answer：Alright, so Ingrid baked a circular pie with an 8-inch diameter and then trimmed it to make the largest possible regular dodecagon. I need to find the area she trimmed off, which is given as ( api - b ), and then find ( a + b ).First, let me visualize the problem. A regular dodecagon has 12 equal sides and angles. The largest regular dodecagon that can fit inside a circle will have all its vertices touching the circumference of the circle. So, the circle is the circumcircle of the dodecagon.Given the diameter of the pie is 8 inches, so the radius ( r ) is half of that, which is 4 inches. So, the radius ( r = 4 ) inches.The area of the original circular pie is ( pi r^2 = pi (4)^2 = 16pi ) square inches.Now, I need to find the area of the regular dodecagon inscribed in this circle. The area of a regular polygon with ( n ) sides inscribed in a circle of radius ( r ) is given by the formula:[text{Area} = frac{1}{2} n r^2 sinleft( frac{2pi}{n} right)]Here, ( n = 12 ) because it's a dodecagon. Plugging in the values:[text{Area}_{text{dodecagon}} = frac{1}{2} times 12 times 4^2 times sinleft( frac{2pi}{12} right)]Simplifying step by step:First, ( frac{1}{2} times 12 = 6 ).Then, ( 4^2 = 16 ).So, multiplying those together: ( 6 times 16 = 96 ).Now, the angle inside the sine function is ( frac{2pi}{12} = frac{pi}{6} ) radians.So, ( sinleft( frac{pi}{6} right) ) is a known value. ( sinleft( frac{pi}{6} right) = frac{1}{2} ).Therefore, the area becomes:[96 times frac{1}{2} = 48]So, the area of the dodecagon is 48 square inches.Wait, hold on. That seems too straightforward. Let me double-check the formula.Yes, the formula for the area of a regular polygon is ( frac{1}{2} n r^2 sinleft( frac{2pi}{n} right) ). So, plugging in ( n = 12 ), ( r = 4 ), and ( sin(pi/6) = 1/2 ), it does compute to 48.So, the area of the dodecagon is 48 square inches.The area of the original pie is ( 16pi ) square inches.Therefore, the area trimmed off is the difference between the two:[16pi - 48]So, in the form ( api - b ), ( a = 16 ) and ( b = 48 ).Therefore, ( a + b = 16 + 48 = 64 ).Wait, but I have a doubt here. Is the area of the dodecagon really 48? Because 48 is a whole number, but I remember that regular dodecagons can have more complex areas. Let me think again.Wait, maybe I made a mistake in the formula. Let me check the formula for the area of a regular polygon.Yes, the formula is ( frac{1}{2} times text{perimeter} times text{apothem} ). Alternatively, when you know the radius, it's ( frac{1}{2} n r^2 sinleft( frac{2pi}{n} right) ). So, that should be correct.But let me compute ( sin(pi/6) ) again. ( pi/6 ) is 30 degrees, and sine of 30 degrees is indeed 1/2. So, that part is correct.So, 12 sides, radius 4, so each of the 12 triangles that make up the dodecagon has an area of ( frac{1}{2} r^2 sin(theta) ), where ( theta = 2pi/12 = pi/6 ). So, each triangle has area ( frac{1}{2} times 16 times frac{1}{2} = 4 ). Then, 12 triangles would be 48. So, that seems correct.Therefore, the area of the dodecagon is indeed 48, and the area of the circle is ( 16pi ). So, the area trimmed off is ( 16pi - 48 ).Hence, ( a = 16 ) and ( b = 48 ), so ( a + b = 64 ).But wait, let me think again. I remember that sometimes when dealing with regular polygons, especially with a high number of sides, the area approaches the area of the circle, but in this case, the dodecagon is significantly smaller? Wait, no, a dodecagon is a 12-sided polygon, which is quite close to a circle. So, the area should be close to the circle's area.Wait, but 48 is approximately 15.278 (since ( 16pi ) is approximately 50.265). So, 48 is just slightly less than the circle's area. That seems reasonable because a dodecagon is close to a circle but still a polygon.Wait, but 48 is actually less than 16π, which is approximately 50.265. So, the area trimmed off is about 2.265 square inches, which is a small area, as expected.But let me confirm the formula once more.Yes, the formula is correct. The area of a regular polygon with n sides is ( frac{1}{2} n r^2 sin(2pi/n) ). So, for n=12, r=4:Area = 0.5 * 12 * 16 * sin(π/6) = 6 * 16 * 0.5 = 6 * 8 = 48. So, that's correct.Therefore, the area trimmed off is 16π - 48, so a=16, b=48, a+b=64.Wait, but just to be thorough, let me compute the area another way.Another way to compute the area of a regular polygon is using the formula:[text{Area} = frac{1}{2} n s a]Where ( s ) is the side length and ( a ) is the apothem.But I don't know the side length or the apothem here. Maybe I can compute them.First, the side length ( s ) of a regular polygon inscribed in a circle of radius ( r ) is given by:[s = 2 r sinleft( frac{pi}{n} right)]So, for n=12, r=4:[s = 2 * 4 * sinleft( frac{pi}{12} right) = 8 sinleft( 15^circ right)]We know that ( sin(15^circ) = sin(45^circ - 30^circ) = sin45 cos30 - cos45 sin30 ).Calculating:( sin45 = frac{sqrt{2}}{2} approx 0.7071 )( cos30 = frac{sqrt{3}}{2} approx 0.8660 )( cos45 = frac{sqrt{2}}{2} approx 0.7071 )( sin30 = 0.5 )So,( sin15 = 0.7071 * 0.8660 - 0.7071 * 0.5 )Calculating:First term: 0.7071 * 0.8660 ≈ 0.6124Second term: 0.7071 * 0.5 ≈ 0.3536So, ( sin15 ≈ 0.6124 - 0.3536 ≈ 0.2588 )Therefore, ( s ≈ 8 * 0.2588 ≈ 2.0704 ) inches.Now, the apothem ( a ) is the distance from the center to the midpoint of a side. It can be calculated as:[a = r cosleft( frac{pi}{n} right)]So, for n=12, r=4:[a = 4 cosleft( frac{pi}{12} right) = 4 cos(15^circ)]We know that ( cos(15^circ) = cos(45^circ - 30^circ) = cos45 cos30 + sin45 sin30 )Calculating:( cos45 = frac{sqrt{2}}{2} ≈ 0.7071 )( cos30 = frac{sqrt{3}}{2} ≈ 0.8660 )( sin45 = frac{sqrt{2}}{2} ≈ 0.7071 )( sin30 = 0.5 )So,( cos15 = 0.7071 * 0.8660 + 0.7071 * 0.5 )Calculating:First term: 0.7071 * 0.8660 ≈ 0.6124Second term: 0.7071 * 0.5 ≈ 0.3536So, ( cos15 ≈ 0.6124 + 0.3536 ≈ 0.9660 )Therefore, ( a ≈ 4 * 0.9660 ≈ 3.864 ) inches.Now, using the area formula ( frac{1}{2} n s a ):Plugging in the values:( n = 12 ), ( s ≈ 2.0704 ), ( a ≈ 3.864 )So,Area ≈ 0.5 * 12 * 2.0704 * 3.864Calculating step by step:0.5 * 12 = 66 * 2.0704 ≈ 12.422412.4224 * 3.864 ≈ let's compute 12 * 3.864 = 46.368, and 0.4224 * 3.864 ≈ 1.632So, total ≈ 46.368 + 1.632 ≈ 48So, that confirms the area is indeed approximately 48 square inches.Therefore, the area trimmed off is ( 16pi - 48 ), so ( a = 16 ), ( b = 48 ), and ( a + b = 64 ).I think that's solid. I used two different methods to compute the area of the dodecagon and both gave me 48. So, I feel confident that the area trimmed off is ( 16pi - 48 ), leading to ( a + b = 64 ).**Final Answer**boxed{64}

answer：Alright, so I have this problem about product inspection. Let me read it again: There are 100 products, and among them, 3 are defective. We need to randomly select 4 products for inspection. The question is asking for the number of ways to exactly select 2 defective products. Hmm, okay.I think this is a combinatorics problem. It involves combinations because the order in which we select the products doesn't matter. So, I need to figure out how many ways we can choose 2 defective products out of the 3 available, and then pair that with choosing the remaining 2 non-defective products from the rest.Let me break it down. First, there are 3 defective products. We want exactly 2 of them in our sample. So, how many ways can we choose 2 defective products from 3? That should be a combination problem, right? The formula for combinations is C(n, k) = n! / (k! * (n - k)!), where n is the total number of items, and k is the number we want to choose.So, for the defective products, it's C(3, 2). Let me calculate that. 3! is 6, and 2! is 2, so 6 / (2 * (3 - 2)!) = 6 / (2 * 1) = 3. So, there are 3 ways to choose 2 defective products from 3.Now, for the non-defective products. There are 100 total products, and 3 are defective, so 100 - 3 = 97 non-defective products. We need to choose the remaining 2 products from these 97. So, that would be C(97, 2).Let me compute that. 97! / (2! * (97 - 2)!) = (97 * 96) / (2 * 1) because the factorials cancel out beyond that. So, 97 * 96 is... let's see, 97 * 96. Well, 100 * 96 is 9600, so subtract 3 * 96 which is 288, so 9600 - 288 = 9312. Then, divide by 2: 9312 / 2 = 4656. So, C(97, 2) is 4656.Now, to find the total number of ways to select exactly 2 defective and 2 non-defective products, we multiply the two combinations together. So, 3 * 4656. Let me calculate that.3 * 4000 is 12,000, and 3 * 656 is 1,968. So, adding those together: 12,000 + 1,968 = 13,968. So, the total number of ways is 13,968.Wait, let me double-check my calculations to make sure I didn't make a mistake. For C(97, 2), 97*96 is indeed 9312, and dividing by 2 gives 4656. Then, 4656 * 3 is 13,968. Yeah, that seems right.Alternatively, I can think of it as the number of ways to choose 2 defective and 2 non-defective. So, yes, it's the product of the two combinations. So, I think that's correct.I don't think I missed anything here. The problem is straightforward once you break it down into the two separate combinations and then multiply them. So, I feel confident that 13,968 is the right answer.**Final Answer**The number of ways to exactly select 2 defective products is boxed{13968}.

answer：Alright, so I have this equation to solve: (3^{n} + 55 = m^{2}), where (n) and (m) are natural numbers. Hmm, okay. Let me think about how to approach this.First off, natural numbers are positive integers, right? So (n) and (m) have to be 1, 2, 3, and so on. The equation is an exponential equation, which can sometimes be tricky because exponents can grow really quickly. But maybe I can find some small values of (n) that satisfy the equation.Let me start by plugging in some small natural numbers for (n) and see if (m) comes out as an integer.Starting with (n = 1):(3^{1} + 55 = 3 + 55 = 58). Is 58 a perfect square? Hmm, 7^2 is 49 and 8^2 is 64, so 58 is between them. Not a perfect square. So (n = 1) doesn't work.Next, (n = 2):(3^{2} + 55 = 9 + 55 = 64). Wait, 64 is 8 squared, right? So (m = 8). That works! Okay, so (n = 2) and (m = 8) is a solution.But before I get too excited, I should check if there are more solutions. Maybe (n = 3):(3^{3} + 55 = 27 + 55 = 82). 82 isn't a perfect square. 9^2 is 81 and 10^2 is 100, so no luck there.How about (n = 4):(3^{4} + 55 = 81 + 55 = 136). 136 is between 11^2 (121) and 12^2 (144). Not a square.Moving on to (n = 5):(3^{5} + 55 = 243 + 55 = 298). 298 is between 17^2 (289) and 18^2 (324). Not a square.Trying (n = 6):(3^{6} + 55 = 729 + 55 = 784). Wait, 784 is 28 squared, right? 28^2 is 784. So (m = 28). That works too! So (n = 6) and (m = 28) is another solution.Hmm, okay, so now I have two solutions: (n = 2, m = 8) and (n = 6, m = 28). Let me check (n = 7) just to see if there's another one.(n = 7):(3^{7} + 55 = 2187 + 55 = 2242). 2242 is between 47^2 (2209) and 48^2 (2304). Not a square.What about (n = 8):(3^{8} + 55 = 6561 + 55 = 6616). 6616 is between 81^2 (6561) and 82^2 (6724). Not a square.Hmm, seems like the numbers are getting pretty big, and the squares are spaced further apart. Maybe I should check if there are any more solutions beyond (n = 6). But before I go on testing more values, perhaps there's a smarter way to approach this.Let me think about the equation modulo some number to see if I can find constraints on (n) or (m). Maybe modulo 3 or modulo 4?Starting with modulo 3. Let's see:The equation is (3^{n} + 55 equiv m^{2} mod 3).First, (3^{n}) modulo 3 is 0 for any (n geq 1), because 3 is a multiple of 3.55 modulo 3: 55 divided by 3 is 18 with a remainder of 1. So 55 ≡ 1 mod 3.So the equation becomes (0 + 1 equiv m^{2} mod 3), which simplifies to (1 equiv m^{2} mod 3).Now, what are the quadratic residues modulo 3? That is, what are the possible values of (m^2 mod 3)?Squares modulo 3 can be:- If (m ≡ 0 mod 3), then (m^2 ≡ 0 mod 3).- If (m ≡ 1 mod 3), then (m^2 ≡ 1 mod 3).- If (m ≡ 2 mod 3), then (m^2 ≡ 4 ≡ 1 mod 3).So, (m^2) modulo 3 is either 0 or 1. In our case, the equation tells us (m^2 ≡ 1 mod 3). So (m) cannot be a multiple of 3. That's a useful piece of information.Now, let's try modulo 4.Looking at the equation (3^{n} + 55 = m^{2}) modulo 4.First, 3 modulo 4 is 3, so (3^{n}) modulo 4:- If (n = 1): 3 mod 4 = 3- If (n = 2): 9 mod 4 = 1- If (n = 3): 27 mod 4 = 3- If (n = 4): 81 mod 4 = 1- So, it alternates between 3 and 1 depending on whether (n) is odd or even.55 modulo 4: 55 divided by 4 is 13 with remainder 3. So 55 ≡ 3 mod 4.So, depending on whether (n) is odd or even, the left-hand side is:- If (n) is odd: 3 + 3 = 6 ≡ 2 mod 4- If (n) is even: 1 + 3 = 4 ≡ 0 mod 4Now, squares modulo 4 can be:- If (m) is even: (m = 2k), so (m^2 = 4k^2 ≡ 0 mod 4)- If (m) is odd: (m = 2k + 1), so (m^2 = 4k^2 + 4k + 1 ≡ 1 mod 4)So, (m^2) modulo 4 is either 0 or 1.Looking back at our equation:- If (n) is odd: LHS ≡ 2 mod 4, but RHS can only be 0 or 1 mod 4. So this is impossible. Therefore, (n) cannot be odd.- If (n) is even: LHS ≡ 0 mod 4, which is possible because RHS can be 0 mod 4. So (n) must be even.So, from this, we can conclude that (n) must be even. That's a helpful constraint. So, (n = 2, 4, 6, 8, ldots)We already checked (n = 2) and (n = 6), which worked, and (n = 4) and (n = 8) didn't. Maybe I should check (n = 10):(3^{10} + 55 = 59049 + 55 = 59104). Is 59104 a perfect square?Let me see, 243^2 is 59049, so 243^2 = 59049, and 244^2 = 59536. So 59104 is between them. 243^2 = 59049, so 59104 - 59049 = 55. So 59104 is 243^2 + 55. Wait, that's the original equation. So 243^2 + 55 = 59104, which is not a perfect square because 244^2 is 59536, which is larger. So, no solution at (n = 10).Hmm, maybe I should try another modulus. Let's try modulo 5.So, equation is (3^{n} + 55 ≡ m^{2} mod 5).First, 3^n modulo 5 cycles every 4 because 3^1=3, 3^2=9≡4, 3^3=12≡2, 3^4=6≡1, then repeats.55 modulo 5 is 0, since 55 is a multiple of 5.So, the equation becomes (3^{n} + 0 ≡ m^{2} mod 5), which is (3^{n} ≡ m^{2} mod 5).So, let's see what (3^{n}) can be modulo 5:- If (n ≡ 1 mod 4): 3- If (n ≡ 2 mod 4): 4- If (n ≡ 3 mod 4): 2- If (n ≡ 0 mod 4): 1Quadratic residues modulo 5 are:- 0^2 = 0- 1^2 = 1- 2^2 = 4- 3^2 = 9 ≡ 4- 4^2 = 16 ≡ 1So, quadratic residues modulo 5 are 0, 1, and 4.Therefore, (m^2) modulo 5 can be 0, 1, or 4.So, (3^{n}) must be congruent to 0, 1, or 4 modulo 5.But (3^{n}) modulo 5 is 1, 3, 4, or 2, depending on (n). So, (3^{n}) can be 1, 3, 4, or 2 mod 5.But (m^2) can only be 0, 1, or 4 mod 5.Therefore, (3^{n}) must be congruent to 1, 4, or 0 mod 5. But (3^{n}) can never be 0 mod 5 because 3 and 5 are coprime. So, (3^{n}) must be 1 or 4 mod 5.Looking back, (3^{n}) is 1 mod 5 when (n ≡ 0 mod 4), and 4 mod 5 when (n ≡ 2 mod 4). So, (n) must be congruent to 0 or 2 mod 4.But earlier, we saw that (n) must be even. So, combining both, (n) must be even, and specifically, (n ≡ 0) or (2 mod 4). So, (n) can be 2, 4, 6, 8, etc.But we saw that for (n = 4), the result wasn't a square. Maybe I can find another modulus to get more constraints.Let me try modulo 8.Equation: (3^{n} + 55 ≡ m^{2} mod 8).First, compute (3^{n}) modulo 8:3^1 = 3 mod 83^2 = 9 ≡ 1 mod 83^3 = 3*1 = 3 mod 83^4 = 3^2 * 3^2 = 1*1 = 1 mod 8So, it cycles every 2: 3, 1, 3, 1,...55 modulo 8: 55 / 8 = 6*8=48, 55-48=7. So 55 ≡ 7 mod 8.So, depending on whether (n) is odd or even:- If (n) is odd: 3 + 7 = 10 ≡ 2 mod 8- If (n) is even: 1 + 7 = 8 ≡ 0 mod 8Now, squares modulo 8 can be:0^2 = 01^2 = 12^2 = 43^2 = 14^2 = 05^2 = 16^2 = 47^2 = 1So, possible quadratic residues modulo 8 are 0, 1, and 4.So, for our equation:- If (n) is odd: LHS ≡ 2 mod 8, but RHS can only be 0,1,4. So impossible. So (n) cannot be odd, which we already knew.- If (n) is even: LHS ≡ 0 mod 8, so RHS must be ≡ 0 mod 8. Therefore, (m^2 ≡ 0 mod 8), which implies that (m) must be even, specifically, divisible by 4 because if (m) is even, (m = 2k), then (m^2 = 4k^2). For (m^2 ≡ 0 mod 8), (k^2) must be even, so (k) must be even, so (m = 4k).So, (m) must be a multiple of 4. That's another constraint.So, putting it all together:- (n) must be even.- (n ≡ 0) or (2 mod 4).- (m) must be a multiple of 4.We already found solutions at (n = 2) and (n = 6). Let's see if we can find another one.Wait, let's see (n = 10) didn't work, as we saw earlier. Let me check (n = 12):(3^{12} + 55 = 531441 + 55 = 531496). Is this a perfect square?Let me see, 729^2 = 531441, so 729^2 = 531441, then 730^2 = 532900. So 531496 is between them. 531496 - 531441 = 55. So, 531496 = 729^2 + 55, which is not a perfect square because 730^2 is 532900, which is much larger. So, no solution at (n = 12).Hmm, maybe I need a different approach. Let's think about the equation (3^{n} = m^{2} - 55). So, (m^2 - 3^{n} = 55). This resembles a difference of squares, but one term is a power of 3.Alternatively, maybe I can factor this as (m^2 - 55 = 3^{n}). Hmm, not sure if that helps.Wait, since (m^2 - 55 = 3^{n}), and (3^{n}) is a power of 3, that means (m^2 - 55) must also be a power of 3. So, (m^2 = 3^{n} + 55). So, (m^2) is 55 more than a power of 3.I wonder if there's a way to bound (n) or (m) so that I can limit the possible values.Let me consider the growth rates. (3^{n}) grows exponentially, while (m^2) grows quadratically. So, for large (n), (3^{n}) will dominate, and (m^2) will be much smaller. But since (m^2 = 3^{n} + 55), (m) is roughly (3^{n/2}). So, the difference between (m) and (3^{n/2}) is small compared to (3^{n/2}) itself.But maybe I can write (m = 3^{k} + t), where (k) is some integer and (t) is a small integer. Let me try that.Let me suppose that (m = 3^{k} + t), where (t) is small compared to (3^{k}). Then, (m^2 = (3^{k} + t)^2 = 3^{2k} + 2*3^{k}*t + t^2).But we know that (m^2 = 3^{n} + 55). So, equating:(3^{2k} + 2*3^{k}*t + t^2 = 3^{n} + 55).Hmm, maybe if (2k = n), then (3^{2k} = 3^{n}), so that term cancels out. Then, we have:(2*3^{k}*t + t^2 = 55).So, (t(2*3^{k} + t) = 55).Since (t) is a positive integer (because (m > 3^{k})), we can factor 55 into pairs of positive integers:55 = 1 * 5555 = 5 * 11So, possible pairs for (t) and (2*3^{k} + t) are (1,55) and (5,11).Let me check each case.Case 1: (t = 1), (2*3^{k} + t = 55)So, (2*3^{k} + 1 = 55)Subtract 1: (2*3^{k} = 54)Divide by 2: (3^{k} = 27)So, (k = 3), since (3^3 = 27).Therefore, (m = 3^{3} + 1 = 27 + 1 = 28). Then, (n = 2k = 6). So, that's the solution we found earlier: (n = 6), (m = 28).Case 2: (t = 5), (2*3^{k} + t = 11)So, (2*3^{k} + 5 = 11)Subtract 5: (2*3^{k} = 6)Divide by 2: (3^{k} = 3)So, (k = 1).Therefore, (m = 3^{1} + 5 = 3 + 5 = 8). Then, (n = 2k = 2). That's the other solution we found: (n = 2), (m = 8).So, these are the only two cases where (t) is positive and (2*3^{k} + t) is a factor of 55. Therefore, the only solutions are (n = 2) and (n = 6).Wait, but just to make sure, is there a possibility that (t) is negative? Because (t) could be negative, but since (m) is a natural number, (m = 3^{k} + t) must be positive. So, (t) can't be so negative that (3^{k} + t) becomes non-positive. But since (t) is multiplied by (2*3^{k} + t) to get 55, which is positive, both factors must be positive. Therefore, (t) must be positive. So, we don't have to consider negative (t).Therefore, the only solutions are (n = 2) and (n = 6), with corresponding (m = 8) and (m = 28).Just to recap, we started by testing small values of (n), found two solutions, then used modular arithmetic to constrain (n) to be even and (m) to be a multiple of 4. Then, by expressing (m) as (3^{k} + t) and solving for (t), we found that the only possible values of (k) that satisfy the equation lead to the solutions we already found. Therefore, these are the only solutions.I think that's thorough enough. I don't see any other approaches that would yield more solutions, especially since the equation grows so quickly, and the modular constraints limit the possibilities.**Final Answer**The natural number solutions are (boxed{n = 2}) and (boxed{n = 6}).Wait, hold on. The question asks to solve the equation in natural numbers, so it's about finding all pairs ((n, m)). So, actually, the solutions are pairs: ((2, 8)) and ((6, 28)). But the user instruction says "put your final answer within boxed{}", and in the initial problem statement, it's just to solve the equation, so maybe they expect the values of (n). Hmm.Looking back, the original equation is (3^{n} + 55 = m^{2}), so the solutions are (n = 2) and (n = 6). So, I think it's appropriate to box both values of (n). Alternatively, if they expect both (n) and (m), maybe box both pairs, but the instruction wasn't entirely clear. Since the problem is to solve for (n) and (m), but the user instruction says "put your final answer within boxed{}", perhaps they expect both solutions. But in the initial problem, it's just to solve the equation, so maybe listing both solutions.But in the think process, I concluded that the only solutions are (n = 2) and (n = 6), with corresponding (m = 8) and (m = 28). So, perhaps the answer should be both pairs. But the user might just want the values of (n). Hmm.Wait, looking back at the problem statement: "Solve the equation (3^{n} + 55 = m^{2}) in natural numbers." So, it's to find all natural numbers (n) and (m) that satisfy the equation. So, the solutions are the pairs ((n, m)). So, perhaps the answer should be both pairs.But in the initial problem, the user didn't specify whether to provide both (n) and (m), but since it's an equation with two variables, the solution is the set of pairs. However, in the instruction, they say "put your final answer within boxed{}", which is typically used for single answers. Maybe they expect both (n) values? Or both pairs?Alternatively, maybe they expect both solutions written as (n = 2) and (n = 6), each boxed separately. But in the think process, I concluded both (n = 2) and (n = 6) are solutions.Wait, actually, in the problem statement, it's to solve the equation in natural numbers, so the solutions are the pairs ((n, m)). So, perhaps the answer should be both pairs. But how to box that?Alternatively, since the question is in the format "Solve the equation...", and the user instruction is to put the final answer in a box, maybe they expect the values of (n). So, I think it's safer to box both (n = 2) and (n = 6).But in the think process, I concluded both solutions, so perhaps the final answer is both (n = 2) and (n = 6). So, I'll present them both boxed.**Final Answer**The natural number solutions for (n) are (boxed{2}) and (boxed{6}).

answer：Okay, so I need to calculate the sum ( s(1) + s(2) + s(3) + ldots + s(321) ), where ( s(n) ) is the sum of all odd digits of the number ( n ). Hmm, let me think about how to approach this.First, let me understand what ( s(n) ) does. For any number ( n ), it looks at each digit, checks if it's odd, and if it is, adds it to the sum. For example, ( s(173) = 1 + 7 + 3 = 11 ), and ( s(1623) = 1 + 3 = 4 ). So, if a digit is even, it doesn't contribute to the sum. Got it.Now, I need to compute this sum from 1 to 321. That's 321 numbers. Calculating each ( s(n) ) individually and adding them up sounds tedious, especially since 321 is a relatively large number. There must be a smarter way to do this without having to compute each term separately.Maybe I can break down the problem by looking at each digit place separately—units, tens, hundreds—and calculate the contribution of each place to the total sum. That is, for each digit place, figure out how many times each odd digit (1, 3, 5, 7, 9) appears, multiply that by the digit, and then sum all those contributions. That sounds promising.Let me structure this. I'll consider the numbers from 000 to 321, padding the numbers with leading zeros to make them all three-digit numbers. This way, I can handle each digit place uniformly. So, numbers like 1 become 001, 10 becomes 010, and so on. This padding won't affect the sum ( s(n) ) because leading zeros don't contribute anything (they're even). So, the total sum will remain the same.Now, I can think of each number as a three-digit number ( ABC ), where ( A ) is the hundreds digit, ( B ) is the tens digit, and ( C ) is the units digit. Each of these digits can range from 0 to 9, except that ( A ) can only go up to 3 because we're only going up to 321.So, I need to calculate the contribution of each digit place (hundreds, tens, units) separately, considering the range from 000 to 321.Let me start with the units place.**Units Place Contribution:**In the units place, each digit from 0 to 9 cycles every 10 numbers. So, in each block of 10 numbers, each digit 0-9 appears exactly once. Since we're dealing with numbers from 000 to 321, which is 322 numbers (including 000), let's see how many complete blocks of 10 there are and how many extra numbers there are.Total numbers: 322Number of complete blocks of 10: 322 / 10 = 32.2, so 32 complete blocks and 2 extra numbers.In each complete block, the units digit cycles through 0-9, so each odd digit (1,3,5,7,9) appears exactly once per block. Therefore, in 32 blocks, each odd digit appears 32 times.Additionally, in the extra 2 numbers (320 and 321), the units digits are 0 and 1. So, only the digit 1 appears once in the units place.Therefore, total contributions from the units place:Each odd digit (1,3,5,7,9) appears 32 times, plus 1 more time for the digit 1.So, total count for each odd digit:- 1: 32 + 1 = 33- 3: 32- 5: 32- 7: 32- 9: 32Now, the contribution to the total sum from the units place is:( 1 times 33 + 3 times 32 + 5 times 32 + 7 times 32 + 9 times 32 )Let me compute this:First, compute the sum for digits 3,5,7,9:Each of these digits appears 32 times, so:( (3 + 5 + 7 + 9) times 32 = (24) times 32 = 768 )Then, add the contribution from digit 1:( 1 times 33 = 33 )Total units place contribution: 768 + 33 = 801Okay, so the units place contributes 801 to the total sum.**Tens Place Contribution:**Now, moving on to the tens place. Similar to the units place, but the cycle is every 100 numbers. Each digit 0-9 appears 10 times in each block of 100 numbers.Total numbers: 322Number of complete blocks of 100: 322 / 100 = 3.22, so 3 complete blocks and 22 extra numbers.In each complete block of 100, each digit in the tens place cycles through 0-9 ten times. So, each odd digit appears 10 times per block.Therefore, in 3 complete blocks, each odd digit appears 30 times.In the extra 22 numbers (300 to 321), the tens digits are:Numbers 300-309: tens digit 0Numbers 310-319: tens digit 1Numbers 320-321: tens digit 2So, in the extra 22 numbers, the tens digits are:- 0 appears 10 times (300-309)- 1 appears 10 times (310-319)- 2 appears 2 times (320-321)So, among these, the odd digits in the tens place are only 1, which appears 10 times.Therefore, total contributions from the tens place:Each odd digit (1,3,5,7,9) appears 30 times, plus digit 1 appears 10 more times.So, total count for each odd digit:- 1: 30 + 10 = 40- 3: 30- 5: 30- 7: 30- 9: 30Now, the contribution to the total sum from the tens place is:( 1 times 40 + 3 times 30 + 5 times 30 + 7 times 30 + 9 times 30 )Compute this:First, compute the sum for digits 3,5,7,9:Each of these digits appears 30 times, so:( (3 + 5 + 7 + 9) times 30 = (24) times 30 = 720 )Then, add the contribution from digit 1:( 1 times 40 = 40 )Total tens place contribution: 720 + 40 = 760So, the tens place contributes 760.**Hundreds Place Contribution:**Now, onto the hundreds place. This is a bit trickier because our range only goes up to 321, so the hundreds digit doesn't go beyond 3.Total numbers: 322In the hundreds place, digits can be 0, 1, 2, or 3.But we need to consider how many times each odd digit (1,3) appears in the hundreds place.Let's break it down:- Numbers 000-099: hundreds digit is 0 (even)- Numbers 100-199: hundreds digit is 1 (odd)- Numbers 200-299: hundreds digit is 2 (even)- Numbers 300-321: hundreds digit is 3 (odd)So, let's calculate how many numbers fall into each category:- 000-099: 100 numbers- 100-199: 100 numbers- 200-299: 100 numbers- 300-321: 22 numbersSo, hundreds digit is 1 for 100 numbers and 3 for 22 numbers.Therefore, the odd digits in the hundreds place are:- 1 appears 100 times- 3 appears 22 timesDigits 5,7,9 never appear in the hundreds place in this range.Therefore, the contribution from the hundreds place is:( 1 times 100 + 3 times 22 )Compute this:( 100 + 66 = 166 )So, the hundreds place contributes 166.**Total Sum:**Now, adding up the contributions from each place:- Units place: 801- Tens place: 760- Hundreds place: 166Total sum ( S = 801 + 760 + 166 )Let me compute this:First, 801 + 760 = 1561Then, 1561 + 166 = 1727Wait, that seems straightforward, but let me double-check my calculations.Wait, 801 + 760: 800 + 760 is 1560, plus 1 is 1561. Then, 1561 + 166: 1561 + 100 = 1661, 1661 + 66 = 1727. Yes, that's correct.But hold on, I feel like 1727 might be too high. Let me verify my reasoning.Wait, when I considered the units place, I included 000 to 321, which is 322 numbers. But the original problem is from 1 to 321, which is 321 numbers. So, does including 000 affect the sum? Because 000 would have s(000) = 0, which doesn't contribute anything. So, actually, the total from 000 to 321 is the same as from 0 to 321, but since s(0) = 0, it's the same as from 1 to 321. So, my calculation is correct.But let me double-check the units place:Total numbers: 322Units digit cycles every 10. So, 322 / 10 = 32.2, so 32 full cycles, each contributing 5 odd digits (1,3,5,7,9) each appearing once. So, 32 * 5 = 160, but wait, that's the number of odd digits, but each odd digit contributes its value.Wait, no. Wait, in each cycle of 10, each odd digit appears once. So, for 32 cycles, each odd digit appears 32 times. Then, the last 2 numbers: 320 and 321. Their units digits are 0 and 1, so only 1 appears once. So, total counts:1: 32 + 1 = 333: 325: 327: 329: 32So, total contribution is 1*33 + 3*32 + 5*32 + 7*32 + 9*32Which is 33 + (3+5+7+9)*32 = 33 + 24*32 = 33 + 768 = 801. That seems correct.Similarly, for the tens place:322 numbers. Each block of 100 has 10 occurrences of each digit in the tens place. So, 3 blocks: 30 occurrences each for digits 0-9. Then, the last 22 numbers: 300-321.In the tens place, 300-309: tens digit 0 (10 numbers)310-319: tens digit 1 (10 numbers)320-321: tens digit 2 (2 numbers)So, in the tens place, the odd digits are only 1, which appears 10 times in the extra 22 numbers.So, total counts:1: 30 + 10 = 403: 305: 307: 309: 30Thus, contribution is 1*40 + 3*30 + 5*30 + 7*30 + 9*30 = 40 + (24)*30 = 40 + 720 = 760. That seems correct.Hundreds place:Numbers 100-199: hundreds digit 1 (100 numbers)Numbers 300-321: hundreds digit 3 (22 numbers)So, 1 appears 100 times, 3 appears 22 times.Thus, contribution is 1*100 + 3*22 = 100 + 66 = 166.So, adding up: 801 + 760 = 1561; 1561 + 166 = 1727.Wait, 1727 seems high, but let me think about it. Each digit place contributes significantly because each odd digit is being counted multiple times across all numbers.Alternatively, maybe I can compute the sum by considering each digit place separately, but perhaps I made a mistake in the hundreds place.Wait, in the hundreds place, numbers 100-199: 100 numbers, so hundreds digit is 1, which is odd, so each of these contributes 1 to the sum. So, total contribution is 100*1 = 100.Similarly, numbers 300-321: 22 numbers, hundreds digit is 3, which is odd, so each contributes 3. So, total contribution is 22*3 = 66.So, total hundreds place contribution: 100 + 66 = 166. That seems correct.Wait, another way to think about it: for each digit place, the total contribution is the number of times each odd digit appears multiplied by the digit. So, for units and tens, we have 801 and 760, which are both over 700, and hundreds is 166. So, total is 1727.But let me cross-verify with another approach.Alternative Approach:Instead of breaking it down by digit places, maybe I can compute the sum for each number from 1 to 321 by considering each digit.But that would be time-consuming. Alternatively, perhaps I can compute the average contribution per digit and multiply by the number of digits.Wait, but that might not be straightforward.Alternatively, let me think about how many times each digit appears in each place.Wait, that's essentially what I did earlier.Wait, but perhaps I can compute the total number of odd digits across all numbers, and then compute the sum.But no, because each digit contributes its value, not just 1 for being odd.Wait, so actually, my initial approach is correct.Wait, another way: let's compute the sum contributed by each digit place.For units place:Each digit 1,3,5,7,9 appears a certain number of times, each contributing their value.Similarly for tens and hundreds.So, my calculation seems correct.Wait, but let me compute the total number of odd digits across all places.Wait, units place: 33 + 32 + 32 + 32 + 32 = 161 odd digitsTens place: 40 + 30 + 30 + 30 + 30 = 160 odd digitsHundreds place: 100 + 22 = 122 odd digitsTotal odd digits: 161 + 160 + 122 = 443Wait, but the total sum is 1727, which is the sum of all these odd digits. So, 443 digits, each contributing their value. So, 1727 is the sum.But let me see: if each digit is contributing its value, then 1727 is the sum.Wait, but let me compute the average value per odd digit.Total sum: 1727Number of odd digits: 443Average value: 1727 / 443 ≈ 3.9Which makes sense, since the digits are 1,3,5,7,9, which average around 5, but since 1 appears more times (due to the way numbers are distributed), the average is a bit lower.Wait, actually, 1 appears 33 + 40 + 100 = 173 times3 appears 32 + 30 + 22 = 84 times5 appears 32 + 30 = 62 times7 appears 32 + 30 = 62 times9 appears 32 + 30 = 62 timesWait, hold on, that doesn't add up.Wait, in units place:1:33, 3:32,5:32,7:32,9:32Tens place:1:40,3:30,5:30,7:30,9:30Hundreds place:1:100,3:22So, total counts:1:33 + 40 + 100 = 1733:32 + 30 + 22 = 845:32 + 30 = 627:32 + 30 = 629:32 + 30 = 62So, total counts:1:1733:845:627:629:62Total digits: 173 + 84 + 62 + 62 + 62 = 443Now, total sum:1*173 + 3*84 + 5*62 + 7*62 + 9*62Compute each term:1*173 = 1733*84 = 2525*62 = 3107*62 = 4349*62 = 558Now, sum these up:173 + 252 = 425425 + 310 = 735735 + 434 = 11691169 + 558 = 1727Yes, that matches the total sum I calculated earlier. So, that seems consistent.Therefore, I think my initial calculation is correct, and the total sum is 1727.But just to be thorough, let me test with a smaller range to see if my method works.Suppose I calculate s(1) + s(2) + ... + s(10).Compute manually:s(1)=1, s(2)=0, s(3)=3, s(4)=0, s(5)=5, s(6)=0, s(7)=7, s(8)=0, s(9)=9, s(10)=1.Total sum: 1 + 0 + 3 + 0 + 5 + 0 + 7 + 0 + 9 + 1 = 26.Using my method:Numbers from 000 to 010 (11 numbers).Units place:Each digit 0-9 appears once in units place for 10 numbers, plus the 11th number is 010, units digit 0.So, units digits:0:2, 1:1, 2:1, ..., 9:1But wait, actually, numbers 000-009: units digits 0-9, then 010: units digit 0.So, in units place:Digits 0:2, 1:1, 2:1, ..., 9:1.So, odd digits in units place:1,3,5,7,9 each appear once, and digit 1 appears once more in 010? Wait, no, 010 has units digit 0, so only the first 10 numbers have units digits 0-9, and 010 adds another 0.So, units digits:0:2, 1:1, 2:1, 3:1, 4:1, 5:1, 6:1, 7:1, 8:1, 9:1.So, odd digits in units place: 1,3,5,7,9 each appear once.Thus, units place contribution: 1+3+5+7+9=25.Wait, but in reality, from 1 to 10, the units digits are 1,2,...,9,0. So, s(10)=1, which is units digit 0 (even) and tens digit 1 (odd). So, in my method, I considered units digits only, but actually, s(n) includes all odd digits, so in 10, the tens digit is 1, which is odd, so s(10)=1.Wait, so my earlier approach of breaking down by digit places is correct because I considered each digit place separately.Wait, but in the smaller example, using my method:Units place: from 000 to 010, units digits:0:2, 1:1, 3:1,5:1,7:1,9:1.So, units place contribution: 1+3+5+7+9=25.Tens place: from 000 to 010, tens digits:0:10 (numbers 000-009), and 1:1 (number 010).So, tens digits:0:10, 1:1.Odd digits in tens place: only 1 appears once.Thus, tens place contribution:1*1=1.Hundreds place: all numbers have hundreds digit 0, which is even, so no contribution.Total sum:25 +1=26, which matches the manual calculation.So, my method works for the smaller case.Therefore, confident that my method is correct, the total sum from 1 to 321 is 1727.**Final Answer**boxed{1727}

answer：Okay, so I have this problem about a tennis tournament in Moscow. Each team has 3 players, and every team plays against every other team. When two teams play against each other, each player from one team plays exactly one game against each player from the other team. The total number of games that can be played is 200, and I need to figure out the maximum number of teams that can participate so that all games are played without exceeding 200.Hmm, let me break this down. First, each team has 3 players. So, if there are, say, n teams, each team will play against n-1 other teams. Now, when two teams play each other, how many games are played? Since each player on one team plays each player on the other team once, that would be 3 players times 3 players, right? So that's 9 games per match between two teams.Wait, so for each pair of teams, there are 9 games. So if there are n teams, the total number of games would be the number of pairs of teams multiplied by 9. The number of pairs of teams is given by the combination formula C(n, 2), which is n(n-1)/2. So the total number of games is 9 * C(n, 2) = 9 * [n(n-1)/2].So, putting that into an equation: Total games = (9/2) * n(n - 1). And we know that this total can't exceed 200. So, (9/2) * n(n - 1) ≤ 200.Let me write that down:(9/2) * n(n - 1) ≤ 200I need to solve for n here. Let me rearrange this inequality.First, multiply both sides by 2 to eliminate the denominator:9 * n(n - 1) ≤ 400Then, divide both sides by 9:n(n - 1) ≤ 400 / 9Calculating 400 divided by 9, that's approximately 44.444...So, n(n - 1) ≤ 44.444...Hmm, so n squared minus n minus 44.444 is less than or equal to zero. Let me write that as a quadratic inequality:n² - n - 44.444 ≤ 0To solve this, I can find the roots of the equation n² - n - 44.444 = 0 and then determine the interval where the quadratic is less than or equal to zero.Using the quadratic formula, n = [1 ± sqrt(1 + 4 * 44.444)] / 2Calculating the discriminant:sqrt(1 + 4 * 44.444) = sqrt(1 + 177.776) = sqrt(178.776)What's the square root of 178.776? Let me approximate.13² is 169, 14² is 196, so it's between 13 and 14.13.5² is 182.25, which is higher than 178.776.So, let me try 13.4²: 13 * 13 is 169, 0.4² is 0.16, and cross term is 2*13*0.4=10.4, so total is 169 + 10.4 + 0.16 = 179.56. Hmm, that's still higher than 178.776.Wait, 13.3²: 13² is 169, 0.3² is 0.09, cross term is 2*13*0.3=7.8, so total is 169 + 7.8 + 0.09 = 176.89. That's lower than 178.776.So, between 13.3 and 13.4.Let me compute 13.35²:13.35² = (13 + 0.35)² = 13² + 2*13*0.35 + 0.35² = 169 + 9.1 + 0.1225 = 178.2225Still a bit lower than 178.776.13.36²: Let's compute 13.36².13.36 * 13.36:First, 13 * 13 = 16913 * 0.36 = 4.680.36 * 13 = 4.680.36 * 0.36 = 0.1296So, adding up:169 + 4.68 + 4.68 + 0.1296 = 169 + 9.36 + 0.1296 = 178.4896Still lower than 178.776.13.37²: Let's compute 13.37².Similarly, 13.37 * 13.37:13 * 13 = 16913 * 0.37 = 4.810.37 * 13 = 4.810.37 * 0.37 = 0.1369Adding up:169 + 4.81 + 4.81 + 0.1369 = 169 + 9.62 + 0.1369 = 178.7569That's very close to 178.776. So, it's approximately 13.37.So, sqrt(178.776) ≈ 13.37.So, going back to the quadratic formula:n = [1 ± 13.37] / 2We can ignore the negative root because the number of teams can't be negative, so:n = (1 + 13.37)/2 ≈ 14.37 / 2 ≈ 7.185So, n ≈ 7.185Since n has to be an integer, and the quadratic is less than or equal to zero between the two roots, which are approximately -6.185 and 7.185. Since n can't be negative, the upper bound is approximately 7.185. So, the maximum integer n is 7.But wait, let me verify this because sometimes approximations can lead to errors.Let me compute the total number of games when n=7:Total games = (9/2)*7*6 = (9/2)*42 = 9*21 = 189 games.And for n=8:Total games = (9/2)*8*7 = (9/2)*56 = 9*28 = 252 games.252 is more than 200, so n=8 is too many. So, n=7 gives 189 games, which is under 200, and n=8 gives 252, which is over.Wait, but the question says "so that all games are played." So, we need to have the total games not exceeding 200. So, n=7 gives 189, which is under, but maybe we can have more teams if the total is exactly 200? But 200 isn't a multiple of 9/2 times n(n-1). Let me check if 200 can be achieved.Wait, 200 divided by 9/2 is 200 * 2 /9 ≈ 44.444, which is the same as before. So, n(n-1) must be less than or equal to 44.444. Since n must be integer, n(n-1) must be less than or equal to 44. So, n=7: 7*6=42 ≤44, which is okay. n=8: 8*7=56>44, which is too big.So, n=7 is the maximum number of teams.Wait, but hold on, is there a way to have more teams without exceeding 200 games? Maybe if some teams don't play each other? But the problem says each team plays against every other team. So, no, all teams must play each other, so n(n-1)/2 matches, each contributing 9 games.So, n=7 is the maximum.But let me double-check the total games for n=7: 7 teams, each plays 6 others, each match has 9 games, so total games: 7*6/2 *9=21*9=189.Yes, that's correct. For n=8, it's 28*9=252, which is over 200.So, 7 teams can participate, and all their games can be played without exceeding 200.Wait, but the problem says "so that all games are played." So, it's possible that 7 teams only play 189 games, but maybe we can have more teams if some games are not played? But the problem says each team plays against every other team, and each player plays exactly one game against each player of the other team. So, it's required that all games are played, so we can't have fewer games.Therefore, 7 teams is the maximum number where the total games (189) are less than or equal to 200. If we have 8 teams, it's 252, which exceeds 200.Wait, but 200 is the maximum number of games allowed. So, 189 is under, but maybe we can have more teams if the total games are exactly 200? But 200 isn't a multiple of 9/2 times n(n-1). Let me see.Wait, 200 divided by 9 is approximately 22.222. So, n(n-1)/2 must be approximately 22.222. So, n(n-1) ≈44.444, as before. So, n is around 7, as before.So, 7 teams give 189 games, which is under 200. So, can we have 7 teams and have some extra games? But the problem says each team must play every other team, so you can't have extra games beyond that. So, 7 teams is the maximum where all required games can be played without exceeding 200.Wait, but what if we have 7 teams, but some teams don't play all their matches? But no, the problem states that each team plays against every other team, so all matches must be played. So, 7 teams is the maximum.Alternatively, maybe the problem allows for the total games to be up to 200, so 7 teams give 189, which is under, but maybe 8 teams would require 252, which is over. So, 7 is the maximum.Wait, but let me think again. Maybe I made a mistake in calculating the total number of games.Each team has 3 players, so when two teams play, each of the 3 players from the first team plays each of the 3 players from the second team. So, 3*3=9 games per pair of teams.So, for n teams, the number of pairs is C(n,2)=n(n-1)/2. So, total games=9*C(n,2)=9*(n(n-1)/2).So, total games= (9/2)*n(n-1). So, yes, that's correct.So, setting (9/2)*n(n-1) ≤200.Multiply both sides by 2: 9n(n-1) ≤400.Divide by 9: n(n-1) ≤44.444...So, n(n-1) must be less than or equal to 44.444.So, n=7: 7*6=42 ≤44.444, okay.n=8:8*7=56>44.444, too big.So, n=7 is the maximum.Therefore, the answer is 7 teams.**Final Answer**The maximum number of teams that can participate is boxed{7}.

answer：Alright, so I have this problem here: "The sum of the coefficients of all the terms in the expansion of ((x - 2y)^{18}) is..." and then it gives me options from A to E. Hmm, okay. I need to figure out what the sum of the coefficients is. Let me think about how to approach this.First, I remember that when you expand a binomial like ((a + b)^n), the coefficients can be found using the binomial theorem. The binomial theorem states that ((a + b)^n = sum_{k=0}^{n} binom{n}{k} a^{n - k} b^k). So, in this case, our binomial is ((x - 2y)^{18}). That means (a = x) and (b = -2y), right?But wait, the question isn't asking for the expansion itself or specific coefficients. It's asking for the sum of all the coefficients. Hmm, how do I find that? I think there's a trick for finding the sum of coefficients in a polynomial. If I substitute specific values for the variables, I can get the sum of coefficients.Let me recall. If I set all variables to 1, then the value of the polynomial becomes the sum of its coefficients. For example, if I have a polynomial like (3x^2 + 4x + 5), and I set (x = 1), then the polynomial becomes (3(1)^2 + 4(1) + 5 = 3 + 4 + 5 = 12), which is the sum of the coefficients. So, applying this idea to the given expression, I can substitute (x = 1) and (y = 1) into ((x - 2y)^{18}) to find the sum of the coefficients.Let me write that down. If I substitute (x = 1) and (y = 1), then the expression becomes ((1 - 2(1))^{18}). Simplifying inside the parentheses first: (1 - 2 = -1). So, it becomes ((-1)^{18}). Hmm, what is ((-1)) raised to the 18th power? Since 18 is an even number, the result should be positive 1. So, ((-1)^{18} = 1).Therefore, the sum of the coefficients is 1. Let me check the answer choices: (A) 0, (B) 1, (C) 19, (D) -1, (E) -19. So, option B is 1. That must be the correct answer.Wait, just to make sure I didn't make a mistake. Let me go through the steps again. The sum of coefficients is found by plugging in 1 for all variables. So, substituting (x = 1) and (y = 1) into ((x - 2y)^{18}) gives ((1 - 2)^{18} = (-1)^{18}). Since any negative number raised to an even power is positive, that's 1. Yep, that seems right.I don't think I made any mistakes here. It's a straightforward substitution. So, yeah, the answer should be 1. Option B.**Final Answer**boxed{B}