answer：Okay, so I have this problem here: If the arc length corresponding to a central angle of 1 radian is 6, then the area of the sector formed by this central angle is _______. Hmm, let me think about how to approach this.First, I remember that the arc length (s) in a circle is related to the radius (r) and the central angle (θ) in radians by the formula s = rθ. Since the central angle here is 1 radian, that should make things simpler.Given that s = 6 and θ = 1 radian, I can plug these values into the formula to find the radius. So, 6 = r * 1, which means r = 6. Okay, so the radius of the circle is 6 units. That seems straightforward.Now, the problem is asking for the area of the sector formed by this central angle. I recall that the area (A) of a sector is given by the formula A = (1/2) * r² * θ. Let me make sure I remember that correctly. Yes, it's similar to the formula for the area of a circle, but scaled by the angle over the full angle of 2π radians. So, for a sector, it's (θ / 2π) * πr², which simplifies to (1/2) * r² * θ. That makes sense.So, plugging in the values I have, r = 6 and θ = 1 radian. Let me compute that step by step. First, square the radius: 6 squared is 36. Then, multiply that by the angle: 36 * 1 = 36. Finally, take half of that: (1/2) * 36 = 18. So, the area of the sector should be 18.Wait, let me double-check that. If the arc length is 6, which equals r * θ, and θ is 1, then r is indeed 6. Then, the area formula is (1/2) * r² * θ, which is (1/2) * 36 * 1, which is 18. Yeah, that seems right.Is there another way to think about this? Maybe using proportions. The entire circle has an area of πr², which would be π*36 = 36π. Since the central angle is 1 radian, and the full circle is 2π radians, the sector is (1 / 2π) of the entire circle. So, the area would be (1 / 2π) * 36π = 36π / 2π = 18. Yep, same result. That confirms it.I think I'm confident that the area is 18. I don't see any mistakes in my reasoning. The key was remembering the correct formulas for arc length and sector area, and then plugging in the known values correctly. It's always good to check using a different method to ensure consistency.So, to recap: Arc length formula gives radius, which is 6. Then, sector area formula with radius 6 and angle 1 radian gives 18. Alternatively, proportion of the circle's area gives the same result. So, the answer should be 18.**Final Answer**The area of the sector is boxed{18}.

answer：Okay, so I have this problem where there's a 100x100 table, and each cell has a natural number. The condition is that each number is either greater than all its neighbors or less than all its neighbors. Neighbors are cells that share a common side, so each cell can have up to four neighbors. The question is asking for the smallest possible sum of all these numbers.First, I need to understand the problem better. Each cell is either a local maximum or a local minimum. That means every cell is either higher than all its adjacent cells or lower than all of them. So, in other words, the table is made up entirely of peaks and valleys, with no flat areas or slopes.I wonder if there's a pattern or a configuration that can minimize the total sum. Since we're dealing with natural numbers, the smallest possible number is 1. So, maybe using 1s and 2s in some alternating pattern could work? Let's think about smaller grids first to get an idea.Let's consider a 2x2 grid. If I place 1s and 2s in a checkerboard pattern, each 1 would be surrounded by 2s, and each 2 would be surrounded by 1s. So, in this case, each 1 is a local minimum, and each 2 is a local maximum. The sum would be 1+2+1+2=6. Alternatively, if I use all 1s, then each cell would be equal to its neighbors, which doesn't satisfy the condition because they need to be either greater or less than all neighbors. So, a checkerboard pattern seems necessary.Extending this idea to a 3x3 grid. If I try a checkerboard pattern, the center cell would have four neighbors. If the center is 1, then all four surrounding cells must be 2. But then, each of those 2s would have neighbors: one 1 in the center and two or three other 2s. Wait, that's a problem because a 2 cannot be both greater than the center 1 and less than or equal to its other neighbors. So, in a 3x3 grid, a simple checkerboard might not work because the corner cells would have two 2s and one 1, which violates the condition for the 2s.Hmm, so maybe the checkerboard pattern works for even-sized grids but not for odd-sized ones? Or perhaps there's a different pattern needed. Let's think about a 3x3 grid. Maybe using 1s and 2s in a way that each 1 is surrounded by 2s, and each 2 is surrounded by 1s. But in a 3x3 grid, the center cell is surrounded by four cells. If the center is 1, then all four surrounding cells must be 2. But then, each of those 2s is adjacent to two other 2s, which would make them not strictly greater or less than all neighbors. So that doesn't work.Wait, maybe the 3x3 grid can't have a checkerboard pattern because of the odd size. So perhaps the minimal configuration is different. Maybe using 1s and 2s in a way that every 2 is only adjacent to 1s. So, in a 3x3 grid, maybe place 1s on the corners and center, and 2s on the edges. Let's see:1 2 12 1 21 2 1In this case, each 1 is surrounded by 2s, so they are local minima. Each 2 is surrounded by 1s and other 2s. Wait, the edge 2s are adjacent to two 1s and two 2s. So, the 2s are not all greater than their neighbors because they are equal to some neighbors. So, that doesn't satisfy the condition either.Hmm, maybe in a 3x3 grid, it's impossible to have such a configuration with only 1s and 2s? Or perhaps we need to use higher numbers. Let's try:1 3 13 2 31 3 1Here, the center is 2, which is less than all its neighbors (3s). The 3s are greater than their 1 and 2 neighbors. The 1s are less than all their neighbors (3s). So, this works. The sum is 1+3+1 +3+2+3 +1+3+1 = 1*4 + 3*4 + 2 = 4 + 12 + 2 = 18.Alternatively, if I use 1s and 2s, but arrange them differently. Maybe:1 2 12 1 21 2 1But as before, the 2s are adjacent to other 2s, which violates the condition. So, perhaps in odd-sized grids, we need to introduce a higher number in the center to satisfy the condition.So, for the 3x3 grid, the minimal sum is 18 with numbers 1, 2, and 3. Maybe in larger grids, a similar approach is needed?Wait, but the original problem is a 100x100 grid, which is even-sized. So, maybe a checkerboard pattern works here without issues. Let's test that idea.In a checkerboard pattern, each cell alternates between 1 and 2. So, in a 100x100 grid, half of the cells would be 1s and half would be 2s. Since 100 is even, there are exactly 5000 cells of each. So, the total sum would be 5000*1 + 5000*2 = 5000 + 10000 = 15000.But wait, let's verify if this satisfies the condition. Each 1 is surrounded by 2s, so each 1 is a local minimum. Each 2 is surrounded by 1s, so each 2 is a local maximum. Yes, that works. So, the sum is 15000.But is this the minimal possible? Or can we do better?Wait, perhaps using more 1s and fewer 2s? But in a checkerboard pattern, every 1 is adjacent only to 2s, and every 2 is adjacent only to 1s. So, we can't have more 1s without violating the condition because two 1s would be adjacent, making them neither greater nor less than each other.Alternatively, maybe using a different pattern where some cells are higher than their neighbors but not necessarily alternating. But I think the checkerboard pattern is the most efficient way to minimize the sum because it uses the smallest possible numbers, 1 and 2, in a way that satisfies the condition.Wait, but in the 3x3 grid, we saw that a checkerboard pattern doesn't work because of the center cell. But in an even-sized grid, like 2x2 or 100x100, the checkerboard pattern works perfectly because every cell has neighbors of the opposite color, so no two same numbers are adjacent.Therefore, for the 100x100 grid, the minimal sum is achieved by a checkerboard pattern of 1s and 2s, resulting in a total sum of 15000.But hold on, let me think again. In the 3x3 grid, the minimal sum was 18, which is 4*1 + 4*3 + 1*2. But if I use a checkerboard pattern, I can't get a lower sum because of the constraints. So, perhaps in the 100x100 grid, the checkerboard is indeed the minimal.Wait, another thought: maybe using 1s and 0s? But the problem states natural numbers, which typically start at 1. So, 0 is not allowed. So, the smallest number we can use is 1.Therefore, the minimal sum is 15000.But let me check if there's a way to have more 1s without violating the condition. For example, if I have a 1, then all its neighbors must be greater than 1, so at least 2. But if I have a 2, its neighbors can be 1 or higher. But if a 2 is adjacent to another 2, then that 2 is not greater than all its neighbors, so that's not allowed. Therefore, 2s cannot be adjacent to each other. So, the only way to arrange 1s and 2s without having two 2s adjacent is the checkerboard pattern.Therefore, the minimal sum is indeed 15000.Wait, but in the 3x3 grid, the checkerboard pattern doesn't work because of the center cell. So, maybe in the 100x100 grid, the checkerboard works because it's even-sized, but in odd-sized grids, it doesn't. So, perhaps in even-sized grids, the checkerboard is the minimal configuration.Yes, that makes sense. So, for a 100x100 grid, the minimal sum is 15000.But let me think again. Is there a way to have some cells with 1, some with 2, and some with higher numbers, but overall resulting in a lower total sum? For example, if some cells are 3, but that might require more 1s or 2s, but perhaps the overall sum is lower.Wait, no, because 3 is larger than 2, so replacing a 2 with a 3 would increase the sum. So, using higher numbers would only increase the total sum, not decrease it. Therefore, the minimal sum is achieved by using the smallest possible numbers, which are 1 and 2, arranged in a checkerboard pattern.Therefore, the minimal sum is 15000.But wait, let me think about the 1x1 grid. If it's just one cell, it has no neighbors, so it can be 1. The sum is 1.For a 2x2 grid, checkerboard: 1,2,2,1. Sum is 6.For a 3x3 grid, as we saw, it's 18.So, for even n, the sum is (n^2)/2 * (1 + 2) = (n^2)/2 * 3.Wait, for 2x2, that would be (4)/2 *3=6, which matches.For 100x100, it's (10000)/2 *3=15000, which matches.So, that seems consistent.Therefore, the minimal sum is 15000.**Final Answer**The smallest possible sum is boxed{15000}.

answer：Alright, so I've got this problem here where I need to divide the set {1, 2, 3, ..., 1998} into 999 disjoint pairs {a_i, b_i}, and for each pair, the absolute difference |a_i - b_i| is either 1 or 6. Then, I have to prove that the last digit of the sum of all these differences is 9. Hmm, okay, let's break this down step by step.First, let me understand the problem. We have numbers from 1 to 1998, which is 1998 numbers in total. We need to pair them up into 999 pairs, each pair consisting of two numbers where their difference is either 1 or 6. So, each pair is either consecutive numbers (difference 1) or numbers that are 6 apart (difference 6). Then, we sum up all these differences and look at the last digit of that total sum. The claim is that this last digit must be 9, regardless of how we pair the numbers, as long as the differences are 1 or 6.Wait, is that true? Let me think. Maybe I need to consider the properties of the numbers and how the differences contribute to the total sum modulo 10.So, let's denote S as the sum of all |a_i - b_i|. We need to find S mod 10 and show that it's 9.Each term in the sum is either 1 or 6, so each term contributes either 1 or 6 to the total sum. Therefore, S is equal to the number of pairs with difference 1 multiplied by 1 plus the number of pairs with difference 6 multiplied by 6.Let me denote x as the number of pairs with difference 1, and y as the number of pairs with difference 6. Then, we have x + y = 999, since there are 999 pairs in total.Therefore, S = x*1 + y*6 = x + 6y.But since x + y = 999, we can express x as 999 - y. Substituting that into S:S = (999 - y) + 6y = 999 + 5y.So, S = 999 + 5y. Therefore, S mod 10 is equal to (999 mod 10) + (5y mod 10).Calculating 999 mod 10: 999 divided by 10 is 99 with a remainder of 9, so 999 mod 10 is 9.So, S mod 10 = (9 + (5y mod 10)) mod 10.Now, 5y mod 10 can be either 0 or 5, depending on whether y is even or odd. Because 5 multiplied by an even number is 0 mod 10, and 5 multiplied by an odd number is 5 mod 10.Therefore, S mod 10 is either (9 + 0) mod 10 = 9 or (9 + 5) mod 10 = 14 mod 10 = 4.Hmm, so depending on whether y is even or odd, the last digit of S is either 9 or 4. But the problem states that the last digit must be 9. So, that suggests that y must be even in any such pairing. Therefore, I need to show that y must be even.So, how can I show that y is even? Maybe by considering the entire set and how the numbers are paired with differences of 1 or 6.Let me think about the numbers modulo something, maybe modulo 2 or modulo 5 or something else. Since the differences are 1 or 6, which are both congruent to 1 mod 5 (since 6 mod 5 is 1). So, each difference is 1 mod 5. Therefore, each term in the sum S is 1 mod 5, so S is equal to 999 mod 5, which is 999 divided by 5 is 199 with a remainder of 4, so 999 mod 5 is 4. Therefore, S mod 5 is 4.But wait, S is also equal to 999 + 5y, so S mod 5 is equal to (999 mod 5) + (5y mod 5) = 4 + 0 = 4. So that's consistent.But how does this help me? Maybe not directly. Let me think differently.Perhaps I can model this as a graph problem. Each number from 1 to 1998 is a vertex, and edges connect numbers that are 1 or 6 apart. Then, the problem reduces to finding a perfect matching in this graph, and then the sum of the edge weights (which are 1 or 6) modulo 10 is 9.But maybe that's overcomplicating it. Let me think about the parity of the numbers.Each pair {a_i, b_i} consists of two numbers. If their difference is 1, then one is even and the other is odd. If their difference is 6, then both are even or both are odd because 6 is even. So, for difference 1, the pair consists of one even and one odd number. For difference 6, the pair consists of two even or two odd numbers.Now, in the set {1, 2, 3, ..., 1998}, there are 999 even numbers and 999 odd numbers. Because 1998 is even, so half of them are even, half are odd.Each pair with difference 1 takes one even and one odd number. Each pair with difference 6 takes either two even or two odd numbers.Let me denote the number of pairs with difference 1 as x, which we already defined. Each such pair removes one even and one odd number from the set. The remaining pairs, which are y in number, must pair the remaining numbers. Since we started with 999 even and 999 odd, after x pairs of difference 1, we have 999 - x even numbers and 999 - x odd numbers left.But the remaining y pairs must pair these remaining numbers. Each of these y pairs can either pair two even numbers or two odd numbers. So, the number of remaining even numbers must be even, because each pair of even numbers removes two evens, and similarly for odds.Therefore, 999 - x must be even, because we have 999 - x even numbers left, and each pair removes two. Similarly, 999 - x must be even for the odd numbers as well.So, 999 - x is even. Therefore, x must be odd, because 999 is odd. Odd minus x is even, so x must be odd.Therefore, x is odd. Since x + y = 999, and x is odd, y must be even because 999 is odd. Odd plus even is odd, so y must be even.Therefore, y is even. So, going back to our earlier expression, S = 999 + 5y. Since y is even, 5y is divisible by 10, so 5y mod 10 is 0. Therefore, S mod 10 is 999 mod 10 + 0 = 9.Therefore, the last digit of S is 9.Wait, that seems to make sense. Let me recap:1. The set has 999 even and 999 odd numbers.2. Each pair with difference 1 uses one even and one odd.3. The remaining pairs must pair the remaining evens and odds, which requires that the remaining count of evens and odds is even, because each pair uses two of the same parity.4. Therefore, 999 - x must be even, so x must be odd.5. Since x + y = 999, and x is odd, y must be even.6. Then, S = 999 + 5y, and since y is even, 5y is 0 mod 10.7. So, S mod 10 is 999 mod 10, which is 9.Therefore, the last digit is 9.I think that's a solid argument. Let me check if I missed anything.- The key point is that after pairing x pairs with difference 1, the remaining numbers must be paired in such a way that the counts of evens and odds are even, which forces x to be odd, hence y even.- Since each difference 6 pair preserves the parity (both even or both odd), the remaining numbers after x pairs must have an even count, so x must be odd.- Therefore, y is even, so 5y is 0 mod 10, and 999 mod 10 is 9, so total sum mod 10 is 9.Yes, that seems correct. I don't see any flaws in this reasoning. So, the last digit must be 9.**Final Answer**boxed{9}

answer：First, I note that the total number of students in the class is 45.Out of these, 19 students did not participate in either competition. This means that the number of students who participated in at least one competition is 45 minus 19, which equals 26 students.Next, I consider the number of students who participated in each competition. There were 12 students in the volleyball competition and 20 students in the track and field competition. Adding these gives a total of 32 participants.However, since only 26 students participated in at least one competition, the overlap between the two groups must account for the difference. The number of students who participated in both competitions is 32 minus 26, which equals 6 students.

answer：Okay, so I have this problem: I need to find the sum of 1/n over all positive integers n such that the decimal representation of 1/n terminates. Hmm, interesting. I remember something about terminating decimals from my math classes, but let me try to recall the exact conditions.First, I think a decimal terminates if the denominator, after simplifying the fraction, has no prime factors other than 2 or 5. So, for example, 1/2 = 0.5, which terminates, and 1/5 = 0.2, which also terminates. But something like 1/3 = 0.333... doesn't terminate because 3 isn't a factor of 2 or 5.So, if that's the case, then n must be a number whose only prime factors are 2 and 5. That means n can be written in the form 2^a * 5^b, where a and b are non-negative integers. For example, n could be 1 (which is 2^0 * 5^0), 2, 4, 5, 8, 10, 16, 20, 25, and so on.Therefore, the set of all such n is the set of all numbers of the form 2^a * 5^b. So, to solve the problem, I need to compute the sum of 1/n for all n = 2^a * 5^b where a and b are non-negative integers.Let me write that out more formally. The sum S is:S = sum_{a=0}^{∞} sum_{b=0}^{∞} 1/(2^a * 5^b)Hmm, that looks like a double sum. I think I can separate this into two separate geometric series because the terms are multiplicative. So, maybe I can rewrite it as:S = (sum_{a=0}^{∞} 1/2^a) * (sum_{b=0}^{∞} 1/5^b)Is that right? Let me check. If I expand the product, I get terms like (1 + 1/2 + 1/4 + ...) multiplied by (1 + 1/5 + 1/25 + ...), which when multiplied together give all possible 1/(2^a * 5^b). Yeah, that seems correct.So, now I just need to compute each of these geometric series separately and then multiply them together.First, let's compute the sum over a:sum_{a=0}^{∞} 1/2^aI know that the sum of a geometric series sum_{k=0}^{∞} r^k is 1/(1 - r) when |r| < 1. Here, r = 1/2, which is less than 1, so the sum is 1/(1 - 1/2) = 1/(1/2) = 2.Similarly, the sum over b:sum_{b=0}^{∞} 1/5^bAgain, using the same formula, r = 1/5, so the sum is 1/(1 - 1/5) = 1/(4/5) = 5/4.So, putting it all together, the total sum S is:S = 2 * (5/4) = 10/4 = 5/2.Wait, that seems straightforward. So, the sum is 5/2, which is 2.5. Let me just make sure I didn't make a mistake.Let me test with some small values of a and b and see if the partial sums approach 5/2.For example, take a from 0 to 2 and b from 0 to 2.Compute the sum:For a=0: 1/(2^0 * 5^0) = 1For a=0, b=1: 1/(1*5) = 1/5For a=0, b=2: 1/(1*25) = 1/25For a=1: 1/(2*1) = 1/2For a=1, b=1: 1/(2*5) = 1/10For a=1, b=2: 1/(2*25) = 1/50For a=2: 1/(4*1) = 1/4For a=2, b=1: 1/(4*5) = 1/20For a=2, b=2: 1/(4*25) = 1/100Adding all these up:1 + 1/5 + 1/25 + 1/2 + 1/10 + 1/50 + 1/4 + 1/20 + 1/100Let me compute this:Convert all to 100 denominators:1 = 100/1001/5 = 20/1001/25 = 4/1001/2 = 50/1001/10 = 10/1001/50 = 2/1001/4 = 25/1001/20 = 5/1001/100 = 1/100Adding them up:100 + 20 + 4 + 50 + 10 + 2 + 25 + 5 + 1 = let's compute step by step:100 + 20 = 120120 + 4 = 124124 + 50 = 174174 + 10 = 184184 + 2 = 186186 + 25 = 211211 + 5 = 216216 + 1 = 217So, total is 217/100 = 2.17Hmm, 2.17 is less than 5/2 = 2.5. But if I take more terms, it should approach 2.5.Let me try adding a=3 and b=3.Compute the additional terms:For a=3: 1/(8*1) = 1/8For a=3, b=1: 1/(8*5) = 1/40For a=3, b=2: 1/(8*25) = 1/200For a=3, b=3: 1/(8*125) = 1/1000Similarly, for b=3 and a=0,1,2:For a=0, b=3: 1/(1*125) = 1/125For a=1, b=3: 1/(2*125) = 1/250For a=2, b=3: 1/(4*125) = 1/500So, adding these:1/8 = 0.1251/40 = 0.0251/200 = 0.0051/1000 = 0.0011/125 = 0.0081/250 = 0.0041/500 = 0.002Adding these decimals:0.125 + 0.025 = 0.150.15 + 0.005 = 0.1550.155 + 0.001 = 0.1560.156 + 0.008 = 0.1640.164 + 0.004 = 0.1680.168 + 0.002 = 0.17So, adding 0.17 to the previous total of 2.17 gives 2.34.Still less than 2.5, but getting closer. If I keep adding more terms, it should approach 2.5.So, that seems to confirm that the sum converges to 5/2.Therefore, the sum of 1/n over all positive integers n where 1/n has a terminating decimal is 5/2.But just to make sure, let me think about another way to approach this.I remember that the set of numbers n with terminating decimals is the multiplicative semigroup generated by 2 and 5. So, every n can be written as 2^a * 5^b, and each such n is unique for each pair (a, b). Therefore, the sum over all such n is the product of the sums over powers of 2 and powers of 5.Which is exactly what I did earlier. So, that seems consistent.Alternatively, I can think of the sum as a double sum, and since the terms are independent in a and b, the product of the two geometric series is the correct approach.So, yeah, I think 5/2 is the right answer.**Final Answer**The sum is boxed{dfrac{5}{2}}.

answer：Alright, so I have this problem here about triangle ABC. It says that the dot product of vectors AB and AC is 10, and the dot product of vectors BA and BC is 6. I need to find the magnitude of vector AB. Hmm, okay, let me think about how to approach this.First, let me recall what the dot product means. The dot product of two vectors is equal to the product of their magnitudes and the cosine of the angle between them. So, if I have two vectors, say **u** and **v**, then **u** ⋅ **v** = |**u**||**v**|cosθ, where θ is the angle between them. That might come in handy here.Now, in triangle ABC, the vectors AB and AC start from point A, right? So, vector AB goes from A to B, and vector AC goes from A to C. Similarly, vector BA goes from B to A, and vector BC goes from B to C. So, the first dot product is between AB and AC, and the second is between BA and BC.Let me denote the lengths of the sides as follows: let |AB| = c, |BC| = a, and |AC| = b. That's the standard notation, I think. So, in triangle ABC, side opposite to A is a, opposite to B is c, and opposite to C is b. Wait, no, actually, standard notation is usually |BC| = a, |AC| = b, and |AB| = c. Yeah, that's right.So, |AB| = c, |BC| = a, |AC| = b.Given that, let's write down the given dot products.First, AB ⋅ AC = 10. Using the dot product formula, that would be |AB||AC|cosθ, where θ is the angle between AB and AC. But in triangle ABC, the angle at point A is between sides AB and AC. So, that angle is angle A. Therefore, AB ⋅ AC = |AB||AC|cos A = c * b * cos A = 10.Similarly, the second dot product is BA ⋅ BC = 6. Vector BA is just the negative of vector AB, right? So, BA = -AB. So, BA ⋅ BC = (-AB) ⋅ BC = - (AB ⋅ BC). So, that equals 6. Therefore, AB ⋅ BC = -6.But let's see, AB ⋅ BC is another dot product. Let me think about what angle that corresponds to. Vector AB starts at A and goes to B, vector BC starts at B and goes to C. So, the angle between AB and BC is actually the angle at point B, but since AB is from A to B and BC is from B to C, the angle between them is 180 degrees minus angle B. Because if you imagine vectors AB and BC, they form a sort of "V" shape at point B, but AB is coming into B and BC is going out from B. So, the angle between AB and BC is actually supplementary to angle B.Wait, maybe I should draw a diagram to visualize this. But since I can't draw, I'll try to imagine it.So, at point B, the angle inside the triangle is angle B, which is between sides AB and BC. But vector AB is from A to B, and vector BC is from B to C. So, the angle between AB and BC is actually 180° - angle B. Because if you have vector AB pointing towards B and vector BC pointing away from B, the angle between them is the external angle at B, which is 180° minus the internal angle.Therefore, the dot product AB ⋅ BC would be |AB||BC|cos(180° - B). But cos(180° - B) is equal to -cos B. So, AB ⋅ BC = |AB||BC|(-cos B) = - |AB||BC|cos B.But from earlier, we have AB ⋅ BC = -6. So, substituting, we get:- |AB||BC|cos B = -6Which simplifies to |AB||BC|cos B = 6.But |AB| is c, |BC| is a, so that's c * a * cos B = 6.So, now, from the first dot product, we have c * b * cos A = 10.So, we have two equations:1. c * b * cos A = 102. c * a * cos B = 6Hmm, okay. Now, in a triangle, we have the Law of Cosines, which relates the sides and the angles. The Law of Cosines states that:a² = b² + c² - 2bc cos ASimilarly,b² = a² + c² - 2ac cos BAnd,c² = a² + b² - 2ab cos CSo, maybe I can use these to relate the given dot products to the sides.From equation 1: c * b * cos A = 10From the Law of Cosines, cos A = (b² + c² - a²)/(2bc)So, substituting into equation 1:c * b * [(b² + c² - a²)/(2bc)] = 10Simplify this:The b and c in the numerator and denominator cancel out, so we get:(b² + c² - a²)/2 = 10Multiply both sides by 2:b² + c² - a² = 20Similarly, from equation 2: c * a * cos B = 6From the Law of Cosines, cos B = (a² + c² - b²)/(2ac)Substitute into equation 2:c * a * [(a² + c² - b²)/(2ac)] = 6Simplify:The a and c cancel out, so we get:(a² + c² - b²)/2 = 6Multiply both sides by 2:a² + c² - b² = 12So, now we have two equations:1. b² + c² - a² = 202. a² + c² - b² = 12Hmm, interesting. Let me write them down:Equation 1: b² + c² - a² = 20Equation 2: a² + c² - b² = 12Now, if I add these two equations together, what do I get?Adding equation 1 and equation 2:(b² + c² - a²) + (a² + c² - b²) = 20 + 12Simplify:b² - b² + c² + c² - a² + a² = 32Which simplifies to:2c² = 32Divide both sides by 2:c² = 16Therefore, c = sqrt(16) = 4Since c is the length of a side, it must be positive, so c = 4.So, |AB| = c = 4.Wait, so that's the answer? Let me just verify to make sure I didn't make any mistakes.So, starting from the two dot products, I expressed them in terms of sides and angles, then used the Law of Cosines to relate them to the sides. Then, I set up two equations and added them to eliminate a² and b², resulting in an equation for c², which gave me c = 4.Let me double-check the steps.First, AB ⋅ AC = |AB||AC|cos A = 10, which is c * b * cos A = 10.Then, BA ⋅ BC = (-AB) ⋅ BC = - (AB ⋅ BC) = 6, so AB ⋅ BC = -6.But AB ⋅ BC = |AB||BC|cos(theta), where theta is the angle between AB and BC, which is 180° - angle B, so cos(theta) = -cos B. Therefore, AB ⋅ BC = - |AB||BC|cos B = -6, which gives |AB||BC|cos B = 6.So, that's correct.Then, using Law of Cosines, cos A = (b² + c² - a²)/(2bc), so plugging into the first equation, we get c * b * (b² + c² - a²)/(2bc) = 10, which simplifies to (b² + c² - a²)/2 = 10, so b² + c² - a² = 20.Similarly, for the second equation, cos B = (a² + c² - b²)/(2ac), so plugging into the second equation, we get c * a * (a² + c² - b²)/(2ac) = 6, which simplifies to (a² + c² - b²)/2 = 6, so a² + c² - b² = 12.Adding these two equations: (b² + c² - a²) + (a² + c² - b²) = 20 + 12 => 2c² = 32 => c² = 16 => c = 4.Yes, that seems solid. So, I think the answer is 4.**Final Answer**The magnitude of vector AB is boxed{4}.