💡First, identify the range of four-digit numbers, which is from 1000 to 9999.Next, consider that the number must be even, so the units digit must be 0, 2, 4, 6, or 8.For each possible units digit, determine the corresponding tens digit such that their sum is 12:- If the units digit is 0, the tens digit would be 12 (invalid since it's not a single digit).- If the units digit is 2, the tens digit would be 10 (invalid for the same reason).- If the units digit is 4, the tens digit is 8.- If the units digit is 6, the tens digit is 6.- If the units digit is 8, the tens digit is 4.This results in three valid combinations for the tens and units digits: 48, 66, and 84.Finally, calculate the total number of valid four-digit numbers by multiplying the number of valid thousands digits (1-9, giving 9 options) by the number of valid tens-units combinations (3 options):9 (thousands digits) × 3 (tens-units pairs) = 27.

💡First, identify the total number of courses Xiao Ming can choose from, which is 4.Since Xiao Ming selects one course randomly, each course has an equal probability of being chosen.To find the probability of selecting "cooking," divide the number of ways to select "cooking" by the total number of courses.This results in a probability of 1/4.

💡First, I need to determine how many sticks of gum are in one brown box. Since there are 480 sticks of gum in 8 brown boxes, I can divide 480 by 8 to find the number of sticks in one box.Next, I'll calculate how many sticks of gum are in one carton. Each carton contains 5 packs of gum, and each pack has 3 sticks. Multiplying 5 packs by 3 sticks per pack gives the total number of sticks per carton.Finally, to find out how many cartons are in each brown box, I'll divide the number of sticks per box by the number of sticks per carton. This will give me the number of cartons in each brown box.

💡Alright, so I have this inequality: (-x^2 + 6a x - 3a^2 geq 0). I need to find the minimum value of (x_1 + x_2 + frac{3a}{x_1 x_2}), where (x_1) and (x_2) are the endpoints of the solution set for (x), and (a > 0). Hmm, okay, let me try to break this down step by step.First, I remember that when dealing with quadratic inequalities, it's often helpful to rewrite them in standard form. The given inequality is (-x^2 + 6a x - 3a^2 geq 0). Maybe I can multiply both sides by (-1) to make the coefficient of (x^2) positive, but I have to remember that multiplying by a negative number reverses the inequality sign. So, that would give me:(x^2 - 6a x + 3a^2 leq 0)Okay, so now the inequality is (x^2 - 6a x + 3a^2 leq 0). Since this is a quadratic equation, I can find its roots to determine the intervals where the inequality holds. The quadratic is (x^2 - 6a x + 3a^2), so let me set it equal to zero:(x^2 - 6a x + 3a^2 = 0)To find the roots, I can use the quadratic formula: (x = frac{-b pm sqrt{b^2 - 4ac}}{2a}), where (a = 1), (b = -6a), and (c = 3a^2). Plugging these into the formula:(x = frac{6a pm sqrt{(6a)^2 - 4 cdot 1 cdot 3a^2}}{2 cdot 1})Simplifying inside the square root:((6a)^2 = 36a^2)(4 cdot 1 cdot 3a^2 = 12a^2)So, the discriminant is (36a^2 - 12a^2 = 24a^2). Therefore, the roots are:(x = frac{6a pm sqrt{24a^2}}{2})Simplify (sqrt{24a^2}):(sqrt{24a^2} = sqrt{24} cdot a = 2sqrt{6}a)So, the roots become:(x = frac{6a pm 2sqrt{6}a}{2})Factor out (2a) from the numerator:(x = frac{2a(3 pm sqrt{6})}{2} = a(3 pm sqrt{6}))Therefore, the roots are (x_1 = a(3 - sqrt{6})) and (x_2 = a(3 + sqrt{6})). Since (a > 0), (x_1 < x_2), so the solution set for the inequality (x^2 - 6a x + 3a^2 leq 0) is the interval ([x_1, x_2]), which is ([a(3 - sqrt{6}), a(3 + sqrt{6})]).Now, I need to compute (x_1 + x_2 + frac{3a}{x_1 x_2}). Let's compute each part step by step.First, (x_1 + x_2):(x_1 + x_2 = a(3 - sqrt{6}) + a(3 + sqrt{6}) = a cdot 3 - a cdot sqrt{6} + a cdot 3 + a cdot sqrt{6})Simplify:The (-a sqrt{6}) and (+a sqrt{6}) cancel out, so:(x_1 + x_2 = 3a + 3a = 6a)Okay, that's straightforward. Now, let's compute (x_1 x_2):(x_1 x_2 = [a(3 - sqrt{6})][a(3 + sqrt{6})] = a^2 (3 - sqrt{6})(3 + sqrt{6}))Notice that ((3 - sqrt{6})(3 + sqrt{6})) is a difference of squares:((3)^2 - (sqrt{6})^2 = 9 - 6 = 3)So, (x_1 x_2 = a^2 cdot 3 = 3a^2)Now, compute (frac{3a}{x_1 x_2}):(frac{3a}{x_1 x_2} = frac{3a}{3a^2} = frac{1}{a})So, putting it all together:(x_1 + x_2 + frac{3a}{x_1 x_2} = 6a + frac{1}{a})Now, I need to find the minimum value of this expression with respect to (a), given that (a > 0). So, essentially, I need to minimize (6a + frac{1}{a}).This looks like a problem where I can apply the AM-GM inequality. The AM-GM inequality states that for non-negative real numbers, the arithmetic mean is greater than or equal to the geometric mean. Specifically, for two positive numbers (u) and (v):(frac{u + v}{2} geq sqrt{uv})Equality holds when (u = v).In this case, let me set (u = 6a) and (v = frac{1}{a}). Then, applying AM-GM:(frac{6a + frac{1}{a}}{2} geq sqrt{6a cdot frac{1}{a}})Simplify the right side:(sqrt{6a cdot frac{1}{a}} = sqrt{6})So, multiplying both sides by 2:(6a + frac{1}{a} geq 2sqrt{6})Therefore, the minimum value of (6a + frac{1}{a}) is (2sqrt{6}), achieved when (6a = frac{1}{a}). Let's solve for (a) to confirm:(6a = frac{1}{a})Multiply both sides by (a):(6a^2 = 1)Divide both sides by 6:(a^2 = frac{1}{6})Take the square root (since (a > 0)):(a = frac{1}{sqrt{6}} = frac{sqrt{6}}{6})So, when (a = frac{sqrt{6}}{6}), the expression (6a + frac{1}{a}) reaches its minimum value of (2sqrt{6}).Therefore, the minimum value of (x_1 + x_2 + frac{3a}{x_1 x_2}) is (2sqrt{6}).**Final Answer**The minimum value is boxed{2sqrt{6}}.

💡Alright, so I came across this problem about "magic numbers." It says that for a positive integer ( n ), if the sum of its digits ( s(n) ) plus the product of its digits ( p(n) ) equals ( n ), then ( n ) is called a magic number. The task is to find the sum of all such magic numbers.Hmm, okay. Let me try to break this down. First, I need to understand what ( s(n) ) and ( p(n) ) represent. For example, if ( n = 23 ), then ( s(n) = 2 + 3 = 5 ) and ( p(n) = 2 times 3 = 6 ). So, ( s(n) + p(n) = 5 + 6 = 11 ), which is not equal to 23. Therefore, 23 is not a magic number.I think the key here is to find all numbers ( n ) such that when you add the sum of their digits and the product of their digits, you get the number itself. So, ( n = s(n) + p(n) ).Let me consider single-digit numbers first. If ( n ) is a single-digit number, say ( n = 5 ), then ( s(n) = 5 ) and ( p(n) = 5 ). So, ( s(n) + p(n) = 5 + 5 = 10 ), which is not equal to 5. Therefore, no single-digit numbers are magic numbers.Okay, moving on to two-digit numbers. Let me denote a two-digit number as ( n = 10a + b ), where ( a ) and ( b ) are the tens and units digits respectively. Then, ( s(n) = a + b ) and ( p(n) = a times b ). The condition becomes:[10a + b = (a + b) + (a times b)]Simplifying this equation:[10a + b = a + b + ab][10a = a + ab][10a - a = ab][9a = ab][9 = b]So, ( b = 9 ). That means, for any two-digit number where the units digit is 9, it will satisfy the condition. Therefore, the two-digit magic numbers are 19, 29, 39, 49, 59, 69, 79, 89, and 99.Let me verify this with an example. Take ( n = 19 ). Then, ( s(n) = 1 + 9 = 10 ) and ( p(n) = 1 times 9 = 9 ). Adding them together, ( 10 + 9 = 19 ), which equals ( n ). So, 19 is indeed a magic number. Similarly, checking 29: ( s(n) = 2 + 9 = 11 ), ( p(n) = 2 times 9 = 18 ), and ( 11 + 18 = 29 ). Perfect.What about three-digit numbers? Let me test a three-digit number, say 119. ( s(n) = 1 + 1 + 9 = 11 ), ( p(n) = 1 times 1 times 9 = 9 ), so ( s(n) + p(n) = 11 + 9 = 20 ), which is not equal to 119. Hmm, so 119 isn't a magic number. Maybe I should try another one. Let's try 129: ( s(n) = 1 + 2 + 9 = 12 ), ( p(n) = 1 times 2 times 9 = 18 ), so ( 12 + 18 = 30 ), which is not 129.Wait, maybe three-digit numbers don't work? Let me see if there's a way to generalize this. Suppose ( n ) is a ( k )-digit number, ( n = a_1a_2...a_k ). Then, ( s(n) = a_1 + a_2 + ... + a_k ) and ( p(n) = a_1 times a_2 times ... times a_k ). The condition is ( n = s(n) + p(n) ).For a three-digit number, ( n = 100a + 10b + c ). Then, ( s(n) = a + b + c ) and ( p(n) = a times b times c ). So, the equation becomes:[100a + 10b + c = (a + b + c) + (a times b times c)]Simplifying:[100a + 10b + c - a - b - c = a times b times c][99a + 9b = a times b times c]Hmm, this seems more complicated. Let me see if I can find any solutions here. Let's try small values for ( a ), starting with ( a = 1 ):For ( a = 1 ):[99(1) + 9b = 1 times b times c][99 + 9b = b times c][c = frac{99 + 9b}{b}][c = frac{99}{b} + 9]Since ( c ) must be an integer between 0 and 9, let's see for which ( b ) this holds.- ( b = 1 ): ( c = 99 + 9 = 108 ) → Not possible, since ( c ) must be a single digit.- ( b = 3 ): ( c = 33 + 9 = 42 ) → Not possible.- ( b = 9 ): ( c = 11 + 9 = 20 ) → Not possible.No solutions for ( a = 1 ).Trying ( a = 2 ):[99(2) + 9b = 2 times b times c][198 + 9b = 2bc][2bc = 198 + 9b][c = frac{198 + 9b}{2b}][c = frac{198}{2b} + frac{9b}{2b} = frac{99}{b} + frac{9}{2}]Again, ( c ) must be an integer between 0 and 9.- ( b = 3 ): ( c = 33 + 4.5 = 37.5 ) → Not integer.- ( b = 9 ): ( c = 11 + 4.5 = 15.5 ) → Not integer.No solutions here either.Trying ( a = 3 ):[99(3) + 9b = 3bc][297 + 9b = 3bc][3bc = 297 + 9b][c = frac{297 + 9b}{3b} = frac{99}{b} + 3]Testing ( b ):- ( b = 9 ): ( c = 11 + 3 = 14 ) → Not possible.No solutions.It seems like for three-digit numbers, it's difficult to find any magic numbers. Maybe there are none? Let me check another approach.Suppose ( n ) is a three-digit number, ( n = 100a + 10b + c ). Then, ( s(n) = a + b + c ) and ( p(n) = a times b times c ). The equation is:[100a + 10b + c = a + b + c + abc][99a + 9b = abc]So, ( abc = 99a + 9b ). Let's factor out ( a ):[a(bc - 99) = 9b]Since ( a ) is at least 1, ( bc - 99 ) must be positive because ( 9b ) is positive. Therefore, ( bc > 99 ). But ( b ) and ( c ) are single digits, so ( b leq 9 ) and ( c leq 9 ). Therefore, ( bc leq 81 ). But 81 is less than 99, which means ( bc - 99 ) is negative. This leads to a contradiction because the left side would be negative or zero, while the right side is positive. Therefore, there are no three-digit magic numbers.Okay, so three-digit numbers don't work. What about four-digit numbers? Let me see if the same logic applies.For a four-digit number ( n = 1000a + 100b + 10c + d ), ( s(n) = a + b + c + d ), ( p(n) = a times b times c times d ). The equation becomes:[1000a + 100b + 10c + d = a + b + c + d + abcd][999a + 99b + 9c = abcd]This seems even more complicated. Let me try to analyze it similarly.Factor out ( a ):[a(999 + b times c times d) = 99b + 9c]Wait, that doesn't seem straightforward. Alternatively, since ( abcd ) is the product, and ( a, b, c, d ) are digits, the maximum product is ( 9 times 9 times 9 times 9 = 6561 ). However, ( 999a + 99b + 9c ) is at least 999 (when ( a=1, b=0, c=0 )) and can be much larger. So, it's unlikely that ( abcd ) can catch up with ( 999a + 99b + 9c ), especially since ( a ) is at least 1.Therefore, it's probable that there are no four-digit magic numbers either.Given this pattern, it seems that only two-digit numbers can satisfy the condition ( n = s(n) + p(n) ). Earlier, I found that for two-digit numbers, ( b = 9 ), so the magic numbers are 19, 29, 39, 49, 59, 69, 79, 89, and 99.Let me list them out:- 19: ( 1 + 9 + 1 times 9 = 1 + 9 + 9 = 19 )- 29: ( 2 + 9 + 2 times 9 = 2 + 9 + 18 = 29 )- 39: ( 3 + 9 + 3 times 9 = 3 + 9 + 27 = 39 )- 49: ( 4 + 9 + 4 times 9 = 4 + 9 + 36 = 49 )- 59: ( 5 + 9 + 5 times 9 = 5 + 9 + 45 = 59 )- 69: ( 6 + 9 + 6 times 9 = 6 + 9 + 54 = 69 )- 79: ( 7 + 9 + 7 times 9 = 7 + 9 + 63 = 79 )- 89: ( 8 + 9 + 8 times 9 = 8 + 9 + 72 = 89 )- 99: ( 9 + 9 + 9 times 9 = 9 + 9 + 81 = 99 )All of these satisfy the condition. Now, I need to find the sum of all these magic numbers.Calculating the sum:19 + 29 + 39 + 49 + 59 + 69 + 79 + 89 + 99Let me add them step by step:19 + 29 = 4848 + 39 = 8787 + 49 = 136136 + 59 = 195195 + 69 = 264264 + 79 = 343343 + 89 = 432432 + 99 = 531So, the total sum is 531.I think that's all. I don't see any other numbers beyond two digits that satisfy the condition, and single-digit numbers don't work either. Therefore, the sum of all magic numbers is 531.**Final Answer**The sum of all magic numbers is boxed{531}.

💡First, I need to determine the original volume of the ice before any melting occurred. Let's denote this original volume as ( V ) cubic inches.In the first hour, the ice loses (frac{3}{4}) of its volume. This means that only (frac{1}{4}) of the original volume remains after the first hour. Therefore, the volume after the first hour is:[frac{1}{4}V]In the second hour, the ice loses (frac{3}{4}) of its remaining volume. So, it retains (frac{1}{4}) of the volume from the first hour. Thus, the volume after the second hour is:[frac{1}{4} times frac{1}{4}V = frac{1}{16}V]We are given that after two hours, the volume of the ice is 0.25 cubic inches. This gives us the equation:[frac{1}{16}V = 0.25]To find the original volume ( V ), we solve for ( V ):[V = 0.25 times 16 = 4]Therefore, the original volume of the ice was 4 cubic inches.