answer：First, I need to determine the universal set ( U ) by finding the union of sets ( A ) and ( B ). Combining all unique elements from both sets, ( U = {3, 4, 5, 7, 8, 9} ).Next, I'll find the intersection of sets ( A ) and ( B ), which consists of the elements common to both sets. Thus, ( A cap B = {4, 7, 9} ).Finally, to find the complement of ( A cap B ) with respect to the universal set ( U ), I'll identify all elements in ( U ) that are not in ( A cap B ). This gives ( complement_U (A cap B) = {3, 5, 8} ). Therefore, the number of elements in this complement set is 3.

answer：First, I'll list out the given relationships:1. All Arogs are Brafs.2. All Crups are Brafs.3. All Dramps are Arogs.4. All Crups are Dramps.Next, I'll analyze each statement to see which one is implied by these facts.**Statement A: All Dramps are Brafs and are Crups.**- Since all Dramps are Arogs and all Arogs are Brafs, it follows that all Dramps are Brafs.- Additionally, since all Crups are Dramps, and all Dramps are Arogs, it means all Crups are also Arogs and Brafs.- Therefore, all Dramps are Brafs and Crups.**Statement B: All Brafs are Crups and are Dramps.**- This is not necessarily true because the given information only states that all Arogs and Crups are Brafs, but it doesn't imply that all Brafs are Crups or Dramps.**Statement C: All Arogs are Crups and are Dramps.**- While all Dramps are Arogs, the reverse isn't necessarily true. Not all Arogs need to be Dramps or Crups.**Statement D: All Crups are Arogs and are Brafs.**- This is true because all Crups are Dramps, and all Dramps are Arogs, making all Crups Arogs and Brafs.**Statement E: All Arogs are Dramps and some Arogs may not be Crups.**- This is incorrect because not all Arogs are necessarily Dramps; only all Dramps are Arogs.After evaluating all statements, both A and D are true based on the given relationships. However, since the question asks for the statement implied by the facts, the most comprehensive and directly supported answer is **Statement A**.

answer：Okay, so I have this problem here where I need to find the dot product of two vectors. The vectors given are (2a - b) and a. The angle between vectors a and b is 120 degrees, and the magnitudes of a and b are 2 and 5 respectively. Hmm, let me think about how to approach this.First, I remember that the dot product of two vectors can be calculated in two ways. One way is by using the formula involving the magnitudes and the cosine of the angle between them, which is |a||b|cosθ. The other way is by expanding the expression if there are coefficients or operations involved, like in this case where we have 2a - b.So, maybe I should start by expanding the dot product (2a - b) · a. Let me write that out:(2a - b) · a = 2a · a - b · aOkay, that makes sense. I can distribute the dot product over the subtraction. Now, I need to compute each term separately.First, let's compute 2a · a. I know that a · a is equal to the magnitude of a squared, right? Because the dot product of a vector with itself is just |a|². So, a · a = |a|². Given that |a| is 2, that would be 2² = 4. Therefore, 2a · a = 2 * 4 = 8.Alright, that's the first term. Now, the second term is -b · a. Hmm, I remember that the dot product is commutative, so b · a is the same as a · b. So, I can write this as - (a · b). Now, I need to compute a · b.To compute a · b, I can use the formula |a||b|cosθ. We know |a| is 2, |b| is 5, and the angle θ between them is 120 degrees. So, plugging those values in:a · b = |a||b|cos(120°) = 2 * 5 * cos(120°)I should compute cos(120°). I remember that 120 degrees is in the second quadrant, and its cosine is negative. Also, 120° is equal to 180° - 60°, so cos(120°) is equal to -cos(60°). Since cos(60°) is 0.5, cos(120°) is -0.5.So, substituting that back in:a · b = 2 * 5 * (-0.5) = 10 * (-0.5) = -5Therefore, b · a is also -5, so - (a · b) becomes - (-5) = 5.Now, putting it all together, the first term was 8 and the second term is 5, so:(2a - b) · a = 8 + 5 = 13Wait, hold on, let me double-check my steps to make sure I didn't make a mistake. So, expanding the dot product, I had 2a · a - b · a. Then, a · a is |a|², which is 4, so 2*4 is 8. Then, a · b is |a||b|cosθ, which is 2*5*cos(120°), which is 10*(-0.5) = -5. So, -b · a is -(-5) = 5. Adding 8 and 5 gives 13. That seems correct.Alternatively, maybe I can compute it another way to verify. Let's think about the vector 2a - b. Its magnitude isn't directly needed, but perhaps I can compute the dot product using components. Maybe that would be a good way to check.To do that, I need to express vectors a and b in component form. Let me choose a coordinate system where vector a lies along the positive x-axis. So, vector a would be (2, 0). Then, vector b makes a 120° angle with a. So, the components of vector b would be |b|cosθ and |b|sinθ, where θ is 120°.Calculating the components of b:x-component: 5 * cos(120°) = 5 * (-0.5) = -2.5y-component: 5 * sin(120°) = 5 * (√3/2) ≈ 5 * 0.866 ≈ 4.33So, vector b is (-2.5, 4.33). Vector a is (2, 0).Now, let's compute 2a - b. 2a would be (4, 0). Subtracting b from that:2a - b = (4 - (-2.5), 0 - 4.33) = (6.5, -4.33)Now, vector a is (2, 0). So, the dot product of (6.5, -4.33) and (2, 0) is:6.5 * 2 + (-4.33) * 0 = 13 + 0 = 13Hmm, that's the same result as before. So, that confirms that the answer is indeed 13.Wait, but let me just make sure that I didn't make any calculation errors in the component method. Let's recalculate the components.Vector a is (2, 0). Vector b is (5cos120°, 5sin120°). Cos120° is -0.5, so 5*(-0.5) = -2.5. Sin120° is √3/2, which is approximately 0.866, so 5*0.866 is approximately 4.33. So, vector b is (-2.5, 4.33). Then, 2a is (4, 0). So, 2a - b is (4 - (-2.5), 0 - 4.33) = (6.5, -4.33). Then, the dot product with a, which is (2, 0), is 6.5*2 + (-4.33)*0 = 13 + 0 = 13. Yep, that's correct.Alternatively, maybe I can think about the projection of vectors or something else, but I think both methods I tried gave me the same answer, so I feel confident that 13 is the correct result.Just to recap, the steps were:1. Expand the dot product (2a - b) · a into 2(a · a) - (b · a).2. Compute a · a as |a|², which is 4, so 2*4 = 8.3. Compute a · b as |a||b|cosθ, which is 2*5*cos120° = -5.4. Therefore, - (a · b) is 5.5. Adding 8 and 5 gives 13.6. Verified using component method, got the same result.So, I think that's solid. I don't see any mistakes in my reasoning, so I'm pretty sure the answer is 13.**Final Answer**The value of (2overrightarrow{a}-overrightarrow{b})cdotoverrightarrow{a} is boxed{13}.

answer：Okay, so I have this problem here where the volume of a cylinder is exactly equal to its surface area, and I need to find the minimum possible value for the volume. The options are given as A. (48 pi), B. (50 pi), C. (54 pi), and D. (66 pi). Hmm, let me try to figure this out step by step.First, I remember that the volume (V) of a cylinder is given by the formula (V = pi r^2 h), where (r) is the radius and (h) is the height. The surface area (S) of a cylinder, on the other hand, is calculated as (S = 2pi r^2 + 2pi r h). This includes the areas of the two circular bases and the lateral surface area.The problem states that the volume is equal to the surface area, so I can set up the equation:[pi r^2 h = 2pi r^2 + 2pi r h]Okay, so now I have this equation. I need to solve for either (r) or (h) and then find the minimum volume. Let me try to simplify this equation.First, I can divide both sides by (pi) to make it simpler:[r^2 h = 2 r^2 + 2 r h]Hmm, that's better. Now, let's see. I can subtract (2 r^2 + 2 r h) from both sides to get all terms on one side:[r^2 h - 2 r^2 - 2 r h = 0]Looking at this, I can factor out an (r) from each term:[r (r h - 2 r - 2 h) = 0]So, either (r = 0) or (r h - 2 r - 2 h = 0). But (r = 0) doesn't make sense for a cylinder because it would mean the cylinder has no radius, which isn't possible. So, we can ignore that solution and focus on:[r h - 2 r - 2 h = 0]Let me rearrange this equation to solve for one variable in terms of the other. Let's try to solve for (h):[r h - 2 r - 2 h = 0]Bring the terms with (h) to one side:[r h - 2 h = 2 r]Factor out (h) from the left side:[h (r - 2) = 2 r]Now, solve for (h):[h = frac{2 r}{r - 2}]Okay, so now I have (h) expressed in terms of (r). Since the volume is equal to the surface area, and we need to find the minimum volume, I can substitute this expression for (h) back into the volume formula and then find the minimum value.So, the volume (V) is:[V = pi r^2 h = pi r^2 left( frac{2 r}{r - 2} right ) = pi frac{2 r^3}{r - 2}]So, (V(r) = frac{2 pi r^3}{r - 2}). Now, I need to find the minimum value of this function. To find the minimum, I should take the derivative of (V) with respect to (r) and set it equal to zero.Let me compute the derivative (V'(r)). First, let me write (V(r)) as:[V(r) = 2 pi cdot frac{r^3}{r - 2}]So, to take the derivative, I can use the quotient rule. The quotient rule states that if I have a function (f(r) = frac{u(r)}{v(r)}), then the derivative (f'(r)) is:[f'(r) = frac{u'(r) v(r) - u(r) v'(r)}{[v(r)]^2}]Applying this to (V(r)), let me set (u(r) = r^3) and (v(r) = r - 2). Then, (u'(r) = 3 r^2) and (v'(r) = 1).So, plugging into the quotient rule:[V'(r) = 2 pi cdot frac{(3 r^2)(r - 2) - (r^3)(1)}{(r - 2)^2}]Let me compute the numerator:First term: (3 r^2 (r - 2) = 3 r^3 - 6 r^2)Second term: (- r^3)So, combining these:[3 r^3 - 6 r^2 - r^3 = (3 r^3 - r^3) - 6 r^2 = 2 r^3 - 6 r^2]Therefore, the derivative is:[V'(r) = 2 pi cdot frac{2 r^3 - 6 r^2}{(r - 2)^2}]Simplify the numerator:Factor out a (2 r^2):[2 r^3 - 6 r^2 = 2 r^2 (r - 3)]So, the derivative becomes:[V'(r) = 2 pi cdot frac{2 r^2 (r - 3)}{(r - 2)^2} = frac{4 pi r^2 (r - 3)}{(r - 2)^2}]Now, to find critical points, set (V'(r) = 0):[frac{4 pi r^2 (r - 3)}{(r - 2)^2} = 0]The denominator ((r - 2)^2) is always positive for (r neq 2), so we can ignore it for setting the equation to zero. The numerator must be zero:[4 pi r^2 (r - 3) = 0]Since (4 pi) is never zero, we have:Either (r^2 = 0) or (r - 3 = 0)(r^2 = 0) implies (r = 0), but as before, that's not possible for a cylinder.(r - 3 = 0) implies (r = 3).So, the critical point is at (r = 3). Now, we need to check if this is a minimum.To confirm whether this critical point is a minimum, we can use the second derivative test or analyze the behavior of the first derivative around (r = 3).Alternatively, since it's the only critical point in the domain (r > 2) (because the height (h = frac{2 r}{r - 2}) must be positive, so (r - 2 > 0) implies (r > 2)), we can test values around (r = 3) to see if the derivative changes from negative to positive, indicating a minimum.Let me pick a value slightly less than 3, say (r = 2.5):Compute (V'(2.5)):Numerator: (4 pi (2.5)^2 (2.5 - 3) = 4 pi (6.25)(-0.5) = 4 pi (-3.125) = -12.5 pi)Denominator: ((2.5 - 2)^2 = (0.5)^2 = 0.25)So, (V'(2.5) = -12.5 pi / 0.25 = -50 pi), which is negative.Now, pick a value slightly more than 3, say (r = 4):Compute (V'(4)):Numerator: (4 pi (4)^2 (4 - 3) = 4 pi (16)(1) = 64 pi)Denominator: ((4 - 2)^2 = (2)^2 = 4)So, (V'(4) = 64 pi / 4 = 16 pi), which is positive.Therefore, the derivative changes from negative to positive at (r = 3), indicating that (r = 3) is indeed a point of local minimum. Since this is the only critical point in the domain, it must be the absolute minimum.So, the minimum volume occurs when (r = 3). Now, let's compute the volume at (r = 3).First, find (h):[h = frac{2 r}{r - 2} = frac{2 times 3}{3 - 2} = frac{6}{1} = 6]So, the height (h) is 6 when the radius (r) is 3.Now, compute the volume (V):[V = pi r^2 h = pi (3)^2 (6) = pi times 9 times 6 = 54 pi]Therefore, the minimum possible volume is (54 pi), which corresponds to option C.Wait, just to make sure, let me verify the surface area as well to ensure that it's equal to the volume.Compute the surface area (S):[S = 2 pi r^2 + 2 pi r h = 2 pi (9) + 2 pi (3)(6) = 18 pi + 36 pi = 54 pi]Yes, that's equal to the volume (54 pi). So, everything checks out.I think that's solid. I don't see any mistakes in my calculations. So, the answer should be C, (54 pi).**Final Answer**boxed{C}

answer：Okay, so I need to figure out the largest number Kolya can get by rearranging the digits of 2015, inserting a multiplication sign between any two digits, and then calculating the result. Hmm, let me break this down step by step.First, let's list out the digits we have: 2, 0, 1, and 5. Kolya can rearrange these digits in any order, so the first thing I should do is consider all possible permutations of these four digits. However, since we have four digits, there are 4! = 24 possible arrangements. That's a lot, but maybe we can find a smarter way than checking all 24.But before that, let me think about the multiplication. When we insert a multiplication sign between two digits, we're essentially splitting the four-digit number into two numbers and multiplying them. The goal is to maximize the product. So, for example, if we have digits arranged as ABCD, we can split it as AB * CD, A * BCD, ABC * D, or A * B * C * D? Wait, no, actually, the problem says putting a multiplication sign between any two digits, so it's only one multiplication sign, meaning we split the four-digit number into two parts, each part being a number, and then multiply them.So, for example, if the digits are arranged as 1502, we can split it as 15 * 02 = 30, or 1 * 502 = 502, or 150 * 2 = 300. So, each arrangement can be split in three different ways, depending on where we place the multiplication sign.Therefore, for each permutation of the digits, there are three possible products, and we need to find the maximum among all these possibilities.But 24 permutations times 3 splits each is 72 calculations. That's a bit time-consuming, but maybe manageable. Alternatively, maybe we can find a pattern or a way to maximize the product without checking all possibilities.I remember that for two numbers with a fixed sum, their product is maximized when they are as close as possible to each other. But in this case, the sum isn't fixed; instead, we have a fixed number of digits, and we're trying to split them into two numbers whose product is maximized.Wait, actually, the number of digits is fixed, but their arrangement can vary. So, to maximize the product, we need to arrange the digits such that the two numbers we get after splitting are as large as possible, but also balanced in terms of their magnitudes.For example, if one number is too large and the other is too small, the product might not be as big as when both numbers are moderately large.So, perhaps the optimal split is when the two numbers are of similar lengths. Since we have four digits, splitting them into two two-digit numbers might give a better product than splitting into a one-digit and a three-digit number.Let me test this idea.If we split into two two-digit numbers, each number can be as large as possible, given the digits. For example, using the digits 5, 2, 1, 0, the largest two-digit numbers we can make are 52 and 10, but 52 * 10 = 520. Alternatively, 51 * 20 = 1020, which is larger. Hmm, so maybe arranging the digits to make two two-digit numbers with higher values can give a better product.Alternatively, if we split into a one-digit and a three-digit number, the three-digit number can be quite large, but the one-digit number is limited to 5, 2, 1, or 0. For example, 5 * 210 = 1050, which is actually larger than 520. So, in this case, splitting into a one-digit and a three-digit number gives a larger product.Wait, so which is better? It depends on the arrangement.So, perhaps we need to consider both possibilities: splitting into two two-digit numbers and splitting into a one-digit and a three-digit number. Then, for each permutation, calculate both possibilities and see which gives the maximum.But that's going to be a lot of calculations. Maybe I can find a way to maximize each case separately.First, let's consider splitting into two two-digit numbers.To maximize the product of two two-digit numbers, we should assign the largest digits to the tens place of both numbers. So, the two largest digits are 5 and 2. So, the two two-digit numbers would be 5X and 2X, where X are the remaining digits, which are 1 and 0.So, arranging 5 and 2 in the tens place, and 1 and 0 in the units place. So, possible combinations:51 * 20 = 102050 * 21 = 1050So, 50 * 21 = 1050 is larger.Alternatively, if we arrange the digits differently, say 52 * 10 = 520, which is smaller.So, 50 * 21 = 1050 seems better.Alternatively, if we arrange the digits as 25 * 10 = 250, which is worse.So, in the two two-digit case, the maximum product is 1050.Now, let's consider splitting into a one-digit and a three-digit number.To maximize the product, we should assign the largest digit as the one-digit number, which is 5, and then arrange the remaining digits to form the largest possible three-digit number.The remaining digits are 2, 1, 0. The largest three-digit number we can make is 210. So, 5 * 210 = 1050.Alternatively, if we take 2 as the one-digit number, the remaining digits are 5, 1, 0. The largest three-digit number is 510, so 2 * 510 = 1020.If we take 1 as the one-digit number, the remaining digits are 5, 2, 0. The largest three-digit number is 520, so 1 * 520 = 520.If we take 0 as the one-digit number, the remaining digits are 5, 2, 1. The largest three-digit number is 521, but 0 * 521 = 0.So, the maximum product in this case is 1050.Wait, so both splitting into two two-digit numbers and splitting into a one-digit and a three-digit number can give us 1050 as the maximum.But is there a way to get higher than 1050?Wait, let's think again. Maybe if we arrange the digits differently.What if we split the number into a three-digit and a one-digit number, but the one-digit number isn't necessarily the largest digit? For example, arranging the digits as 201 * 5 = 1005, which is less than 1050.Alternatively, 210 * 5 = 1050, which is the same as before.Alternatively, 251 * 0 = 0, which is worse.Alternatively, 250 * 1 = 250, which is worse.So, no improvement there.What about splitting into two two-digit numbers but arranging the digits differently?Wait, earlier we saw that 50 * 21 = 1050, but what if we arrange the digits as 52 * 10 = 520, which is worse.Alternatively, 15 * 20 = 300, which is worse.Alternatively, 12 * 50 = 600, which is worse.Alternatively, 10 * 52 = 520, which is worse.So, seems like 50 * 21 = 1050 is the maximum in the two two-digit case.Wait, but is there a way to get a higher product?Wait, let's think about the digits again. We have 2, 0, 1, 5.If we split into a one-digit and a three-digit number, the maximum is 5 * 210 = 1050.If we split into two two-digit numbers, the maximum is 50 * 21 = 1050.But is there a way to arrange the digits such that both numbers are larger?Wait, for example, can we have a three-digit number and a one-digit number where the three-digit number is larger than 210?Wait, the digits are 2, 0, 1, 5. The largest three-digit number we can make is 521, but if we use 5 as the one-digit number, the three-digit number is 210. If we use 2 as the one-digit number, the three-digit number is 510, which is larger than 210, but 2 * 510 = 1020, which is less than 5 * 210 = 1050.Alternatively, if we use 1 as the one-digit number, the three-digit number is 520, but 1 * 520 = 520, which is worse.So, 5 * 210 = 1050 is still the maximum.Wait, but what if we arrange the digits as 2015, and split it as 20 * 15 = 300. That's worse.Alternatively, 201 * 5 = 1005, which is less than 1050.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.Alternatively, 21 * 05 = 105, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.Wait, so maybe 1050 is the maximum.But hold on, let's think about another arrangement.What if we arrange the digits as 2501, and split it as 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.Alternatively, 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 2501 split as 25 * 01 = 25, which is worse.Alternatively, 2501 split as 2 * 501 = 1002, which is less than 1050.Alternatively, 2501 split as 250 * 1 = 250, which is worse.So, that's not better.Wait, another arrangement: 1520.Split as 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.Alternatively, 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.Alternatively, 1520 split as 15 * 20 = 300, which is worse.Alternatively, 1520 split as 1 * 520 = 520, which is worse.Alternatively, 1520 split as 152 * 0 = 0, which is worse.So, that's not better.Wait, let's try another arrangement: 5120.Split as 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.Alternatively, 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.Alternatively, 5120 split as 51 * 20 = 1020, which is less than 1050.Alternatively, 5120 split as 5 * 120 = 600, which is worse.Alternatively, 5120 split as 512 * 0 = 0, which is worse.So, still 1050 is higher.Wait, let's try another arrangement: 5210.Split as 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.Alternatively, 521 * 0 = 0, which is worse.Alternatively, 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.Alternatively, 521 * 0 = 0, which is worse.So, same result.Wait, another arrangement: 5021.Split as 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.Alternatively, 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.So, same maximum.Wait, another arrangement: 5012.Split as 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.Alternatively, 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.So, still 1050 is higher.Wait, let's try 5102.Split as 51 * 02 = 102, which is worse.Alternatively, 5 * 102 = 510, which is worse.Alternatively, 510 * 2 = 1020, which is less than 1050.Alternatively, 51 * 02 = 102, which is worse.Alternatively, 5 * 102 = 510, which is worse.Alternatively, 510 * 2 = 1020, which is less than 1050.So, same result.Wait, another arrangement: 2501.Split as 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.Alternatively, 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.So, same.Wait, let's try 2150.Split as 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.Alternatively, 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.So, same maximum.Wait, another arrangement: 2105.Split as 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.Alternatively, 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.So, same result.Wait, let's try 2015.Split as 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.Alternatively, 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.So, same.Wait, another arrangement: 1520.Wait, I think I already tried that earlier, and it gave 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.So, same.Wait, another arrangement: 1502.Split as 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.Alternatively, 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.So, same.Wait, another arrangement: 1052.Split as 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.Alternatively, 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.So, same.Wait, another arrangement: 1025.Split as 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.Alternatively, 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.So, same.Wait, another arrangement: 1205.Split as 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.Alternatively, 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.So, same.Wait, another arrangement: 1250.Split as 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.Alternatively, 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.So, same.Wait, another arrangement: 1502.Wait, I think I did that earlier.Wait, let me think differently. Maybe I can consider all possible splits and see which gives the maximum.But since I've gone through many permutations and the maximum I can get is 1050, maybe that's the answer.But wait, let me think again. Is there a way to get higher than 1050?Wait, let's consider splitting into a three-digit and a one-digit number, but maybe arranging the digits differently.Wait, for example, if we arrange the digits as 5210, and split as 521 * 0 = 0, which is worse.Alternatively, 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.Alternatively, 521 * 0 = 0, which is worse.Alternatively, 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.So, same result.Wait, another arrangement: 5120.Split as 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.Alternatively, 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.So, same.Wait, another arrangement: 5021.Split as 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.Alternatively, 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.So, same.Wait, another arrangement: 5012.Split as 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.Alternatively, 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.So, same.Wait, another arrangement: 2510.Split as 25 * 10 = 250, which is worse.Alternatively, 2 * 510 = 1020, which is less than 1050.Alternatively, 251 * 0 = 0, which is worse.Alternatively, 25 * 10 = 250, which is worse.Alternatively, 2 * 510 = 1020, which is less than 1050.Alternatively, 251 * 0 = 0, which is worse.So, same.Wait, another arrangement: 2501.Split as 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.Alternatively, 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.So, same.Wait, another arrangement: 2150.Split as 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.Alternatively, 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.So, same.Wait, another arrangement: 2105.Split as 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.Alternatively, 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.So, same.Wait, another arrangement: 2015.Split as 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.Alternatively, 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.So, same.Wait, another arrangement: 1520.Split as 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.Alternatively, 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.So, same.Wait, another arrangement: 1502.Split as 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.Alternatively, 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.So, same.Wait, another arrangement: 1052.Split as 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.Alternatively, 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.So, same.Wait, another arrangement: 1025.Split as 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.Alternatively, 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.So, same.Wait, another arrangement: 1205.Split as 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.Alternatively, 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.So, same.Wait, another arrangement: 1250.Split as 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.Alternatively, 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.So, same.Wait, another arrangement: 1520.Wait, I think I did that earlier.Wait, another arrangement: 1502.Wait, same as before.Wait, another arrangement: 1052.Same as before.Wait, another arrangement: 1025.Same as before.Wait, another arrangement: 1205.Same as before.Wait, another arrangement: 1250.Same as before.So, after going through all these permutations, it seems that the maximum product Kolya can get is 1050.But wait, let me think again. Is there a way to arrange the digits such that the product is higher?Wait, for example, if we arrange the digits as 5210, and split it as 521 * 0 = 0, which is worse.Alternatively, 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.Alternatively, 521 * 0 = 0, which is worse.Alternatively, 52 * 10 = 520, which is worse.Alternatively, 5 * 210 = 1050, which is the same as before.So, same result.Wait, another arrangement: 5120.Split as 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.Alternatively, 51 * 20 = 1020, which is less than 1050.Alternatively, 5 * 120 = 600, which is worse.Alternatively, 512 * 0 = 0, which is worse.So, same.Wait, another arrangement: 5021.Split as 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.Alternatively, 50 * 21 = 1050, which is the same as before.Alternatively, 5 * 021 = 105, which is worse.Alternatively, 502 * 1 = 502, which is worse.So, same.Wait, another arrangement: 5012.Split as 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.Alternatively, 50 * 12 = 600, which is worse.Alternatively, 5 * 012 = 60, which is worse.Alternatively, 501 * 2 = 1002, which is less than 1050.So, same.Wait, another arrangement: 2510.Split as 25 * 10 = 250, which is worse.Alternatively, 2 * 510 = 1020, which is less than 1050.Alternatively, 251 * 0 = 0, which is worse.Alternatively, 25 * 10 = 250, which is worse.Alternatively, 2 * 510 = 1020, which is less than 1050.Alternatively, 251 * 0 = 0, which is worse.So, same.Wait, another arrangement: 2501.Split as 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.Alternatively, 25 * 01 = 25, which is worse.Alternatively, 2 * 501 = 1002, which is less than 1050.Alternatively, 250 * 1 = 250, which is worse.So, same.Wait, another arrangement: 2150.Split as 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.Alternatively, 21 * 50 = 1050, which is the same as before.Alternatively, 2 * 150 = 300, which is worse.Alternatively, 215 * 0 = 0, which is worse.So, same.Wait, another arrangement: 2105.Split as 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.Alternatively, 21 * 05 = 105, which is worse.Alternatively, 2 * 105 = 210, which is worse.Alternatively, 210 * 5 = 1050, which is the same as before.So, same.Wait, another arrangement: 2015.Split as 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.Alternatively, 20 * 15 = 300, which is worse.Alternatively, 2 * 015 = 30, which is worse.Alternatively, 201 * 5 = 1005, which is less than 1050.So, same.Wait, another arrangement: 1520.Split as 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.Alternatively, 15 * 20 = 300, which is worse.Alternatively, 1 * 520 = 520, which is worse.Alternatively, 152 * 0 = 0, which is worse.So, same.Wait, another arrangement: 1502.Split as 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.Alternatively, 15 * 02 = 30, which is worse.Alternatively, 1 * 502 = 502, which is worse.Alternatively, 150 * 2 = 300, which is worse.So, same.Wait, another arrangement: 1052.Split as 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.Alternatively, 10 * 52 = 520, which is worse.Alternatively, 1 * 052 = 52, which is worse.Alternatively, 105 * 2 = 210, which is worse.So, same.Wait, another arrangement: 1025.Split as 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.Alternatively, 10 * 25 = 250, which is worse.Alternatively, 1 * 025 = 25, which is worse.Alternatively, 102 * 5 = 510, which is worse.So, same.Wait, another arrangement: 1205.Split as 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.Alternatively, 12 * 05 = 60, which is worse.Alternatively, 1 * 205 = 205, which is worse.Alternatively, 120 * 5 = 600, which is worse.So, same.Wait, another arrangement: 1250.Split as 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.Alternatively, 12 * 50 = 600, which is worse.Alternatively, 1 * 250 = 250, which is worse.Alternatively, 125 * 0 = 0, which is worse.So, same.Wait, I think I've exhausted all possible permutations, and the maximum product I can get is 1050.But just to be thorough, let me consider another approach.Since we have four digits, and we can split them into two numbers in different ways, let's consider all possible splits:1. Split after first digit: one-digit * three-digit2. Split after second digit: two-digit * two-digit3. Split after third digit: three-digit * one-digitFor each permutation, we can calculate these three splits and take the maximum.But since we have 24 permutations, it's a lot, but maybe we can find the maximum for each split type.For split after first digit (one-digit * three-digit):To maximize the product, we need the one-digit number to be as large as possible, and the three-digit number to be as large as possible.The largest one-digit number is 5, and the largest three-digit number from the remaining digits (2,1,0) is 210. So, 5 * 210 = 1050.Alternatively, if we take 2 as the one-digit number, the three-digit number is 510, so 2 * 510 = 1020.If we take 1 as the one-digit number, the three-digit number is 520, so 1 * 520 = 520.If we take 0 as the one-digit number, the three-digit number is 521, but 0 * 521 = 0.So, the maximum for this split is 1050.For split after second digit (two-digit * two-digit):To maximize the product, we need both two-digit numbers to be as large as possible.The two largest digits are 5 and 2, so we can form 52 and 10, but 52 * 10 = 520.Alternatively, arranging the digits as 51 and 20, which gives 51 * 20 = 1020.Alternatively, arranging the digits as 50 and 21, which gives 50 * 21 = 1050.Alternatively, arranging the digits as 25 and 10, which gives 25 * 10 = 250.Alternatively, arranging the digits as 21 and 50, which gives 21 * 50 = 1050.So, the maximum for this split is 1050.For split after third digit (three-digit * one-digit):This is similar to the first split, just reversed. The maximum would be the same as the first split, which is 1050.Therefore, regardless of the split, the maximum product is 1050.So, after considering all possible splits and permutations, the largest number Kolya can get is 1050.**Final Answer**The largest number Kolya can get is boxed{1050}.

answer：Alright, so I have this problem here where I need to find all the "4-inclusive numbers" between 0 and 100. These are numbers that are either multiples of 4 or contain the digit 4. Then, I have to arrange them in ascending order and find their sum. Hmm, okay, let me break this down step by step.First, I need to understand what exactly constitutes a "4-inclusive number." The problem says it's a number that is a multiple of 4 or contains the digit 4. So, for example, 20 is a multiple of 4, and 34 contains the digit 4. Got it. So, I need to consider both conditions: divisibility by 4 and the presence of the digit 4.Now, my range is from 0 to 100. That includes all integers from 0 up to and including 100. So, I need to check each number in this range and see if it meets either of the two conditions.I think the best approach is to separate the problem into two parts: first, find all multiples of 4 between 0 and 100, and second, find all numbers that contain the digit 4 in the same range. Then, I can combine these two sets, making sure to remove any duplicates (since some numbers might be both multiples of 4 and contain the digit 4), and finally sum them all up.Let me start with the multiples of 4. To find these, I can use the formula for an arithmetic sequence. The first multiple of 4 in the range is 0, and the last one is 100 since 100 divided by 4 is 25. So, the sequence is 0, 4, 8, 12, ..., 100.The formula for the nth term of an arithmetic sequence is a_n = a_1 + (n - 1)d, where a_1 is the first term, d is the common difference, and n is the number of terms. Here, a_1 is 0, d is 4, and a_n is 100.So, plugging in the values: 100 = 0 + (n - 1)*4. Solving for n: (n - 1)*4 = 100 => n - 1 = 25 => n = 26. So, there are 26 multiples of 4 between 0 and 100.Now, let me list these multiples to make sure I have them all:0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.Yep, that's 26 numbers. Okay, so that's my first set.Next, I need to find all numbers between 0 and 100 that contain the digit 4. This includes numbers where 4 is in the tens place or the units place. Let's break this down.First, numbers from 0 to 99. For the units place, numbers ending with 4 are 4, 14, 24, 34, 44, 54, 64, 74, 84, 94. That's 10 numbers.For the tens place, numbers from 40 to 49. That's 10 numbers: 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Wait, but 44 is already included in both lists, so I need to be careful not to double-count it. So, the total numbers containing the digit 4 from 0 to 99 are 10 (units) + 10 (tens) - 1 (overlap) = 19 numbers.But wait, hold on. Let me list them all to make sure:Units place: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94.Tens place: 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.So, combining these, we have:4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 40, 41, 42, 43, 45, 46, 47, 48, 49.Wait, that's 19 numbers. Let me count them:1. 42. 143. 244. 345. 446. 547. 648. 749. 8410. 9411. 4012. 4113. 4214. 4315. 4516. 4617. 4718. 4819. 49Yes, 19 numbers. Okay, so that's my second set.But wait, 100 is also in the range, right? So, does 100 contain the digit 4? No, it doesn't. So, 100 is only a multiple of 4, not a number containing the digit 4. So, we don't have to worry about 100 in this set.So, now I have two sets:Set A: Multiples of 4: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.Set B: Numbers containing digit 4: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 40, 41, 42, 43, 45, 46, 47, 48, 49.Now, I need to combine these two sets, removing duplicates, and then sort them in ascending order.First, let's identify the duplicates. That is, numbers that are in both Set A and Set B.Looking at Set A: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.Looking at Set B: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 40, 41, 42, 43, 45, 46, 47, 48, 49.So, comparing these, the common numbers are: 4, 24, 40, 44, 48, 64, 84.Let me list them:- 4 is in both.- 24 is in both.- 40 is in both.- 44 is in both.- 48 is in both.- 64 is in both.- 84 is in both.So, that's 7 duplicates.Therefore, the total number of unique "4-inclusive numbers" is |Set A| + |Set B| - |Duplicates| = 26 + 19 - 7 = 38 numbers.Wait, let me verify that. 26 multiples of 4, 19 numbers with digit 4, overlapping 7 numbers. So, 26 + 19 = 45, minus 7 duplicates gives 38 unique numbers. Okay, that seems right.Now, I need to list all these 38 numbers in ascending order. Hmm, that might be a bit tedious, but let's try to do it systematically.First, let's list all numbers from Set A and Set B, then sort them.But maybe a better approach is to merge the two sorted lists, removing duplicates as we go.Set A is already sorted: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.Set B is also sorted: 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94.Wait, actually, Set B isn't entirely sorted. Let me sort Set B:From the earlier list, Set B is: 4, 14, 24, 34, 44, 54, 64, 74, 84, 94, 40, 41, 42, 43, 45, 46, 47, 48, 49.Wait, that's not in order. So, to sort Set B, let's arrange them properly:4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94.Yes, that's sorted.So, now, to merge Set A and Set B, we can use a two-pointer technique, similar to merging two sorted arrays.Let me write down both sets:Set A: 0, 4, 8, 12, 16, 20, 24, 28, 32, 36, 40, 44, 48, 52, 56, 60, 64, 68, 72, 76, 80, 84, 88, 92, 96, 100.Set B: 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94.Now, let's go through each number in order, adding them to the merged list if they haven't been added already.Starting with the smallest number, which is 0 from Set A. Add 0.Next, 4 is in both Set A and Set B. Add 4.Then, 8 is only in Set A. Add 8.Next, 12 is only in Set A. Add 12.14 is only in Set B. Add 14.16 is only in Set A. Add 16.20 is only in Set A. Add 20.24 is in both. Add 24.28 is only in Set A. Add 28.32 is only in Set A. Add 32.34 is only in Set B. Add 34.36 is only in Set A. Add 36.40 is in both. Add 40.41 is only in Set B. Add 41.42 is only in Set B. Add 42.43 is only in Set B. Add 43.44 is in both. Add 44.45 is only in Set B. Add 45.46 is only in Set B. Add 46.47 is only in Set B. Add 47.48 is in both. Add 48.49 is only in Set B. Add 49.52 is only in Set A. Add 52.54 is only in Set B. Add 54.56 is only in Set A. Add 56.60 is only in Set A. Add 60.64 is in both. Add 64.68 is only in Set A. Add 68.72 is only in Set A. Add 72.74 is only in Set B. Add 74.76 is only in Set A. Add 76.80 is only in Set A. Add 80.84 is in both. Add 84.88 is only in Set A. Add 88.92 is only in Set A. Add 92.94 is only in Set B. Add 94.96 is only in Set A. Add 96.100 is only in Set A. Add 100.Okay, so compiling all these, the merged list is:0, 4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 52, 54, 56, 60, 64, 68, 72, 74, 76, 80, 84, 88, 92, 94, 96, 100.Let me count these to make sure we have 38 numbers.Counting them one by one:1. 02. 43. 84. 125. 146. 167. 208. 249. 2810. 3211. 3412. 3613. 4014. 4115. 4216. 4317. 4418. 4519. 4620. 4721. 4822. 4923. 5224. 5425. 5626. 6027. 6428. 6829. 7230. 7431. 7632. 8033. 8434. 8835. 9236. 9437. 9638. 100Yes, that's 38 numbers. Perfect.Now, the next step is to find the sum of all these numbers. That might take some time, but I can try to do it step by step.Alternatively, maybe I can find a smarter way to compute the sum without adding each number individually. Let me think.First, let's note that the merged list is a combination of two sets: multiples of 4 and numbers containing the digit 4. However, since some numbers are in both sets, we have to be careful not to double-count them. But since we already merged them into a unique list, we can just sum all the numbers in the merged list.So, perhaps the easiest way is to add them one by one, but let's see if we can group them or find a pattern to make the addition easier.Looking at the list:0, 4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 52, 54, 56, 60, 64, 68, 72, 74, 76, 80, 84, 88, 92, 94, 96, 100.I can try to pair numbers that add up to 100 or some round number to make the addition easier, but given the range, it might not be straightforward.Alternatively, I can split the list into two parts: numbers from 0 to 49 and numbers from 50 to 100, and compute the sum for each part separately.Let me try that.First, numbers from 0 to 49:0, 4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.That's 22 numbers.Numbers from 50 to 100:52, 54, 56, 60, 64, 68, 72, 74, 76, 80, 84, 88, 92, 94, 96, 100.That's 16 numbers.Wait, let me verify the counts:From 0-49: 22 numbers.From 50-100: 16 numbers.Total: 22 + 16 = 38, which matches our earlier count. Good.Now, let's compute the sum for 0-49 first.List: 0, 4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Let me add them step by step:Start with 0.0 + 4 = 44 + 8 = 1212 + 12 = 2424 + 14 = 3838 + 16 = 5454 + 20 = 7474 + 24 = 9898 + 28 = 126126 + 32 = 158158 + 34 = 192192 + 36 = 228228 + 40 = 268268 + 41 = 309309 + 42 = 351351 + 43 = 394394 + 44 = 438438 + 45 = 483483 + 46 = 529529 + 47 = 576576 + 48 = 624624 + 49 = 673.So, the sum from 0 to 49 is 673.Now, moving on to numbers from 50 to 100:52, 54, 56, 60, 64, 68, 72, 74, 76, 80, 84, 88, 92, 94, 96, 100.Let me add these as well.Starting with 52.52 + 54 = 106106 + 56 = 162162 + 60 = 222222 + 64 = 286286 + 68 = 354354 + 72 = 426426 + 74 = 500500 + 76 = 576576 + 80 = 656656 + 84 = 740740 + 88 = 828828 + 92 = 920920 + 94 = 10141014 + 96 = 11101110 + 100 = 1210.So, the sum from 50 to 100 is 1210.Therefore, the total sum of all "4-inclusive numbers" from 0 to 100 is 673 + 1210.Let me compute that: 673 + 1210.673 + 1200 = 18731873 + 10 = 1883.So, the total sum is 1883.Wait, let me verify my addition steps because sometimes when adding sequentially, errors can creep in.First, for the 0-49 range:0, 4, 8, 12, 14, 16, 20, 24, 28, 32, 34, 36, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49.Adding them step by step:Start with 0.0 + 4 = 44 + 8 = 1212 + 12 = 2424 + 14 = 3838 + 16 = 5454 + 20 = 7474 + 24 = 9898 + 28 = 126126 + 32 = 158158 + 34 = 192192 + 36 = 228228 + 40 = 268268 + 41 = 309309 + 42 = 351351 + 43 = 394394 + 44 = 438438 + 45 = 483483 + 46 = 529529 + 47 = 576576 + 48 = 624624 + 49 = 673.Yes, that seems correct.Now, for the 50-100 range:52, 54, 56, 60, 64, 68, 72, 74, 76, 80, 84, 88, 92, 94, 96, 100.Adding them step by step:Start with 52.52 + 54 = 106106 + 56 = 162162 + 60 = 222222 + 64 = 286286 + 68 = 354354 + 72 = 426426 + 74 = 500500 + 76 = 576576 + 80 = 656656 + 84 = 740740 + 88 = 828828 + 92 = 920920 + 94 = 10141014 + 96 = 11101110 + 100 = 1210.Yes, that also seems correct.So, adding 673 and 1210:673 + 1210.Let me compute this again:673 + 1210.Adding the hundreds: 600 + 1200 = 1800Adding the tens: 70 + 10 = 80Adding the units: 3 + 0 = 3Total: 1800 + 80 + 3 = 1883.So, the total sum is 1883.Wait, but let me cross-verify this with another method to be sure. Maybe using the formula for the sum of an arithmetic series for Set A and then adding the sum of Set B, subtracting the sum of duplicates.Set A is the multiples of 4 from 0 to 100. The sum of an arithmetic series is given by S = n/2 * (a_1 + a_n), where n is the number of terms, a_1 is the first term, and a_n is the last term.We already found that n = 26, a_1 = 0, a_n = 100.So, S_A = 26/2 * (0 + 100) = 13 * 100 = 1300.Set B is the numbers containing the digit 4 from 0 to 99, which we found to be 19 numbers. Let me compute their sum.List of Set B: 4, 14, 24, 34, 40, 41, 42, 43, 44, 45, 46, 47, 48, 49, 54, 64, 74, 84, 94.Let me add these:4 + 14 = 1818 + 24 = 4242 + 34 = 7676 + 40 = 116116 + 41 = 157157 + 42 = 199199 + 43 = 242242 + 44 = 286286 + 45 = 331331 + 46 = 377377 + 47 = 424424 + 48 = 472472 + 49 = 521521 + 54 = 575575 + 64 = 639639 + 74 = 713713 + 84 = 797797 + 94 = 891.So, the sum of Set B is 891.Now, the sum of duplicates is the sum of numbers that are in both Set A and Set B. We identified them as: 4, 24, 40, 44, 48, 64, 84.Let me compute their sum:4 + 24 = 2828 + 40 = 6868 + 44 = 112112 + 48 = 160160 + 64 = 224224 + 84 = 308.So, the sum of duplicates is 308.Therefore, the total sum of "4-inclusive numbers" is S_A + S_B - duplicates_sum = 1300 + 891 - 308.Compute that:1300 + 891 = 21912191 - 308 = 1883.Yes, that matches the previous result. So, the total sum is indeed 1883.Therefore, after carefully going through the process, cross-verifying the sums, I'm confident that the sum of all "4-inclusive numbers" from 0 to 100 is 1883.**Final Answer**The sum of all "4-inclusive numbers" in the range [0, 100] is boxed{1883}.