answer：Okay, so I have this geometry problem here about triangle ABC. Points X and Y are on sides AB and AC respectively, and Z is the intersection of lines BY and CX. I need to prove that the sum of the areas of triangles BZX and CZY is greater than twice the area of triangle XYZ. Hmm, that sounds a bit tricky, but let me try to break it down.First, I should probably draw a diagram to visualize what's going on. So, triangle ABC, with AB and AC having points X and Y somewhere along them. Then lines BY and CX intersect at point Z. So, Z is inside the triangle, right? Now, I need to consider the areas of these smaller triangles: BZX, CZY, and XYZ.Maybe I can use some area ratio properties here. I remember that when two lines intersect inside a triangle, the areas of the resulting smaller triangles can be related through ratios of segments. Let me think about that.If I consider triangles that share the same base or height, their areas are proportional to their corresponding heights or bases. So, maybe I can express the areas [BZX] and [CZY] in terms of some ratios involving segments around point Z.Let me denote some variables. Let's say the area of triangle ABC is S. Then, the areas of the smaller triangles would be fractions of S. But I'm not sure if that's the right approach. Maybe I need to look at the ratios of the areas directly.Wait, another thought: perhaps using Ceva's Theorem or Menelaus's Theorem could help here. These theorems relate the ratios of lengths when lines are drawn from vertices intersecting opposite sides. But I'm not sure if they directly apply to areas. Maybe I need a different approach.Alternatively, I could use coordinate geometry. Assign coordinates to the points and calculate the areas algebraically. That might be a bit involved, but it could work. Let me try that.Let's place triangle ABC in a coordinate system. Let me set point A at (0, 0), point B at (1, 0), and point C at (0, 1). So, AB is from (0,0) to (1,0), and AC is from (0,0) to (0,1). Then, point X is somewhere on AB, say at (x, 0) where 0 < x < 1, and point Y is somewhere on AC, say at (0, y) where 0 < y < 1.Now, lines BY and CX intersect at Z. Let me find the coordinates of Z. The equation of line BY: it goes from B(1,0) to Y(0,y). The slope of BY is (y - 0)/(0 - 1) = -y. So, the equation is y = -y(x - 1), which simplifies to y = -yx + y.Similarly, the equation of line CX: it goes from C(0,1) to X(x,0). The slope is (0 - 1)/(x - 0) = -1/x. So, the equation is y = (-1/x)(x - 0) + 1, which simplifies to y = -x/x + 1 = -1 + 1 = 0? Wait, that can't be right. Let me recalculate.Wait, the equation of line CX: it goes from (0,1) to (x,0). The slope is (0 - 1)/(x - 0) = -1/x. So, the equation is y - 1 = (-1/x)(x - 0), which simplifies to y - 1 = -1, so y = 0. That can't be right because that would mean the line CX is the x-axis, but point C is at (0,1), so that can't be. I must have made a mistake.Wait, no, the equation should be y - 1 = (-1/x)(x - 0). So, y - 1 = -1, so y = 0. Hmm, that suggests that line CX is the x-axis, but point C is at (0,1), which is not on the x-axis. That doesn't make sense. I think I messed up the equation.Let me try again. The general equation of a line through two points (x1, y1) and (x2, y2) is (y - y1) = m(x - x1), where m is the slope. So, for line CX, which goes from (0,1) to (x,0), the slope m is (0 - 1)/(x - 0) = -1/x. So, the equation is y - 1 = (-1/x)(x - 0), which simplifies to y - 1 = -1, so y = 0. Wait, that still gives y = 0, which is the x-axis. But point C is at (0,1), which isn't on the x-axis. That must mean I made a mistake in setting up the coordinates.Wait, no, point C is at (0,1), and point X is at (x,0). So, the line CX connects (0,1) to (x,0). So, when x is not zero, the line should have a slope of -1/x, but when x approaches zero, the slope becomes very steep. But when I plug in x = 0, it's undefined, which makes sense because it's a vertical line. Hmm, maybe I need to use a different approach for the equation.Alternatively, I can write the parametric equations for lines BY and CX and find their intersection point Z.For line BY: starting at B(1,0) and going towards Y(0,y). So, parametric equations can be written as:x = 1 - ty = 0 + t*ywhere t ranges from 0 to 1.For line CX: starting at C(0,1) and going towards X(x,0). So, parametric equations:x = 0 + s*xy = 1 - s*1where s ranges from 0 to 1.Now, to find point Z, we need to solve for t and s where the two parametric equations meet:1 - t = s*xandt*y = 1 - sSo, from the first equation: s = (1 - t)/xSubstitute into the second equation:t*y = 1 - (1 - t)/xMultiply both sides by x:t*y*x = x - (1 - t)t*y*x = x - 1 + tBring terms with t to one side:t*y*x - t = x - 1t*(y*x - 1) = x - 1So, t = (x - 1)/(y*x - 1)Hmm, that seems a bit messy. Let me see if I can simplify it.Wait, maybe I can express t in terms of s or vice versa. Alternatively, maybe using mass point geometry could help here. But I'm not too familiar with that. Maybe I should stick to coordinate geometry.Alternatively, perhaps using vectors could help. Let me try that.Let me assign vectors to the points. Let me set point A as the origin, so A = (0,0). Then, vector AB is (1,0), and vector AC is (0,1). Point X is on AB, so its position vector is x*(1,0) = (x,0). Similarly, point Y is on AC, so its position vector is y*(0,1) = (0,y).Now, line BY goes from B(1,0) to Y(0,y). The parametric equation for BY can be written as B + t(Y - B) = (1,0) + t*(-1, y). So, any point on BY is (1 - t, 0 + t*y).Similarly, line CX goes from C(0,1) to X(x,0). The parametric equation for CX is C + s(X - C) = (0,1) + s*(x, -1). So, any point on CX is (0 + s*x, 1 - s).To find point Z, we set the two parametric equations equal:1 - t = s*xandt*y = 1 - sFrom the first equation, s = (1 - t)/xSubstitute into the second equation:t*y = 1 - (1 - t)/xMultiply both sides by x:t*y*x = x - (1 - t)t*y*x = x - 1 + tBring terms with t to one side:t*y*x - t = x - 1t*(y*x - 1) = x - 1So, t = (x - 1)/(y*x - 1)Hmm, that's the same result as before. Let me see if I can simplify this expression.Let me factor out a negative sign in the numerator and denominator:t = (-(1 - x))/( - (1 - y*x)) = (1 - x)/(1 - y*x)So, t = (1 - x)/(1 - y*x)Similarly, s = (1 - t)/x = [1 - (1 - x)/(1 - y*x)] / xLet me compute that:s = [ (1 - y*x - 1 + x ) / (1 - y*x) ] / x= [ (x - y*x) / (1 - y*x) ] / x= [ x(1 - y) / (1 - y*x) ] / x= (1 - y)/(1 - y*x)So, s = (1 - y)/(1 - y*x)Okay, so now I have t and s in terms of x and y.Now, the coordinates of point Z are:From BY: (1 - t, t*y) = (1 - (1 - x)/(1 - y*x), (1 - x)/(1 - y*x)*y )Simplify the x-coordinate:1 - (1 - x)/(1 - y*x) = [ (1 - y*x) - (1 - x) ] / (1 - y*x )= (1 - y*x - 1 + x ) / (1 - y*x )= (x - y*x ) / (1 - y*x )= x(1 - y ) / (1 - y*x )Similarly, the y-coordinate:(1 - x)/(1 - y*x)*y = y(1 - x)/(1 - y*x )So, point Z has coordinates ( x(1 - y)/(1 - y*x ), y(1 - x)/(1 - y*x ) )Hmm, that's a bit complicated, but maybe I can use these coordinates to compute the areas.Let me recall that the area of a triangle with vertices at (x1,y1), (x2,y2), (x3,y3) is given by:Area = | (x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)) / 2 |So, let's compute the areas [BZX], [CZY], and [XYZ].First, let's find [BZX]. The vertices are B(1,0), Z(x(1 - y)/(1 - y*x ), y(1 - x)/(1 - y*x )), and X(x,0).Using the area formula:[BZX] = | (1*(y_z - 0) + x_z*(0 - 0) + x*(0 - y_z )) / 2 |= | (1*y_z + 0 + x*(-y_z )) / 2 |= | (y_z - x*y_z ) / 2 |= | y_z (1 - x ) / 2 |Similarly, [CZY] is the area of triangle C(0,1), Z(x_z, y_z ), Y(0,y).Using the area formula:[CZY] = | (0*(y_z - y) + x_z*(y - 1) + 0*(1 - y_z )) / 2 |= | 0 + x_z*(y - 1) + 0 | / 2= | x_z*(y - 1) / 2 |= | x_z*(y - 1) / 2 |But since y < 1 (because Y is inside AC), (y - 1) is negative, so the absolute value makes it positive:= | x_z*(1 - y ) / 2 |Now, [XYZ] is the area of triangle X(x,0), Y(0,y), Z(x_z, y_z ).Using the area formula:[XYZ] = | x*(y - y_z ) + 0*(y_z - 0 ) + x_z*(0 - y ) / 2 |= | x*(y - y_z ) + 0 + x_z*(-y ) / 2 |= | x*y - x*y_z - x_z*y / 2 |So, now I have expressions for [BZX], [CZY], and [XYZ].Let me write them again:[BZX] = | y_z (1 - x ) / 2 |[CZY] = | x_z (1 - y ) / 2 |[XYZ] = | x*y - x*y_z - x_z*y / 2 |But since all areas are positive, I can drop the absolute value signs.So,[BZX] = y_z (1 - x ) / 2[CZY] = x_z (1 - y ) / 2[XYZ] = (x*y - x*y_z - x_z*y ) / 2Now, I need to compute [BZX] + [CZY] and compare it to 2*[XYZ].So,[BZX] + [CZY] = (y_z (1 - x ) + x_z (1 - y )) / 2And 2*[XYZ] = (x*y - x*y_z - x_z*y )So, the inequality to prove is:(y_z (1 - x ) + x_z (1 - y )) / 2 > (x*y - x*y_z - x_z*y )Multiply both sides by 2:y_z (1 - x ) + x_z (1 - y ) > 2x*y - 2x*y_z - 2x_z*yBring all terms to the left side:y_z (1 - x ) + x_z (1 - y ) - 2x*y + 2x*y_z + 2x_z*y > 0Simplify term by term:y_z (1 - x ) + 2x*y_z = y_z (1 - x + 2x ) = y_z (1 + x )Similarly, x_z (1 - y ) + 2x_z*y = x_z (1 - y + 2y ) = x_z (1 + y )So, the left side becomes:y_z (1 + x ) + x_z (1 + y ) - 2x*y > 0So, we have:y_z (1 + x ) + x_z (1 + y ) > 2x*yNow, recall that from earlier, we have expressions for x_z and y_z in terms of x and y:x_z = x(1 - y ) / (1 - y*x )y_z = y(1 - x ) / (1 - y*x )So, let's substitute these into the inequality:[ y(1 - x ) / (1 - y*x ) ]*(1 + x ) + [ x(1 - y ) / (1 - y*x ) ]*(1 + y ) > 2x*yFactor out 1/(1 - y*x ):[ y(1 - x )(1 + x ) + x(1 - y )(1 + y ) ] / (1 - y*x ) > 2x*yCompute the numerator:First term: y(1 - x )(1 + x ) = y(1 - x^2 )Second term: x(1 - y )(1 + y ) = x(1 - y^2 )So, numerator = y(1 - x^2 ) + x(1 - y^2 )So, the inequality becomes:[ y(1 - x^2 ) + x(1 - y^2 ) ] / (1 - y*x ) > 2x*yMultiply both sides by (1 - y*x ), which is positive because y*x < 1 (since x and y are between 0 and 1). So, the inequality direction remains the same.So,y(1 - x^2 ) + x(1 - y^2 ) > 2x*y*(1 - y*x )Expand the right side:2x*y - 2x^2*y^2So, the inequality is:y - y x^2 + x - x y^2 > 2x y - 2x^2 y^2Bring all terms to the left:y - y x^2 + x - x y^2 - 2x y + 2x^2 y^2 > 0Let me rearrange the terms:= y + x - y x^2 - x y^2 - 2x y + 2x^2 y^2Hmm, this seems complicated. Maybe I can factor it somehow.Let me group terms:= (y + x ) - (y x^2 + x y^2 ) - 2x y + 2x^2 y^2Notice that y x^2 + x y^2 = x y (x + y )So,= (x + y ) - x y (x + y ) - 2x y + 2x^2 y^2Factor out (x + y ):= (x + y )(1 - x y ) - 2x y + 2x^2 y^2Hmm, still not obvious. Maybe I can factor further.Let me write it as:= (x + y )(1 - x y ) - 2x y (1 - x y )Because 2x^2 y^2 = 2x y * x y, but that might not help.Wait, let me see:= (x + y )(1 - x y ) - 2x y + 2x^2 y^2= (x + y )(1 - x y ) - 2x y (1 - x y )Because 2x^2 y^2 = 2x y * x y, but 1 - x y is a common factor.Wait, let me factor out (1 - x y ):= (1 - x y )(x + y - 2x y ) + 2x^2 y^2 - 2x^2 y^2 ?Wait, no, that doesn't seem right. Maybe I need a different approach.Alternatively, let me consider that 1 - x y is positive, so maybe I can divide both sides by (1 - x y ) to simplify.Wait, but I already multiplied both sides by (1 - x y ), so maybe that's not helpful.Alternatively, maybe I can consider the entire expression as a quadratic in terms of x or y.Let me try to see if the expression can be written as a square or something similar.Alternatively, perhaps I can use the AM-GM inequality somewhere. Let me think.Wait, another approach: maybe instead of coordinate geometry, I can use area ratios directly.Let me recall that in triangle ABC, with points X on AB and Y on AC, and Z the intersection of BY and CX, the areas of the smaller triangles can be related through the ratios of the segments.I remember that the ratio of areas can be expressed in terms of the ratios of the lengths.Specifically, the ratio [BZX]/[XYZ] is equal to BZ/ZY, and similarly [CZY]/[XYZ] is equal to CZ/ZX.So, if I can show that BZ/ZY + CZ/ZX > 2, then multiplying both sides by [XYZ] would give [BZX] + [CZY] > 2 [XYZ], which is what I need to prove.So, let me denote:Let’s let’s denote the ratio BZ/ZY = m and CZ/ZX = n.So, I need to show that m + n > 2.If I can show that m + n > 2, then I'm done.But how can I relate m and n?Wait, perhaps using the fact that in triangle ABC, the cevians BY and CX intersect at Z, so by Ceva's theorem, the product of the ratios is 1.Wait, Ceva's theorem states that for concurrent cevians, (AX/XB) * (BY/YC) * (CZ/ZA) = 1.But in this case, we have cevians BY and CX intersecting at Z, but we don't have the third cevian. So, maybe Ceva's theorem isn't directly applicable here.Alternatively, maybe using Menelaus's theorem.Wait, Menelaus's theorem relates the ratios of lengths when a transversal crosses the sides of a triangle. Maybe I can apply it here.Alternatively, perhaps using the concept of similar triangles or area ratios.Wait, another idea: since Z is the intersection of BY and CX, maybe I can express the ratios BZ/ZY and CZ/ZX in terms of the areas of other triangles.Wait, let me consider triangles that share the same base or height.For example, triangles BZX and XYZ share the base XZ, so their areas are proportional to their heights from B and Y respectively.Similarly, triangles CZY and XYZ share the base ZY, so their areas are proportional to their heights from C and X respectively.Hmm, maybe that's a way to express the ratios.Alternatively, perhaps using the concept of mass point geometry, which assigns weights to the vertices based on the ratios of the segments.But I'm not too familiar with mass point geometry, so maybe I should stick to area ratios.Wait, let me try to express the ratios BZ/ZY and CZ/ZX in terms of areas.Let me denote [BZX] = A, [CZY] = B, and [XYZ] = C.We need to show that A + B > 2C.From earlier, I have:A/C = BZ/ZY = mB/C = CZ/ZX = nSo, A = mC and B = nC.Thus, the inequality becomes mC + nC > 2C, which simplifies to (m + n)C > 2C. Since C > 0, we can divide both sides by C, giving m + n > 2.So, I need to show that m + n > 2.Now, how can I relate m and n?Wait, perhaps using the fact that in triangle ABC, the cevians BY and CX intersect at Z, so the ratios of the segments are related.Wait, let me consider the areas of triangles ABZ and ACZ.Wait, maybe not. Alternatively, perhaps using the fact that the sum of the ratios is greater than 2 due to some inequality.Wait, another idea: perhaps using the AM-GM inequality on m and n.If I can show that m + n >= 2*sqrt(mn), and if I can show that sqrt(mn) > 1, then m + n > 2.But I need to find a relationship between m and n.Wait, let me recall that in triangle ABC, the cevians BY and CX intersect at Z, so the product of the ratios is 1.Wait, no, Ceva's theorem requires three cevians, but here we have only two. So, maybe not directly applicable.Alternatively, perhaps using the ratio of areas to express m and n.Wait, let me consider the areas of triangles ABZ and ACZ.Wait, maybe not. Alternatively, perhaps using the fact that the ratio of areas can be expressed in terms of the ratios of the segments.Wait, let me think about the areas in terms of the heights.Let me denote h_B as the height from B to AZ, and h_Y as the height from Y to AZ.Similarly, the area of triangle BZX is (1/2)*XZ*h_B, and the area of triangle XYZ is (1/2)*XZ*h_Y.So, the ratio [BZX]/[XYZ] = h_B / h_Y = BZ / ZY = m.Similarly, for triangle CZY and XYZ, the ratio [CZY]/[XYZ] = h_C / h_X = CZ / ZX = n.So, m = h_B / h_Y and n = h_C / h_X.Now, perhaps I can relate h_B, h_Y, h_C, and h_X.Wait, since h_B and h_Y are heights from B and Y to AZ, and h_C and h_X are heights from C and X to AZ.But I'm not sure how to relate these heights directly.Wait, another idea: perhaps using the fact that the heights are proportional to the distances from the points to the line AZ.So, if I denote d_B as the distance from B to AZ, d_Y as the distance from Y to AZ, d_C as the distance from C to AZ, and d_X as the distance from X to AZ, then:m = d_B / d_Yn = d_C / d_XSo, I need to show that d_B / d_Y + d_C / d_X > 2.Hmm, maybe I can use the AM-GM inequality here.By AM-GM, (d_B / d_Y + d_C / d_X ) / 2 >= sqrt( (d_B / d_Y )*(d_C / d_X ) )So, if I can show that sqrt( (d_B / d_Y )*(d_C / d_X ) ) > 1, then (d_B / d_Y + d_C / d_X ) / 2 > 1, which implies d_B / d_Y + d_C / d_X > 2.So, I need to show that (d_B / d_Y )*(d_C / d_X ) > 1.Which is equivalent to (d_B * d_C ) / (d_Y * d_X ) > 1.So, d_B * d_C > d_Y * d_X.Is this true?Well, since points X and Y are inside AB and AC respectively, and Z is inside the triangle, perhaps the distances from B and C to AZ are greater than the distances from Y and X to AZ.Wait, let me think about that.Since Y is on AC, which is closer to C than B is, so the distance from Y to AZ might be less than the distance from B to AZ. Similarly, X is on AB, which is closer to B than C is, so the distance from X to AZ might be less than the distance from C to AZ.So, perhaps d_B > d_Y and d_C > d_X.If that's the case, then d_B * d_C > d_Y * d_X.Because d_B > d_Y and d_C > d_X, multiplying them would give d_B * d_C > d_Y * d_X.So, if that's true, then (d_B * d_C ) / (d_Y * d_X ) > 1, which implies that sqrt( (d_B / d_Y )*(d_C / d_X ) ) > 1, and thus by AM-GM, d_B / d_Y + d_C / d_X > 2.Therefore, m + n > 2, which implies [BZX] + [CZY] > 2 [XYZ].So, that seems to work.Wait, but I need to make sure that d_B > d_Y and d_C > d_X.Is that necessarily true?Well, since Y is on AC, which is closer to C, and AZ is a line from A to Z inside the triangle, the distance from Y to AZ should be less than the distance from B to AZ, because B is farther away from AZ than Y is.Similarly, X is on AB, which is closer to B, so the distance from X to AZ should be less than the distance from C to AZ.Therefore, d_B > d_Y and d_C > d_X.Thus, d_B * d_C > d_Y * d_X.Therefore, the product (d_B / d_Y )*(d_C / d_X ) > 1, so by AM-GM, the sum m + n > 2.Hence, [BZX] + [CZY] > 2 [XYZ].So, that seems to be the way to go.Alternatively, I could have used the concept of similar triangles or other area ratio properties, but this approach using heights and AM-GM seems to work.I think this is a valid proof. I should probably write it up more formally, but the key idea is using the ratio of areas as ratios of heights, then applying AM-GM inequality to show that the sum of the ratios is greater than 2.

answer：First, I need to determine the total number of possible six-digit numbers that can be formed using the digits 1, 2, 3, 0, 5, and 8. Since the number cannot start with 0, I'll calculate the permutations accordingly.Next, I'll check the conditions for a number to be divisible by 15. A number is divisible by 15 if it is divisible by both 3 and 5. For divisibility by 5, the number must end with either 0 or 5. For divisibility by 3, the sum of its digits must be divisible by 3.I'll calculate the sum of the given digits: 1 + 2 + 3 + 0 + 5 + 8 = 19. Since 19 is not divisible by 3, none of the permutations of these digits will satisfy the divisibility rule for 3. Therefore, it's impossible for any arrangement of these digits to be divisible by 15.Finally, since there are no valid numbers that meet the criteria, the probability is 0.

answer：First, I need to find the greatest common divisor (GCD) of 459 and 357. I'll use the Euclidean Algorithm for this.I'll start by dividing the larger number, 459, by the smaller number, 357. This gives me a quotient of 1 and a remainder of 102.Next, I'll take the previous divisor, 357, and divide it by the remainder, 102. This results in a quotient of 3 and a remainder of 51.Now, I'll divide the last remainder, 102, by the new remainder, 51. This time, there's no remainder, which means I've found the GCD.Therefore, the greatest common divisor of 459 and 357 is 51.

answer：Okay, so I have this problem where I'm given a function f(x) = sin x + cos x. Then, they define a sequence of functions where each subsequent function is the derivative of the previous one. Specifically, f₁(x) is the first derivative of f(x), f₂(x) is the second derivative, and so on, up to f_{2016}(x). I need to find the value of f_{2016}(π/3) and choose the correct answer from the given options.First, let me write down what I know. The function is f(x) = sin x + cos x. The first derivative, f₁(x), would be the derivative of sin x plus the derivative of cos x. I remember that the derivative of sin x is cos x, and the derivative of cos x is -sin x. So, f₁(x) = cos x - sin x.Now, moving on to the second derivative, f₂(x). That would be the derivative of f₁(x). So, I need to differentiate cos x and -sin x. The derivative of cos x is -sin x, and the derivative of -sin x is -cos x. Therefore, f₂(x) = -sin x - cos x.Next, the third derivative, f₃(x), is the derivative of f₂(x). Differentiating -sin x gives -cos x, and differentiating -cos x gives sin x. So, f₃(x) = -cos x + sin x.Then, the fourth derivative, f₄(x), is the derivative of f₃(x). Differentiating -cos x gives sin x, and differentiating sin x gives cos x. So, f₄(x) = sin x + cos x. Wait a minute, that's the original function f(x)! So, f₄(x) = f(x).This suggests that the derivatives of f(x) cycle every four derivatives. In other words, f_{n+4}(x) = f_n(x) for any positive integer n. So, the sequence of derivatives is periodic with a period of 4.Given that, to find f_{2016}(x), I can figure out where 2016 falls in the cycle. Since the period is 4, I can divide 2016 by 4 and find the remainder. If the remainder is 0, it means f_{2016}(x) is the same as f₄(x), which is f(x). If the remainder is 1, it would be f₁(x), and so on.Let me calculate 2016 divided by 4. 2016 ÷ 4 = 504 with a remainder of 0. So, 2016 is exactly 504 times 4, meaning f_{2016}(x) = f₄(x) = f(x) = sin x + cos x.Therefore, f_{2016}(π/3) is equal to sin(π/3) + cos(π/3). I need to compute these values.I remember that sin(π/3) is √3/2 and cos(π/3) is 1/2. So, adding them together: √3/2 + 1/2 = (√3 + 1)/2.Looking back at the options, option B is (√3 + 1)/2, which matches my result. So, the answer should be option B.Wait, let me double-check my steps to make sure I didn't make any mistakes. Starting from f(x), taking derivatives step by step, I got f₁(x) = cos x - sin x, f₂(x) = -sin x - cos x, f₃(x) = -cos x + sin x, and f₄(x) = sin x + cos x, which is back to f(x). So, the cycle is indeed every four derivatives.Calculating 2016 divided by 4 gives 504 with no remainder, so f_{2016}(x) is f(x). Plugging in π/3, I correctly identified sin(π/3) and cos(π/3). So, the sum is (√3 + 1)/2, which is option B. I think that's correct.I don't see any mistakes in my reasoning, so I feel confident that option B is the right answer.**Final Answer**boxed{B}

answer：First, I need to determine how many more milkshakes were sold compared to ice cream cones.I know that the shop sold 82 milkshakes and 67 ice cream cones.To find the difference, I subtract the number of ice cream cones from the number of milkshakes.So, 82 milkshakes minus 67 ice cream cones equals 15.Therefore, 15 more milkshakes were sold than ice cream cones.

answer：Okay, so I have this problem where two numbers are selected independently from the set of positive integers less than or equal to 6. The condition is that exactly one of the numbers must be even. I need to find the probability that the sum of these two numbers is less than their product. Hmm, let me break this down step by step.First, let me understand the set we're dealing with. The positive integers less than or equal to 6 are {1, 2, 3, 4, 5, 6}. So, we're selecting two numbers from this set. The numbers are selected independently, which I think means that the selection of one doesn't affect the selection of the other, and they can be the same number. But wait, the problem says "independently selected," so I guess that means they can be the same number. So, for example, selecting 2 and 2 is allowed.But hold on, the problem also specifies that exactly one of the numbers must be even. So, in other words, one number has to be even, and the other has to be odd. That means we can't have both numbers even or both numbers odd. So, our possible pairs are either (even, odd) or (odd, even). Since the selection is independent, these are two separate cases.Let me list out the even and odd numbers in the set. The even numbers are {2, 4, 6}, and the odd numbers are {1, 3, 5}. So, there are 3 even numbers and 3 odd numbers.Now, since exactly one of the numbers must be even, the total number of possible pairs is the number of ways to choose an even number and an odd number. Since the selection is independent, the total number of such pairs is 3 (even) * 3 (odd) for (even, odd) and another 3 (odd) * 3 (even) for (odd, even). So, in total, that's 9 + 9 = 18 possible pairs where exactly one number is even.Wait, but hold on, is that correct? Because if we're selecting two numbers independently, does order matter? For example, is (2, 3) different from (3, 2)? I think in probability, unless specified otherwise, we usually consider ordered pairs. So, yes, (2, 3) and (3, 2) are different. So, the total number of ordered pairs where exactly one is even is indeed 18.Now, the next part is to find the probability that the sum of the two numbers is less than their product. So, mathematically, we need to find the number of pairs (a, b) such that a + b < a * b. Let me write that inequality down:a + b < a * bHmm, I can rearrange this inequality to make it easier to analyze. Let me subtract a + b from both sides:0 < a * b - a - bWhich simplifies to:a * b - a - b > 0Hmm, I can factor this expression. Let me see:a * b - a - b = a(b - 1) - b = a(b - 1) - bWait, that doesn't seem helpful. Maybe I can factor it differently. Let me try adding 1 to both sides:a * b - a - b + 1 > 1Now, the left side can be factored as:(a - 1)(b - 1) > 1Oh, that's a neat trick! So, the inequality a + b < a * b is equivalent to (a - 1)(b - 1) > 1. That might be easier to work with.So, now I need to find all ordered pairs (a, b) where exactly one of a or b is even, and (a - 1)(b - 1) > 1.Let me list all possible ordered pairs where exactly one is even. Since there are 18 such pairs, I can go through them one by one and check the condition.But before I do that, maybe I can find a pattern or a smarter way to count the number of pairs that satisfy (a - 1)(b - 1) > 1.Let me think about the possible values of a and b. Since a and b are from 1 to 6, let's consider the possible values of (a - 1) and (b - 1). These will range from 0 to 5.So, (a - 1)(b - 1) > 1 means that the product of (a - 1) and (b - 1) must be greater than 1. So, let's see when this product is greater than 1.If either (a - 1) or (b - 1) is 0, then the product is 0, which is not greater than 1. So, we need both (a - 1) and (b - 1) to be at least 1. That means a and b must be at least 2.Wait, but if a is 2, then (a - 1) is 1, and similarly for b. So, if both a and b are at least 2, then (a - 1)(b - 1) is at least 1*1 = 1. But we need it to be greater than 1, so we need at least one of (a - 1) or (b - 1) to be greater than 1.So, in other words, either a > 2 or b > 2, or both.But wait, let's think again. If both a and b are 2, then (a - 1)(b - 1) = 1*1 = 1, which is not greater than 1. So, even if both are 2, it doesn't satisfy the inequality. So, we need at least one of a or b to be greater than 2.But in our case, since exactly one of a or b is even, let's consider the cases where a is even and b is odd, and vice versa.Let me first consider the case where a is even and b is odd.Possible even a's: 2, 4, 6Possible odd b's: 1, 3, 5Now, for each even a, let's see which odd b's satisfy (a - 1)(b - 1) > 1.Let's start with a = 2:(a - 1) = 1So, (1)(b - 1) > 1 => b - 1 > 1 => b > 2So, b must be greater than 2. Since b is odd, the possible values are 3 and 5.So, for a = 2, b can be 3 or 5. That's 2 possibilities.Next, a = 4:(a - 1) = 3So, (3)(b - 1) > 1 => b - 1 > 1/3Since b is an integer, b - 1 must be at least 1, so b > 1.But b is odd, so b can be 1, 3, 5. But b must be greater than 1, so b = 3 or 5.Wait, but (3)(b - 1) > 1. If b = 1, then (3)(0) = 0, which is not greater than 1. So, b must be 3 or 5.So, for a = 4, b can be 3 or 5. That's 2 possibilities.Similarly, a = 6:(a - 1) = 5So, (5)(b - 1) > 1 => b - 1 > 1/5Again, since b is an integer, b - 1 must be at least 1, so b > 1.So, b can be 3 or 5. So, for a = 6, b can be 3 or 5. That's 2 possibilities.So, in total, for a even and b odd, we have 2 + 2 + 2 = 6 valid pairs.Now, let's consider the case where a is odd and b is even.Possible odd a's: 1, 3, 5Possible even b's: 2, 4, 6Again, for each odd a, let's see which even b's satisfy (a - 1)(b - 1) > 1.Starting with a = 1:(a - 1) = 0So, (0)(b - 1) = 0, which is not greater than 1. So, no valid b's for a = 1.Next, a = 3:(a - 1) = 2So, (2)(b - 1) > 1 => b - 1 > 1/2Since b is an integer, b - 1 must be at least 1, so b > 1.But b is even, so b can be 2, 4, 6. But b must be greater than 1, so b = 2, 4, 6.Wait, let's check:If b = 2:(2 - 1) = 1, so (2)(1) = 2 > 1. So, valid.If b = 4:(4 - 1) = 3, so (2)(3) = 6 > 1. Valid.If b = 6:(6 - 1) = 5, so (2)(5) = 10 > 1. Valid.So, for a = 3, all even b's are valid. That's 3 possibilities.Next, a = 5:(a - 1) = 4So, (4)(b - 1) > 1 => b - 1 > 1/4Since b is an integer, b - 1 must be at least 1, so b > 1.So, b can be 2, 4, 6.Again, let's check:b = 2:(2 - 1) = 1, so (4)(1) = 4 > 1. Valid.b = 4:(4 - 1) = 3, so (4)(3) = 12 > 1. Valid.b = 6:(6 - 1) = 5, so (4)(5) = 20 > 1. Valid.So, for a = 5, all even b's are valid. That's 3 possibilities.So, in total, for a odd and b even, we have 0 (from a=1) + 3 (from a=3) + 3 (from a=5) = 6 valid pairs.Wait, but earlier, when a was even and b was odd, we had 6 valid pairs, and now when a is odd and b is even, we have 6 valid pairs. So, in total, 6 + 6 = 12 valid pairs where (a - 1)(b - 1) > 1.But wait, let me double-check. When a is even and b is odd, we had 2 possibilities for each a (2,4,6), so 2*3=6. When a is odd and b is even, for a=3 and a=5, we had 3 possibilities each, so 3*2=6. But for a=1, we had 0. So, total is 6 + 6 = 12.Now, the total number of ordered pairs where exactly one is even is 18, as we calculated earlier.So, the probability is the number of valid pairs divided by the total number of pairs, which is 12/18. Simplifying that, we get 2/3.Wait, but let me make sure I didn't miss any cases or count anything incorrectly.Let me list all the valid pairs explicitly.When a is even and b is odd:a=2: b=3,5 → (2,3), (2,5)a=4: b=3,5 → (4,3), (4,5)a=6: b=3,5 → (6,3), (6,5)So, that's 6 pairs.When a is odd and b is even:a=1: no valid b'sa=3: b=2,4,6 → (3,2), (3,4), (3,6)a=5: b=2,4,6 → (5,2), (5,4), (5,6)So, that's 6 pairs.Total valid pairs: 6 + 6 = 12.Total possible pairs with exactly one even: 18.So, probability is 12/18 = 2/3.Wait, but let me check if all these pairs actually satisfy a + b < a * b.Let's take a few examples:(2,3): 2+3=5, 2*3=6 → 5 < 6 → valid.(2,5): 2+5=7, 2*5=10 → 7 < 10 → valid.(4,3): 4+3=7, 4*3=12 → 7 < 12 → valid.(4,5): 4+5=9, 4*5=20 → 9 < 20 → valid.(6,3): 6+3=9, 6*3=18 → 9 < 18 → valid.(6,5): 6+5=11, 6*5=30 → 11 < 30 → valid.Now, for the odd a's:(3,2): 3+2=5, 3*2=6 → 5 < 6 → valid.(3,4): 3+4=7, 3*4=12 → 7 < 12 → valid.(3,6): 3+6=9, 3*6=18 → 9 < 18 → valid.(5,2): 5+2=7, 5*2=10 → 7 < 10 → valid.(5,4): 5+4=9, 5*4=20 → 9 < 20 → valid.(5,6): 5+6=11, 5*6=30 → 11 < 30 → valid.So, all 12 pairs satisfy the condition. Therefore, the probability is indeed 12/18, which simplifies to 2/3.I think that's correct. So, the probability is 2/3.