💡Okay, so I have this problem where I need to figure out how many three-digit numbers can be formed using the digits 1 through 9, such that the sum of the three digits is 17. Hmm, let me think about how to approach this.First, I know that a three-digit number has a hundreds place, a tens place, and a ones place. Each of these places can be filled with any digit from 1 to 9, right? So, I need to find all possible combinations of three digits (each from 1 to 9) that add up to 17.This sounds like a problem where I can use combinatorics, specifically something called "stars and bars." I remember that stars and bars is a method used to find the number of ways to distribute indistinguishable objects into distinguishable bins. In this case, the "objects" are the total sum (17), and the "bins" are the three digits.But wait, there's a catch. Each digit has to be at least 1 and at most 9. So, I can't just use the standard stars and bars formula because it allows for zeros and doesn't have an upper limit. I need to adjust for these constraints.Let me recall the formula for the number of solutions to the equation (x_1 + x_2 + x_3 = n) where each (x_i geq 1) and (x_i leq 9). The standard stars and bars formula without restrictions is (binom{n-1}{k-1}), where (k) is the number of variables. But since each (x_i) has a maximum value of 9, I need to subtract the cases where any (x_i) exceeds 9.This is similar to the inclusion-exclusion principle. So, the formula becomes:[text{Number of solutions} = binom{n-1}{k-1} - binom{k}{1}binom{n - m - 1}{k - 1} + binom{k}{2}binom{n - 2m - 1}{k - 1} - dots]Where (m) is the maximum value each variable can take, which is 9 in this case. But this might get complicated because we have three variables, and the number of terms can increase.Alternatively, maybe I can adjust the variables to account for the minimum value. If each digit is at least 1, let me define new variables (y_i = x_i - 1). Then each (y_i) is at least 0, and the equation becomes:[(y_1 + 1) + (y_2 + 1) + (y_3 + 1) = 17][y_1 + y_2 + y_3 = 14]Now, I need to find the number of non-negative integer solutions to this equation where each (y_i leq 8) (since (x_i leq 9) implies (y_i leq 8)).Using the stars and bars formula without restrictions, the number of solutions is (binom{14 + 3 - 1}{3 - 1} = binom{16}{2} = 120).But now I need to subtract the cases where any (y_i > 8). Let's use inclusion-exclusion for this.First, calculate the number of solutions where (y_1 > 8). If (y_1 geq 9), let (y_1' = y_1 - 9), so the equation becomes:[y_1' + y_2 + y_3 = 14 - 9 = 5]The number of solutions is (binom{5 + 3 - 1}{3 - 1} = binom{7}{2} = 21).Similarly, the number of solutions where (y_2 > 8) is also 21, and the same for (y_3 > 8). So, for each variable exceeding 8, there are 21 solutions. Since there are 3 variables, we subtract (3 times 21 = 63).But wait, we might have subtracted too much if two variables exceed 8 simultaneously. Let's check if that's possible. If two variables, say (y_1) and (y_2), are both greater than 8, then (y_1 geq 9) and (y_2 geq 9). The equation becomes:[y_1' + y_2' + y_3 = 14 - 9 - 9 = -4]This is impossible because the sum can't be negative. So, there are no solutions where two variables exceed 8. Therefore, we don't need to add anything back in.So, the total number of valid solutions is (120 - 63 = 57).But wait, does this account for all the restrictions? Each (y_i) is at most 8, so each (x_i) is at most 9, which is exactly what we needed. So, yes, 57 should be the correct number of solutions.Therefore, the number of three-digit numbers that can be formed using digits 1 through 9 with the sum of digits equal to 17 is 57.

💡Alright, so I have this inequality to prove: ( a^3 + b^3 > a^2b + ab^2 ), where ( a ) and ( b ) are both positive numbers and ( a ) is not equal to ( b ). Hmm, okay. Let me think about how to approach this.First off, I remember that for inequalities involving polynomials, sometimes factoring can help simplify things. Maybe I can factor both sides or rearrange the terms to see if something cancels out or becomes more manageable.Let me write down the inequality again:( a^3 + b^3 > a^2b + ab^2 )Hmm, I notice that both sides have similar terms. Maybe I can subtract ( a^2b + ab^2 ) from both sides to get everything on one side:( a^3 + b^3 - a^2b - ab^2 > 0 )Now, let's see if I can factor this expression. I recall that ( a^3 + b^3 ) can be factored as ( (a + b)(a^2 - ab + b^2) ). Let me try that:( (a + b)(a^2 - ab + b^2) - a^2b - ab^2 > 0 )Hmm, not sure if that helps directly. Maybe I should factor the entire expression ( a^3 + b^3 - a^2b - ab^2 ). Let me group the terms:( (a^3 - a^2b) + (b^3 - ab^2) )Factor out ( a^2 ) from the first group and ( b^2 ) from the second group:( a^2(a - b) + b^2(b - a) )Wait, ( (b - a) ) is the same as ( -(a - b) ), so I can rewrite this as:( a^2(a - b) - b^2(a - b) )Now, factor out ( (a - b) ):( (a - b)(a^2 - b^2) )Oh, nice! Now, ( a^2 - b^2 ) can be factored further as ( (a - b)(a + b) ). So, substituting that in:( (a - b)(a - b)(a + b) )Which simplifies to:( (a - b)^2(a + b) )So, putting it all together, the original expression ( a^3 + b^3 - a^2b - ab^2 ) factors to ( (a - b)^2(a + b) ). Therefore, our inequality becomes:( (a - b)^2(a + b) > 0 )Now, let's analyze this expression. Since ( a ) and ( b ) are both positive, ( a + b ) is definitely positive. Also, ( (a - b)^2 ) is a square of a real number, which is always non-negative. But since ( a neq b ), ( (a - b)^2 ) is actually positive. So, we have the product of two positive numbers, which is positive. Therefore, ( (a - b)^2(a + b) > 0 ) is true.Thus, the original inequality ( a^3 + b^3 > a^2b + ab^2 ) holds true for all positive numbers ( a ) and ( b ) where ( a neq b ).Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from the beginning, I subtracted ( a^2b + ab^2 ) from both sides, which seems correct. Then, I factored the expression by grouping, which also seems valid. Factoring out ( (a - b) ) and then recognizing ( a^2 - b^2 ) as a difference of squares was a good move. The final expression ( (a - b)^2(a + b) ) is indeed positive because both ( (a - b)^2 ) and ( (a + b) ) are positive. So, I think my reasoning holds up.I guess another way to think about it is by using the AM-GM inequality or maybe calculus, but factoring seems straightforward here. I don't see any errors in my logic, so I feel confident about this proof.**Final Answer**The inequality is proved, so the final result is boxed{a^3 + b^3 > a^2b + ab^2}.**Final Answer**boxed{a^3 + b^3 > a^2b + ab^2}

💡Okay, so I need to find the molecular weight of a compound with the formula C4H8O2. The problem already gives me the molecular weights for each part: carbon is 48, hydrogen is 8, and oxygen is 32. Hmm, that seems straightforward, but I want to make sure I understand everything correctly.First, let me recall what molecular weight means. Molecular weight is the sum of the atomic weights of all the atoms in a molecule. So, for C4H8O2, I need to add up the weights of all the carbon, hydrogen, and oxygen atoms in the molecule.The formula C4H8O2 tells me there are 4 carbon atoms, 8 hydrogen atoms, and 2 oxygen atoms in one molecule of this compound. The problem already provides the total molecular weights for each element part: carbon part is 48, hydrogen part is 8, and oxygen part is 32. So, I think I just need to add these three numbers together to get the total molecular weight.Let me write that down:Total molecular weight = Carbon part + Hydrogen part + Oxygen partTotal molecular weight = 48 + 8 + 32Now, adding those numbers: 48 plus 8 is 56, and 56 plus 32 is 88. So, the total molecular weight should be 88.Wait, but just to double-check, maybe I should verify the individual atomic weights to ensure that the given molecular weights make sense. I know that the atomic weight of carbon is approximately 12, hydrogen is about 1, and oxygen is around 16.So, for 4 carbon atoms: 4 * 12 = 48. That matches the given carbon part.For 8 hydrogen atoms: 8 * 1 = 8. That also matches the given hydrogen part.For 2 oxygen atoms: 2 * 16 = 32. That matches the given oxygen part.Okay, so the individual calculations check out. Adding them together gives 48 + 8 + 32 = 88. That seems correct.I don't see any mistakes in my reasoning. The total molecular weight of C4H8O2 is indeed 88.

💡Okay, so I need to figure out how much was spent on beads for each bracelet. Let's see, I start by understanding the problem. I'm selling charm bracelets. For each bracelet, I spend 1 on the string and some amount on beads. I sell each bracelet for 6. If I sell 25 bracelets and make a profit of 50, how much did I spend on beads for each bracelet?First, I need to recall what profit means. Profit is the money I make after subtracting the costs from the revenue. So, profit equals revenue minus costs.Let me break it down step by step.1. **Calculate Total Revenue:** I sold 25 bracelets at 6 each. So, total revenue is 25 multiplied by 6. [ text{Total Revenue} = 25 times 6 = 150 text{ dollars} ]2. **Understand Total Costs:** For each bracelet, I spend 1 on the string and some amount on beads. Let's call the cost of beads for each bracelet ( B ) dollars. So, the total cost per bracelet is ( 1 + B ) dollars. Since I made 25 bracelets, the total cost is 25 multiplied by ( (1 + B) ). [ text{Total Cost} = 25 times (1 + B) ]3. **Set Up the Profit Equation:** Profit is total revenue minus total cost, and we know the profit is 50. [ 50 = 150 - 25 times (1 + B) ]4. **Solve for ( B ):** Let's simplify the equation step by step. First, expand the total cost: [ 50 = 150 - 25 - 25B ] Combine like terms: [ 50 = 125 - 25B ] Now, subtract 125 from both sides: [ 50 - 125 = -25B ] [ -75 = -25B ] Finally, divide both sides by -25 to solve for ( B ): [ B = frac{-75}{-25} = 3 ]So, the amount spent on beads for each bracelet is 3.Wait, let me double-check my calculations to make sure I didn't make any mistakes.- Total Revenue: 25 bracelets × 6 = 150. That seems right.- Total Cost: 25 bracelets × (1 + 3) = 25 × 4 = 100.- Profit: 150 - 100 = 50. Yes, that matches the given profit.Everything checks out. So, I'm confident that the amount spent on beads per bracelet is 3.

💡First, identify the digits in the number 3003, which are 3, 0, 0, and 3.Since we're forming a four-digit number, the thousands place cannot be 0. Therefore, it must be occupied by one of the two 3s.After placing a 3 in the thousands place, we're left with the digits 3, 0, and 0 to arrange in the remaining three positions.The number of unique permutations of these three digits can be calculated using the formula for permutations of a multiset:frac{3!}{1!2!} = frac{6}{2} = 3Thus, there are 3 different four-digit numbers that can be formed by arranging the digits in 3003.

💡Okay, so I have this problem where a line intersects a parabola at two points, A and B. The line is given by the equation y = kx - 2, and the parabola is y² = 8x. The x-coordinate of the midpoint of segment AB is 2, and I need to find the value of k. Hmm, let me try to figure this out step by step.First, I know that to find the points of intersection between the line and the parabola, I can substitute the expression for y from the line equation into the parabola equation. That should give me a quadratic equation in terms of x, which I can then solve to find the x-coordinates of points A and B.So, substituting y = kx - 2 into y² = 8x, I get:(kx - 2)² = 8xExpanding the left side, that becomes:k²x² - 4kx + 4 = 8xNow, I need to rearrange this equation to form a standard quadratic equation. Let me move all terms to one side:k²x² - 4kx + 4 - 8x = 0Combining like terms, the equation becomes:k²x² - (4k + 8)x + 4 = 0Okay, so this is a quadratic in x. Let's denote this as:ax² + bx + c = 0Where:a = k²b = -(4k + 8)c = 4Now, since the line intersects the parabola at two points, this quadratic equation should have two real roots. The x-coordinates of points A and B are these roots. Let's denote them as x₁ and x₂.From Vieta's formulas, I remember that the sum of the roots of a quadratic equation ax² + bx + c = 0 is given by -b/a. So, the sum of x₁ and x₂ is:x₁ + x₂ = -b/a = (4k + 8)/k²The problem states that the x-coordinate of the midpoint of AB is 2. The midpoint's x-coordinate is the average of x₁ and x₂, which is (x₁ + x₂)/2. So,(x₁ + x₂)/2 = 2Multiplying both sides by 2 gives:x₁ + x₂ = 4But from Vieta's formula, we have:x₁ + x₂ = (4k + 8)/k²So, setting these equal:(4k + 8)/k² = 4Now, let's solve for k. Multiply both sides by k² to eliminate the denominator:4k + 8 = 4k²Bring all terms to one side to form a quadratic equation:4k² - 4k - 8 = 0I can simplify this equation by dividing all terms by 4:k² - k - 2 = 0Now, let's factor this quadratic equation. Looking for two numbers that multiply to -2 and add to -1. Hmm, those numbers would be -2 and +1.So, factoring:(k - 2)(k + 1) = 0Setting each factor equal to zero gives the possible solutions:k - 2 = 0 => k = 2k + 1 = 0 => k = -1Now, I need to check if both solutions are valid. Remember, the quadratic equation in x must have two distinct real roots for the line to intersect the parabola at two points. Let's check the discriminant for each value of k.The discriminant D of a quadratic equation ax² + bx + c = 0 is given by D = b² - 4ac.First, for k = 2:a = (2)² = 4b = -(4*2 + 8) = -(8 + 8) = -16c = 4So, D = (-16)² - 4*4*4 = 256 - 64 = 192Since D > 0, there are two distinct real roots. So, k = 2 is valid.Now, for k = -1:a = (-1)² = 1b = -(4*(-1) + 8) = -(-4 + 8) = -4c = 4So, D = (-4)² - 4*1*4 = 16 - 16 = 0Since D = 0, there is exactly one real root, meaning the line is tangent to the parabola. But the problem states that the line intersects the parabola at two points, so k = -1 would result in only one point of contact, which doesn't satisfy the condition. Therefore, k = -1 is not a valid solution.Thus, the only valid solution is k = 2.I think that's it. I followed the steps: substituted the line into the parabola, formed a quadratic equation, used Vieta's formula to relate the sum of roots to the midpoint, solved for k, and checked the validity of each solution by ensuring the discriminant was positive. Everything seems to add up, so I'm confident that k = 2 is the correct answer.