🧠:<think>First, I need to determine the increase in water level caused by placing the stone in the bowl. The water level rose from 4 cm to 9 cm, so the rise in water level is 9 cm minus 4 cm, which equals 5 cm.Next, I'll calculate the volume of water displaced by the stone. This can be done by multiplying the width, length, and the rise in water level. The width of the bowl is 16 cm, the length is 14 cm, and the rise in water level is 5 cm. So, the volume of the stone is 16 cm multiplied by 14 cm, and then multiplied by 5 cm. Calculating this gives 16 times 14 equals 224, and then 224 multiplied by 5 equals 1120.Therefore, the volume of the stone is 1120 cubic centimeters.</think>

🧠:<think>Okay, so I have this problem here where I need to find the number of 6-digit positive integers. The digits have to be in increasing order, and none of these integers can start with the digit 6. The digits we can use are from 1 to 6, inclusive. Hmm, let me try to wrap my head around this.First, let me understand what it means for digits to be in increasing order. That means each digit is strictly larger than the one before it, right? So, for example, a number like 123456 would be valid because each digit is one more than the previous. But something like 112345 wouldn't be because the first two digits are the same, and 123345 also wouldn't be because the third and fourth digits are the same. So, all digits must be unique and each subsequent digit must be larger than the previous one.Now, since we're dealing with 6-digit numbers, each number must have exactly 6 digits, and the first digit can't be 6. The digits we can use are 1 through 6, so each digit in the number must be one of these. Also, since the digits are in increasing order, once we choose the digits, their order is fixed. That simplifies things a bit because we don't have to worry about permutations; we just need to choose the right combination of digits.So, the problem boils down to selecting 6 distinct digits from 1 to 6, arranged in increasing order, without the first digit being 6. Wait, but hold on, if we're selecting 6 digits from 1 to 6, and all digits must be unique, that means we have to use each digit exactly once, right? Because there are only 6 digits available, and we need 6 digits for the number. So, the only possible 6-digit number with digits in increasing order from 1 to 6 is 123456. But that starts with 1, not 6, so it should be valid.But wait, that seems too straightforward. Maybe I'm misunderstanding the problem. Let me read it again: "the number of 6-digit positive integers where digits are in increasing order and none of these integers start with the digit 6." So, digits are from 1 to 6, inclusive, and we need 6-digit numbers where each digit is larger than the previous one, and the first digit isn't 6.Hmm, so if we have to use 6 digits, each from 1 to 6, in increasing order, the only possible number is 123456. But that starts with 1, so it's allowed. So, is the answer just 1? That seems too simple, and the problem mentions digits that can be used are from 1 to 6, so maybe repetition is allowed? Wait, no, because if digits are in increasing order, they must be strictly increasing, so repetition isn't allowed. So, each digit must be unique and strictly increasing.Wait, but if each digit must be unique and strictly increasing, and we have to use 6 digits from 1 to 6, then the only possible number is 123456. So, that would mean there's only one such number. But the problem says "none of these integers start with the digit 6," so 123456 is fine because it starts with 1. So, is the answer 1?But that seems too straightforward, and the initial problem might be more complex. Maybe I'm misinterpreting the problem. Let me think again.Wait, maybe the digits don't have to be strictly increasing? No, the problem says "digits are in increasing order," which usually means strictly increasing. So, each digit must be larger than the previous one. So, with digits from 1 to 6, the only 6-digit number possible is 123456.But then, why does the problem mention "none of these integers start with the digit 6"? Because if we have to use all 6 digits, the first digit has to be 1, right? So, maybe the problem is not requiring all digits to be used? Maybe it's just that each digit is from 1 to 6, and the digits are in increasing order, but not necessarily using all 6 digits.Wait, the problem says "6-digit positive integers," so each number must have exactly 6 digits. So, each digit is from 1 to 6, and digits are in increasing order, meaning each digit is strictly larger than the previous one, and none of these integers start with 6.So, if we have to have 6 digits, each from 1 to 6, in increasing order, and the first digit can't be 6, then how many such numbers are there?Wait, but if we have to use 6 digits from 1 to 6, and each digit is unique and in increasing order, then the only possible number is 123456. So, that's the only one. But that seems too simple, and the answer is 1. But maybe I'm missing something.Alternatively, maybe the digits can be repeated? But no, because if digits are in increasing order, they can't be repeated. So, each digit must be unique and strictly increasing.Wait, but if we have to form a 6-digit number with digits from 1 to 6, in increasing order, then the only possible number is 123456. So, that's the only one. Therefore, the number of such integers is 1.But the problem says "none of these integers start with the digit 6." So, if we had a number starting with 6, it would be invalid. But in this case, the only possible number is 123456, which starts with 1, so it's valid. So, the answer is 1.But I feel like this is too straightforward, and maybe I'm misinterpreting the problem. Let me read it again: "Let N denote the number of 6-digit positive integers where digits are in increasing order and none of these integers start with the digit 6. The digits that can be used are from 1 to 6 (inclusive). Determine the number of such integers."Wait, maybe the digits don't have to be strictly increasing? Maybe non-decreasing? But the problem says "increasing order," which usually means strictly increasing. So, each digit must be larger than the previous one.Alternatively, maybe the digits can be in non-decreasing order, allowing repeats. But the problem says "increasing order," which typically means strictly increasing. So, I think it's safe to assume that each digit must be strictly larger than the previous one.So, with that in mind, since we have to use 6 digits from 1 to 6, each exactly once, in increasing order, the only possible number is 123456. Therefore, N is 1.But wait, that seems too simple, and the problem is presented as a more complex combinatorial problem. Maybe I'm missing something. Let me think differently.Perhaps the problem is not requiring all 6 digits to be used, but just that the digits are in increasing order, and each digit is from 1 to 6, and the number is 6 digits long. So, for example, a number like 112345 would be invalid because digits are not strictly increasing, but 123456 is valid.Wait, but if we have to form a 6-digit number with digits from 1 to 6, in increasing order, then each digit must be unique and strictly increasing. So, the only possible number is 123456. So, N is 1.But again, that seems too simple. Maybe the problem allows for digits to be repeated, but in increasing order, meaning non-decreasing. So, digits can be the same or increasing. But the problem says "increasing order," which usually means strictly increasing. So, I think it's safe to assume that each digit must be strictly larger than the previous one.Wait, but if that's the case, then the only possible 6-digit number is 123456, so N is 1. But the problem also says "none of these integers start with the digit 6." So, if we had a number starting with 6, it would be invalid. But in this case, the only possible number is 123456, which starts with 1, so it's valid. So, N is 1.But I'm still not sure. Maybe the problem is allowing for digits to be in non-decreasing order, meaning repeats are allowed. So, for example, 112345 would be allowed, but digits must be in non-decreasing order. In that case, the number of such 6-digit numbers would be equal to the number of combinations with repetition.So, the formula for combinations with repetition is C(n + k - 1, k), where n is the number of types of items, and k is the number of items to choose. In this case, n is 6 (digits 1 to 6), and k is 6 (digits in the number). So, the number of such numbers would be C(6 + 6 - 1, 6) = C(11, 6) = 462.But then, the problem says "none of these integers start with the digit 6." So, we need to subtract the number of 6-digit numbers that start with 6. So, how many 6-digit numbers start with 6 and have digits in non-decreasing order?If the first digit is 6, then the remaining 5 digits must be from 6 to 6, since they have to be in non-decreasing order and can't be less than 6. So, all the remaining digits must be 6. So, the only number starting with 6 is 666666. So, there's only 1 such number.Therefore, the total number of 6-digit numbers with digits in non-decreasing order from 1 to 6 is 462, and subtracting the 1 number that starts with 6, we get 461.But wait, the problem says "digits are in increasing order," which I thought meant strictly increasing, but if it's non-decreasing, then the answer would be 461. But I'm not sure if "increasing order" allows for repeats.Wait, in mathematics, "increasing" can sometimes mean non-decreasing, but often it's used to mean strictly increasing. So, I'm a bit confused here. Let me check the problem statement again: "digits are in increasing order." Hmm, it doesn't specify strictly, so maybe it's non-decreasing.But in the initial problem, the user wrote: "digits are in increasing order and none of these integers start with the digit 6." So, if it's non-decreasing, then the number of such numbers is C(11,6) = 462, and subtracting the 1 number that starts with 6, we get 461.But wait, in the initial problem, the user wrote: "the digits that can be used are from 1 to 6 (inclusive)." So, digits can be repeated? Or not? If digits are in increasing order, does that allow repeats?Wait, in standard terminology, "increasing" usually means strictly increasing, meaning each digit is larger than the previous one. So, in that case, the number of such numbers would be C(6,6) = 1, which is 123456. But that seems too simple.Alternatively, if "increasing" allows for non-decreasing, then the number is C(11,6) = 462, and subtracting the 1 number that starts with 6, we get 461.But the problem is a bit ambiguous. Let me think about how the problem is presented. It says "digits are in increasing order," which could be interpreted as strictly increasing. So, in that case, the only possible number is 123456, so N is 1.But the problem also mentions "none of these integers start with the digit 6," which suggests that there might be numbers starting with 6, but we have to exclude them. So, if we have to exclude numbers starting with 6, then perhaps there are multiple numbers, and we have to count them accordingly.Wait, if we consider strictly increasing digits, then the first digit can be from 1 to 5, because if the first digit is 6, then the remaining digits would have to be greater than 6, but we only have digits up to 6. So, actually, if we have strictly increasing digits, the first digit can't be 6, because there are no digits after 6 to complete the 6-digit number.Wait, that makes sense. So, if we have strictly increasing digits, the first digit can be at most 1, because if the first digit is 1, the next digits can be 2,3,4,5,6. If the first digit is 2, then the next digits would have to be 3,4,5,6, but that's only 4 digits, so we can't form a 6-digit number. Similarly, if the first digit is 3, we can't form a 6-digit number because we only have 3,4,5,6 left, which is 4 digits. So, actually, the only possible 6-digit number with strictly increasing digits from 1 to 6 is 123456.Therefore, N is 1.But then, why does the problem mention "none of these integers start with the digit 6"? Because in this case, the only possible number doesn't start with 6, so it's automatically excluded. So, maybe the problem is intended to have more numbers, and perhaps the digits can be repeated.Wait, maybe the problem is allowing for digits to be in non-decreasing order, meaning repeats are allowed. So, for example, 112345 would be allowed, as long as each digit is not less than the previous one.In that case, the number of such 6-digit numbers would be equal to the number of combinations with repetition, which is C(n + k - 1, k), where n is 6 (digits 1 to 6) and k is 6 (digits in the number). So, that would be C(11,6) = 462.But then, we need to exclude the numbers that start with 6. So, how many 6-digit numbers start with 6 and have digits in non-decreasing order?If the first digit is 6, then the remaining 5 digits must be 6 or higher, but since we only have digits up to 6, all the remaining digits must be 6. So, the only number starting with 6 is 666666. Therefore, there's only 1 such number.So, subtracting that from the total, we get 462 - 1 = 461.But wait, the problem says "digits are in increasing order," which I thought meant strictly increasing, but if it's non-decreasing, then the answer is 461.But I'm still confused because the problem didn't specify whether increasing means strictly or non-decreasing. In mathematics, "increasing" can sometimes mean non-decreasing, but often it's used to mean strictly increasing. So, I'm not sure.Wait, let's think about the problem again. It says "digits are in increasing order." If it meant strictly increasing, then the only possible number is 123456, so N is 1. But if it meant non-decreasing, then N is 461.But the problem also mentions "none of these integers start with the digit 6." If we consider strictly increasing, then the first digit can't be 6, as we saw earlier, because we can't have a 6-digit number starting with 6 and having strictly increasing digits. So, in that case, the only number is 123456, which doesn't start with 6, so N is 1.But if we consider non-decreasing, then we have 462 numbers, and subtracting the 1 that starts with 6, we get 461.So, which interpretation is correct? I think the key is in the problem statement: "digits are in increasing order." If it's strictly increasing, then the answer is 1. If it's non-decreasing, the answer is 461.But the problem also says "none of these integers start with the digit 6." So, if we consider strictly increasing, then the first digit can't be 6, so we don't have to subtract anything, because the only number is 123456. So, N is 1.But if we consider non-decreasing, then we have to subtract the 1 number that starts with 6, giving us 461.So, which one is it? I think the problem is more likely to be about non-decreasing digits because otherwise, the answer is trivially 1, which seems too simple. Also, the mention of "none of these integers start with the digit 6" suggests that there might be some numbers starting with 6 that need to be excluded, which would only make sense if we're considering non-decreasing digits.Therefore, I think the correct interpretation is that digits can be in non-decreasing order, allowing repeats, and we need to count all such 6-digit numbers from 1 to 6, excluding those that start with 6.So, using the stars and bars method, the number of non-decreasing 6-digit numbers with digits from 1 to 6 is C(11,6) = 462. Then, subtract the number of such numbers that start with 6, which is 1 (666666). Therefore, the total number is 462 - 1 = 461.But wait, is that correct? Let me think again. If the first digit is 6, then all the remaining digits must be 6 as well, because they have to be greater than or equal to 6, but we can't have digits higher than 6. So, the only number starting with 6 is 666666. So, subtracting 1 from 462 gives us 461.But wait, is that the only number starting with 6? Let me think. If the first digit is 6, then the remaining 5 digits can be 6 or higher, but since the maximum digit is 6, they all have to be 6. So, yes, only one number starts with 6.Therefore, the total number of such numbers is 462 - 1 = 461.But wait, let me double-check the stars and bars calculation. The formula for combinations with repetition is C(n + k - 1, k), where n is the number of types and k is the number of items. Here, n = 6 (digits 1-6) and k = 6 (digits in the number). So, C(6 + 6 - 1, 6) = C(11,6) = 462. That seems correct.And the number of numbers starting with 6 is 1, as we established. So, 462 - 1 = 461.But wait, the problem says "none of these integers start with the digit 6." So, does that mean we have to exclude all numbers where the first digit is 6, regardless of the rest? Yes, that's what it means. So, we have to subtract all such numbers.But in the case of non-decreasing digits, the only number starting with 6 is 666666. So, subtracting 1 gives us 461.Therefore, the answer should be 461.But wait, let me think again. If the digits are in non-decreasing order, then the first digit can be from 1 to 6, but if it's 6, the rest have to be 6. So, only one number starts with 6.Therefore, the total number of numbers is 462 - 1 = 461.But wait, another way to think about it is to fix the first digit to be less than 6, and then count the number of non-decreasing sequences of length 6 where the first digit is from 1 to 5.So, for each possible first digit d (where d is from 1 to 5), the remaining 5 digits must be greater than or equal to d and less than or equal to 6.So, for each d, the number of such sequences is C(6 - d + 5, 5). Wait, no, that's not quite right.Wait, the number of non-decreasing sequences of length 5 where each digit is at least d and at most 6 is equal to the number of combinations with repetition, which is C((6 - d + 1) + 5 - 1, 5) = C(11 - d, 5).Wait, that might not be the right way to think about it.Alternatively, for each d from 1 to 5, the number of non-decreasing sequences of length 6 starting with d is equal to the number of non-decreasing sequences of length 5 with digits from d to 6.So, for each d, the number is C(6 - d + 5, 5) = C(11 - d, 5).Wait, let me test this with d=1. Then, the number of sequences is C(11 - 1,5) = C(10,5) = 252.But the total number of non-decreasing sequences starting with 1 is C(10,5) = 252.Similarly, for d=2, it's C(9,5) = 126.For d=3, C(8,5) = 56.For d=4, C(7,5) = 21.For d=5, C(6,5) = 6.For d=6, C(5,5) = 1.So, adding these up: 252 + 126 + 56 + 21 + 6 + 1 = 462, which matches the total number of non-decreasing sequences.Therefore, if we want to exclude the numbers starting with 6, we just subtract the 1 number starting with 6, giving us 462 - 1 = 461.So, the answer is 461.But wait, let me think again. If we fix the first digit to be less than 6, then the number of such numbers is the sum from d=1 to d=5 of C(11 - d,5). Which is 252 + 126 + 56 + 21 + 6 = 461.Yes, that's correct.Therefore, the number of 6-digit positive integers with digits in non-decreasing order from 1 to 6, none starting with 6, is 461.But wait, the problem says "digits are in increasing order." So, if "increasing" means strictly increasing, then the answer is 1. If it means non-decreasing, the answer is 461.I think the problem is more likely to mean non-decreasing because otherwise, the answer is trivial. Also, the mention of excluding numbers starting with 6 suggests that there are multiple numbers to exclude, but in the strictly increasing case, there are none.Therefore, I think the correct interpretation is non-decreasing, so the answer is 461.But wait, let me check the initial problem again. It says "digits are in increasing order." Hmm, in some contexts, "increasing" can mean non-decreasing, but in others, it's strictly increasing. Since the problem is about forming numbers, which are sequences, "increasing" usually means strictly increasing in the context of digits. So, maybe the answer is 1.But that seems too simple, and the problem is presented as a combinatorial problem, suggesting a more complex answer.Wait, maybe the problem is allowing for digits to be in non-decreasing order, but not necessarily strictly increasing. So, the answer is 461.But I'm still not sure. Let me think about the problem again.If the digits are in increasing order, and we have to form a 6-digit number, with digits from 1 to 6, and none starting with 6.If strictly increasing, only 123456 is possible, so N=1.If non-decreasing, then N=461.But the problem says "digits are in increasing order," which is ambiguous. However, in combinatorics, when we talk about "increasing sequences," it often means strictly increasing. So, maybe the answer is 1.But then, why mention excluding numbers starting with 6? Because in the strictly increasing case, the first digit can't be 6, so we don't have to exclude anything. So, the answer is 1.Alternatively, maybe the problem is allowing for digits to be in non-decreasing order, and the mention of "increasing order" is a bit loose.Given the ambiguity, I think the problem is more likely to be about non-decreasing digits, so the answer is 461.But to be thorough, let me consider both interpretations.Case 1: Strictly increasing digits.In this case, each digit must be unique and in increasing order. So, the only possible 6-digit number is 123456. Since it doesn't start with 6, N=1.Case 2: Non-decreasing digits.In this case, digits can repeat, but each digit is greater than or equal to the previous one. The number of such numbers is C(11,6)=462. Subtracting the 1 number that starts with 6, we get 461.Given that the problem is presented as a combinatorial problem, and the mention of excluding numbers starting with 6, I think the intended interpretation is non-decreasing digits, so the answer is 461.Therefore, the number of such integers is 461.But wait, let me double-check the stars and bars calculation.The number of non-decreasing sequences of length k with elements from a set of size n is C(n + k - 1, k). Here, n=6 (digits 1-6), k=6. So, C(6+6-1,6)=C(11,6)=462.Yes, that's correct.Then, the number of sequences starting with 6 is 1, as established earlier.Therefore, 462 - 1 = 461.Yes, that seems correct.So, the answer is 461.But wait, let me think about it another way. If we fix the first digit to be less than 6, then the remaining 5 digits can be anything from that first digit to 6, in non-decreasing order.So, for each possible first digit d (1 to 5), the number of sequences is C(6 - d + 5,5). Wait, no, that's not quite right.Wait, for each d, the number of non-decreasing sequences of length 5 with digits from d to 6 is C(6 - d + 5,5). Wait, that doesn't seem correct.Wait, the number of non-decreasing sequences of length 5 with digits from d to 6 is equal to the number of combinations with repetition, which is C(6 - d + 5,5). Wait, no, that's not correct.Wait, the formula is C(n + k - 1, k), where n is the number of types and k is the number of items. Here, n=6 - d + 1 (digits from d to 6), and k=5 (remaining digits). So, the number is C((6 - d + 1) + 5 - 1,5) = C(11 - d,5).So, for d=1, it's C(10,5)=252.For d=2, C(9,5)=126.For d=3, C(8,5)=56.For d=4, C(7,5)=21.For d=5, C(6,5)=6.Adding these up: 252 + 126 + 56 + 21 + 6 = 461.Yes, that's correct.Therefore, the total number of such numbers is 461.So, I think that's the answer.</think>

🧠:<think>Alright, so I have these five problems to solve. Let me take them one by one and try to figure them out step by step.Problem 1: Toba-n FractionsOkay, so a Toba-n fraction is just 1/n, right? And I need to sum these from n=101 to n=200. The problem wants me to show that this sum, which they call A, is between 7/12 and 5/6. Hmm, okay.First, I remember that the sum of reciprocals of numbers from 1 to N is called the harmonic series, and it grows kind of slowly. But here, I'm only summing from 101 to 200. Maybe I can approximate this sum using integrals? I think the integral of 1/x from a to b is ln(b) - ln(a), which gives an approximation for the sum.Let me try that. So, the sum from 101 to 200 of 1/n should be approximately ln(200) - ln(100). Wait, but actually, since the sum starts at 101, maybe it's better to use ln(201) - ln(100). Because the integral from 100 to 200 would be ln(200) - ln(100), but since we're starting at 101, it's more accurate to go from 101 to 200, which would be ln(200) - ln(101). Hmm, okay.Let me calculate ln(200) and ln(101). I know ln(100) is about 4.605, so ln(200) would be ln(2*100) = ln(2) + ln(100) ≈ 0.693 + 4.605 ≈ 5.298. And ln(101) is a bit more than ln(100), so maybe around 4.615.So, ln(200) - ln(101) ≈ 5.298 - 4.615 ≈ 0.683. Now, 7/12 is approximately 0.583 and 5/6 is approximately 0.833. So, 0.683 is between 0.583 and 0.833, which means A is between 7/12 and 5/6. That seems to work.But wait, is this approximation accurate enough? Maybe I should consider that the integral is a lower bound for the sum, and the actual sum is a bit higher. So, perhaps the sum is actually a little more than 0.683, but still less than 0.833. Yeah, that makes sense. So, I think this works.Problem 2: System of EquationsAlright, this one looks tricky. We have three equations:1. (ab)/(a + b) = 1/22. (bc)/(b + c) = 1/33. (ac)/(a + c) = 1/7And we need to find (a - c)^b.Hmm, okay. Let me see. Maybe I can rewrite these equations in terms of reciprocals. Let me denote x = 1/a, y = 1/b, z = 1/c. Then, the equations become:1. (1/x + 1/y) = 22. (1/y + 1/z) = 33. (1/x + 1/z) = 7Wait, no. Let me check. If (ab)/(a + b) = 1/2, then 1/(1/a + 1/b) = 1/2, so 1/a + 1/b = 2. Similarly, 1/b + 1/c = 3, and 1/a + 1/c = 7.So, we have:1. x + y = 22. y + z = 33. x + z = 7Where x = 1/a, y = 1/b, z = 1/c.Now, I can solve this system of equations. Let me write them down:1. x + y = 22. y + z = 33. x + z = 7Let me subtract equation 1 from equation 3: (x + z) - (x + y) = 7 - 2 ⇒ z - y = 5.From equation 2, y + z = 3. So, we have:z - y = 5z + y = 3Adding these two equations: 2z = 8 ⇒ z = 4.Then, from z + y = 3 ⇒ y = 3 - z = 3 - 4 = -1.From equation 1, x + y = 2 ⇒ x = 2 - y = 2 - (-1) = 3.So, x = 3, y = -1, z = 4.Therefore, a = 1/x = 1/3, b = 1/y = -1, c = 1/z = 1/4.So, a = 1/3, b = -1, c = 1/4.Now, we need to find (a - c)^b.Compute a - c: 1/3 - 1/4 = (4 - 3)/12 = 1/12.Then, (1/12)^b = (1/12)^{-1} = 12.So, the value is 12.Wait, but b is -1, so (a - c)^b = (1/12)^{-1} = 12. Yeah, that seems right.Problem 3: Area of Quadrilateral PMCNGiven triangle ABC, with M as the midpoint of AC, N as the midpoint of BC, and P any point on AB. We need to find the area of quadrilateral PMCN.Hmm, okay. Since M and N are midpoints, maybe we can use properties of midsegments in triangles.First, let me recall that the midsegment in a triangle is parallel to the third side and half its length. So, MN is a midsegment, so MN is parallel to AB and MN = (1/2)AB.But P is any point on AB. So, the quadrilateral PMCN is formed by points P, M, C, N.Wait, let me visualize this. Triangle ABC, with M midpoint of AC, N midpoint of BC. So, connecting P on AB to M and N.Hmm, maybe I can express the area in terms of the area of ABC.Let me denote the area of ABC as Δ.Since M and N are midpoints, the area of triangle MNC is 1/4 of Δ, because connecting midpoints divides the triangle into four smaller triangles of equal area.But PMCN is a quadrilateral. Maybe it's the area of ABC minus the areas of triangles APM and BPN.Wait, not sure. Alternatively, since M and N are midpoints, maybe PMCN is a trapezoid or something.Alternatively, perhaps the area is always half of the area of ABC, regardless of where P is on AB.Wait, let me think. If P is at A, then quadrilateral PMCN becomes AMCN, which is half of ABC. Similarly, if P is at B, then PMCN becomes BMNC, which is also half of ABC. So, maybe regardless of where P is, the area is always half.So, the area of PMCN is (1/2)Δ.Yeah, that seems plausible.Problem 4: Number of Tracks on the PlaneWe have a particle moving on the xy-plane with two rules:N: (m, n) → (m + 1, n + 1)T: (m, n) → (m + 1, n - 1)We need to find the number of different tracks from (0, 3) to (7, 2).So, each move increases m by 1 and either increases or decreases n by 1. So, from (0, 3), after 7 moves, we need to reach (7, 2). So, the total change in m is +7, which is achieved by 7 moves. The total change in n is from 3 to 2, so a change of -1.Each N move increases n by 1, each T move decreases n by 1. So, if we have x N moves and y T moves, then:x + y = 7 (total moves)x - y = -1 (net change in n)Solving these equations:From x + y = 7 and x - y = -1, adding them gives 2x = 6 ⇒ x = 3, so y = 4.So, we need 3 N moves and 4 T moves. The number of different tracks is the number of ways to arrange these moves, which is the combination C(7, 3) = 35.So, there are 35 different tracks.Problem 5: Marble Game StrategyAndra and Dedi are playing a game with 30 marbles. They take turns, each can take 1 to 6 marbles. The person who takes the last marble wins. Andra starts by taking 3 marbles. Dedi needs to find a strategy to win.So, the key here is to force Andra into a position where she cannot avoid losing. The standard strategy for such games is to leave the opponent with a multiple of (1 + maximum take) marbles, which in this case is 7.So, after Andra takes 3 marbles, there are 27 left. Dedi should take 6 marbles, leaving 21, which is a multiple of 7. Then, whatever Andra takes (say y marbles), Dedi takes 7 - y marbles, maintaining the multiple of 7. This way, Dedi can always force Andra to take the last marble.So, Dedi should take 6 marbles first, then mirror Andra's moves to keep the total taken each round to 7.Final AnswerProblem 1: boxed{frac{7}{12} < A < frac{5}{6}}Problem 2: boxed{12}Problem 3: boxed{frac{1}{2} Delta}Problem 4: boxed{35}Problem 5: Dedi should take boxed{6} marbles first.</think>

🧠:<think>Okay, so I need to prove that when n is greater than or equal to 3, and n is a natural number, then 2 to the power of n is greater than or equal to 2 times (n plus 1). Hmm, let's see. I think I should start by understanding what this inequality is saying. It's comparing an exponential function, 2^n, with a linear function, 2(n+1). Since exponential functions grow much faster than linear ones, I suspect this inequality will hold true for n >= 3.First, maybe I should test the inequality for n = 3 to see if it's true. Let's calculate both sides:For n = 3:Left side: 2^3 = 8Right side: 2*(3 + 1) = 2*4 = 8So, 8 >= 8, which is true. Okay, that's a good start.Now, what about n = 4?Left side: 2^4 = 16Right side: 2*(4 + 1) = 2*5 = 1016 >= 10, which is also true.n = 5:Left side: 2^5 = 32Right side: 2*(5 + 1) = 2*6 = 1232 >= 12, definitely true.So, it seems like the inequality holds for these values. But I need to prove it for all natural numbers n >= 3, not just specific cases. Maybe I can use mathematical induction for this.Mathematical induction has two main steps: the base case and the inductive step. I've already checked the base case for n = 3, and it holds true. Now, I need to assume that the inequality holds for some arbitrary natural number k >= 3, and then show that it must also hold for k + 1.So, let's assume that for n = k, 2^k >= 2(k + 1). This is my inductive hypothesis. Now, I need to show that 2^(k + 1) >= 2((k + 1) + 1) = 2(k + 2).Starting from the left side of the inequality for n = k + 1:2^(k + 1) = 2 * 2^kFrom the inductive hypothesis, I know that 2^k >= 2(k + 1). So, substituting that in:2^(k + 1) = 2 * 2^k >= 2 * 2(k + 1) = 4(k + 1)Now, I need to compare 4(k + 1) with 2(k + 2). Let's see:4(k + 1) = 4k + 42(k + 2) = 2k + 4Subtracting 2k + 4 from both sides:4k + 4 - (2k + 4) = 2kSince k >= 3, 2k >= 6, which is positive. Therefore, 4(k + 1) >= 2(k + 2) because 4(k + 1) - 2(k + 2) = 2k >= 6 > 0.Thus, 2^(k + 1) >= 4(k + 1) >= 2(k + 2). Therefore, the inequality holds for n = k + 1.By the principle of mathematical induction, since the base case holds and the inductive step has been proven, the inequality 2^n >= 2(n + 1) holds for all natural numbers n >= 3.Wait, let me double-check my inductive step. I assumed that 2^k >= 2(k + 1), then multiplied both sides by 2 to get 2^(k + 1) >= 4(k + 1). Then I compared 4(k + 1) with 2(k + 2). Is that correct?Yes, because I need to show that 2^(k + 1) >= 2(k + 2). So, if I can show that 4(k + 1) >= 2(k + 2), then since 2^(k + 1) >= 4(k + 1), it follows that 2^(k + 1) >= 2(k + 2).Simplifying 4(k + 1) >= 2(k + 2):Divide both sides by 2: 2(k + 1) >= (k + 2)Which simplifies to: 2k + 2 >= k + 2Subtract k + 2 from both sides: k >= 0Since k >= 3, this inequality holds true. Therefore, my inductive step is valid.So, putting it all together, the base case holds for n = 3, and the inductive step has been proven, so by induction, the inequality 2^n >= 2(n + 1) is true for all natural numbers n >= 3.</think>