answer：Okay, so I have this number: 32 ※ 35717 ※. It's supposed to be divisible by 72. Hmm, I need to find the digits represented by the asterisks. Let me think about how to approach this.First, I remember that if a number is divisible by 72, it must be divisible by both 8 and 9 because 72 is 8 multiplied by 9, and 8 and 9 are co-prime. So, I can check the divisibility rules for 8 and 9 separately.Starting with the divisibility rule for 8: A number is divisible by 8 if its last three digits form a number that's divisible by 8. In this case, the number is 32 ※ 35717 ※. Let me write it out with placeholders for clarity: 3 2 * 3 5 7 1 7 *. So, the last three digits would be 7, 1, and the second asterisk. Wait, actually, let me count the digits properly.Wait, the number is 32 ※ 35717 ※. Let's break it down:- The first two digits are 3 and 2.- Then there's an asterisk, which is the third digit.- Followed by 3, 5, 7, 1, 7.- And then another asterisk as the last digit.So, the number is 8 digits long: 3 2 * 3 5 7 1 7 *. So, the last three digits are 7, 1, and the last asterisk. So, the number formed by the last three digits is 71*. That needs to be divisible by 8.Okay, so 71* must be divisible by 8. Let's figure out what digit * can be. Let me list the numbers from 710 to 719 and see which are divisible by 8.710 ÷ 8 = 88.75 → Not divisible711 ÷ 8 = 88.875 → Not divisible712 ÷ 8 = 89 → Exactly, so 712 is divisible by 8.713 ÷ 8 = 89.125 → Not divisible714 ÷ 8 = 89.25 → Not divisible715 ÷ 8 = 89.375 → Not divisible716 ÷ 8 = 89.5 → Not divisible717 ÷ 8 = 89.625 → Not divisible718 ÷ 8 = 89.75 → Not divisible719 ÷ 8 = 89.875 → Not divisibleSo, the only number in that range divisible by 8 is 712. Therefore, the last asterisk must be 2. So, the last digit is 2.Alright, so now we know the last digit is 2. So, the number is 32 ※ 357172. Now, let's move on to the divisibility rule for 9.Divisibility by 9: A number is divisible by 9 if the sum of its digits is a multiple of 9. So, let's calculate the sum of the known digits and then figure out what the missing digit (the first asterisk) should be.The number is: 3, 2, *, 3, 5, 7, 1, 7, 2.Wait, hold on, earlier I thought the number was 8 digits, but now it seems like it's 9 digits. Let me check again.Wait, the original number is 32 ※ 35717 ※. So, that's 3, 2, *, 3, 5, 7, 1, 7, *. So, yes, it's 9 digits. So, the number is 3 2 * 3 5 7 1 7 *.So, the digits are: 3, 2, a, 3, 5, 7, 1, 7, b, where a and b are the asterisks. We already found that b is 2.So, the digits are: 3, 2, a, 3, 5, 7, 1, 7, 2.Now, let's compute the sum of these digits:3 + 2 + a + 3 + 5 + 7 + 1 + 7 + 2.Let me add them step by step:3 + 2 = 55 + a = 5 + a5 + a + 3 = 8 + a8 + a + 5 = 13 + a13 + a + 7 = 20 + a20 + a + 1 = 21 + a21 + a + 7 = 28 + a28 + a + 2 = 30 + a.So, the total sum is 30 + a.For the number to be divisible by 9, 30 + a must be a multiple of 9.What's the next multiple of 9 after 30? 9*3=27, 9*4=36. So, 36 is the next multiple.So, 30 + a = 36 → a = 6.Alternatively, if 30 + a could be 27, but 30 is already larger than 27, so the next possible is 36.So, a = 6.Therefore, the first asterisk is 6, and the last asterisk is 2.Let me double-check:Number is 3 2 6 3 5 7 1 7 2.Let's verify divisibility by 8: last three digits are 712. 712 ÷ 8 = 89, which is exact. Good.Divisibility by 9: sum of digits is 3 + 2 + 6 + 3 + 5 + 7 + 1 + 7 + 2.Calculating: 3+2=5, 5+6=11, 11+3=14, 14+5=19, 19+7=26, 26+1=27, 27+7=34, 34+2=36. 36 is divisible by 9. Perfect.So, both conditions are satisfied. Therefore, the digits are 6 and 2.**Final Answer**The digits are boxed{6} and boxed{2}.

answer：First, I need to determine the time it takes to travel each distance at a constant speed of 50 miles per hour.For the 450-mile trip, I'll divide the distance by the speed: 450 miles ÷ 50 mph = 9 hours.For the 475-mile trip, I'll do the same: 475 miles ÷ 50 mph = 9.5 hours.Next, I'll find the difference in travel times: 9.5 hours - 9 hours = 0.5 hours.To convert this time difference into minutes, I'll multiply by 60: 0.5 hours × 60 minutes/hour = 30 minutes.Therefore, the 450-mile trip takes 30 minutes less than the 475-mile trip.

answer：Alright, so I have this problem here: I need to find the coefficient of the ( x^4 ) term in the expansion of the binomial ( (4x^2 - 2x + 1)(2x + 1)^5 ). Hmm, okay. Let me think about how to approach this.First, I remember that when you have a product of two polynomials, the coefficient of a specific term can be found by considering all the ways that term can be formed by multiplying terms from each polynomial. So, in this case, I have two polynomials: ( 4x^2 - 2x + 1 ) and ( (2x + 1)^5 ). I need to find the coefficient of ( x^4 ) in their product.Let me denote the first polynomial as ( P(x) = 4x^2 - 2x + 1 ) and the second polynomial as ( Q(x) = (2x + 1)^5 ). So, I need to compute ( P(x) times Q(x) ) and find the coefficient of ( x^4 ).But expanding ( Q(x) ) directly might be tedious since it's a fifth power. Maybe there's a smarter way to do this without expanding the entire expression. I recall that the binomial theorem can help here. The binomial theorem states that ( (a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k ). So, applying this to ( Q(x) = (2x + 1)^5 ), I can write it as:[Q(x) = sum_{k=0}^{5} binom{5}{k} (2x)^{5 - k} (1)^k = sum_{k=0}^{5} binom{5}{k} 2^{5 - k} x^{5 - k}]Simplifying that, each term in the expansion is ( binom{5}{k} 2^{5 - k} x^{5 - k} ). So, the exponents of ( x ) in ( Q(x) ) range from 0 to 5.Now, since ( P(x) ) is a quadratic polynomial, when I multiply it by ( Q(x) ), the resulting polynomial will have terms with exponents ranging from ( 0 + 2 = 2 ) up to ( 5 + 2 = 7 ). But I specifically need the coefficient of ( x^4 ). So, I need to figure out which combinations of terms from ( P(x) ) and ( Q(x) ) will multiply together to give an ( x^4 ) term.Let me denote ( P(x) = a x^2 + b x + c ), where ( a = 4 ), ( b = -2 ), and ( c = 1 ). Then, when I multiply ( P(x) ) by ( Q(x) ), the ( x^4 ) term will come from:1. The ( x^2 ) term in ( P(x) ) multiplied by the ( x^2 ) term in ( Q(x) ).2. The ( x ) term in ( P(x) ) multiplied by the ( x^3 ) term in ( Q(x) ).3. The constant term in ( P(x) ) multiplied by the ( x^4 ) term in ( Q(x) ).So, in other words, the coefficient of ( x^4 ) in the product will be:[a times text{(coefficient of } x^2 text{ in } Q(x)) + b times text{(coefficient of } x^3 text{ in } Q(x)) + c times text{(coefficient of } x^4 text{ in } Q(x))]So, I need to find the coefficients of ( x^2 ), ( x^3 ), and ( x^4 ) in ( Q(x) ), then multiply each by the corresponding coefficients from ( P(x) ) and add them up.Let me compute each of these coefficients one by one.First, the coefficient of ( x^2 ) in ( Q(x) ). From the expansion above, each term is ( binom{5}{k} 2^{5 - k} x^{5 - k} ). So, for the exponent ( 5 - k = 2 ), we have ( k = 3 ). Therefore, the coefficient is ( binom{5}{3} 2^{5 - 3} ).Calculating that:[binom{5}{3} = frac{5!}{3!2!} = 10][2^{5 - 3} = 2^2 = 4]So, the coefficient is ( 10 times 4 = 40 ).Next, the coefficient of ( x^3 ) in ( Q(x) ). Similarly, ( 5 - k = 3 ) implies ( k = 2 ). So, the coefficient is ( binom{5}{2} 2^{5 - 2} ).Calculating that:[binom{5}{2} = frac{5!}{2!3!} = 10][2^{5 - 2} = 2^3 = 8]So, the coefficient is ( 10 times 8 = 80 ).Now, the coefficient of ( x^4 ) in ( Q(x) ). Here, ( 5 - k = 4 ) implies ( k = 1 ). So, the coefficient is ( binom{5}{1} 2^{5 - 1} ).Calculating that:[binom{5}{1} = 5][2^{5 - 1} = 2^4 = 16]So, the coefficient is ( 5 times 16 = 80 ).Alright, so now I have the coefficients from ( Q(x) ):- Coefficient of ( x^2 ): 40- Coefficient of ( x^3 ): 80- Coefficient of ( x^4 ): 80Now, recalling the coefficients from ( P(x) ):- ( a = 4 ) (coefficient of ( x^2 ))- ( b = -2 ) (coefficient of ( x ))- ( c = 1 ) (constant term)So, plugging these into the formula for the coefficient of ( x^4 ) in the product:[text{Coefficient} = a times 40 + b times 80 + c times 80]Substituting the values:[text{Coefficient} = 4 times 40 + (-2) times 80 + 1 times 80]Let me compute each term:1. ( 4 times 40 = 160 )2. ( -2 times 80 = -160 )3. ( 1 times 80 = 80 )Now, adding these together:[160 - 160 + 80 = (160 - 160) + 80 = 0 + 80 = 80]Wait, so the coefficient is 80? Hmm, that seems straightforward, but let me double-check my calculations to make sure I didn't make a mistake.First, coefficients from ( Q(x) ):- ( x^2 ): 40 (from ( k = 3 ))- ( x^3 ): 80 (from ( k = 2 ))- ( x^4 ): 80 (from ( k = 1 ))Coefficients from ( P(x) ):- ( x^2 ): 4- ( x ): -2- Constant: 1So, the contributions:- ( 4 times 40 = 160 )- ( -2 times 80 = -160 )- ( 1 times 80 = 80 )Adding up: 160 - 160 + 80 = 80.Hmm, that seems correct. Wait, let me think again. Is there another way to compute this?Alternatively, maybe I can compute the entire product and see if the coefficient of ( x^4 ) is indeed 80.But that might be time-consuming, but perhaps for verification, let me try.First, let me compute ( (2x + 1)^5 ) using the binomial theorem.As above, ( (2x + 1)^5 = sum_{k=0}^{5} binom{5}{k} (2x)^{5 - k} (1)^k ).So, expanding term by term:- ( k = 0 ): ( binom{5}{0} (2x)^5 = 1 times 32x^5 = 32x^5 )- ( k = 1 ): ( binom{5}{1} (2x)^4 = 5 times 16x^4 = 80x^4 )- ( k = 2 ): ( binom{5}{2} (2x)^3 = 10 times 8x^3 = 80x^3 )- ( k = 3 ): ( binom{5}{3} (2x)^2 = 10 times 4x^2 = 40x^2 )- ( k = 4 ): ( binom{5}{4} (2x)^1 = 5 times 2x = 10x )- ( k = 5 ): ( binom{5}{5} (2x)^0 = 1 times 1 = 1 )So, ( Q(x) = 32x^5 + 80x^4 + 80x^3 + 40x^2 + 10x + 1 ).Now, ( P(x) = 4x^2 - 2x + 1 ). So, multiplying these two polynomials:( (4x^2 - 2x + 1)(32x^5 + 80x^4 + 80x^3 + 40x^2 + 10x + 1) ).But since I only need the ( x^4 ) term, I can focus on the products that result in ( x^4 ).Let me think about it:- The ( 4x^2 ) term in ( P(x) ) will multiply terms in ( Q(x) ) with exponent ( 2 ) to give ( x^4 ). So, ( 4x^2 times 40x^2 = 160x^4 ).- The ( -2x ) term in ( P(x) ) will multiply terms in ( Q(x) ) with exponent ( 3 ) to give ( x^4 ). So, ( -2x times 80x^3 = -160x^4 ).- The ( 1 ) term in ( P(x) ) will multiply terms in ( Q(x) ) with exponent ( 4 ) to give ( x^4 ). So, ( 1 times 80x^4 = 80x^4 ).Adding these up: ( 160x^4 - 160x^4 + 80x^4 = (160 - 160 + 80)x^4 = 80x^4 ).So, the coefficient is indeed 80. That matches my earlier calculation. So, I think 80 is the correct answer.But just to be thorough, let me compute another term to see if my method is correct. For example, the constant term. The constant term in the product should come from multiplying the constant terms of ( P(x) ) and ( Q(x) ). The constant term in ( P(x) ) is 1, and in ( Q(x) ) it is 1. So, 1 * 1 = 1. Let me check in the full expansion:Looking at ( Q(x) ), the constant term is 1, and in ( P(x) ), it's 1. So, 1 * 1 = 1. That seems correct.Another check: the ( x^5 ) term. It should come from ( 4x^2 times 32x^5 ), but wait, that would be ( x^7 ). Wait, no, actually, the highest degree term in ( P(x) ) is ( x^2 ) and in ( Q(x) ) is ( x^5 ), so the highest degree in the product is ( x^7 ). So, the ( x^5 ) term would come from:- ( 4x^2 times 10x = 40x^3 ) (Wait, no, 4x^2 * 10x is 40x^3, which is lower than x^5. Wait, maybe I'm getting confused.Wait, no, actually, to get an ( x^5 ) term, you need to multiply terms whose exponents add up to 5.So, in ( P(x) ), the exponents are 2, 1, and 0. In ( Q(x) ), the exponents are 5, 4, 3, 2, 1, 0.So, to get ( x^5 ), you can have:- ( x^2 times x^3 ): 4x^2 * 80x^3 = 320x^5- ( x^1 times x^4 ): -2x * 80x^4 = -160x^5- ( x^0 times x^5 ): 1 * 32x^5 = 32x^5Adding these up: 320 - 160 + 32 = 192x^5. So, the coefficient is 192.But I don't know if that's correct, but at least it's a way to check my method. Alternatively, if I compute the entire product, I can see if the ( x^5 ) term is indeed 192, but that might take too long.Alternatively, let me compute the coefficient of ( x^5 ) using the same method as before.Wait, actually, I think my method is correct because when I multiplied ( P(x) ) and ( Q(x) ), the ( x^4 ) term came out to 80, which matched both methods. So, I think my approach is solid.Therefore, I can be confident that the coefficient of ( x^4 ) is 80.**Final Answer**The coefficient of the ( x^4 ) term is boxed{80}.

answer：Okay, so I have this problem here where there's a group of 3 female students and 4 male students. I need to choose 3 representatives, but the catch is that I have to include at least one female and one male. Hmm, okay. Let me think about how to approach this.First, I remember that when dealing with combinations, the formula is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number of items we want to choose. So, maybe I can use combinations here.The total number of ways to choose 3 representatives without any restrictions would be the combination of all 7 students taken 3 at a time. Let me calculate that. So, C(7, 3) = 7! / (3! * (7 - 3)!) = (7 * 6 * 5) / (3 * 2 * 1) = 35. Okay, so there are 35 ways if there are no restrictions.But the problem says I need at least one female and one male. That means I can't have all females or all males. So, maybe I should subtract the cases where all are female or all are male from the total.Let me calculate the number of ways to choose all females. There are 3 females, so choosing 3 females would be C(3, 3) = 1. That makes sense because there's only one way to choose all three females.Now, the number of ways to choose all males. There are 4 males, so choosing 3 males would be C(4, 3). Let me compute that. C(4, 3) = 4! / (3! * (4 - 3)!) = (4 * 3 * 2) / (3 * 2 * 1 * 1) = 4. So, there are 4 ways to choose all males.Therefore, the number of invalid combinations (all female or all male) is 1 + 4 = 5. So, subtracting that from the total number of combinations, which was 35, gives me 35 - 5 = 30. So, there are 30 valid ways to choose the representatives with at least one female and one male.Wait, let me verify this another way to make sure I didn't make a mistake. Another approach is to consider the different possible gender distributions in the group of 3 representatives. Since we need at least one of each, the possible distributions are:1. 1 female and 2 males2. 2 females and 1 maleThere are no other possibilities because 3 females or 3 males are invalid, and we can't have 0 females or 0 males.So, let's calculate each case separately.Case 1: 1 female and 2 males.Number of ways to choose 1 female from 3 is C(3, 1) = 3.Number of ways to choose 2 males from 4 is C(4, 2). Let me compute that: 4! / (2! * 2!) = (4 * 3) / (2 * 1) = 6.So, the total number of ways for this case is 3 * 6 = 18.Case 2: 2 females and 1 male.Number of ways to choose 2 females from 3 is C(3, 2) = 3.Number of ways to choose 1 male from 4 is C(4, 1) = 4.So, the total number of ways for this case is 3 * 4 = 12.Adding both cases together: 18 + 12 = 30. Okay, so that matches my previous result. So, that gives me more confidence that 30 is the correct answer.Just to recap, the total number of unrestricted combinations is 35. The invalid combinations are 1 (all female) + 4 (all male) = 5. So, 35 - 5 = 30. Alternatively, breaking it down into valid cases: 18 + 12 = 30. Both methods give the same answer, so I think that's solid.I don't think I missed any cases here. The only possibilities are 1F2M and 2F1M, and I accounted for both. So, yeah, 30 is the number of different ways to select the representatives with at least one female and one male.**Final Answer**The number of different ways to select the representatives is boxed{30}.

answer：First, I need to determine the number of ways to take one book from any of the three shelves. There are 4 Chinese books on the first shelf, 5 math books on the second shelf, and 6 sports books on the third shelf. Since these are independent choices, I can add the number of books on each shelf to find the total number of ways to take one book from any shelf. This gives me 4 + 5 + 6 = 15 ways.Next, I need to find the number of ways to take one book from each of the first, second, and third shelves. For each shelf, I have a specific number of choices: 4 Chinese books, 5 math books, and 6 sports books. Since these choices are independent, I multiply the number of options for each shelf together to find the total number of ways. This results in 4 × 5 × 6 = 120 ways.

answer：First, I recognize that the sum of the interior angles in any triangle is 180 degrees. Given the ratio of the angles is 1:2:3, I can represent the angles as x, 2x, and 3x.Adding these together: x + 2x + 3x = 6x = 180 degrees. Solving for x gives x = 30 degrees. Therefore, the angles are 30°, 60°, and 90°, indicating that the triangle is a right-angled triangle.In a right-angled triangle, the sides are related by the Pythagorean theorem: a² + b² = c², where c is the hypotenuse. The longest side, which is the hypotenuse, is given as 6 units.Using the properties of a 30-60-90 triangle, the sides are in the ratio 1 : √3 : 2. Here, the hypotenuse corresponds to 2 parts, so each part is 3 units. Therefore, the shortest side, which corresponds to 1 part, is 3 units long.