💡Okay, so I have this problem where I need to find the smallest possible value of ( c + d ) where both ( c ) and ( d ) are positive real numbers. The conditions given are that each of the quadratic equations ( x^2 + cx + 3d = 0 ) and ( x^2 + 3dx + c = 0 ) has real roots. First, I remember that for a quadratic equation ( ax^2 + bx + c = 0 ) to have real roots, the discriminant must be non-negative. The discriminant is given by ( b^2 - 4ac ). So, I need to apply this condition to both equations.Starting with the first equation ( x^2 + cx + 3d = 0 ), the discriminant is ( c^2 - 4 times 1 times 3d ). This simplifies to ( c^2 - 12d ). For real roots, this discriminant must be greater than or equal to zero. So, I have the inequality:[ c^2 - 12d geq 0 ][ Rightarrow c^2 geq 12d ]Okay, that's the first condition. Now moving on to the second equation ( x^2 + 3dx + c = 0 ). The discriminant here is ( (3d)^2 - 4 times 1 times c ), which simplifies to ( 9d^2 - 4c ). Again, for real roots, this discriminant must be non-negative:[ 9d^2 - 4c geq 0 ][ Rightarrow 9d^2 geq 4c ][ Rightarrow c leq frac{9d^2}{4} ]So now I have two inequalities:1. ( c^2 geq 12d )2. ( c leq frac{9d^2}{4} )My goal is to find the minimum value of ( c + d ). Since both ( c ) and ( d ) are positive, I can try to express one variable in terms of the other and then minimize the sum.From the second inequality, I can express ( c ) as ( c leq frac{9d^2}{4} ). Since I want to minimize ( c + d ), I should consider the case where ( c ) is as small as possible, but it still has to satisfy the first inequality ( c^2 geq 12d ). So, substituting ( c = frac{9d^2}{4} ) into the first inequality might give me the boundary condition.Let me try that:Substitute ( c = frac{9d^2}{4} ) into ( c^2 geq 12d ):[ left( frac{9d^2}{4} right)^2 geq 12d ][ frac{81d^4}{16} geq 12d ]Hmm, that looks a bit complicated, but let's see. I can multiply both sides by 16 to eliminate the denominator:[ 81d^4 geq 192d ]Now, I can divide both sides by ( d ) (since ( d ) is positive, this is allowed):[ 81d^3 geq 192 ][ d^3 geq frac{192}{81} ]Simplify ( frac{192}{81} ). Let me see, both are divisible by 3:[ frac{192 ÷ 3}{81 ÷ 3} = frac{64}{27} ]So, ( d^3 geq frac{64}{27} ). Taking the cube root of both sides:[ d geq sqrt[3]{frac{64}{27}} ][ d geq frac{4}{3} ]Okay, so ( d ) must be at least ( frac{4}{3} ). Now, let's find the corresponding ( c ) when ( d = frac{4}{3} ). Using the second inequality ( c leq frac{9d^2}{4} ):[ c leq frac{9 times left( frac{4}{3} right)^2}{4} ][ c leq frac{9 times frac{16}{9}}{4} ][ c leq frac{16}{4} ][ c leq 4 ]So, ( c ) can be at most 4 when ( d = frac{4}{3} ). But wait, from the first inequality ( c^2 geq 12d ), let's check if ( c = 4 ) satisfies this:[ 4^2 = 16 ][ 12d = 12 times frac{4}{3} = 16 ][ 16 geq 16 ]Yes, it does. So, when ( d = frac{4}{3} ) and ( c = 4 ), both discriminants are zero, meaning both quadratics have exactly one real root (a repeated root). Therefore, the smallest possible value of ( c + d ) is ( 4 + frac{4}{3} = frac{12}{3} + frac{4}{3} = frac{16}{3} ).I should double-check my steps to make sure I didn't make any mistakes. Starting from the discriminants, I correctly set up the inequalities. Then, substituting ( c ) in terms of ( d ) led me to the equation involving ( d^3 ). Solving that gave me ( d geq frac{4}{3} ), which seems correct. Plugging ( d = frac{4}{3} ) back into the expression for ( c ) gave me ( c = 4 ), and verifying with the first inequality confirmed it. So, I think my solution is correct, and the minimal value of ( c + d ) is indeed ( frac{16}{3} ).**Final Answer**The smallest possible value of ( c + d ) is boxed{dfrac{16}{3}}.

💡First, I need to simplify the given expression ([left(a-2bright)^{2}-left(a-2bright)left(a+2bright)-4b]div left(-2bright)) before substituting the values of (a = 1) and (b = -2).I'll start by expanding the terms in the numerator:1. Expand ((a - 2b)^2):[(a - 2b)^2 = a^2 - 4ab + 4b^2]2. Expand ((a - 2b)(a + 2b)) using the difference of squares formula:[(a - 2b)(a + 2b) = a^2 - (2b)^2 = a^2 - 4b^2]Now, substitute these expansions back into the original expression:[(a^2 - 4ab + 4b^2) - (a^2 - 4b^2) - 4b]Simplify the expression by combining like terms:[a^2 - 4ab + 4b^2 - a^2 + 4b^2 - 4b = -4ab + 8b^2 - 4b]Next, divide the simplified numerator by (-2b):[frac{-4ab + 8b^2 - 4b}{-2b} = frac{-4ab}{-2b} + frac{8b^2}{-2b} - frac{4b}{-2b} = 2a - 4b + 2]Finally, substitute (a = 1) and (b = -2) into the simplified expression:[2(1) - 4(-2) + 2 = 2 + 8 + 2 = 12]Therefore, the evaluated result of the expression is 12.

💡Alright, so I've got this problem here about a sequence where the first two terms are both 1, and each subsequent term is the sum of the two before it. That sounds familiar—it must be the Fibonacci sequence! So, f₁ = 1, f₂ = 1, f₃ = 2, f₄ = 3, f₅ = 5, and so on. The problem is asking me to prove that the sum of the reciprocals of the products of terms two apart, from f₁f₃ up to f₉₈f₁₀₀, is less than 1. That is, I need to show:1/(f₁f₃) + 1/(f₂f₄) + 1/(f₃f₅) + ... + 1/(f₉₈f₁₀₀) < 1.Hmm, okay. So, first, let me write out the first few terms to see if I can spot a pattern or figure out a way to approach this.Calculating the first few terms:1/(f₁f₃) = 1/(1*2) = 1/2,1/(f₂f₄) = 1/(1*3) = 1/3,1/(f₃f₅) = 1/(2*5) = 1/10,1/(f₄f₆) = 1/(3*8) = 1/24,and so on.Adding these up: 1/2 + 1/3 + 1/10 + 1/24 + ... It seems like each term is getting smaller, which is good because it might converge to something less than 1.But how do I sum this series? It's not a geometric series or anything I recognize immediately. Maybe I can find a telescoping series or some kind of pattern that allows the terms to cancel out when summed.I remember that in some series involving Fibonacci numbers, there are identities that can help. Maybe I can express each term as a difference of fractions that telescope when summed. Let me think about that.Suppose I can write 1/(fₖfₖ₊₂) as something like A/fₖ - B/fₖ₊₁ or something similar. If I can find constants A and B such that:1/(fₖfₖ₊₂) = A/fₖ - B/fₖ₊₁,then when I sum over k, the terms might telescope. Let's try to find such A and B.Starting with the right-hand side:A/fₖ - B/fₖ₊₁.I want this to equal 1/(fₖfₖ₊₂). Let's express 1/(fₖfₖ₊₂) in terms of A and B.First, note that fₖ₊₂ = fₖ₊₁ + fₖ. So, fₖ₊₂ = fₖ₊₁ + fₖ.Let me try to manipulate A/fₖ - B/fₖ₊₁ to get 1/(fₖfₖ₊₂).Let me write A/fₖ - B/fₖ₊₁ as (A fₖ₊₁ - B fₖ)/(fₖ fₖ₊₁).I want this to equal 1/(fₖ fₖ₊₂). So,(A fₖ₊₁ - B fₖ)/(fₖ fₖ₊₁) = 1/(fₖ fₖ₊₂).Multiplying both sides by fₖ fₖ₊₁:A fₖ₊₁ - B fₖ = fₖ₊₁ / fₖ₊₂.Hmm, that seems a bit complicated. Maybe I need a different approach.Wait, maybe I can use the identity that relates Fibonacci numbers and their reciprocals. I recall that there's an identity involving the reciprocal of Fibonacci numbers that can telescope.Alternatively, maybe I can use the fact that fₖ₊₂ = fₖ₊₁ + fₖ, and try to express 1/(fₖfₖ₊₂) in terms of fₖ and fₖ₊₁.Let me try expressing 1/(fₖfₖ₊₂) as (1/fₖ - 1/fₖ₊₁)/something.Let me compute 1/fₖ - 1/fₖ₊₁:1/fₖ - 1/fₖ₊₁ = (fₖ₊₁ - fₖ)/(fₖ fₖ₊₁) = fₖ₋₁/(fₖ fₖ₊₁).Wait, because fₖ₊₁ - fₖ = fₖ₋₁. That's a Fibonacci identity: fₖ₊₁ = fₖ + fₖ₋₁, so fₖ₊₁ - fₖ = fₖ₋₁.So, 1/fₖ - 1/fₖ₊₁ = fₖ₋₁/(fₖ fₖ₊₁).Hmm, but I need to relate this to 1/(fₖfₖ₊₂). Maybe I can manipulate this expression further.Let me see:1/(fₖfₖ₊₂) = ?Since fₖ₊₂ = fₖ₊₁ + fₖ, maybe I can write 1/(fₖfₖ₊₂) = 1/(fₖ(fₖ + fₖ₊₁)).Hmm, not sure if that helps. Maybe I can write it as:1/(fₖfₖ₊₂) = [1/fₖ - 1/fₖ₊₂]/(fₖ₊₂ - fₖ).But fₖ₊₂ - fₖ = fₖ₊₁ + fₖ - fₖ = fₖ₊₁.So,1/(fₖfₖ₊₂) = [1/fₖ - 1/fₖ₊₂]/fₖ₊₁.Hmm, that might be useful. Let me write that down:1/(fₖfₖ₊₂) = (1/fₖ - 1/fₖ₊₂)/fₖ₊₁.But I'm not sure if that helps me telescope the series. Maybe I need a different identity.Wait, going back to the earlier expression:1/fₖ - 1/fₖ₊₁ = fₖ₋₁/(fₖ fₖ₊₁).If I can relate this to 1/(fₖfₖ₊₂), maybe I can find a telescoping pattern.Let me try to express 1/(fₖfₖ₊₂) in terms of differences of reciprocals.Suppose I have:1/(fₖfₖ₊₂) = A(1/fₖ - 1/fₖ₊₁).From earlier, we have:1/fₖ - 1/fₖ₊₁ = fₖ₋₁/(fₖ fₖ₊₁).So, if I set A = 1/fₖ₊₂, then:A(1/fₖ - 1/fₖ₊₁) = (1/fₖ₊₂)(fₖ₋₁/(fₖ fₖ₊₁)).But fₖ₊₂ = fₖ₊₁ + fₖ, and fₖ₋₁ = fₖ₊₁ - fₖ.Wait, maybe I can relate fₖ₋₁ and fₖ₊₂.Alternatively, maybe I can find a constant such that:1/(fₖfₖ₊₂) = (1/fₖ - 1/fₖ₊₁)/fₖ₊₂.Wait, let's test this:(1/fₖ - 1/fₖ₊₁)/fₖ₊₂ = (fₖ₊₁ - fₖ)/(fₖ fₖ₊₁ fₖ₊₂) = fₖ₋₁/(fₖ fₖ₊₁ fₖ₊₂).But 1/(fₖfₖ₊₂) = 1/(fₖ(fₖ + fₖ₊₁)).Not sure if that's helpful.Maybe I need to think differently. Perhaps instead of trying to express each term as a difference, I can look for a telescoping product or another approach.Alternatively, maybe I can use induction to prove the sum up to n terms is less than 1.Wait, the problem is to sum up to 98 terms, but maybe I can generalize it.Let me consider the sum Sₙ = Σₖ=1ⁿ 1/(fₖfₖ₊₂).I need to show that S₉₈ < 1.Perhaps I can find a closed-form expression for Sₙ and then show that it's less than 1.Let me try to compute Sₙ for small n and see if I can spot a pattern.For n=1: S₁ = 1/(1*2) = 1/2.For n=2: S₂ = 1/2 + 1/(1*3) = 1/2 + 1/3 = 5/6 ≈ 0.833.For n=3: S₃ = 5/6 + 1/(2*5) = 5/6 + 1/10 = 25/30 + 3/30 = 28/30 = 14/15 ≈ 0.933.For n=4: S₄ = 14/15 + 1/(3*8) = 14/15 + 1/24 = (14*24 + 15)/(15*24) = (336 + 15)/360 = 351/360 = 117/120 = 39/40 = 0.975.For n=5: S₅ = 39/40 + 1/(5*13) = 39/40 + 1/65 = (39*65 + 40)/(40*65) = (2535 + 40)/2600 = 2575/2600 = 103/104 ≈ 0.9904.Hmm, interesting. It seems like Sₙ is approaching 1 as n increases, but never reaching it. So, for n=5, S₅ ≈ 0.9904, which is still less than 1.If this pattern continues, then S₉₈ would be very close to 1 but still less than 1. So, perhaps I can find a general formula for Sₙ and show that it's always less than 1.Looking back at the small n cases, I notice that Sₙ seems to be approaching 1 as n increases. Maybe Sₙ = 1 - something that goes to zero as n increases.Let me try to find a general expression for Sₙ.Suppose Sₙ = 1 - 1/(fₙ₊₁fₙ₊₂).Let me test this for n=1:S₁ = 1 - 1/(f₂f₃) = 1 - 1/(1*2) = 1 - 1/2 = 1/2. Which matches.For n=2:S₂ = 1 - 1/(f₃f₄) = 1 - 1/(2*3) = 1 - 1/6 = 5/6. Which matches.For n=3:S₃ = 1 - 1/(f₄f₅) = 1 - 1/(3*5) = 1 - 1/15 = 14/15. Which matches.For n=4:S₄ = 1 - 1/(f₅f₆) = 1 - 1/(5*8) = 1 - 1/40 = 39/40. Which matches.For n=5:S₅ = 1 - 1/(f₆f₇) = 1 - 1/(8*13) = 1 - 1/104 = 103/104. Which matches.Okay, so it seems that Sₙ = 1 - 1/(fₙ₊₁fₙ₊₂).If this is true, then for n=98, S₉₈ = 1 - 1/(f₉₉f₁₀₀). Since f₉₉ and f₁₀₀ are positive integers greater than 1, 1/(f₉₉f₁₀₀) is positive, so S₉₈ = 1 - something positive < 1.Therefore, S₉₈ < 1, which is what we needed to prove.But wait, how did I come up with Sₙ = 1 - 1/(fₙ₊₁fₙ₊₂)? I just noticed the pattern from small n, but I need to prove it in general.Let me try to prove by induction that Sₙ = 1 - 1/(fₙ₊₁fₙ₊₂).Base case: n=1.S₁ = 1/(1*2) = 1/2.1 - 1/(f₂f₃) = 1 - 1/(1*2) = 1 - 1/2 = 1/2. So, base case holds.Inductive step: Assume that for some k ≥ 1, Sₖ = 1 - 1/(fₖ₊₁fₖ₊₂).We need to show that Sₖ₊₁ = 1 - 1/(fₖ₊₂fₖ₊₃).Sₖ₊₁ = Sₖ + 1/(fₖ₊₁fₖ₊₃).By the inductive hypothesis:Sₖ₊₁ = [1 - 1/(fₖ₊₁fₖ₊₂)] + 1/(fₖ₊₁fₖ₊₃).Let me combine these terms:= 1 - 1/(fₖ₊₁fₖ₊₂) + 1/(fₖ₊₁fₖ₊₃).Factor out 1/fₖ₊₁:= 1 + (1/fₖ₊₁)[ -1/fₖ₊₂ + 1/fₖ₊₃ ].Now, let's compute the expression in the brackets:-1/fₖ₊₂ + 1/fₖ₊₃ = ( -fₖ₊₃ + fₖ₊₂ ) / (fₖ₊₂fₖ₊₃).But fₖ₊₃ = fₖ₊₂ + fₖ₊₁, so:= ( - (fₖ₊₂ + fₖ₊₁) + fₖ₊₂ ) / (fₖ₊₂fₖ₊₃)= ( -fₖ₊₂ - fₖ₊₁ + fₖ₊₂ ) / (fₖ₊₂fₖ₊₃)= ( -fₖ₊₁ ) / (fₖ₊₂fₖ₊₃).So, plugging back into Sₖ₊₁:= 1 + (1/fₖ₊₁)( -fₖ₊₁ / (fₖ₊₂fₖ₊₃) )= 1 - 1/(fₖ₊₂fₖ₊₃).Which is exactly what we wanted to show. Therefore, by induction, Sₙ = 1 - 1/(fₙ₊₁fₙ₊₂) for all n ≥ 1.Thus, for n=98:S₉₈ = 1 - 1/(f₉₉f₁₀₀).Since f₉₉ and f₁₀₀ are positive integers greater than 1, 1/(f₉₉f₁₀₀) is positive, so S₉₈ = 1 - something positive < 1.Therefore, the sum is less than 1.I think that wraps it up. I used induction to generalize the sum and showed that it's always less than 1 because we're subtracting a positive term from 1.

💡Okay, so I have this problem about splitting piles of stones. Let me try to understand it step by step. We start with a pile of n stones, where n is at least 2. On the first step, we divide this pile approximately in half. If n is even, it's split into two equal piles of n/2 each. If n is odd, it's split into two piles of (n-1)/2 and (n+1)/2. Then, on each subsequent step, every pile that has at least two stones is split in the same way. The process continues until every pile has exactly one stone. We need to prove that the number of steps m required to reach this state satisfies the inequality m - 1 < log₂(n) ≤ m.Alright, so first, let me try to get an intuition about what's happening here. Each time we split a pile, we're roughly halving it. So, if we start with n stones, after one step, we have two piles, each about n/2. Then, each of those piles is split again, and so on. This seems similar to a binary tree where each node splits into two children, and the depth of the tree would correspond to the number of steps m.In a binary tree, the depth is related to the logarithm base 2 of the number of leaves. Here, the number of leaves would be the number of piles, which is n since each pile has one stone. So, the depth of the tree, which is m, should be roughly log₂(n). But the problem states that m - 1 < log₂(n) ≤ m. That suggests that m is the ceiling of log₂(n). So, m is the smallest integer greater than or equal to log₂(n).Let me test this with some small examples to see if it holds.Take n = 2. Log₂(2) = 1. So, m should be 1. Indeed, we split the pile into two piles of 1 each in one step. That works.Take n = 3. Log₂(3) ≈ 1.58496. So, m should be 2. Let's see: first step, split into 1 and 2. Then, in the second step, split the pile of 2 into two piles of 1 each. So, total steps are 2. That matches.n = 4. Log₂(4) = 2. So, m should be 2. Split into two piles of 2 each. Then, each of those splits into two piles of 1. So, two steps. Correct.n = 5. Log₂(5) ≈ 2.32193. So, m should be 3. Let's see: first step, split into 2 and 3. Then, split the 2 into 1 and 1, and split the 3 into 1 and 2. Now, we have three piles: 1, 1, and 2. Then, split the 2 into 1 and 1. So, total steps: 3. Correct.n = 6. Log₂(6) ≈ 2.58496. So, m should be 3. Let's see: first step, split into 3 and 3. Then, each 3 splits into 1 and 2. So, now we have four piles: 1, 2, 1, 2. Then, each 2 splits into 1 and 1. So, total steps: 3. Correct.n = 7. Log₂(7) ≈ 2.80735. So, m should be 3. Let's see: first step, split into 3 and 4. Then, split the 3 into 1 and 2, and the 4 into 2 and 2. Now, we have piles: 1, 2, 2, 2. Then, split each 2 into 1 and 1. So, total steps: 3. Correct.n = 8. Log₂(8) = 3. So, m should be 3. Split into 4 and 4. Then, each 4 splits into 2 and 2. Then, each 2 splits into 1 and 1. So, three steps. Correct.Okay, so from these examples, it seems that m is indeed the ceiling of log₂(n). So, m - 1 < log₂(n) ≤ m. That makes sense because log₂(n) is not necessarily an integer, and m has to be an integer number of steps.Now, how can we prove this in general? Maybe by induction.Let's try induction on n.Base case: n = 2. As we saw, m = 1. Log₂(2) = 1. So, 1 - 1 < 1 ≤ 1. Wait, that would be 0 < 1 ≤ 1, which is true. So, base case holds.Inductive step: Assume that for all k with 2 ≤ k < n, the number of steps m satisfies m - 1 < log₂(k) ≤ m.Now, consider a pile of n stones. If n is even, we split it into two piles of n/2. If n is odd, we split it into (n-1)/2 and (n+1)/2. In either case, the two resulting piles are as equal as possible.Now, the number of steps m(n) required to split n stones would be 1 plus the maximum number of steps required for each of the two resulting piles. Because we have to process both piles, and the total number of steps is determined by the pile that takes longer to split.So, m(n) = 1 + max(m(a), m(b)), where a and b are the sizes of the two resulting piles after the first split.If n is even, a = b = n/2, so m(n) = 1 + m(n/2).If n is odd, a = (n-1)/2 and b = (n+1)/2. So, m(n) = 1 + max(m((n-1)/2), m((n+1)/2)).But since (n+1)/2 is larger than (n-1)/2, m(n) = 1 + m((n+1)/2).Wait, but actually, both a and b are less than n, so by induction, we can apply the inductive hypothesis to them.But maybe it's better to think in terms of the binary tree depth. Each split corresponds to a level in the tree, and the number of steps is the depth of the tree.Alternatively, perhaps we can relate this to the binary representation of n.Wait, but maybe another approach is to consider that each step roughly halves the size of the piles, so the number of steps needed is proportional to log₂(n).But to formalize this, let's consider that after each step, the size of the largest pile is at most half of the original size (rounded up or down). So, the number of steps needed to reduce a pile of size n to piles of size 1 is the smallest m such that n ≤ 2^m.Wait, that's similar to the definition of ceiling of log₂(n). Because 2^m is the smallest power of 2 greater than or equal to n. So, m = ceiling(log₂(n)).Which would mean that m - 1 < log₂(n) ≤ m.Yes, that seems to fit.So, perhaps we can argue that the number of steps m is the smallest integer such that 2^m ≥ n. Therefore, m = ceiling(log₂(n)), which implies m - 1 < log₂(n) ≤ m.But let's make this more precise.Suppose that after m steps, all piles have size 1. Then, in the first step, we split n into two piles, each of size at most ceiling(n/2). Then, in the second step, each of those piles splits into two piles of size at most ceiling(n/4), and so on. After m steps, each pile has size at most ceiling(n/(2^m)).But we need ceiling(n/(2^m)) = 1. So, n/(2^m) ≤ 1, which implies 2^m ≥ n. Therefore, m ≥ log₂(n). Since m must be an integer, m is the smallest integer greater than or equal to log₂(n), which is ceiling(log₂(n)).But wait, in our earlier examples, for n=3, log₂(3)≈1.58496, ceiling is 2, which matches m=2.Similarly, for n=5, log₂(5)≈2.32193, ceiling is 3, which matches m=3.So, this seems to hold.But we need to be careful because the splitting isn't exactly halving each time, especially when n is odd. For example, when n is odd, we split into (n-1)/2 and (n+1)/2, which are not equal. So, the larger pile is (n+1)/2, which is still less than n/2 + 1.Wait, actually, for n odd, (n+1)/2 = (n-1)/2 + 1, so it's just one more than the other pile. So, the maximum pile size after splitting is at most ceiling(n/2).Therefore, regardless of whether n is even or odd, after each split, the maximum pile size is at most ceiling(n/2).So, the number of steps m needed to reduce n to 1 is the smallest m such that ceiling(n/2^m) = 1.Which is equivalent to n/2^m ≤ 1, so 2^m ≥ n, hence m ≥ log₂(n). Since m must be an integer, m = ceiling(log₂(n)).Therefore, m - 1 < log₂(n) ≤ m.Wait, let me check this with n=7.log₂(7)≈2.80735, so ceiling is 3. Indeed, m=3.Similarly, n=8, log₂(8)=3, ceiling is 3, m=3.n=9, log₂(9)≈3.16993, ceiling is 4. Let's see: first split into 4 and 5. Then, split 4 into 2 and 2, and 5 into 2 and 3. Now, we have piles: 2,2,2,3. Then, split each 2 into 1 and 1, and split 3 into 1 and 2. Now, we have: 1,1,1,1,1,2. Then, split the 2 into 1 and 1. So, total steps: 4. Correct.So, this seems consistent.Therefore, the number of steps m is the ceiling of log₂(n), which satisfies m - 1 < log₂(n) ≤ m.Hence, the proof is complete.

💡First, I need to determine the number of years between 2010 and 2013, inclusive. This gives me a total of 4 years.Next, I must identify which of these years are leap years. A leap year occurs every 4 years, and 2012 falls within this range, making it a leap year.A common year has 365 days, while a leap year has 366 days. Therefore, for the 3 common years (2010, 2011, and 2013), there are 3 × 365 = 1,095 days. Adding the 366 days from the leap year 2012, the total number of days is 1,095 + 366 = 1,461 days.

💡Okay, so I have this math problem here that I need to solve. It's a fraction where the numerator is the sum of seven cubes, and the denominator is the sum of four products. Let me write it down to get a clearer picture:[frac{65533^3 + 65534^3 + 65535^3 + 65536^3 + 65537^3 + 65538^3 + 65539^3}{32765 cdot 32766 + 32767 cdot 32768 + 32768 cdot 32769 + 32770 cdot 32771}]Hmm, that's a bit intimidating with all those large numbers. But maybe there's a pattern or a way to simplify this. Let me start by looking at the numerator and the denominator separately.**Numerator Analysis:**The numerator is the sum of cubes from 65533 to 65539. That's seven consecutive numbers. I remember that sometimes when you have sums of cubes, especially symmetrically around a central number, there might be a way to simplify the expression.Let me note that 65536 is a familiar number—it's 2^16, which is 65536. So, 65533 is 65536 minus 3, 65534 is 65536 minus 2, and so on, up to 65539, which is 65536 plus 3. So, if I let N = 65536, then the numerator becomes:[(N - 3)^3 + (N - 2)^3 + (N - 1)^3 + N^3 + (N + 1)^3 + (N + 2)^3 + (N + 3)^3]That looks symmetric! Maybe I can pair the terms to simplify. Let's pair (N - 3)^3 with (N + 3)^3, (N - 2)^3 with (N + 2)^3, and (N - 1)^3 with (N + 1)^3, leaving N^3 in the middle.I recall that the sum of cubes can be expressed using the formula:[a^3 + b^3 = (a + b)(a^2 - ab + b^2)]But since these pairs are symmetric around N, maybe there's a simpler way. Let me compute each pair:1. ((N - 3)^3 + (N + 3)^3)2. ((N - 2)^3 + (N + 2)^3)3. ((N - 1)^3 + (N + 1)^3)Let's compute the first pair:[(N - 3)^3 + (N + 3)^3 = [N^3 - 9N^2 + 27N - 27] + [N^3 + 9N^2 + 27N + 27] = 2N^3 + 54N]Wait, that simplifies nicely! The N^2 terms cancel out, and the constants also cancel. So, it's 2N^3 + 54N.Similarly, let's compute the second pair:[(N - 2)^3 + (N + 2)^3 = [N^3 - 6N^2 + 12N - 8] + [N^3 + 6N^2 + 12N + 8] = 2N^3 + 24N]Again, the N^2 and constant terms cancel out, leaving 2N^3 + 24N.Now, the third pair:[(N - 1)^3 + (N + 1)^3 = [N^3 - 3N^2 + 3N - 1] + [N^3 + 3N^2 + 3N + 1] = 2N^3 + 6N]Same pattern here: 2N^3 + 6N.So, adding all these together along with the middle term N^3:[(2N^3 + 54N) + (2N^3 + 24N) + (2N^3 + 6N) + N^3 = 7N^3 + 84N]Okay, so the numerator simplifies to 7N^3 + 84N, where N = 65536.**Denominator Analysis:**Now, the denominator is the sum of four products:[32765 cdot 32766 + 32767 cdot 32768 + 32768 cdot 32769 + 32770 cdot 32771]Looking at these numbers, I notice that 32768 is half of 65536, which is N. So, let me set M = 32768. Then:- 32765 = M - 3- 32766 = M - 2- 32767 = M - 1- 32768 = M- 32769 = M + 1- 32770 = M + 2- 32771 = M + 3So, the denominator can be rewritten as:[(M - 3)(M - 2) + (M - 1)M + M(M + 1) + (M + 2)(M + 3)]Let me compute each term:1. ((M - 3)(M - 2) = M^2 - 5M + 6)2. ((M - 1)M = M^2 - M)3. (M(M + 1) = M^2 + M)4. ((M + 2)(M + 3) = M^2 + 5M + 6)Now, let's add all these together:[(M^2 - 5M + 6) + (M^2 - M) + (M^2 + M) + (M^2 + 5M + 6)]Combine like terms:- M^2 terms: 4M^2- M terms: (-5M - M + M + 5M) = 0M- Constants: 6 + 0 + 0 + 6 = 12So, the denominator simplifies to 4M^2 + 12.Since M = 32768, and N = 65536 = 2M, we can express the denominator in terms of N:[4M^2 + 12 = 4left(frac{N}{2}right)^2 + 12 = frac{N^2}{4} + 12]**Putting It All Together:**Now, the original expression becomes:[frac{7N^3 + 84N}{frac{N^2}{4} + 12}]Let me factor out 7N from the numerator:[frac{7N(N^2 + 12)}{frac{N^2}{4} + 12}]Notice that the denominator is (frac{N^2}{4} + 12), which can be written as (frac{N^2 + 48}{4}). So, the expression becomes:[frac{7N(N^2 + 12)}{frac{N^2 + 48}{4}} = 7N cdot frac{N^2 + 12}{frac{N^2 + 48}{4}} = 7N cdot frac{4(N^2 + 12)}{N^2 + 48}]Simplify the fraction:[frac{4(N^2 + 12)}{N^2 + 48} = frac{4N^2 + 48}{N^2 + 48} = 4 cdot frac{N^2 + 12}{N^2 + 48}]Wait, that doesn't seem right. Let me check my steps again.Actually, when I factor out 7N, it's:[frac{7N(N^2 + 12)}{frac{N^2}{4} + 12} = 7N cdot frac{N^2 + 12}{frac{N^2 + 48}{4}} = 7N cdot frac{4(N^2 + 12)}{N^2 + 48}]So, it's 7N multiplied by 4(N^2 + 12)/(N^2 + 48). Let me compute this:[7N cdot frac{4(N^2 + 12)}{N^2 + 48} = 28N cdot frac{N^2 + 12}{N^2 + 48}]Hmm, this seems complicated. Maybe I made a mistake earlier. Let me go back.Wait, when I simplified the numerator, I had 7N^3 + 84N, which is 7N(N^2 + 12). The denominator was 4M^2 + 12, which is 4*(N/2)^2 + 12 = N^2/4 + 12.So, the expression is:[frac{7N(N^2 + 12)}{N^2/4 + 12}]Let me write the denominator as (N^2 + 48)/4:[frac{7N(N^2 + 12)}{(N^2 + 48)/4} = 7N cdot frac{4(N^2 + 12)}{N^2 + 48}]So, it's 28N*(N^2 + 12)/(N^2 + 48). Hmm, this still doesn't seem to simplify easily. Maybe I need to factor N^2 + 12 and N^2 + 48.But N^2 + 12 and N^2 + 48 don't have common factors. Wait, but N is 65536, which is a specific number. Maybe plugging in N = 65536 will help.Let me compute N^2:N = 65536, so N^2 = (65536)^2. That's a huge number, but maybe I don't need to compute it directly.Wait, let's see:The expression is 28N*(N^2 + 12)/(N^2 + 48). Let me factor out N^2 from numerator and denominator:[28N cdot frac{N^2(1 + 12/N^2)}{N^2(1 + 48/N^2)} = 28N cdot frac{1 + 12/N^2}{1 + 48/N^2}]Since N is very large (65536), 12/N^2 and 48/N^2 are very small, almost negligible. So, approximately, this is 28N*(1/1) = 28N. But that's an approximation. However, maybe the exact value is 28N because the terms cancel out.Wait, let's test with smaller numbers to see if this pattern holds.Let me take N = 4, so M = 2.Numerator: (1)^3 + 2^3 + 3^3 + 4^3 + 5^3 + 6^3 + 7^3Wait, no, actually, when N=4, the numbers would be N-3=1, N-2=2, N-1=3, N=4, N+1=5, N+2=6, N+3=7.Sum of cubes: 1 + 8 + 27 + 64 + 125 + 216 + 343 = 784.Denominator: (M-3)(M-2) + (M-1)M + M(M+1) + (M+2)(M+3)M=2, so:(-1)(0) + (1)(2) + (2)(3) + (4)(5) = 0 + 2 + 6 + 20 = 28.So, the expression is 784 / 28 = 28.Which is 7*M, since M=2, 7*2=14. Wait, but 784/28=28, which is 7*4=28, where N=4.Wait, so in this case, the result is 7*N, since N=4, 7*4=28.So, perhaps in the general case, the expression simplifies to 7*N.Let me check with another small N.Let N=2, M=1.Numerator: (-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3Wait, but N=2, so N-3=-1, which is negative. Cubes of negative numbers are negative, so:(-1)^3 + 0^3 + 1^3 + 2^3 + 3^3 + 4^3 + 5^3 = -1 + 0 + 1 + 8 + 27 + 64 + 125 = 224.Denominator: (M-3)(M-2) + (M-1)M + M(M+1) + (M+2)(M+3)M=1, so:(-2)(-1) + (0)(1) + (1)(2) + (3)(4) = 2 + 0 + 2 + 12 = 16.So, 224 / 16 = 14, which is 7*M, since M=1, 7*1=7, but 224/16=14=7*2=7*N, since N=2.Wait, so in both cases, the result is 7*N.So, perhaps the general formula is 7*N.Therefore, in our original problem, N=65536, so the result is 7*65536.Let me compute that:7 * 65536 = 458752.Wait, but earlier when I tried to simplify, I got 28N*(N^2 + 12)/(N^2 + 48). If N is 65536, then N^2 is 4294967296.So, N^2 + 12 = 4294967296 + 12 = 4294967308N^2 + 48 = 4294967296 + 48 = 4294967344So, the fraction is 4294967308 / 4294967344 ≈ 0.999999986So, 28N * 0.999999986 ≈ 28N - 28N*(1 - 0.999999986) = 28N - 28N*0.000000014 ≈ 28N - negligible.But since 28N is 7*4N, and N=65536, 4N=262144, so 7*262144=1835008.Wait, that doesn't match the earlier result of 7*N=458752.Hmm, I must have made a mistake in my earlier steps.Wait, let's go back.When I simplified the numerator, I had 7N^3 + 84N.Denominator was 4M^2 + 12, which is 4*(N/2)^2 + 12 = N^2/4 + 12.So, the expression is:(7N^3 + 84N) / (N^2/4 + 12)Let me factor numerator and denominator:Numerator: 7N(N^2 + 12)Denominator: (N^2 + 48)/4So, the expression becomes:7N(N^2 + 12) / [(N^2 + 48)/4] = 7N * 4(N^2 + 12)/(N^2 + 48) = 28N*(N^2 + 12)/(N^2 + 48)Now, let's see if N^2 + 12 and N^2 + 48 have a common factor. They don't, but maybe we can write it as:28N * [ (N^2 + 48) - 36 ] / (N^2 + 48) = 28N * [1 - 36/(N^2 + 48)]But that might not help. Alternatively, let's perform polynomial division.Divide N^2 + 12 by N^2 + 48:(N^2 + 12)/(N^2 + 48) = 1 - 36/(N^2 + 48)So, the expression becomes:28N * [1 - 36/(N^2 + 48)] = 28N - 28N*36/(N^2 + 48)But since N is very large, 36/(N^2 + 48) is negligible, so approximately 28N.But earlier, when I tested with N=4, the result was 28, which is 7*N=28. Similarly, with N=2, the result was 14=7*N.So, perhaps the exact value is 7*N, and the extra terms cancel out.Wait, let me compute 28N*(N^2 + 12)/(N^2 + 48) with N=4:28*4*(16 + 12)/(16 + 48) = 112*(28)/(64) = 112*(7/16) = 112*(0.4375) = 49.But earlier, when N=4, the result was 28. So, that contradicts. Therefore, my assumption that it's 7*N is incorrect.Wait, but when I computed with N=4, the numerator was 784 and denominator was 28, giving 28, which is 7*N=28.But according to the formula 28N*(N^2 + 12)/(N^2 + 48), with N=4, it's 28*4*(16+12)/(16+48)=112*28/64=112*(7/16)=49, which is not 28.So, there must be a mistake in my earlier steps. Let me re-examine.Wait, when I simplified the numerator, I had 7N^3 + 84N.Denominator was 4M^2 + 12, which is 4*(N/2)^2 + 12 = N^2/4 + 12.So, the expression is:(7N^3 + 84N)/(N^2/4 + 12)Let me factor N from numerator and denominator:Numerator: N(7N^2 + 84)Denominator: (N^2 + 48)/4So, the expression becomes:N(7N^2 + 84) / [(N^2 + 48)/4] = 4N(7N^2 + 84)/(N^2 + 48)Factor 7 from numerator:4N*7(N^2 + 12)/(N^2 + 48) = 28N(N^2 + 12)/(N^2 + 48)Now, let's see if (N^2 + 12)/(N^2 + 48) simplifies. It doesn't, but maybe we can write it as:(N^2 + 12)/(N^2 + 48) = 1 - 36/(N^2 + 48)So, the expression becomes:28N[1 - 36/(N^2 + 48)] = 28N - 28N*36/(N^2 + 48)But with N=65536, 36/(N^2 + 48) is extremely small, so the second term is negligible, making the expression approximately 28N.But earlier, when N=4, the exact result was 28, which is 7*N=28, not 28N=112.This inconsistency suggests that my earlier approach might be flawed. Let me try a different method.**Alternative Approach:**Let me consider that both numerator and denominator might be related to N and M in a way that the ratio simplifies to 7*N.Given that in the small cases, the result was 7*N, perhaps it's a general formula.Given N=65536, the result would be 7*65536=458752.But let me verify this with the exact computation.Compute numerator: 7N^3 + 84NN=65536, so N^3=65536^3=2814749767106567N^3=7*281474976710656=197032483697459284N=84*65536=5505024So, numerator=1970324836974592 + 5505024=1970324842479616Denominator: N^2/4 + 12= (65536^2)/4 + 12= (4294967296)/4 +12=1073741824 +12=1073741836So, the expression is 1970324842479616 / 1073741836Let me compute this division:1970324842479616 ÷ 1073741836First, note that 1073741836 * 1835008 = ?Wait, 1073741836 * 1835008 = ?But 1073741836 * 1835008 is a huge number, but let's see:1073741836 * 1835008 = 1073741836 * (1800000 + 35008) = 1073741836*1800000 + 1073741836*35008But this is getting too complicated. Alternatively, let's see if 1073741836 * 1835008 equals the numerator.Wait, 1073741836 * 1835008 = ?But 1073741836 * 1835008 = 1073741836 * (1835000 + 8) = 1073741836*1835000 + 1073741836*8This is still too cumbersome. Maybe I can use the fact that 1073741836 = 4*268435459Wait, 1073741836 ÷ 4 = 268435459So, 1073741836 = 4*268435459Similarly, the numerator is 1970324842479616Let me divide numerator by denominator:1970324842479616 ÷ 1073741836Let me compute how many times 1073741836 fits into 1970324842479616.Compute 1970324842479616 ÷ 1073741836 ≈ ?Let me approximate:1073741836 * 1835008 = ?But 1073741836 * 1835008 = 1073741836 * 1835008Wait, 1073741836 * 1835008 = 1073741836 * (1835000 + 8) = 1073741836*1835000 + 1073741836*8But this is not helpful. Alternatively, let's see:1073741836 * 1835008 = ?Wait, 1073741836 * 1835008 = 1073741836 * (1835000 + 8) = 1073741836*1835000 + 1073741836*8But this is still too big. Maybe I can use the fact that 1073741836 * 1835008 = 1073741836 * (1835000 + 8) = 1073741836*1835000 + 1073741836*8But I'm stuck here. Maybe I should accept that the result is 7*N=458752, as suggested by the small N cases.Therefore, the final answer is 7*65536=458752.