and A3. From each of these, the ant again has two choices, because it can't go back to A. Wait, is that right? If the ant is at A1, it can go to two other vertices, but one of them might lead closer to B, and the other might not. Hmm, maybe.Alternatively, perhaps it's better to model this as a graph problem. The cube can be represented as a graph where each vertex is connected to three others. We need to find the number of shortest paths from A to B.In graph theory, the number of shortest paths between two nodes can be found using breadth-first search (BFS), which explores all nodes at the present depth level before moving on to nodes at the next depth level. Since we're dealing with a cube, which is a regular graph, maybe there's a pattern or formula we can use.Let me try to break it down. The cube has 8 vertices. Let's label them as follows for clarity:- Let’s say vertex A is at (0,0,0).- Then, vertex B, being the opposite corner, would be at (1,1,1).- The other vertices can be labeled with coordinates (x,y,z) where each of x, y, z is either 0 or 1.So, moving from A to B along the edges means changing one coordinate at a time. Each move flips one coordinate from 0 to 1 or from 1 to 0. But since we're starting at (0,0,0) and need to get to (1,1,1), each move must flip a 0 to a 1 because we can't go back once we've moved along an edge.Wait, is that correct? Actually, no, because the ant can move along any edge, so it can go back and forth. But since we're looking for the shortest path, the ant can't revisit any vertex, so it can't go back along the same edge it just came from. So, each move must be towards a new vertex that hasn't been visited before.Therefore, each path is a sequence of three moves, each time moving to a new vertex, until reaching B. So, starting from A, the ant has three choices. From each of those, it has two choices (since it can't go back to A). Then, from the third vertex, it has only one choice to reach B.Wait, but does that hold? Let me test it with an example. Suppose A is (0,0,0). The ant can go to (1,0,0), (0,1,0), or (0,0,1). Let's say it goes to (1,0,0). From there, it can't go back to A, so it can go to (1,1,0) or (1,0,1). Suppose it goes to (1,1,0). Then, from there, it can go to (1,1,1), which is B. Alternatively, from (1,0,0), it could have gone to (1,0,1), and then to (1,1,1). So, from (1,0,0), there are two paths.Similarly, if the ant had gone from A to (0,1,0), then from there, it can go to (1,1,0) or (0,1,1). Each of those would lead to B in one move. So, again, two paths. Same with starting from (0,0,1): from there, it can go to (1,0,1) or (0,1,1), each leading to B.So, for each of the three initial choices, there are two paths. So, 3 * 2 = 6. Therefore, the total number of shortest paths is 6.Wait, but is that all? Let me see. Is there a way the ant can take a different route that still only takes three moves? For example, is there a path that goes through a different sequence of vertices?Suppose the ant goes from A to (1,0,0), then to (1,1,0), then to (1,1,1). That's one path. Another is A to (1,0,0) to (1,0,1) to (1,1,1). Similarly, from A to (0,1,0) to (1,1,0) to B, and A to (0,1,0) to (0,1,1) to B. Same with starting from (0,0,1): two paths each.So, 3 starting edges, each leading to two paths, so 6 in total. That seems consistent.But let me think again. Maybe I'm missing something. Is there a way to traverse different edges without repeating vertices? For example, is there a path that goes through a different vertex that's not directly adjacent to A?Wait, but in three moves, starting from A, you can only reach vertices that are three edges away. Since the cube is a regular graph, all the vertices at distance three from A are the ones that are opposite, which is just B. So, all paths of length three must end at B. So, all such paths are valid.But how many such paths are there? Each path is a sequence of three edges, each time moving to a new vertex, not revisiting any.Alternatively, we can model this as permutations. Since the ant has to change each coordinate exactly once, from 0 to 1. So, the order in which the ant changes the x, y, and z coordinates can vary.So, for example, the ant can first change x, then y, then z; or x, then z, then y; or y, then x, then z; and so on. There are 3! = 6 possible orders. Each order corresponds to a unique path.So, that also gives us 6 paths. That seems to confirm the earlier result.Wait, but let me make sure. Let's enumerate all possible paths.1. A -> (1,0,0) -> (1,1,0) -> B2. A -> (1,0,0) -> (1,0,1) -> B3. A -> (0,1,0) -> (1,1,0) -> B4. A -> (0,1,0) -> (0,1,1) -> B5. A -> (0,0,1) -> (1,0,1) -> B6. A -> (0,0,1) -> (0,1,1) -> BYes, that's six distinct paths. Each one corresponds to a different order of changing the coordinates. So, the number of shortest paths is 6.But just to be thorough, let me consider if there's any other way. For example, is there a path that goes through a different vertex not listed above? Let's see.Suppose the ant goes from A to (1,0,0) to (0,0,0). Wait, no, that would be going back to A, which isn't allowed in a shortest path because it would create a cycle, making the path longer. So, the ant can't revisit A.Similarly, from (1,0,0), the ant can't go back to A, so it has to go to either (1,1,0) or (1,0,1). So, no other options there.Same logic applies to the other intermediate vertices. From (1,1,0), the only unvisited vertex is B, so the ant must go there. Similarly, from (1,0,1), it must go to B.Therefore, there are no other paths beyond the six we've already listed. So, the total number of shortest paths is 6.Wait, but hold on a second. I just thought of something. What if the cube is oriented differently? Does that affect the count? Hmm, no, because regardless of the orientation, the cube's structure is the same. So, the number of paths should remain consistent.Another way to think about it is using combinations. Since the ant has to make three moves, each changing one coordinate. The number of ways to arrange these three moves is 3 factorial, which is 6. So, that also gives us 6.Alternatively, using graph theory, the number of shortest paths from A to B can be calculated by considering the number of ways to traverse the cube's edges without revisiting any vertex. Since each step reduces the distance to B by one, and there are three steps, the number of paths is equal to the number of permutations of the three axes, which is 6.So, all these different approaches—enumeration, coordinate permutation, graph traversal—lead me to the same answer: 6.Therefore, I'm pretty confident that the number of different shortest paths the ant can take from A to B is 6.**Final Answer**The ant has boxed{6} different shortest paths from vertex ( A ) to vertex ( B ).

answer：Okay, so I have this problem where an athlete is shooting at a target, and each shot scores either 8, 9, or 10 points. The athlete made more than 11 shots and scored a total of 100 points. I need to figure out how many 8-point shots the athlete made. Hmm, let's break this down step by step.First, let me define some variables to represent the number of each type of shot. Let's say:- Let ( x ) be the number of 8-point shots.- Let ( y ) be the number of 9-point shots.- Let ( z ) be the number of 10-point shots.We know that each of these point values was scored at least once, so ( x geq 1 ), ( y geq 1 ), and ( z geq 1 ).The total number of shots is more than 11, so ( x + y + z > 11 ).The total points scored is 100, so the equation for the total points is:( 8x + 9y + 10z = 100 ).Our goal is to find the value of ( x ).Since we have three variables and only two equations (the total points and the total shots), it might be a bit tricky, but maybe we can find some constraints or express variables in terms of others.Let me try to express the total number of shots as ( S = x + y + z ). We know that ( S > 11 ).So, we have:1. ( 8x + 9y + 10z = 100 )2. ( x + y + z = S ), where ( S > 11 )3. ( x, y, z geq 1 )Maybe I can express ( y ) and ( z ) in terms of ( x ) or something like that.Alternatively, I can consider the total points equation and try to manipulate it.Let me subtract 8 times the total number of shots from the total points equation to see if that helps.So, subtract ( 8(x + y + z) ) from both sides:( 8x + 9y + 10z - 8x - 8y - 8z = 100 - 8S )Simplify:( y + 2z = 100 - 8S )So, ( y + 2z = 100 - 8S ).Hmm, that's interesting. Let me write that as:( y + 2z = 100 - 8S ) [Equation A]Since ( S = x + y + z ), and ( x, y, z geq 1 ), the minimum value of ( S ) is 3 (if each was 1). But we know ( S > 11 ), so ( S geq 12 ).So, ( S ) can be 12, 13, 14, etc. But let's see how high ( S ) can be.The maximum number of shots would be if all shots were 8-pointers, but since each type is at least 1, the maximum ( S ) is when we have as many 8s as possible.Wait, actually, the maximum ( S ) would be when the total points are minimized for that number of shots. Since 8 is the lowest point value, the maximum ( S ) would be when all shots are 8s, but we have to have at least one 9 and one 10.So, the maximum ( S ) is such that ( 8(S) leq 100 - 9 - 10 = 81 ). So, ( 8S leq 81 ) implies ( S leq 10.125 ). But since ( S > 11 ), this seems contradictory.Wait, maybe I messed up. Let me think again.Wait, the total points are fixed at 100. So, if all shots were 8s, the maximum number of shots would be ( 100 / 8 = 12.5 ). But since we have to have at least one 9 and one 10, the maximum number of shots would be less than 12.5.But wait, the problem says the athlete made more than 11 shots, so ( S geq 12 ). But if all shots were 8s, he could have 12 shots (since 12*8=96), but he needs to have at least one 9 and one 10, so total points would be 96 + 1 + 2 = 99, which is less than 100. So, actually, 12 shots can't reach 100 because even with two extra points, it's only 99. So, maybe 13 shots?Wait, 13 shots: 11*8 + 9 + 10 = 88 + 9 + 10 = 107, which is more than 100. Hmm, so maybe 13 shots is too much?Wait, perhaps I need to find the possible values of ( S ) such that ( 8x + 9y + 10z = 100 ) with ( x + y + z = S ) and ( S > 11 ).Alternatively, maybe I can think in terms of Equation A: ( y + 2z = 100 - 8S ).Since ( y ) and ( z ) are positive integers, ( 100 - 8S ) must be positive as well because ( y + 2z ) is positive.So, ( 100 - 8S > 0 ) implies ( S < 12.5 ). Since ( S ) is an integer greater than 11, the possible values of ( S ) are 12.Wait, so ( S ) can only be 12 because ( S ) must be less than 12.5 and greater than 11, so ( S = 12 ).So, the total number of shots is 12.Therefore, plugging back into Equation A:( y + 2z = 100 - 8*12 = 100 - 96 = 4 ).So, ( y + 2z = 4 ).But we know that ( y geq 1 ) and ( z geq 1 ). So, let's find positive integers ( y ) and ( z ) such that ( y + 2z = 4 ).Possible solutions:Case 1: ( z = 1 ). Then ( y = 4 - 2*1 = 2 ).Case 2: ( z = 2 ). Then ( y = 4 - 2*2 = 0 ). But ( y geq 1 ), so this is invalid.So, the only solution is ( z = 1 ) and ( y = 2 ).Therefore, ( x = S - y - z = 12 - 2 - 1 = 9 ).So, the number of 8-point shots is 9.Wait, let me verify this.Total shots: 9 + 2 + 1 = 12.Total points: 9*8 + 2*9 + 1*10 = 72 + 18 + 10 = 100.Yes, that adds up correctly.But wait, let me check if there are other possible values of ( S ). Earlier, I thought ( S ) must be 12 because ( S < 12.5 ). But is that necessarily the case?Wait, if ( S = 13 ), then ( y + 2z = 100 - 8*13 = 100 - 104 = -4 ). That's negative, which isn't possible because ( y ) and ( z ) are positive. So, ( S ) can't be 13.Similarly, ( S = 11 ) is not allowed because the problem states more than 11 shots, so ( S geq 12 ).Therefore, ( S = 12 ) is the only possible value.Thus, the number of 8-point shots is 9.Wait, but let me think again. Is there another way this could be solved?Alternatively, maybe I can express the total points in terms of the number of 8s, 9s, and 10s.Let me consider that each 9 is 1 more than an 8, and each 10 is 2 more than an 8.So, if all shots were 8s, the total points would be ( 8S ). The extra points beyond ( 8S ) come from the 9s and 10s.So, the extra points are ( 100 - 8S ). Each 9 contributes 1 extra point, and each 10 contributes 2 extra points.So, ( y + 2z = 100 - 8S ).Which is the same as Equation A.Since ( S = 12 ), the extra points are 100 - 96 = 4.So, ( y + 2z = 4 ), which as before, gives ( y = 2 ), ( z = 1 ), so ( x = 9 ).Therefore, the answer is 9.I think that's solid. I don't see any other possibilities because ( S ) can't be higher than 12 without making the extra points negative, and ( S ) can't be lower than 12 because the problem states more than 11 shots.So, yeah, the athlete made 9 shots of 8 points each.**Final Answer**The athlete made boxed{9} eight-point shots.

answer：First, I recognize that the set consists of three numbers. Let's denote them as ( a ), ( b ), and ( c ), where ( a leq b leq c ).Given that the smallest number is 1, we have ( a = 1 ).The median of the set is 4, which means the middle number ( b ) is also 4.The mean of the set is also 4. The mean is calculated by dividing the sum of the numbers by the number of numbers. Therefore, the sum of the three numbers is ( 3 times 4 = 12 ).We can now set up the equation:[1 + 4 + c = 12]Solving for ( c ), we find:[c = 12 - 5 = 7]Finally, the range of the set is the difference between the largest and smallest numbers:[text{Range} = c - a = 7 - 1 = 6]

answer：Okay, so I need to calculate the product (left(1-frac{1}{10}right)left(1-frac{1}{10^{2}}right)left(1-frac{1}{10^{3}}right) ldotsleft(1-frac{1}{10^{99}}right)) to five decimal places. Hmm, that seems like a lot of terms multiplied together. Let me think about how to approach this.First, let me write out the product more formally. It's the product from (k = 1) to (k = 99) of (left(1 - frac{1}{10^k}right)). So, mathematically, that's:[P = prod_{k=1}^{99} left(1 - frac{1}{10^k}right)]I remember that products like this can sometimes be related to infinite products, especially when dealing with exponents. Maybe I can express this as an infinite product and then approximate it since we're only going up to (k = 99). Let me recall if there's a known infinite product formula that resembles this.Wait, I think the infinite product for (frac{1}{1 + x}) is related to something like (prod_{k=1}^{infty} (1 - x^k)), but I'm not sure. Let me check my memory. Oh, right! There's a formula for the infinite product representation of the q-Pochhammer symbol. Specifically, the infinite product (prod_{k=1}^{infty} (1 - x^k)) is known as the Euler function, and it converges for (|x| < 1).In our case, (x = frac{1}{10}), which is less than 1, so the infinite product should converge. Therefore, the product (P) we're calculating is actually a partial product of the infinite product up to (k = 99). So, if I can compute the infinite product and then subtract the tail from (k = 100) to infinity, that might give me a good approximation.But wait, calculating the infinite product exactly might not be straightforward. Maybe I can use logarithms to approximate the product? Taking the logarithm of the product turns it into a sum, which might be easier to handle.Let me try that. Taking the natural logarithm of (P):[ln P = sum_{k=1}^{99} lnleft(1 - frac{1}{10^k}right)]Now, I can approximate this sum. Since each term (frac{1}{10^k}) is small for larger (k), I can use the Taylor series expansion of (ln(1 - x)) around (x = 0), which is:[ln(1 - x) = -x - frac{x^2}{2} - frac{x^3}{3} - ldots]But since (x = frac{1}{10^k}) is small, especially for larger (k), maybe just the first term or two would be sufficient for a good approximation. Let me see.So, approximating each term:[lnleft(1 - frac{1}{10^k}right) approx -frac{1}{10^k} - frac{1}{2 cdot 10^{2k}} - ldots]If I take just the first term, the approximation becomes:[ln P approx -sum_{k=1}^{99} frac{1}{10^k}]This is a geometric series. The sum of a geometric series (sum_{k=1}^{infty} ar^{k-1}) is (frac{a}{1 - r}), but in our case, it's (sum_{k=1}^{99} frac{1}{10^k}), which is a finite geometric series with first term (a = frac{1}{10}) and common ratio (r = frac{1}{10}).The sum of the first (n) terms of a geometric series is:[S_n = a frac{1 - r^n}{1 - r}]So, plugging in the values:[S_{99} = frac{1}{10} cdot frac{1 - left(frac{1}{10}right)^{99}}{1 - frac{1}{10}} = frac{1}{10} cdot frac{1 - 10^{-99}}{frac{9}{10}} = frac{1}{9} left(1 - 10^{-99}right)]Since (10^{-99}) is an extremely small number, practically zero for our purposes, we can approximate:[S_{99} approx frac{1}{9}]Therefore, the logarithm approximation becomes:[ln P approx -frac{1}{9}]Exponentiating both sides gives:[P approx e^{-1/9}]Calculating (e^{-1/9}):First, (1/9 approx 0.111111). So, (e^{-0.111111}). Let me compute this using a calculator or Taylor series.Alternatively, I know that (e^{-x} approx 1 - x + frac{x^2}{2} - frac{x^3}{6} + ldots). Let's compute up to the (x^3) term for better accuracy.So, (x = 0.111111):[e^{-0.111111} approx 1 - 0.111111 + frac{(0.111111)^2}{2} - frac{(0.111111)^3}{6}]Calculating each term:1. (1) is just 1.2. (0.111111) is approximately (1/9).3. ((0.111111)^2 = 0.012345679), so divided by 2 is approximately 0.0061728395.4. ((0.111111)^3 = 0.0013717421), divided by 6 is approximately 0.0002286237.Putting it all together:[1 - 0.111111 + 0.0061728395 - 0.0002286237 approx 1 - 0.111111 = 0.888889 + 0.0061728395 = 0.8950618395 - 0.0002286237 approx 0.8948332158]So, (e^{-1/9} approx 0.894833). But wait, this is just an approximation using the first three terms of the Taylor series. Maybe I should compute more terms or use a calculator for better precision.Alternatively, I can use the fact that (e^{-1/9}) is approximately equal to (1 / e^{1/9}). Let me compute (e^{1/9}) first.(1/9 approx 0.111111). So, (e^{0.111111}). Again, using the Taylor series:[e^x = 1 + x + frac{x^2}{2} + frac{x^3}{6} + frac{x^4}{24} + ldots]So, plugging in (x = 0.111111):1. (1)2. (+ 0.111111)3. (+ frac{(0.111111)^2}{2} = 0.0061728395)4. (+ frac{(0.111111)^3}{6} = 0.0002286237)5. (+ frac{(0.111111)^4}{24} approx frac{0.0000152416}{24} approx 0.000000635)6. Higher terms will be negligible.Adding these up:1. (1 + 0.111111 = 1.111111)2. (+ 0.0061728395 = 1.1172838395)3. (+ 0.0002286237 = 1.1175124632)4. (+ 0.000000635 approx 1.1175130982)So, (e^{0.111111} approx 1.117513). Therefore, (e^{-0.111111} approx 1 / 1.117513). Let's compute that.Calculating (1 / 1.117513):Well, (1 / 1.117513) is approximately equal to 0.894833. Wait, that's the same as before. So, that seems consistent.But is this accurate enough? Because we only took a few terms in the Taylor series. Maybe I should use a calculator for a better approximation. But since I don't have a calculator here, perhaps I can use more terms in the expansion.Alternatively, I can use the fact that (e^{-1/9}) is approximately 0.894833, as we calculated. But let me check if this is correct.Wait, actually, I think I made a mistake. Because when I approximated (ln P) as (-1/9), I neglected the higher-order terms in the expansion of (ln(1 - x)). So, perhaps my approximation is not accurate enough.Let me think again. The logarithm of the product is the sum of the logarithms. Each term is (ln(1 - 10^{-k})). For small (x = 10^{-k}), (ln(1 - x) approx -x - x^2/2 - x^3/3 - ldots). So, if I include more terms in the expansion, I can get a better approximation.Therefore, perhaps I can write:[ln P = sum_{k=1}^{99} lnleft(1 - frac{1}{10^k}right) approx -sum_{k=1}^{99} left( frac{1}{10^k} + frac{1}{2 cdot 10^{2k}} + frac{1}{3 cdot 10^{3k}} + ldots right)]But this seems complicated because it's an infinite series inside the sum. Maybe instead, I can compute the sum numerically, term by term, for each (k) from 1 to 99, using a sufficient number of terms in the expansion for each (ln(1 - 10^{-k})).But that would be time-consuming. Alternatively, perhaps I can compute the exact value by recognizing the product as related to a known function.Wait, I recall that the infinite product (prod_{k=1}^{infty} (1 - x^k)) is related to the generating function for partitions, but I don't know if that helps here. Alternatively, maybe I can express the product as a limit or relate it to a known constant.Alternatively, perhaps I can compute the product numerically by multiplying the terms one by one, keeping track of the product as I go. Since each term is less than 1, the product will decrease, but since we're dealing with exponents, it might converge quickly.But since we have 99 terms, it's a bit tedious, but maybe manageable.Alternatively, perhaps I can use the fact that the product up to (k = n) is approximately (e^{-H_n / 10}), where (H_n) is the nth harmonic number? Wait, not sure about that.Wait, let me think again about the logarithm approach. If I take the logarithm:[ln P = sum_{k=1}^{99} lnleft(1 - frac{1}{10^k}right)]Each term (ln(1 - 10^{-k})) can be approximated as (-10^{-k} - frac{1}{2}10^{-2k} - frac{1}{3}10^{-3k} - ldots). So, if I take the first two terms, the approximation becomes:[ln P approx -sum_{k=1}^{99} left( frac{1}{10^k} + frac{1}{2 cdot 10^{2k}} right )]So, let's compute this sum.First, compute (S_1 = sum_{k=1}^{99} frac{1}{10^k}). As before, this is a geometric series with first term (1/10) and ratio (1/10). The sum is:[S_1 = frac{1/10 (1 - (1/10)^{99})}{1 - 1/10} = frac{1/10}{9/10} (1 - 10^{-99}) = frac{1}{9}(1 - 10^{-99}) approx frac{1}{9}]Similarly, compute (S_2 = sum_{k=1}^{99} frac{1}{2 cdot 10^{2k}} = frac{1}{2} sum_{k=1}^{99} frac{1}{100^k}). This is another geometric series with first term (1/100) and ratio (1/100). The sum is:[S_2 = frac{1}{2} cdot frac{1/100 (1 - (1/100)^{99})}{1 - 1/100} = frac{1}{2} cdot frac{1/100}{99/100} (1 - 100^{-99}) approx frac{1}{2} cdot frac{1}{99} = frac{1}{198}]Therefore, the approximation for (ln P) becomes:[ln P approx -left( frac{1}{9} + frac{1}{198} right ) = -left( frac{22}{198} + frac{1}{198} right ) = -frac{23}{198} approx -0.115656]So, exponentiating:[P approx e^{-0.115656}]Calculating (e^{-0.115656}). Let me use the Taylor series again.(x = 0.115656)[e^{-x} approx 1 - x + frac{x^2}{2} - frac{x^3}{6} + frac{x^4}{24} - ldots]Compute each term:1. (1)2. (-0.115656)3. (frac{(0.115656)^2}{2} = frac{0.013371}{2} = 0.0066855)4. (-frac{(0.115656)^3}{6} = -frac{0.001547}{6} approx -0.0002578)5. (frac{(0.115656)^4}{24} approx frac{0.0001787}{24} approx 0.000007445)6. Higher terms will be negligible.Adding these up:1. (1 - 0.115656 = 0.884344)2. (+ 0.0066855 = 0.8910295)3. (- 0.0002578 = 0.8907717)4. (+ 0.000007445 approx 0.890779145)So, (e^{-0.115656} approx 0.890779). Hmm, that's a bit different from the previous approximation of 0.894833. So, including the second term in the expansion gave me a slightly lower value.But I think this is still an approximation. Maybe I should include more terms or find a better way.Alternatively, perhaps I can compute the product numerically by multiplying the terms step by step. Since each term is (1 - 10^{-k}), which is a decimal number, I can compute the product incrementally.Let me try that. I'll start with (P = 1), and then multiply each term from (k = 1) to (k = 99). Since I can't compute all 99 terms manually, maybe I can compute the first few terms and see if the product converges quickly.Compute the first few terms:- (k = 1): (1 - 1/10 = 0.9). So, (P = 0.9)- (k = 2): (1 - 1/100 = 0.99). (P = 0.9 times 0.99 = 0.891)- (k = 3): (1 - 1/1000 = 0.999). (P = 0.891 times 0.999 approx 0.890109)- (k = 4): (1 - 1/10000 = 0.9999). (P approx 0.890109 times 0.9999 approx 0.890020)- (k = 5): (1 - 1/100000 = 0.99999). (P approx 0.890020 times 0.99999 approx 0.890011)- (k = 6): (1 - 1/1000000 = 0.999999). (P approx 0.890011 times 0.999999 approx 0.890010)- (k = 7): (1 - 1/10000000 = 0.9999999). (P approx 0.890010 times 0.9999999 approx 0.890010)Hmm, interesting. After (k = 6), the product doesn't change much. It seems to stabilize around 0.890010. Let me check the next few terms:- (k = 8): (1 - 1/10^8 = 0.99999999). (P approx 0.890010 times 0.99999999 approx 0.890010)- Similarly, for (k = 9) to (k = 99), each term is so close to 1 that multiplying them won't significantly change the product.Therefore, it seems that the product converges to approximately 0.890010 after a few terms. But wait, earlier when I approximated using the logarithm, I got around 0.890779, which is slightly higher. So, which one is more accurate?Well, the numerical multiplication suggests that the product converges to about 0.890010, while the logarithmic approximation with two terms gave me 0.890779. There's a discrepancy here.Wait, perhaps because the logarithmic approximation is missing higher-order terms, leading to a less accurate result. The numerical multiplication, on the other hand, is more precise because it's directly computing the product without approximations.But let me check more carefully. When I multiplied up to (k = 6), I got 0.890010. Let me compute a few more terms to see if it changes.- (k = 7): (0.890010 times 0.9999999 approx 0.890010 - 0.890010 times 0.0000001 = 0.890010 - 0.000000089001 approx 0.890009910999)- (k = 8): (0.890009910999 times 0.99999999 approx 0.890009910999 - 0.890009910999 times 0.00000001 approx 0.890009910999 - 0.0000000089000991 approx 0.890009902099)- (k = 9): (0.890009902099 times 0.999999999 approx 0.890009902099 - 0.890009902099 times 0.000000001 approx 0.890009902099 - 0.000000000890009902 approx 0.890009901209)So, it's decreasing slightly, but the change is minimal. Each subsequent term reduces the product by an extremely small amount. Therefore, the product is converging to approximately 0.8900099, which is roughly 0.890010.But earlier, when I used the logarithmic approximation with two terms, I got approximately 0.890779, which is higher. So, which one is correct?Wait, perhaps the logarithmic approximation is missing higher-order terms, leading to an overestimation. Because when I included more terms in the logarithm, the approximation became more accurate, but still, the numerical multiplication is more precise.Alternatively, maybe I can use more terms in the logarithmic expansion to get a better approximation.Let me try including the third term in the expansion of (ln(1 - x)). So, for each (k), (ln(1 - 10^{-k}) approx -10^{-k} - frac{1}{2}10^{-2k} - frac{1}{3}10^{-3k}).Therefore, the sum becomes:[ln P approx -sum_{k=1}^{99} left( frac{1}{10^k} + frac{1}{2 cdot 10^{2k}} + frac{1}{3 cdot 10^{3k}} right )]Compute each sum separately.First, (S_1 = sum_{k=1}^{99} frac{1}{10^k} approx frac{1}{9})Second, (S_2 = sum_{k=1}^{99} frac{1}{2 cdot 10^{2k}} approx frac{1}{198})Third, (S_3 = sum_{k=1}^{99} frac{1}{3 cdot 10^{3k}} approx frac{1}{3} cdot frac{1/1000}{1 - 1/1000} = frac{1}{3} cdot frac{1}{999} approx frac{1}{2997} approx 0.0003336)Therefore, the total approximation:[ln P approx -left( frac{1}{9} + frac{1}{198} + 0.0003336 right ) approx -left( 0.111111 + 0.0050505 + 0.0003336 right ) approx -0.1165]So, exponentiating:[P approx e^{-0.1165}]Calculating (e^{-0.1165}):Again, using the Taylor series:(x = 0.1165)[e^{-x} approx 1 - x + frac{x^2}{2} - frac{x^3}{6} + frac{x^4}{24} - ldots]Compute each term:1. (1)2. (-0.1165)3. (frac{(0.1165)^2}{2} = frac{0.01357225}{2} = 0.006786125)4. (-frac{(0.1165)^3}{6} = -frac{0.001581328}{6} approx -0.000263555)5. (frac{(0.1165)^4}{24} approx frac{0.0001841}{24} approx 0.00000767)6. Higher terms are negligible.Adding these up:1. (1 - 0.1165 = 0.8835)2. (+ 0.006786125 = 0.890286125)3. (- 0.000263555 = 0.89002257)4. (+ 0.00000767 approx 0.89003024)So, (e^{-0.1165} approx 0.890030). Hmm, that's very close to the numerical multiplication result of approximately 0.890010. The difference is about 0.00002, which is within the expected error margin considering we only took a few terms in the expansion.Therefore, combining both methods, it seems that the product converges to approximately 0.890010 to 0.890030. Given that the numerical multiplication gave me around 0.890010 after several terms, and the logarithmic approximation with three terms gave me 0.890030, it's reasonable to conclude that the product is approximately 0.89001 to 0.89003.But to get it to five decimal places, I need more precision. Let me try to compute more terms in the logarithmic expansion or find another approach.Alternatively, perhaps I can recognize that the infinite product (prod_{k=1}^{infty} (1 - 10^{-k})) is known and can be expressed in terms of the q-Pochhammer symbol. Specifically, it's ((1/10; 1/10)_{infty}), which is a known constant.Looking it up, I find that the infinite product (prod_{k=1}^{infty} (1 - 10^{-k})) is approximately 0.890010089999999... So, it's approximately 0.89001009.Wait, that's very close to our numerical result. Therefore, the product up to (k = 99) is extremely close to the infinite product, since (10^{-100}) is negligible. Therefore, the product is approximately 0.89001009.Rounded to five decimal places, that would be 0.89001.But let me verify this with another method. I can use the formula for the infinite product:[prod_{k=1}^{infty} (1 - x^k) = expleft( -sum_{k=1}^{infty} frac{sigma_{-1}(k)}{k} x^k right )]Where (sigma_{-1}(k)) is the sum of the reciprocals of the divisors of (k). But this might complicate things further.Alternatively, I can use the fact that the infinite product converges to approximately 0.89001009 as per known mathematical constants.Therefore, given that the product up to (k = 99) is practically equal to the infinite product, the value is approximately 0.89001009, which to five decimal places is 0.89001.But let me check with another approach. Let me compute the product numerically using more precise terms.Starting with (P = 1):- (k = 1): (P = 0.9)- (k = 2): (P = 0.9 times 0.99 = 0.891)- (k = 3): (P = 0.891 times 0.999 = 0.890109)- (k = 4): (P = 0.890109 times 0.9999 = 0.89002001)- (k = 5): (P = 0.89002001 times 0.99999 = 0.890011001)- (k = 6): (P = 0.890011001 times 0.999999 = 0.890010001)- (k = 7): (P = 0.890010001 times 0.9999999 = 0.8900099999999)- (k = 8): (P = 0.8900099999999 times 0.99999999 = 0.8900099999999 - 0.8900099999999 times 0.00000001 approx 0.8900099999999 - 0.000000008900099999999 approx 0.8900099999109)- (k = 9): (P approx 0.8900099999109 times 0.999999999 approx 0.8900099999109 - 0.8900099999109 times 0.000000001 approx 0.8900099999109 - 0.000000000890009999109 approx 0.8900099998219)Continuing this, each subsequent term reduces the product by an extremely small amount, but the product is converging to approximately 0.890010.Therefore, after multiplying up to (k = 9), the product is approximately 0.8900099998219, which is roughly 0.890010 when rounded to six decimal places.Given that each subsequent term beyond (k = 10) contributes a negligible amount (since (10^{-k}) becomes very small), it's safe to say that the product converges to approximately 0.890010.But to ensure accuracy to five decimal places, let me consider the error introduced by truncating the product at (k = 99). The tail of the product from (k = 100) to infinity is:[prod_{k=100}^{infty} left(1 - frac{1}{10^k}right)]Taking the logarithm:[ln left( prod_{k=100}^{infty} left(1 - frac{1}{10^k}right) right ) = sum_{k=100}^{infty} lnleft(1 - frac{1}{10^k}right)]Approximating each term as (-10^{-k}), the sum becomes:[approx -sum_{k=100}^{infty} frac{1}{10^k} = -frac{1}{10^{100}} cdot frac{1}{1 - 1/10} = -frac{1}{9 cdot 10^{99}}]Which is an extremely small number, effectively zero for our purposes. Therefore, the tail contributes negligibly to the product, so the product up to (k = 99) is practically equal to the infinite product.Thus, the value of the product is approximately 0.89001009, which to five decimal places is 0.89001.But wait, let me check the exact value of the infinite product. I recall that the infinite product (prod_{k=1}^{infty} (1 - x^k)) for (x = 1/10) is known and can be computed using the formula:[prod_{k=1}^{infty} (1 - x^k) = expleft( -sum_{k=1}^{infty} frac{sigma_{-1}(k)}{k} x^k right )]But calculating this sum is complicated. Alternatively, I can use the fact that the infinite product is approximately 0.89001009 as per known mathematical constants.Therefore, the product up to (k = 99) is approximately 0.89001009, which rounds to 0.89001 when considering five decimal places.However, to ensure that the fifth decimal place is accurate, I need to check if the sixth decimal place is less than 5. The approximate value is 0.89001009, so the sixth decimal place is 0, which is less than 5. Therefore, rounding to five decimal places gives 0.89001.But wait, let me double-check with another method. I can use the formula for the infinite product:[prod_{k=1}^{infty} (1 - x^k) = frac{1}{(x; x)_{infty}}]But I'm not sure if that helps here. Alternatively, I can use the fact that the infinite product is related to the Dedekind eta function, but that might be overcomplicating things.Alternatively, perhaps I can use the known value of the infinite product for (x = 1/10). After a quick search in my mind, I recall that the infinite product (prod_{k=1}^{infty} (1 - 10^{-k})) is approximately 0.89001009, as mentioned earlier.Therefore, given that the product up to (k = 99) is practically equal to the infinite product, the value is approximately 0.89001009, which to five decimal places is 0.89001.However, to ensure that the fifth decimal place is accurate, I need to consider the error introduced by truncating the product at (k = 99). As calculated earlier, the error is on the order of (10^{-100}), which is negligible for our purposes of five decimal places.Therefore, the product is approximately 0.89001009, which rounds to 0.89001 when considering five decimal places.But wait, let me check if the exact value is slightly higher or lower. If the infinite product is approximately 0.89001009, then to five decimal places, it's 0.89001, since the sixth decimal is 0, which doesn't round up.However, if the exact value is slightly higher, say 0.89001009, then to five decimal places, it's still 0.89001.Therefore, after all these considerations, I conclude that the product is approximately 0.89001 when rounded to five decimal places.**Final Answer**The product is approximately boxed{0.89001}.