💡To simplify the product (frac{10}{5} cdot frac{15}{10} cdot frac{20}{15} dotsm frac{5n+5}{5n} dotsm frac{2010}{2005}), I observe that each numerator cancels with the denominator of the next fraction. This pattern continues throughout the entire product.Starting with (frac{10}{5}), the denominator (5) cancels with the numerator of the previous fraction, and this cancellation continues until only the last numerator and the first denominator remain. Specifically, the last numerator is (2010) and the first denominator is (5).Thus, the simplified product is (frac{2010}{5}).Calculating this gives:[frac{2010}{5} = 402]Therefore, the simplified product is (402).

💡Okay, so I'm trying to figure out if André can end up with exactly 2009 pieces of paper after some number of rounds of cutting. Let me start by understanding the problem step by step.At the beginning, André has 7 pieces of paper. In each round, he takes some of these pieces and cuts each one into 7 smaller pieces. Then, he mixes these cut pieces with the ones he didn't cut in that round. So, each time he cuts a piece, it's replaced by 7 new pieces, effectively increasing the total number of pieces by 6 each time he cuts one piece.Let me try to model this mathematically. Suppose in the first round, André decides to cut 'n₁' pieces. Each of these 'n₁' pieces becomes 7 pieces, so the total number of pieces after the first round would be:Original pieces not cut: 7 - n₁Plus the cut pieces: 7 * n₁Total after first round: (7 - n₁) + 7n₁ = 7 + 6n₁Okay, so after the first round, he has 7 + 6n₁ pieces. Now, in the second round, he can choose to cut some of these pieces again. Let's say he cuts 'n₂' pieces in the second round. Each of these 'n₂' pieces becomes 7 pieces, so the total number of pieces after the second round would be:Pieces not cut: (7 + 6n₁) - n₂Plus the cut pieces: 7 * n₂Total after second round: (7 + 6n₁ - n₂) + 7n₂ = 7 + 6n₁ + 6n₂Hmm, I see a pattern here. After each round, the total number of pieces increases by 6 times the number of pieces he cuts in that round. So, after 'k' rounds, the total number of pieces would be:Total pieces = 7 + 6(n₁ + n₂ + ... + nₖ)Now, we want to know if this total can ever be exactly 2009. So, we set up the equation:7 + 6(n₁ + n₂ + ... + nₖ) = 2009Subtracting 7 from both sides gives:6(n₁ + n₂ + ... + nₖ) = 2002So, we need to check if 2002 is divisible by 6. If it is, then it's possible; otherwise, it's not.Let me check the divisibility of 2002 by 6. A number is divisible by 6 if it's divisible by both 2 and 3.First, check divisibility by 2: 2002 is even, so it's divisible by 2.Next, check divisibility by 3: To check if a number is divisible by 3, add its digits and see if the sum is divisible by 3.2 + 0 + 0 + 2 = 44 is not divisible by 3, so 2002 is not divisible by 3, and therefore not divisible by 6.Since 2002 isn't divisible by 6, there's no integer value for (n₁ + n₂ + ... + nₖ) that satisfies the equation 6(n₁ + n₂ + ... + nₖ) = 2002.Therefore, André cannot have exactly 2009 pieces of paper after any number of rounds.

💡Okay, so I have this problem here: Given points A(a, 1) and B(2, a) are on opposite sides of the y-axis, I need to find the range of the real number a. Hmm, let me think about how to approach this.First, I remember that in the Cartesian coordinate system, the y-axis is the line x = 0. Points on one side of the y-axis have positive x-coordinates, and points on the other side have negative x-coordinates. So, if two points are on opposite sides of the y-axis, one must have a positive x-coordinate and the other a negative x-coordinate.Looking at point A(a, 1), its x-coordinate is 'a'. And point B(2, a) has an x-coordinate of 2. So, point B is fixed at x = 2, which is on the right side of the y-axis because 2 is positive. That means point A must be on the opposite side, which is the left side of the y-axis. For point A to be on the left side, its x-coordinate 'a' must be negative.Wait, is that all? So, if point B is on the right, point A must be on the left, so a must be less than zero. Therefore, the range of 'a' is all real numbers less than zero. That seems straightforward.But let me double-check. If a were positive, then both points A and B would be on the right side of the y-axis, which contradicts the condition that they are on opposite sides. If a were zero, point A would be on the y-axis itself, not on the opposite side. So, a must indeed be negative.Is there any other condition I need to consider? The y-coordinates of the points are 1 and a, respectively. Does the y-coordinate affect their position relative to the y-axis? Hmm, no, because the y-axis is determined solely by the x-coordinate. So, regardless of the y-coordinate, the side of the y-axis a point is on depends only on its x-coordinate.Therefore, the only condition we need is that the x-coordinate of point A is negative, which gives us a < 0. So, the range of 'a' is all real numbers less than zero.I think that's it. There's no upper limit specified for 'a', so it can be any negative real number. So, the range is simply a < 0.**Final Answer**The range of the real number ( a ) is boxed{a < 0}.

💡First, I need to calculate the total rainfall on Mondays by multiplying the number of Mondays by the rainfall per Monday.Next, I'll calculate the total rainfall on Tuesdays by multiplying the number of Tuesdays by the rainfall per Tuesday.After obtaining both totals, I'll subtract the total rainfall on Mondays from the total rainfall on Tuesdays to find out how many more centimeters it rained on Tuesdays compared to Mondays.Finally, I'll present the calculated difference in a clear and concise manner.

💡Okay, so I have this problem where I need to find the points where two parabolas intersect. The equations given are ( y = 3x^2 - 12x - 9 ) and ( y = 2x^2 - 8x + 5 ). Hmm, I remember that to find the points of intersection between two graphs, I need to set their equations equal to each other because at the points of intersection, both the x and y values are the same.Alright, so let me set them equal:( 3x^2 - 12x - 9 = 2x^2 - 8x + 5 )Now, I need to solve for x. To do that, I should bring all terms to one side so that I can have a quadratic equation set to zero. Let me subtract ( 2x^2 - 8x + 5 ) from both sides:( 3x^2 - 12x - 9 - (2x^2 - 8x + 5) = 0 )Simplifying that, I distribute the negative sign:( 3x^2 - 12x - 9 - 2x^2 + 8x - 5 = 0 )Now, combine like terms:- ( 3x^2 - 2x^2 = x^2 )- ( -12x + 8x = -4x )- ( -9 - 5 = -14 )So, the equation simplifies to:( x^2 - 4x - 14 = 0 )Alright, now I have a quadratic equation. I need to solve for x. I can use the quadratic formula here, which is ( x = frac{-b pm sqrt{b^2 - 4ac}}{2a} ), where a, b, and c are coefficients from the quadratic equation ( ax^2 + bx + c = 0 ).In this case, a = 1, b = -4, and c = -14.Plugging these into the quadratic formula:( x = frac{-(-4) pm sqrt{(-4)^2 - 4(1)(-14)}}{2(1)} )Simplify step by step:First, calculate the numerator:- ( -(-4) = 4 )- ( (-4)^2 = 16 )- ( 4(1)(-14) = -56 )- So, inside the square root: ( 16 - (-56) = 16 + 56 = 72 )So, now we have:( x = frac{4 pm sqrt{72}}{2} )Simplify the square root of 72. I know that 72 can be broken down into 36*2, and the square root of 36 is 6, so:( sqrt{72} = sqrt{36*2} = 6sqrt{2} )So, substituting back:( x = frac{4 pm 6sqrt{2}}{2} )Now, we can simplify this fraction by dividing both numerator terms by 2:( x = 2 pm 3sqrt{2} )So, the two x-values where the parabolas intersect are ( x = 2 + 3sqrt{2} ) and ( x = 2 - 3sqrt{2} ).Now, I need to find the corresponding y-values for each x to get the complete points of intersection. I can plug these x-values back into either of the original equations. I'll choose the first equation ( y = 3x^2 - 12x - 9 ) because it seems straightforward.First, let's find y when ( x = 2 + 3sqrt{2} ):( y = 3(2 + 3sqrt{2})^2 - 12(2 + 3sqrt{2}) - 9 )I need to expand ( (2 + 3sqrt{2})^2 ). Let's do that step by step:( (2 + 3sqrt{2})^2 = 2^2 + 2*2*3sqrt{2} + (3sqrt{2})^2 )= 4 + 12sqrt{2} + 9*2= 4 + 12sqrt{2} + 18= 22 + 12sqrt{2}Now, plug that back into the equation:( y = 3*(22 + 12sqrt{2}) - 12*(2 + 3sqrt{2}) - 9 )Let's compute each term:- ( 3*(22 + 12sqrt{2}) = 66 + 36sqrt{2} )- ( -12*(2 + 3sqrt{2}) = -24 - 36sqrt{2} )- The last term is -9.Now, combine all these:( 66 + 36sqrt{2} - 24 - 36sqrt{2} - 9 )Simplify term by term:- ( 66 - 24 = 42 )- ( 42 - 9 = 33 )- ( 36sqrt{2} - 36sqrt{2} = 0 )So, y = 33 when ( x = 2 + 3sqrt{2} ).Wait, that seems a bit off because when I calculated earlier, I thought it was 66 - 36√2. Did I make a mistake?Let me double-check the calculation:Starting again with ( x = 2 + 3sqrt{2} ):( y = 3x^2 - 12x - 9 )First, compute ( x^2 ):( (2 + 3sqrt{2})^2 = 4 + 12sqrt{2} + 18 = 22 + 12sqrt{2} )Multiply by 3:( 3*(22 + 12sqrt{2}) = 66 + 36sqrt{2} )Now, compute -12x:( -12*(2 + 3sqrt{2}) = -24 - 36sqrt{2} )Add the constant term -9:So, total y = 66 + 36√2 -24 -36√2 -9Simplify:66 -24 = 4242 -9 = 3336√2 -36√2 = 0So, y = 33. Hmm, that's different from what I thought earlier. Maybe I made a mistake in the initial calculation.Wait, perhaps I should try plugging into the other equation to verify.Let's use the second equation ( y = 2x^2 - 8x + 5 ) with ( x = 2 + 3sqrt{2} ):Compute ( x^2 ):( (2 + 3sqrt{2})^2 = 4 + 12sqrt{2} + 18 = 22 + 12sqrt{2} )Multiply by 2:( 2*(22 + 12sqrt{2}) = 44 + 24sqrt{2} )Compute -8x:( -8*(2 + 3sqrt{2}) = -16 -24sqrt{2} )Add the constant term +5:Total y = 44 +24√2 -16 -24√2 +5Simplify:44 -16 = 2828 +5 = 3324√2 -24√2 = 0So, y = 33. Okay, so both equations give y = 33 when x = 2 + 3√2. That seems correct. So, my initial calculation was wrong because I must have miscalculated earlier.Now, let's find y when ( x = 2 - 3sqrt{2} ).Again, using the first equation ( y = 3x^2 - 12x - 9 ):Compute ( x^2 ):( (2 - 3sqrt{2})^2 = 4 - 12sqrt{2} + 18 = 22 - 12sqrt{2} )Multiply by 3:( 3*(22 - 12sqrt{2}) = 66 - 36sqrt{2} )Compute -12x:( -12*(2 - 3sqrt{2}) = -24 + 36sqrt{2} )Add the constant term -9:Total y = 66 -36√2 -24 +36√2 -9Simplify:66 -24 = 4242 -9 = 33-36√2 +36√2 = 0So, y = 33 again.Wait, that's interesting. Both x-values give the same y-value, 33. So, both points of intersection are at y = 33, but with different x-values.So, the points are ( (2 + 3sqrt{2}, 33) ) and ( (2 - 3sqrt{2}, 33) ).But wait, earlier when I calculated using the first equation, I thought I got different y-values, but now both give y = 33. That seems correct because when I checked with the second equation, I also got y = 33. So, my initial confusion was because I miscalculated earlier.Therefore, the points of intersection are ( (2 + 3sqrt{2}, 33) ) and ( (2 - 3sqrt{2}, 33) ).To list them in order of increasing x-coordinate, since ( 2 - 3sqrt{2} ) is less than ( 2 + 3sqrt{2} ), the points are:( (2 - 3sqrt{2}, 33) ) and ( (2 + 3sqrt{2}, 33) ).I think that's it. I should double-check my calculations to make sure I didn't make any errors.Let me verify the quadratic equation step again:Original equations:( 3x^2 - 12x - 9 = 2x^2 - 8x + 5 )Subtracting ( 2x^2 - 8x + 5 ) from both sides:( 3x^2 - 12x - 9 - 2x^2 + 8x -5 = 0 )Simplify:( x^2 -4x -14 = 0 )Yes, that's correct.Using quadratic formula:( x = [4 ± sqrt(16 + 56)] / 2 = [4 ± sqrt(72)] / 2 = [4 ± 6√2]/2 = 2 ± 3√2 )That's correct.Then, plugging back into the equation, both times I got y = 33, which seems consistent.So, the final points are ( (2 - 3sqrt{2}, 33) ) and ( (2 + 3sqrt{2}, 33) ).

💡Okay, so I have this problem where I need to prove that for any natural number ( n ), the sum of the reciprocals of all possible products of distinct integers from 1 to ( n ) equals ( n ). That is,[sum_{1 leqslant i_{1}<i_{2}<ldots<i_{k} leqslant n} frac{1}{i_{1} i_{2} ldots i_{k}} = n]where the summation is over all possible sets of numbers ( i_{1}<i_{2}<ldots<i_{k} ) for ( k = 1, 2, ldots, n ).Hmm, let me try to understand what this sum actually represents. It seems like we're considering all possible subsets of the set ( {1, 2, ldots, n} ) where the subsets have at least one element, and for each subset, we take the product of its elements and then take the reciprocal of that product. Finally, we sum all these reciprocals.So, for example, if ( n = 2 ), the subsets are ( {1} ), ( {2} ), and ( {1, 2} ). The sum would be:[frac{1}{1} + frac{1}{2} + frac{1}{1 cdot 2} = 1 + 0.5 + 0.5 = 2]Which equals ( n = 2 ). That checks out.Similarly, for ( n = 3 ), the subsets are ( {1} ), ( {2} ), ( {3} ), ( {1, 2} ), ( {1, 3} ), ( {2, 3} ), and ( {1, 2, 3} ). The sum is:[frac{1}{1} + frac{1}{2} + frac{1}{3} + frac{1}{1 cdot 2} + frac{1}{1 cdot 3} + frac{1}{2 cdot 3} + frac{1}{1 cdot 2 cdot 3}]Calculating each term:[1 + 0.5 + 0.333... + 0.5 + 0.333... + 0.1666... + 0.1666... approx 1 + 0.5 + 0.333 + 0.5 + 0.333 + 0.166 + 0.166 approx 3]Which equals ( n = 3 ). So, it seems to hold for small values. But how do I prove it for any ( n )?Maybe I can think about generating functions or some combinatorial identity. Let me recall that the product ( (1 + x)(1 + y) ) expands to ( 1 + x + y + xy ). If I consider the product over all terms ( (1 + frac{1}{i}) ) for ( i ) from 1 to ( n ), then expanding this product would give me all possible sums of reciprocals of products of subsets, similar to the sum in the problem.So, let's define:[P = prod_{i=1}^{n} left(1 + frac{1}{i}right)]Expanding this product, each term in the expansion corresponds to choosing either 1 or ( frac{1}{i} ) from each factor ( (1 + frac{1}{i}) ). If we choose ( frac{1}{i} ) from some subset of the factors, the product of these chosen terms gives the reciprocal of the product of the indices ( i ). Summing all these terms gives exactly the sum we're interested in.Therefore, the sum ( S ) is equal to ( P - 1 ), since ( P ) includes the term 1 (from choosing 1 in all factors), which we need to subtract.So,[S = P - 1 = prod_{i=1}^{n} left(1 + frac{1}{i}right) - 1]Now, let's compute ( P ):[P = prod_{i=1}^{n} left(1 + frac{1}{i}right) = prod_{i=1}^{n} frac{i + 1}{i} = frac{2}{1} times frac{3}{2} times frac{4}{3} times ldots times frac{n + 1}{n}]Notice that in this product, each numerator cancels with the denominator of the next fraction:[frac{2}{1} times frac{3}{2} times frac{4}{3} times ldots times frac{n + 1}{n} = frac{n + 1}{1} = n + 1]Therefore,[S = P - 1 = (n + 1) - 1 = n]So, the sum equals ( n ), which is what we wanted to prove.Wait, let me double-check this reasoning. I considered the product ( P ) which expands to include all possible subset products, subtracted 1 to exclude the empty product, and then simplified the product to ( n + 1 ). Subtracting 1 gives ( n ), which matches the desired result. The telescoping nature of the product seems correct, as each intermediate term cancels out. Is there another way to approach this? Maybe by induction. Let's try that.**Base Case:** For ( n = 1 ), the sum is just ( frac{1}{1} = 1 ), which equals ( n = 1 ). So, the base case holds.**Inductive Step:** Assume that for some ( k geq 1 ), the sum ( S_k = k ). Now, consider ( S_{k+1} ). The sum ( S_{k+1} ) includes all subsets of ( {1, 2, ldots, k+1} ). These subsets can be divided into two categories: those that do not include ( k+1 ) and those that do.- The subsets that do not include ( k+1 ) are exactly the subsets of ( {1, 2, ldots, k} ), whose sum is ( S_k = k ) by the inductive hypothesis.- The subsets that include ( k+1 ) can be formed by taking each subset of ( {1, 2, ldots, k} ) and adding ( k+1 ) to it. For each such subset, the reciprocal of the product becomes ( frac{1}{(k+1) times text{product of the subset}} ). Therefore, the sum over these subsets is ( frac{1}{k+1} times S_k ).Thus,[S_{k+1} = S_k + frac{1}{k+1} times S_k = S_k left(1 + frac{1}{k+1}right) = S_k times frac{k+2}{k+1}]But wait, according to the inductive hypothesis, ( S_k = k ). So,[S_{k+1} = k times frac{k+2}{k+1} = frac{k(k+2)}{k+1}]Hmm, this doesn't immediately simplify to ( k + 1 ). Did I make a mistake?Wait, let's reconsider. The sum over subsets that include ( k+1 ) is not just ( frac{1}{k+1} times S_k ). Because when you add ( k+1 ) to a subset, the reciprocal of the product becomes ( frac{1}{(k+1) times text{product of the subset}} ). However, the empty subset (which doesn't include any elements) when combined with ( k+1 ) gives just ( frac{1}{k+1} ). So, actually, the sum over subsets that include ( k+1 ) is ( frac{1}{k+1} times (S_k + 1) ), where the additional 1 accounts for the empty subset.Wait, no. The empty subset corresponds to the term ( frac{1}{k+1} ) when combined with ( k+1 ). But in our original sum ( S_{k+1} ), we are only considering non-empty subsets. So, actually, the sum over subsets that include ( k+1 ) is ( frac{1}{k+1} times (S_k + 1) - frac{1}{k+1} times 1 ), because the empty subset contributes ( frac{1}{k+1} ), but we need to exclude it since we are only summing over non-empty subsets.Wait, this is getting confusing. Maybe a better approach is to express ( S_{k+1} ) as:[S_{k+1} = S_k + sum_{text{subsets including } k+1} frac{1}{text{product}}]The subsets including ( k+1 ) can be formed by taking each subset of ( {1, 2, ldots, k} ) and adding ( k+1 ) to it. For each subset ( A ) of ( {1, 2, ldots, k} ), the corresponding subset ( A cup {k+1} ) contributes ( frac{1}{(k+1) times prod A} ). Therefore, the sum over these subsets is:[sum_{A subseteq {1, 2, ldots, k}} frac{1}{(k+1) times prod A} = frac{1}{k+1} sum_{A subseteq {1, 2, ldots, k}} frac{1}{prod A}]But ( sum_{A subseteq {1, 2, ldots, k}} frac{1}{prod A} ) is exactly ( S_k + 1 ), where the additional 1 comes from the empty subset. However, in our original sum ( S_{k+1} ), we are only considering non-empty subsets, so we need to subtract the term corresponding to the empty subset, which is ( frac{1}{k+1} ).Wait, no. The sum ( sum_{A subseteq {1, 2, ldots, k}} frac{1}{prod A} ) includes the empty subset, which contributes 1. But in our case, when we form subsets of ( {1, 2, ldots, k+1} ) that include ( k+1 ), the empty subset combined with ( k+1 ) gives just ( frac{1}{k+1} ). However, in ( S_{k+1} ), we are summing over all non-empty subsets, so the term ( frac{1}{k+1} ) is included. Therefore, the sum over subsets including ( k+1 ) is:[frac{1}{k+1} times (S_k + 1)]But wait, ( S_k ) is the sum over non-empty subsets of ( {1, 2, ldots, k} ), so ( S_k + 1 ) would include the empty subset. Therefore, the sum over subsets including ( k+1 ) is:[frac{1}{k+1} times (S_k + 1)]But in ( S_{k+1} ), we need to include all non-empty subsets, so:[S_{k+1} = S_k + frac{1}{k+1} times (S_k + 1)]Simplifying:[S_{k+1} = S_k + frac{S_k}{k+1} + frac{1}{k+1} = S_k left(1 + frac{1}{k+1}right) + frac{1}{k+1}]But this seems complicated. Maybe I'm overcomplicating it. Let's go back to the generating function approach, which seemed straightforward.We have:[S_n = prod_{i=1}^{n} left(1 + frac{1}{i}right) - 1 = (n + 1) - 1 = n]This seems much simpler and avoids the confusion of the inductive step. Therefore, the generating function approach provides a clear and concise proof.Alternatively, another way to think about it is to consider the sum as the sum over all non-empty subsets, and each element ( i ) is included in exactly half of the subsets. However, this line of reasoning might not directly lead to the result, but it's worth exploring.Wait, actually, each element ( i ) is included in ( 2^{n-1} ) subsets, but that might not be directly useful here. Instead, focusing on the generating function seems more effective.In conclusion, the generating function approach provides a straightforward proof by recognizing that the sum is the expansion of a product, which telescopes to ( n + 1 ), and subtracting 1 gives the desired result.