💡Okay, so I have this problem about tetrahedrons and spheres that touch all the planes of their faces. I need to prove that for any tetrahedron, there are at least 5 and no more than 8 such spheres. Hmm, interesting. Let me try to break this down.First, I know that a tetrahedron has four triangular faces. A sphere that touches all the planes of its faces is called a tangent sphere, right? So, these spheres can be either inscribed or exscribed relative to the tetrahedron. I remember that for a tetrahedron, there's one inscribed sphere that touches all four faces from the inside. But what about the exscribed ones?Wait, in 2D, for a triangle, there are four circles: one inscribed and three exscribed. Each exscribed circle touches one side and the extensions of the other two sides. So, maybe in 3D, for a tetrahedron, there are more possibilities?Let me think. If I consider each face of the tetrahedron, maybe I can have spheres that are tangent to that face and the extensions of the other three faces. That would be similar to the exscribed circles in 2D. So, for each face, there might be an exscribed sphere. That would give me four exscribed spheres, right?Adding the one inscribed sphere, that makes five spheres in total. So, that's the minimum number, five. But the problem says there can be up to eight. How does that happen?Hmm, maybe there are more possibilities when considering different combinations of signs for the distances from the sphere's center to each face. Let me recall the formula for the radius of an inscribed sphere. It's related to the volume and the surface areas.The formula is ( r = frac{3V}{S_1 + S_2 + S_3 + S_4} ), where ( V ) is the volume and ( S_i ) are the areas of the faces. But for exscribed spheres, the formula might involve subtracting some areas instead of adding. So, maybe each exscribed sphere corresponds to a different combination of adding and subtracting the areas.Wait, so if I consider each face, I can have a sphere that's tangent to that face and the extensions of the other three. That would mean for each face, I subtract its area and add the others. So, that gives four exscribed spheres. Plus the inscribed one, that's five.But how do we get up to eight? Maybe there are more combinations where we subtract more than one area? Let me think. If I subtract two areas and add the other two, does that give a valid sphere?But wait, if I subtract two areas, then the sum ( sum varepsilon_i S_i ) might not necessarily be positive. The formula for the radius requires that the sum is positive, right? So, if I have ( sum varepsilon_i S_i > 0 ), then the radius is positive, which is meaningful.So, for each combination of signs ( varepsilon_i = pm 1 ), if the sum ( sum varepsilon_i S_i ) is positive, then we have a valid sphere. How many such combinations are there?There are four faces, so each can have a +1 or -1, giving ( 2^4 = 16 ) possible combinations. But not all of these will result in a positive sum. In fact, for each combination, if the sum is positive, then the opposite combination (where all signs are flipped) will have a negative sum, which isn't valid. So, the number of valid spheres is at most half of 16, which is 8.That makes sense. So, the maximum number of spheres is 8. But why is the minimum 5?Well, we always have the inscribed sphere, which corresponds to all ( varepsilon_i = +1 ). Then, for each face, we can have an exscribed sphere where ( varepsilon_i = -1 ) for that face and +1 for the others. That gives four more spheres, totaling five.But why can't we have more than five? Because sometimes, when we try to subtract two areas, the sum might not be positive. For example, if the sum of two areas is greater than the sum of the other two, then subtracting those two would result in a negative sum, which isn't allowed.Wait, so if the areas are such that for any two faces, the sum of their areas isn't equal to the sum of the other two, then all combinations where we subtract one or two areas might still result in a positive sum. But if the areas are such that some combinations lead to a negative sum, then those spheres wouldn't exist.So, the number of spheres depends on the specific areas of the faces. If the areas are such that all possible combinations (except the ones that would result in a negative sum) give a positive sum, then we can have up to eight spheres. But if some combinations lead to a negative sum, then we have fewer spheres.Therefore, the minimum number is five because we always have the inscribed sphere and four exscribed spheres. The maximum is eight because there are eight possible valid combinations where the sum is positive.Wait, but how do we know that there are at least five? What if some of the exscribed spheres don't exist? For example, if subtracting one area makes the sum negative, then that sphere wouldn't exist. But I think in a tetrahedron, each face's area is less than the sum of the other three. So, subtracting one area would still leave a positive sum.Let me check that. Suppose ( S_1 < S_2 + S_3 + S_4 ). Then, ( S_2 + S_3 + S_4 - S_1 > 0 ). So, the exscribed sphere corresponding to subtracting ( S_1 ) exists. Similarly for the other faces. So, each exscribed sphere exists because each face's area is less than the sum of the other three.Therefore, we always have at least five spheres: one inscribed and four exscribed. For the maximum, it's eight, but that requires that all combinations where we subtract one or two areas still result in a positive sum. If the areas are such that subtracting two areas doesn't make the sum negative, then we can have more spheres.But in general, for any tetrahedron, the areas might not satisfy that condition. So, the maximum is eight, but it's not always achieved. However, the problem states "no more than 8," so that's consistent.So, putting it all together, for any tetrahedron, there are at least five and no more than eight spheres that touch all the planes of its faces. The five include the inscribed sphere and four exscribed spheres, while the eight comes from all possible valid combinations of signs in the formula for the radius.I think that makes sense. I should probably write this up more formally, but that's the gist of it.

💡First, I need to determine the sampling ratio used for selecting students. There are 210 freshmen, and 7 were selected. To find the ratio, I divide the total number of freshmen by the number selected:[frac{210}{7} = 30]This means that 1 out of every 30 students is selected.Next, I apply this ratio to the number of seniors. There are 300 seniors, so the number of seniors selected is:[frac{300}{30} = 10]Therefore, 10 seniors should be selected.

💡First, I need to evaluate the inner function ( f(4) ) using the given function ( f(x) = log_{2}x ).Calculating ( f(4) ):[f(4) = log_{2}4 = 2]Next, I use the result from the first step to evaluate ( f(f(4)) ), which is ( f(2) ).Calculating ( f(2) ):[f(2) = log_{2}2 = 1]Therefore, the value of ( f(f(4)) ) is 1.

💡Alright, so I need to figure out how many positive integer multiples of 143 can be expressed in the form (10^{j} - 10^{i}), where (i) and (j) are integers with (0 leq i < j leq 99). Hmm, okay. Let me break this down step by step.First, I know that 143 is a composite number. Let me factorize it to see its prime components. 143 divided by 11 is 13, right? So, (143 = 11 times 13). That means any multiple of 143 must be divisible by both 11 and 13.The expression given is (10^{j} - 10^{i}). I can factor out (10^{i}) from both terms, which gives me (10^{i}(10^{j-i} - 1)). So, the expression simplifies to (10^{i} times (10^{k} - 1)), where (k = j - i). Since (i < j), (k) is a positive integer.Now, since 143 divides (10^{j} - 10^{i}), and (10^{i}) is a power of 10, which is co-prime with 143 (because 10 and 143 share no common factors other than 1), it must be that 143 divides (10^{k} - 1). So, the key here is to find all (k) such that (10^{k} equiv 1 mod 143).To find such (k), I need to determine the order of 10 modulo 143. The order of 10 modulo a number is the smallest positive integer (k) such that (10^{k} equiv 1 mod) that number. Since 143 is the product of two primes, 11 and 13, I can use the Chinese Remainder Theorem to find the order modulo 143 by finding the orders modulo 11 and 13 separately and then taking the least common multiple (LCM) of those orders.Let me find the order of 10 modulo 11 first. I know that 10 modulo 11 is -1, so (10^1 equiv -1 mod 11), and (10^2 equiv 1 mod 11). So, the order of 10 modulo 11 is 2.Next, the order of 10 modulo 13. Let me compute powers of 10 modulo 13 until I get back to 1. - (10^1 mod 13 = 10)- (10^2 mod 13 = 100 mod 13 = 9)- (10^3 mod 13 = 10 times 9 = 90 mod 13 = 12)- (10^4 mod 13 = 10 times 12 = 120 mod 13 = 3)- (10^5 mod 13 = 10 times 3 = 30 mod 13 = 4)- (10^6 mod 13 = 10 times 4 = 40 mod 13 = 1)So, the order of 10 modulo 13 is 6.Now, the order of 10 modulo 143 is the LCM of the orders modulo 11 and 13, which are 2 and 6, respectively. The LCM of 2 and 6 is 6. Therefore, the smallest (k) such that (10^{k} equiv 1 mod 143) is 6. That means (k) must be a multiple of 6 for (10^{k} - 1) to be divisible by 143.So, (k = j - i) must be a multiple of 6. Let me denote (k = 6m), where (m) is a positive integer. Now, I need to find all pairs ((i, j)) such that (j - i = 6m) and (0 leq i < j leq 99).Let me consider the possible values of (m). Since (j) can be at most 99 and (i) is at least 0, the maximum value of (k = j - i) is 99 - 0 = 99. Therefore, the maximum (m) such that (6m leq 99) is (m = lfloor 99 / 6 rfloor = 16), because (6 times 16 = 96), and (6 times 17 = 102) which is greater than 99.So, (m) can range from 1 to 16. For each (m), the number of valid pairs ((i, j)) is equal to the number of integers (i) such that (i + 6m leq 99). That is, (i) can range from 0 to (99 - 6m). Therefore, the number of pairs for each (m) is (100 - 6m).Wait, hold on. Let me double-check that. If (k = 6m), then (j = i + 6m). Since (j) must be less than or equal to 99, (i) must be less than or equal to (99 - 6m). Therefore, the number of possible (i) values is (100 - 6m), because (i) can be 0, 1, ..., up to (99 - 6m). So, yes, the number of pairs for each (m) is (100 - 6m).But wait, when (m = 1), (k = 6), so (i) can be from 0 to 93, which gives 94 pairs. Similarly, for (m = 2), (k = 12), (i) can be from 0 to 87, giving 88 pairs. Continuing this way, for (m = 16), (k = 96), (i) can be from 0 to 3, giving 4 pairs.So, the number of pairs for each (m) is decreasing by 6 each time: 94, 88, 82, ..., 4. This is an arithmetic sequence where the first term (a_1 = 94), the common difference (d = -6), and the last term (a_n = 4).I need to find how many terms are in this sequence. Let me use the formula for the nth term of an arithmetic sequence:(a_n = a_1 + (n - 1)d)Plugging in the known values:(4 = 94 + (n - 1)(-6))Solving for (n):(4 = 94 - 6(n - 1))Subtract 94 from both sides:(-90 = -6(n - 1))Divide both sides by -6:(15 = n - 1)Add 1 to both sides:(n = 16)So, there are 16 terms in this sequence. Now, to find the total number of pairs, I need to sum all these terms. The sum (S) of an arithmetic sequence is given by:(S = frac{n}{2} times (a_1 + a_n))Plugging in the values:(S = frac{16}{2} times (94 + 4) = 8 times 98 = 784)Therefore, there are 784 valid pairs ((i, j)) such that (10^{j} - 10^{i}) is a multiple of 143.Wait a second, let me make sure I didn't make a mistake in calculating the number of terms. When (m = 1), (k = 6), and the number of pairs is 94. Then, for each subsequent (m), the number decreases by 6. So, the sequence is 94, 88, 82, ..., 4. The difference between each term is -6, and the first term is 94, last term is 4.To find the number of terms, I can also think of it as how many times I subtract 6 from 94 until I reach 4. Let me compute:94 - 6*(n-1) = 4So, 94 - 4 = 6*(n-1)90 = 6*(n-1)n - 1 = 15n = 16Yes, that's consistent with what I found earlier. So, 16 terms.Then, the sum is indeed 16/2*(94 + 4) = 8*98 = 784.Therefore, the total number of positive integer multiples of 143 that can be expressed in the form (10^{j} - 10^{i}) with (0 leq i < j leq 99) is 784.I think that's solid. I considered the factorization, found the order, determined the necessary condition on (k), calculated the number of valid pairs for each possible (k), recognized the arithmetic sequence, and summed it up correctly. I don't see any mistakes in my reasoning.**Final Answer**boxed{784}

💡First, I recognize that any non-zero number raised to the power of 0 equals 1. Therefore, ((sqrt{3} - 1)^0 = 1).Next, I understand that (2^{-1}) is the reciprocal of 2, which is (frac{1}{2}).Finally, I add the two results together: (1 + frac{1}{2} = frac{3}{2}).

💡First, I need to determine the total number of towels Maria initially bought by adding the green and white towels together.Next, I will subtract the number of towels she gave to her mother from the total to find out how many towels she has left.Finally, I will present the final answer clearly.