💡Okay, so I have this problem here: there's a theater with 750 seats, and out of those, 218 are empty. I need to figure out how many people are actually watching the movie. Hmm, alright, let's break this down step by step.First, I know that the total number of seats in the theater is 750. That's the maximum number of people that could be there if every seat was occupied. But right now, not all seats are filled. Specifically, 218 seats are empty. So, to find out how many people are watching the movie, I need to subtract the number of empty seats from the total number of seats.Let me write that down to visualize it better:Total seats = 750Empty seats = 218People watching = Total seats - Empty seatsSo, plugging in the numbers:People watching = 750 - 218Now, I need to do this subtraction. Let's see, 750 minus 200 would be 550, and then minus 18 more would be 532. So, 750 - 218 equals 532.Wait a minute, let me double-check that to make sure I didn't make a mistake. If I add 218 to 532, does it give me back 750? Let's see:532 + 200 = 732732 + 18 = 750Yes, that adds up correctly. So, 532 people are watching the movie in the theater.I think that's it. It seems straightforward, but it's always good to verify my calculations to avoid any errors. Especially since sometimes when dealing with numbers, it's easy to misplace a digit or miscalculate. But in this case, everything checks out.Another way to think about it is to consider the theater's capacity and how many seats are being used. If there are 750 seats and 218 are empty, then the number of occupied seats is simply the difference between these two numbers. This makes sense because the total capacity minus the unused seats gives the number of people present.I could also visualize this as a simple equation:Total seats = Occupied seats + Empty seatsWe know the total and the empty seats, so solving for the occupied seats:Occupied seats = Total seats - Empty seatsOccupied seats = 750 - 218Occupied seats = 532This aligns with what I calculated earlier, so I'm confident that 532 is the correct number of people watching the movie.Just to explore it a bit more, if I were to represent this on a number line, starting at 750 and moving back 218 units, I would land at 532. This visual representation helps reinforce the idea that subtracting the empty seats from the total gives the number of people present.Additionally, if I were to use place value to break down the subtraction:750-218------?Starting from the rightmost digit:0 - 8: I can't subtract 8 from 0, so I need to borrow from the tens place. The 5 in the tens place becomes 4, and the 0 becomes 10. Now, 10 - 8 = 2.Next, the tens place: 4 - 1 = 3.Finally, the hundreds place: 7 - 2 = 5.Putting it all together, I get 532.This methodical approach ensures that each digit is handled correctly, preventing any mistakes that might come from rushing through the calculation.In summary, by subtracting the number of empty seats from the total number of seats, I accurately determine that there are 532 people watching the movie in the theater. It's a straightforward problem, but taking the time to verify and explore different methods helps solidify the solution.

💡To determine the three-digit integer that results in the greatest product when multiplied by a two-digit integer using the digits 1, 3, 4, 5, 7, and 8 exactly once, follow these steps:1. **Identify the Largest Digits for the Hundreds and Tens Places:** - The highest digits available are 8 and 7. Assign 8 to the hundreds place of the three-digit number and 7 to the tens place of the two-digit number to maximize the product.2. **Assign the Next Largest Digit to the Units Place of the Two-Digit Number:** - The next largest digit is 5. Assign 5 to the units place of the two-digit number.3. **Distribute the Remaining Digits to the Three-Digit Number:** - The remaining digits are 1, 3, and 4. To maximize the three-digit number, assign the largest remaining digit (4) to the tens place and the next largest (3) to the units place. This forms the three-digit number 843.4. **Form the Two-Digit Number:** - Using the assigned digits, the two-digit number becomes 75.5. **Calculate the Product:** - Multiply the three-digit number (843) by the two-digit number (75): [ 843 times 75 = 63225 ]6. **Verify with Alternative Assignments:** - Consider assigning 7 to the hundreds place and 8 to the tens place. This forms the three-digit number 743 and the two-digit number 85. Calculate the product: [ 743 times 85 = 63155 ] - Compare the two products: 63225 is greater than 63155.7. **Conclusion:** - The configuration with the three-digit number 843 and the two-digit number 75 yields the highest product.

💡Okay, so I have this problem: For which values of ( n ) can the weights of 1 gram, 2 grams, 3 grams, up to ( n ) grams be divided into three equal-mass piles? Hmm, interesting. I need to figure out for which ( n ) this is possible.First, let me understand the problem. I have weights starting from 1 gram and going up to ( n ) grams. I need to split these weights into three groups where each group has the same total mass. So, the sum of all the weights must be divisible by 3, right? Because if the total mass isn't divisible by 3, I can't split it into three equal parts.Let me write down the total mass. The sum of the first ( n ) natural numbers is given by the formula:[S = frac{n(n + 1)}{2}]So, for this sum to be divisible by 3, ( frac{n(n + 1)}{2} ) must be a multiple of 3. That means ( n(n + 1) ) must be divisible by 6 because 2 and 3 are coprime. Since ( n ) and ( n + 1 ) are consecutive integers, one of them is even, so ( n(n + 1) ) is always divisible by 2. Therefore, the key condition is that ( n(n + 1) ) must be divisible by 3.Now, for ( n(n + 1) ) to be divisible by 3, either ( n ) or ( n + 1 ) must be divisible by 3. Because 3 is a prime number, it can only divide one of the consecutive numbers. So, this gives us two cases:1. ( n equiv 0 mod 3 )2. ( n + 1 equiv 0 mod 3 ) which implies ( n equiv 2 mod 3 )So, ( n ) must be congruent to 0 or 2 modulo 3. That is, ( n ) can be written as ( 3k ) or ( 3k + 2 ) for some integer ( k ).But wait, is that all? Let me test this with some small values of ( n ) to see if it holds.Let's try ( n = 3 ):Total mass ( S = frac{3 times 4}{2} = 6 ). Divided by 3, each pile should be 2 grams. But the weights are 1, 2, 3. Can I split them into three piles each of 2 grams? No, because the smallest weight is 1 gram, and the next is 2 grams. If I try to make a pile of 2 grams, I can only use the 2-gram weight. Then I have 1 gram and 3 grams left, which can't be split into two more piles of 2 grams. So, ( n = 3 ) doesn't work.Hmm, so maybe ( n ) has to be greater than 3? Let's try ( n = 4 ):Total mass ( S = frac{4 times 5}{2} = 10 ). Divided by 3, each pile should be approximately 3.333 grams. But since we're dealing with whole grams, this isn't possible. So, ( n = 4 ) doesn't work either.Wait, according to my earlier condition, ( n = 4 ) is ( 1 mod 3 ), which doesn't satisfy ( n equiv 0 ) or ( 2 mod 3 ). So, that's consistent.Let's try ( n = 5 ):Total mass ( S = frac{5 times 6}{2} = 15 ). Divided by 3, each pile should be 5 grams. Let's see if we can split 1, 2, 3, 4, 5 into three piles each of 5 grams.One possible way:- Pile 1: 5 grams (just the 5-gram weight)- Pile 2: 4 + 1 = 5 grams- Pile 3: 3 + 2 = 5 gramsYes, that works! So, ( n = 5 ) works, which is ( 2 mod 3 ).Next, ( n = 6 ):Total mass ( S = frac{6 times 7}{2} = 21 ). Divided by 3, each pile should be 7 grams.Let's try to split 1, 2, 3, 4, 5, 6 into three piles of 7 grams each.One possible way:- Pile 1: 6 + 1 = 7 grams- Pile 2: 5 + 2 = 7 grams- Pile 3: 4 + 3 = 7 gramsYes, that works too! So, ( n = 6 ) works, which is ( 0 mod 3 ).Let's try ( n = 7 ):Total mass ( S = frac{7 times 8}{2} = 28 ). Divided by 3, each pile should be approximately 9.333 grams. Not possible with whole grams, so ( n = 7 ) doesn't work. Which is ( 1 mod 3 ), consistent with our condition.( n = 8 ):Total mass ( S = frac{8 times 9}{2} = 36 ). Divided by 3, each pile should be 12 grams.Let's see if we can split 1 through 8 into three piles of 12 grams each.One possible way:- Pile 1: 8 + 4 = 12 grams- Pile 2: 7 + 5 = 12 grams- Pile 3: 6 + 3 + 2 + 1 = 12 gramsYes, that works! So, ( n = 8 ) works, which is ( 2 mod 3 ).( n = 9 ):Total mass ( S = frac{9 times 10}{2} = 45 ). Divided by 3, each pile should be 15 grams.Let's try to split 1 through 9 into three piles of 15 grams each.One possible way:- Pile 1: 9 + 6 = 15 grams- Pile 2: 8 + 7 = 15 grams- Pile 3: 5 + 4 + 3 + 2 + 1 = 15 gramsYes, that works! So, ( n = 9 ) works, which is ( 0 mod 3 ).So, from these examples, it seems that when ( n ) is congruent to 0 or 2 modulo 3, and ( n > 3 ), it's possible to split the weights into three equal-mass piles.But wait, what about ( n = 2 )?Total mass ( S = frac{2 times 3}{2} = 3 ). Divided by 3, each pile should be 1 gram. But we have weights 1 and 2. We can't split them into three piles each of 1 gram because we only have two weights. So, ( n = 2 ) doesn't work, even though ( 2 equiv 2 mod 3 ). So, we need ( n > 3 ).Similarly, ( n = 1 ) is trivial but doesn't make sense for splitting into three piles.So, putting it all together, the condition is that ( n > 3 ) and ( n equiv 0 ) or ( 2 mod 3 ).I think that's the answer. To summarize, for ( n ) greater than 3, if ( n ) is a multiple of 3 or one less than a multiple of 3, then the weights can be divided into three equal-mass piles.

💡Okay, so I have this problem here: "sinalpha = frac{2}{3}" is a ( ) of "cos2alpha = frac{1}{9}". The options are A, B, C, D, which correspond to different types of conditions. I need to figure out whether "sinalpha = frac{2}{3}" is a necessary and sufficient condition, just sufficient, just necessary, or neither for "cos2alpha = frac{1}{9}".Alright, let's start by recalling some trigonometric identities. I remember that there's a double-angle formula for cosine, which is cos2alpha = 1 - 2sin^2alpha. Maybe that can help me relate sinalpha to cos2alpha.So, if I know that sinalpha = frac{2}{3}, I can plug that into the double-angle formula to find cos2alpha. Let me do that:cos2alpha = 1 - 2left(frac{2}{3}right)^2 = 1 - 2left(frac{4}{9}right) = 1 - frac{8}{9} = frac{1}{9}.Okay, so if sinalpha = frac{2}{3}, then cos2alpha does equal frac{1}{9}. That means that knowing sinalpha = frac{2}{3} is enough to guarantee that cos2alpha = frac{1}{9}. So, in this direction, it's a sufficient condition.But is it necessary? That is, does cos2alpha = frac{1}{9} necessarily mean that sinalpha = frac{2}{3}? Let me check.Starting from cos2alpha = frac{1}{9}, I can use the same double-angle formula to solve for sinalpha:cos2alpha = 1 - 2sin^2alphaSo,frac{1}{9} = 1 - 2sin^2alphaSubtracting 1 from both sides:frac{1}{9} - 1 = -2sin^2alphafrac{1 - 9}{9} = -2sin^2alphafrac{-8}{9} = -2sin^2alphaDivide both sides by -2:frac{4}{9} = sin^2alphaTaking the square root of both sides:sinalpha = pmfrac{2}{3}Hmm, so sinalpha could be either frac{2}{3} or -frac{2}{3}. That means that cos2alpha = frac{1}{9} doesn't uniquely determine sinalpha to be frac{2}{3}; it could also be negative. Therefore, "sinalpha = frac{2}{3}" isn't a necessary condition because there's another possibility for sinalpha that still satisfies cos2alpha = frac{1}{9}.So, putting it all together: "sinalpha = frac{2}{3}" is sufficient to ensure that "cos2alpha = frac{1}{9}", but it's not necessary because there's another value of sinalpha that also satisfies the equation for cos2alpha.Therefore, the correct answer should be that "sinalpha = frac{2}{3}" is a sufficient but not necessary condition for "cos2alpha = frac{1}{9}".**Final Answer**boxed{B}

💡Okay, so I have this problem about a geometric sequence. Let me try to understand it step by step. The first term is 1, and the number of terms is even. The sum of the odd terms is 85, and the sum of the even terms is 170. I need to find the common ratio and the number of terms. The options are given, so maybe I can check them, but I think I should try to solve it myself first.Alright, let's recall what a geometric sequence is. It's a sequence where each term after the first is found by multiplying the previous term by a constant called the common ratio, denoted as r. So, the sequence would look like: 1, r, r², r³, r⁴, ..., up to some number of terms.Since the number of terms is even, let's say there are 2n terms. That way, n is an integer, and the sequence has an even number of terms. So, the terms are: 1 (term 1), r (term 2), r² (term 3), r³ (term 4), ..., up to term 2n.Now, the problem mentions the sum of the odd terms and the sum of the even terms. Let's clarify what that means. The odd terms are the 1st, 3rd, 5th, etc., terms, and the even terms are the 2nd, 4th, 6th, etc., terms. So, in this case, since there are 2n terms, there will be n odd terms and n even terms.Let me write down the sum of the odd terms. The odd terms are: 1, r², r⁴, ..., up to r^{2(n-1)}. So, the sum of the odd terms, let's call it S_odd, is:S_odd = 1 + r² + r⁴ + ... + r^{2(n-1)}Similarly, the even terms are: r, r³, r⁵, ..., up to r^{2n-1}. So, the sum of the even terms, S_even, is:S_even = r + r³ + r⁵ + ... + r^{2n-1}Given that S_odd = 85 and S_even = 170.Hmm, I notice that S_even is exactly r times S_odd. Let me check that:S_even = r + r³ + r⁵ + ... + r^{2n-1} = r(1 + r² + r⁴ + ... + r^{2(n-1)}) = r * S_oddSo, S_even = r * S_odd. Given that S_odd = 85 and S_even = 170, we can write:170 = r * 85So, solving for r:r = 170 / 85 = 2Alright, so the common ratio r is 2. That's straightforward.Now, we need to find the number of terms, which is 2n. We already know r = 2, and S_odd = 85. Let's use the formula for the sum of a geometric series to find n.The sum of the odd terms is a geometric series with first term 1, common ratio r², and number of terms n. The formula for the sum of a geometric series is:S = a * (1 - r^k) / (1 - r)Where a is the first term, r is the common ratio, and k is the number of terms.In this case, for S_odd:S_odd = 1 * (1 - (r²)^n) / (1 - r²) = (1 - r^{2n}) / (1 - r²)We know S_odd = 85 and r = 2, so plugging in:85 = (1 - 2^{2n}) / (1 - 4) = (1 - 2^{2n}) / (-3)Multiply both sides by -3:85 * (-3) = 1 - 2^{2n}-255 = 1 - 2^{2n}Subtract 1 from both sides:-256 = -2^{2n}Multiply both sides by -1:256 = 2^{2n}Now, 256 is 2^8, so:2^8 = 2^{2n}Therefore, 8 = 2nSo, n = 4But wait, the number of terms is 2n, so 2n = 8.So, the number of terms is 8, and the common ratio is 2.Let me double-check to make sure I didn't make any mistakes.First, with r = 2 and n = 4, the number of terms is 8.Sum of odd terms: 1 + 4 + 16 + 64 = 85. That's correct.Sum of even terms: 2 + 8 + 32 + 128 = 170. That's also correct.So, everything checks out.Looking back at the options:A: 8, 2 – That's ratio 8, terms 2. Doesn't match.B: 2, 4 – Ratio 2, terms 4. But we have 8 terms, so no.C: 4, 10 – Ratio 4, terms 10. Doesn't match.D: 2, 8 – Ratio 2, terms 8. That's exactly what we found.So, the correct answer is D.**Final Answer**boxed{D}

💡First, I need to determine how many people fall into each discount category and how many pay the full price.There are 48 people in total. The first 10 people receive a 40% discount, and the next 20 people receive a 15% discount. This leaves 48 - 10 - 20 = 18 people who pay the full price.Next, I'll calculate the revenue from each group:1. **First 10 people with a 40% discount:** - Original ticket price: 20 - Discount amount: 40% of 20 = 8 - Discounted ticket price: 20 - 8 = 12 - Total revenue from this group: 10 people * 12 = 1202. **Next 20 people with a 15% discount:** - Original ticket price: 20 - Discount amount: 15% of 20 = 3 - Discounted ticket price: 20 - 3 = 17 - Total revenue from this group: 20 people * 17 = 3403. **Remaining 18 people paying full price:** - Total revenue from this group: 18 people * 20 = 360Finally, I'll sum the revenues from all groups to find the total revenue:120 (first group) + 340 (second group) + 360 (third group) = 820Therefore, the total revenue from ticket sales is 820.