answer：First, I need to determine the shares of A, B, and C based on the given conditions.Let’s denote B's share as ( x ) rupees.According to the problem, A's share is Rs. 20 more than B's share. Therefore, A's share is ( x + 20 ) rupees.Additionally, A's share is Rs. 20 less than C's share. This means C's share is ( (x + 20) + 20 = x + 40 ) rupees.The total amount shared among A, B, and C is Rs. 120. So, I can set up the equation:[(x + 20) + x + (x + 40) = 120]Simplifying the equation:[3x + 60 = 120]Subtracting 60 from both sides:[3x = 60]Dividing both sides by 3:[x = 20]Therefore, B's share is Rs. 20.

answer：First, I need to determine the total number of dogwood trees that will be planted in the park. The workers are planting 41 trees today and 20 trees tomorrow. Adding these together gives a total of 61 trees to be planted.Next, I will add the number of trees currently in the park to the total number of trees that will be planted. There are already 39 dogwood trees in the park. Adding the 61 trees to be planted results in a total of 100 dogwood trees.Therefore, after the workers finish planting, the park will have 100 dogwood trees.

answer：Okay, so I have this inequality problem to solve. It says that for all real numbers ( x geq -1 ), and for any ( a_1, a_2, ldots, a_n geq 1 ) where ( n geq 2 ), the following inequality should hold:[frac{a_1 + x}{2} cdot frac{a_2 + x}{2} cdot ldots cdot frac{a_n + x}{2} leq frac{a_1 a_2 ldots a_n + x}{2}]Hmm, I need to find all real numbers ( x geq -1 ) such that this inequality is true no matter what values ( a_1, a_2, ldots, a_n ) take, as long as they are at least 1.First, maybe I should try plugging in some specific values for ( a_i ) to see if I can get some constraints on ( x ). A common strategy is to set all ( a_i ) equal to 1 because that's the minimum value they can take. Let me try that.If ( a_1 = a_2 = ldots = a_n = 1 ), then the left-hand side (LHS) of the inequality becomes:[left( frac{1 + x}{2} right)^n]And the right-hand side (RHS) becomes:[frac{1 cdot 1 cdot ldots cdot 1 + x}{2} = frac{1 + x}{2}]So the inequality simplifies to:[left( frac{1 + x}{2} right)^n leq frac{1 + x}{2}]Hmm, okay. Let's denote ( y = frac{1 + x}{2} ). Then the inequality becomes:[y^n leq y]Since ( y ) is a real number, and ( x geq -1 ), so ( 1 + x geq 0 ), which means ( y geq 0 ).So we have ( y^n leq y ). Let's analyze this inequality.Case 1: ( y = 0 ). Then both sides are 0, so the inequality holds.Case 2: ( y > 0 ). Then we can divide both sides by ( y ) (since ( y neq 0 )):[y^{n - 1} leq 1]Since ( n geq 2 ), ( n - 1 geq 1 ). So we have:[y leq 1]But ( y = frac{1 + x}{2} ), so:[frac{1 + x}{2} leq 1 implies 1 + x leq 2 implies x leq 1]So from this case, we get ( x leq 1 ).Case 3: ( y < 0 ). But wait, ( y = frac{1 + x}{2} ). Since ( x geq -1 ), ( 1 + x geq 0 ), so ( y geq 0 ). Therefore, this case doesn't occur.So combining the results, we have ( x leq 1 ) and ( x geq -1 ). So ( x ) must be in the interval ( [-1, 1] ).But wait, is this the only constraint? I mean, I only tested the case where all ( a_i = 1 ). Maybe I need to check other cases where ( a_i ) are larger than 1 to see if ( x ) must be even smaller or something.Let me try another case. Suppose ( n = 2 ) and ( a_1 = a_2 = t ) where ( t geq 1 ). Then the inequality becomes:[left( frac{t + x}{2} right)^2 leq frac{t^2 + x}{2}]Let me expand the left-hand side:[frac{(t + x)^2}{4} = frac{t^2 + 2tx + x^2}{4}]So the inequality is:[frac{t^2 + 2tx + x^2}{4} leq frac{t^2 + x}{2}]Multiply both sides by 4 to eliminate denominators:[t^2 + 2tx + x^2 leq 2t^2 + 2x]Bring all terms to the left:[t^2 + 2tx + x^2 - 2t^2 - 2x leq 0]Simplify:[-t^2 + 2tx + x^2 - 2x leq 0]Let me rearrange terms:[- t^2 + 2tx + x^2 - 2x leq 0]Hmm, this is a quadratic in ( t ). Let me write it as:[- t^2 + 2tx + (x^2 - 2x) leq 0]Multiply both sides by -1 (which reverses the inequality):[t^2 - 2tx - (x^2 - 2x) geq 0]Simplify:[t^2 - 2tx - x^2 + 2x geq 0]Let me factor this expression if possible. Let's see:[t^2 - 2tx - x^2 + 2x = t^2 - 2tx + ( -x^2 + 2x )]Hmm, maybe group terms differently:[(t^2 - 2tx + x^2) - 2x^2 + 2x = (t - x)^2 - 2x^2 + 2x]Wait, that might not help. Alternatively, perhaps complete the square for ( t ).The expression is quadratic in ( t ):[t^2 - 2tx + ( -x^2 + 2x )]The quadratic in ( t ) is ( t^2 - 2x t + (-x^2 + 2x) ). Let's find its discriminant:Discriminant ( D = ( -2x )^2 - 4 cdot 1 cdot (-x^2 + 2x ) = 4x^2 + 4x^2 - 8x = 8x^2 - 8x )So ( D = 8x(x - 1) )For the quadratic to be non-negative for all ( t geq 1 ), the quadratic must be always non-negative, which would require that the discriminant is non-positive (so that the quadratic doesn't cross the t-axis) and the leading coefficient is positive.But the leading coefficient is 1, which is positive. So we need the discriminant ( D leq 0 ):[8x(x - 1) leq 0]Divide both sides by 8:[x(x - 1) leq 0]This inequality holds when ( x ) is between 0 and 1, inclusive. So ( 0 leq x leq 1 ).Wait, but earlier we had ( x leq 1 ) and ( x geq -1 ). Now, with this case, we get ( 0 leq x leq 1 ). So that's a stricter condition.Hmm, so perhaps ( x ) must be between 0 and 1? But wait, let me check another case to see.Suppose ( n = 2 ) and ( a_1 = t geq 1 ), ( a_2 = 1 ). Then the inequality becomes:[frac{t + x}{2} cdot frac{1 + x}{2} leq frac{t cdot 1 + x}{2}]Simplify:[frac{(t + x)(1 + x)}{4} leq frac{t + x}{2}]Multiply both sides by 4:[(t + x)(1 + x) leq 2(t + x)]Assuming ( t + x neq 0 ), we can divide both sides by ( t + x ):[1 + x leq 2]Which simplifies to:[x leq 1]But this doesn't give any new information beyond what we had before.Wait, but earlier when I set both ( a_1 ) and ( a_2 ) to ( t ), I got that ( x ) must be between 0 and 1. But when I set one ( a_i ) to ( t ) and the other to 1, I only get ( x leq 1 ). So which one is correct?Maybe I need to consider that for different configurations of ( a_i ), different constraints on ( x ) emerge. So perhaps the most restrictive condition is ( x in [0, 1] ).But let me test another case. Suppose ( n = 3 ), ( a_1 = a_2 = a_3 = t geq 1 ). Then the inequality becomes:[left( frac{t + x}{2} right)^3 leq frac{t^3 + x}{2}]Let me compute both sides.Left-hand side:[frac{(t + x)^3}{8} = frac{t^3 + 3t^2 x + 3t x^2 + x^3}{8}]Right-hand side:[frac{t^3 + x}{2}]So the inequality is:[frac{t^3 + 3t^2 x + 3t x^2 + x^3}{8} leq frac{t^3 + x}{2}]Multiply both sides by 8:[t^3 + 3t^2 x + 3t x^2 + x^3 leq 4t^3 + 4x]Bring all terms to the left:[t^3 + 3t^2 x + 3t x^2 + x^3 - 4t^3 - 4x leq 0]Simplify:[-3t^3 + 3t^2 x + 3t x^2 + x^3 - 4x leq 0]Hmm, this is a cubic in ( t ). It's getting complicated. Maybe instead of trying to analyze this directly, I should think about the behavior as ( t ) becomes very large.As ( t to infty ), the leading term on the left-hand side is ( -3t^3 ), and on the right-hand side, it's ( 4t^3 ). So the inequality becomes:[-3t^3 leq 4t^3]Which simplifies to:[-3 leq 4]Which is always true. So for large ( t ), the inequality holds regardless of ( x ). Hmm, so maybe the constraints on ( x ) come from smaller values of ( t ).Let me try ( t = 1 ). Then the inequality becomes:[frac{(1 + x)^3}{8} leq frac{1 + x}{2}]Multiply both sides by 8:[(1 + x)^3 leq 4(1 + x)]Assuming ( 1 + x neq 0 ), divide both sides by ( 1 + x ):[(1 + x)^2 leq 4]Which gives:[1 + 2x + x^2 leq 4 implies x^2 + 2x - 3 leq 0]Factor:[(x + 3)(x - 1) leq 0]So the solution is ( -3 leq x leq 1 ). But since ( x geq -1 ), this gives ( -1 leq x leq 1 ).Hmm, so when ( t = 1 ), we get ( x in [-1, 1] ). But earlier, when ( t ) was variable, we got ( x in [0, 1] ). So which one is it?Wait, maybe I made a mistake in interpreting the quadratic case. Let me go back.When I set ( a_1 = a_2 = t ), I ended up with the quadratic inequality:[t^2 - 2tx - x^2 + 2x geq 0]And I found that the discriminant condition gives ( x in [0, 1] ). But when I set ( t = 1 ), I get ( x in [-1, 1] ). So perhaps the quadratic case is too restrictive?Alternatively, maybe I need to ensure that the inequality holds for all ( t geq 1 ), so the quadratic must be non-negative for all ( t geq 1 ). So perhaps I need to ensure that the quadratic is non-negative for all ( t geq 1 ), not just for all ( t ).Wait, that's a different condition. So the quadratic ( t^2 - 2tx - x^2 + 2x geq 0 ) must hold for all ( t geq 1 ).To ensure that, we can consider the minimum of the quadratic in ( t ). Since the quadratic opens upwards (coefficient of ( t^2 ) is positive), the minimum occurs at the vertex.The vertex occurs at ( t = frac{2x}{2} = x ).So the minimum value is:[f(x) = x^2 - 2x cdot x - x^2 + 2x = x^2 - 2x^2 - x^2 + 2x = -2x^2 + 2x]So for the quadratic to be non-negative for all ( t geq 1 ), we need that the minimum value ( f(x) geq 0 ):[-2x^2 + 2x geq 0 implies -2x(x - 1) geq 0 implies 2x(x - 1) leq 0]So ( x(x - 1) leq 0 ), which implies ( 0 leq x leq 1 ).Therefore, in this case, ( x ) must be between 0 and 1.But earlier, when I set ( t = 1 ), I got ( x in [-1, 1] ). So which one is correct?I think the key is that the inequality must hold for all ( a_i geq 1 ), including when some ( a_i ) are 1 and others are larger. So perhaps the most restrictive condition is ( x in [0, 1] ).Wait, but when I set all ( a_i = 1 ), I got ( x leq 1 ), and when I set ( a_i = t geq 1 ), I got ( x in [0, 1] ). So maybe the correct interval is ( x in [0, 1] ).But let me test another case. Suppose ( n = 2 ), ( a_1 = 2 ), ( a_2 = 1 ). Then the inequality becomes:[frac{2 + x}{2} cdot frac{1 + x}{2} leq frac{2 cdot 1 + x}{2}]Simplify:[frac{(2 + x)(1 + x)}{4} leq frac{2 + x}{2}]Multiply both sides by 4:[(2 + x)(1 + x) leq 2(2 + x)]Assuming ( 2 + x neq 0 ), divide both sides by ( 2 + x ):[1 + x leq 2]Which gives ( x leq 1 ). Again, no new information.Wait, but if I set ( a_1 = 2 ), ( a_2 = 2 ), then:[left( frac{2 + x}{2} right)^2 leq frac{4 + x}{2}]Compute LHS:[frac{(2 + x)^2}{4} = frac{4 + 4x + x^2}{4} = 1 + x + frac{x^2}{4}]RHS:[frac{4 + x}{2} = 2 + frac{x}{2}]So the inequality is:[1 + x + frac{x^2}{4} leq 2 + frac{x}{2}]Bring all terms to the left:[1 + x + frac{x^2}{4} - 2 - frac{x}{2} leq 0]Simplify:[-1 + frac{x}{2} + frac{x^2}{4} leq 0]Multiply by 4:[-4 + 2x + x^2 leq 0]Which is:[x^2 + 2x - 4 leq 0]Solve the quadratic inequality:The roots are ( x = frac{-2 pm sqrt{4 + 16}}{2} = frac{-2 pm sqrt{20}}{2} = frac{-2 pm 2sqrt{5}}{2} = -1 pm sqrt{5} )So the inequality ( x^2 + 2x - 4 leq 0 ) holds for ( x in [-1 - sqrt{5}, -1 + sqrt{5}] ). Since ( x geq -1 ), the relevant interval is ( x in [-1, -1 + sqrt{5}] ).But ( -1 + sqrt{5} approx 1.236 ), which is greater than 1. So this suggests that ( x ) can be up to approximately 1.236, which contradicts our earlier result.Wait, that can't be right because when we set all ( a_i = 1 ), we found ( x leq 1 ). So perhaps this case is not restrictive enough?Wait, no. The problem is that when ( a_i ) are larger, the inequality might hold even for larger ( x ), but we need the inequality to hold for all ( a_i geq 1 ). So if for some ( a_i ), the inequality only holds when ( x leq 1 ), then ( x ) must be at most 1.But in this case, when ( a_1 = a_2 = 2 ), the inequality holds for ( x leq -1 + sqrt{5} approx 1.236 ). But since we need the inequality to hold for all ( a_i geq 1 ), including when ( a_i = 1 ), which requires ( x leq 1 ), the stricter condition is ( x leq 1 ).Therefore, combining all these cases, it seems that ( x ) must be in the interval ( [-1, 1] ).But let me try to think of another approach. Maybe using induction or some inequality like AM-GM.Wait, the given inequality resembles the AM-GM inequality, but it's not exactly the same. Let me see.The left-hand side is the product of terms ( frac{a_i + x}{2} ), and the right-hand side is ( frac{prod a_i + x}{2} ).If I take the logarithm of both sides, maybe I can turn the product into a sum, but I'm not sure if that helps.Alternatively, perhaps I can consider the function ( f(a_1, a_2, ldots, a_n) = prod_{i=1}^n frac{a_i + x}{2} - frac{prod_{i=1}^n a_i + x}{2} leq 0 ).We need this function to be non-positive for all ( a_i geq 1 ).Maybe I can analyze the behavior of this function. For example, when all ( a_i = 1 ), we have ( f = left( frac{1 + x}{2} right)^n - frac{1 + x}{2} leq 0 ), which we already solved.When one ( a_i ) increases, how does ( f ) behave? Let's fix all ( a_j = 1 ) except ( a_1 = t geq 1 ). Then:[f(t, 1, 1, ldots, 1) = frac{t + x}{2} cdot left( frac{1 + x}{2} right)^{n - 1} - frac{t + x}{2}]Factor out ( frac{t + x}{2} ):[frac{t + x}{2} left( left( frac{1 + x}{2} right)^{n - 1} - 1 right) leq 0]Since ( t + x geq 1 + x geq 0 ) (because ( x geq -1 )), the sign of ( f ) depends on the term in the parentheses:[left( frac{1 + x}{2} right)^{n - 1} - 1 leq 0]Which implies:[left( frac{1 + x}{2} right)^{n - 1} leq 1]Again, since ( frac{1 + x}{2} geq 0 ), we can take the (n-1)th root:[frac{1 + x}{2} leq 1 implies x leq 1]So this again gives ( x leq 1 ).Therefore, combining all these cases and analyses, it seems that ( x ) must satisfy ( -1 leq x leq 1 ).To ensure that this is indeed the case, let me try to see if ( x = 1 ) satisfies the inequality for arbitrary ( a_i geq 1 ).Set ( x = 1 ). Then the inequality becomes:[prod_{i=1}^n frac{a_i + 1}{2} leq frac{prod_{i=1}^n a_i + 1}{2}]Is this true?Let me consider ( n = 2 ). Then:[frac{a_1 + 1}{2} cdot frac{a_2 + 1}{2} leq frac{a_1 a_2 + 1}{2}]Multiply both sides by 4:[(a_1 + 1)(a_2 + 1) leq 2(a_1 a_2 + 1)]Expand the left-hand side:[a_1 a_2 + a_1 + a_2 + 1 leq 2a_1 a_2 + 2]Bring all terms to the left:[a_1 a_2 + a_1 + a_2 + 1 - 2a_1 a_2 - 2 leq 0]Simplify:[- a_1 a_2 + a_1 + a_2 - 1 leq 0]Factor:[- (a_1 a_2 - a_1 - a_2 + 1) = - (a_1 - 1)(a_2 - 1) leq 0]Since ( a_1, a_2 geq 1 ), ( (a_1 - 1)(a_2 - 1) geq 0 ), so the negative of that is ( leq 0 ). Therefore, the inequality holds.Similarly, for ( x = -1 ), let's check.Set ( x = -1 ). Then the inequality becomes:[prod_{i=1}^n frac{a_i - 1}{2} leq frac{prod_{i=1}^n a_i - 1}{2}]But since ( a_i geq 1 ), ( a_i - 1 geq 0 ), so all terms are non-negative.Let me test ( n = 2 ), ( a_1 = a_2 = 1 ):LHS: ( frac{0}{2} cdot frac{0}{2} = 0 )RHS: ( frac{1 - 1}{2} = 0 )So equality holds.Now, ( a_1 = 2 ), ( a_2 = 1 ):LHS: ( frac{2 - 1}{2} cdot frac{1 - 1}{2} = frac{1}{2} cdot 0 = 0 )RHS: ( frac{2 cdot 1 - 1}{2} = frac{1}{2} )So ( 0 leq frac{1}{2} ), which holds.Another case, ( a_1 = 2 ), ( a_2 = 2 ):LHS: ( frac{2 - 1}{2} cdot frac{2 - 1}{2} = frac{1}{2} cdot frac{1}{2} = frac{1}{4} )RHS: ( frac{4 - 1}{2} = frac{3}{2} )So ( frac{1}{4} leq frac{3}{2} ), which holds.Wait, but what if ( a_1 = 1 ), ( a_2 = 1 ), ( n = 2 ):LHS: 0RHS: 0Equality holds.But what if ( a_1 = 1 ), ( a_2 = t geq 1 ):LHS: ( frac{0}{2} cdot frac{t - 1}{2} = 0 )RHS: ( frac{t - 1}{2} geq 0 )So inequality holds.Similarly, for ( x = -1 ), the inequality seems to hold.Therefore, combining all these observations, it seems that ( x ) must be in the interval ( [-1, 1] ).I think I've considered multiple cases and checked the boundary conditions. It seems consistent that ( x ) must be between -1 and 1, inclusive.

answer：Alright, so I have this problem here: Given the imaginary unit ( i ), find the maximum value of ( |z| ) for a complex number ( z ) satisfying the condition ( |z - i| = |3 - 4i| ). Hmm, okay, let me try to unpack this step by step.First, I remember that for any complex number ( z = x + yi ), the modulus ( |z| ) represents its distance from the origin in the complex plane. So, ( |z| = sqrt{x^2 + y^2} ). Got that down.Now, the condition given is ( |z - i| = |3 - 4i| ). Let me recall that ( |z - a| ) represents the distance from ( z ) to the point ( a ) in the complex plane. So, in this case, ( |z - i| ) is the distance from ( z ) to ( i ), which is the point ( (0, 1) ) in the complex plane.Next, I need to figure out what ( |3 - 4i| ) is. I think this is the modulus of the complex number ( 3 - 4i ), which should be calculated as ( sqrt{3^2 + (-4)^2} = sqrt{9 + 16} = sqrt{25} = 5 ). So, the condition simplifies to ( |z - i| = 5 ). That means all the complex numbers ( z ) that are at a distance of 5 units from the point ( i ) (which is ( (0, 1) )) lie on a circle centered at ( (0, 1) ) with radius 5.Okay, so now I need to find the maximum value of ( |z| ) for all ( z ) on this circle. ( |z| ) is the distance from the origin to the point ( z ). So, essentially, I need to find the point on the circle ( |z - i| = 5 ) that is farthest from the origin.Let me visualize this. The circle is centered at ( (0, 1) ) with radius 5. The origin is at ( (0, 0) ). The distance from the origin to the center of the circle is 1 unit. So, the origin is inside the circle because the radius is 5, which is much larger than 1.To find the farthest point from the origin on this circle, I think I can use the fact that the farthest point will lie along the line connecting the origin and the center of the circle. Since the center is at ( (0, 1) ), the line connecting the origin and the center is the vertical line ( x = 0 ).So, starting from the origin, moving towards the center at ( (0, 1) ), and then extending to the edge of the circle in that direction should give me the farthest point. The distance from the origin to the center is 1, and the radius is 5, so the farthest point should be ( 1 + 5 = 6 ) units away from the origin.Wait, let me verify that. If I move from the origin towards the center, which is 1 unit away, and then go another 5 units in the same direction, I should reach a point that's 6 units from the origin. That makes sense because the circle extends 5 units in all directions from the center, so in the direction away from the origin, it would be 1 + 5 = 6.But just to be thorough, maybe I should set up some equations to confirm this.Let me denote ( z = x + yi ). Then, the condition ( |z - i| = 5 ) becomes ( |x + (y - 1)i| = 5 ), which translates to ( sqrt{x^2 + (y - 1)^2} = 5 ). Squaring both sides, I get ( x^2 + (y - 1)^2 = 25 ). That's the equation of the circle centered at ( (0, 1) ) with radius 5.Now, I want to maximize ( |z| = sqrt{x^2 + y^2} ). To maximize this, I can instead maximize ( x^2 + y^2 ) since the square root is a monotonically increasing function.So, I need to maximize ( x^2 + y^2 ) subject to the constraint ( x^2 + (y - 1)^2 = 25 ). This sounds like a problem of maximizing one function given another constraint, which I think can be approached using the method of Lagrange multipliers or by using geometric intuition.But since I already have a geometric interpretation, let me stick with that for simplicity. The maximum distance from the origin to a point on the circle will be the distance from the origin to the center of the circle plus the radius of the circle. As I calculated earlier, the distance from the origin to the center is 1, and the radius is 5, so the maximum distance should be ( 1 + 5 = 6 ).Just to double-check, let's consider the point on the circle in the direction away from the origin. That would be along the line from the origin through the center of the circle, which is the positive y-axis. So, starting at the origin, moving up to the center at ( (0, 1) ), and then another 5 units up along the y-axis would bring us to ( (0, 6) ). The distance from the origin to ( (0, 6) ) is indeed 6.Alternatively, if I consider the point in the opposite direction, towards the negative y-axis, that point would be ( (0, 1 - 5) = (0, -4) ). The distance from the origin to ( (0, -4) ) is 4, which is less than 6, so that makes sense.Therefore, the maximum value of ( |z| ) is 6.But just to make sure I haven't missed anything, let me try to approach this algebraically as well.We have the constraint ( x^2 + (y - 1)^2 = 25 ) and we want to maximize ( x^2 + y^2 ).Let me expand the constraint equation:( x^2 + y^2 - 2y + 1 = 25 )Simplify:( x^2 + y^2 - 2y = 24 )Now, let me denote ( S = x^2 + y^2 ), which is the quantity we want to maximize.From the constraint, we have:( S - 2y = 24 )So, ( S = 24 + 2y )Therefore, to maximize ( S ), we need to maximize ( y ), since ( S ) is directly proportional to ( y ).So, what's the maximum value of ( y ) on the circle ( x^2 + (y - 1)^2 = 25 )?Well, the circle is centered at ( (0, 1) ) with radius 5, so the maximum y-coordinate occurs at the top of the circle, which is ( y = 1 + 5 = 6 ).Therefore, the maximum value of ( y ) is 6, and substituting back into ( S = 24 + 2y ), we get:( S = 24 + 2*6 = 24 + 12 = 36 )Thus, ( x^2 + y^2 = 36 ), so ( |z| = sqrt{36} = 6 ).This confirms my earlier geometric reasoning. So, the maximum value of ( |z| ) is indeed 6.I think that's solid. I approached it both geometrically and algebraically, and both methods led me to the same answer. I don't see any mistakes in my reasoning, so I'm confident that 6 is the correct maximum value.**Final Answer**The maximum value of ( |z| ) is boxed{6}.

answer：Okay, so I have this problem where I need to find four positive integers whose product is 2002, and their sum is less than 40. Hmm, let me think about how to approach this.First, I know that 2002 is the product of these four numbers. So, maybe I should start by factoring 2002 into its prime factors. That might help me figure out possible combinations of four numbers.Let me try dividing 2002 by small primes. Starting with 2: 2002 divided by 2 is 1001. Okay, so 2 is a factor. Now, 1001 is a number I recognize—it's 7 times 11 times 13. Let me check: 7 times 11 is 77, and 77 times 13 is 1001. Yep, that's correct. So, putting it all together, 2002 factors into 2 × 7 × 11 × 13. So, the prime factors are 2, 7, 11, and 13. That gives me one possible set of four numbers: 2, 7, 11, and 13. Let me check their sum: 2 + 7 is 9, plus 11 is 20, plus 13 is 33. That's 33, which is definitely less than 40. So, that's a valid set.But the problem says "four positive integers," and doesn't specify they have to be distinct or prime. So, maybe there are other combinations where some of the numbers are 1, which is a positive integer. Let me explore that.If I include 1 as one of the numbers, then I need to adjust the other numbers accordingly. Since 1 doesn't affect the product, I can combine some of the prime factors to make larger numbers. For example, if I take 2 and 7, their product is 14. So, another possible set could be 1, 14, 11, and 13. Let me check the sum: 1 + 14 is 15, plus 11 is 26, plus 13 is 39. That's still less than 40, so that works too.What if I try combining different factors? Let's see. If I take 2 and 11, that's 22. So, another set could be 1, 7, 22, and 13. Let me add those up: 1 + 7 is 8, plus 22 is 30, plus 13 is 43. Oh, that's more than 40, so that doesn't work. Okay, so that combination is out.How about combining 2 and 13? That gives me 26. So, the set would be 1, 7, 11, and 26. Let's add them: 1 + 7 is 8, plus 11 is 19, plus 26 is 45. That's way over 40. So, that's not good either.What if I combine 7 and 11? That's 77. So, the set would be 1, 2, 77, and 13. Adding those: 1 + 2 is 3, plus 77 is 80, plus 13 is 93. That's way too high. Definitely not.How about combining 7 and 13? That's 91. So, the set would be 1, 2, 11, and 91. Adding them: 1 + 2 is 3, plus 11 is 14, plus 91 is 105. Again, way over 40.Hmm, maybe combining two different pairs? Let's see. If I take 2 and 7 to make 14, and then 11 and 13 to make 143. So, the set would be 1, 14, and 143. Wait, that's only three numbers. I need four numbers. So, maybe 1, 14, 11, and 13, which I already considered earlier. That's the same as the second set I had.Alternatively, what if I take 2 and 11 to make 22, and then 7 and 13 to make 91. So, the set would be 1, 22, and 91. Again, that's only three numbers. I need four, so maybe 1, 22, 7, and 13. Wait, that's the same as before, which summed to 43. So, that's over.Wait, maybe I can combine more than two factors? Let's see. If I take 2, 7, and 11, that's 154. So, the set would be 1, 154, and 13. That's only three numbers. Not enough.Alternatively, 2, 7, and 13 is 182. So, the set would be 1, 182, and 11. Again, only three numbers.Hmm, maybe I'm overcomplicating this. Let me list out all the possible combinations of four positive integers that multiply to 2002, and then check their sums.Starting with the prime factors: 2, 7, 11, 13. Sum is 33, which is good.Now, if I replace one of the primes with 1 and combine two primes into a composite number:- Replace 2 with 1, combine 2 and 7 into 14: 1, 14, 11, 13. Sum is 39, which is still less than 40.- Replace 7 with 1, combine 7 and 2 into 14: same as above.- Replace 11 with 1, combine 11 and 2 into 22: 1, 22, 7, 13. Sum is 43, which is too high.- Replace 13 with 1, combine 13 and 2 into 26: 1, 26, 7, 11. Sum is 45, too high.Alternatively, combining different pairs:- Combine 7 and 11 into 77: 1, 2, 77, 13. Sum is 93, too high.- Combine 7 and 13 into 91: 1, 2, 11, 91. Sum is 105, too high.- Combine 11 and 13 into 143: 1, 2, 7, 143. Sum is 153, way too high.So, the only valid combinations are the two I found earlier: 2, 7, 11, 13 and 1, 14, 11, 13.Wait, let me double-check if there are any other combinations where I might have more than two numbers greater than 1. For example, could I have three numbers greater than 1 and one 1? Let's see.If I have three numbers greater than 1, then the fourth is 1. So, the product would be 1 × a × b × c = 2002. So, a × b × c = 2002.But 2002 is 2 × 7 × 11 × 13, so if I have three numbers, each would have to be a product of some combination of these primes. Let me see if I can find three numbers whose product is 2002, and then add 1 to make four numbers.But wait, 2002 is already the product of four primes, so if I group them into three numbers, one of them would have to be a product of two primes, and the other two would be single primes. For example:- 2 × 7 = 14, so the three numbers could be 14, 11, 13. Then adding 1, we have 1, 14, 11, 13, which is the same as before.Alternatively, 2 × 11 = 22, so the three numbers would be 22, 7, 13. Adding 1, we get 1, 22, 7, 13, which sums to 43, too high.Similarly, 2 × 13 = 26, so the three numbers are 26, 7, 11. Adding 1, sum is 45, too high.7 × 11 = 77, so three numbers are 77, 2, 13. Adding 1, sum is 93, too high.7 × 13 = 91, so three numbers are 91, 2, 11. Adding 1, sum is 105, too high.11 × 13 = 143, so three numbers are 143, 2, 7. Adding 1, sum is 153, too high.So, the only valid three-number combination that works when adding 1 is 14, 11, 13, which gives a sum of 39.Alternatively, could I have two numbers greater than 1 and two 1s? Let's see.If I have two 1s, then the product would be 1 × 1 × a × b = 2002, so a × b = 2002. Then, a and b would have to be factors of 2002. Let me list the factor pairs of 2002:1 × 20022 × 10017 × 28611 × 18213 × 15414 × 14322 × 9126 × 77So, if I take two 1s and two of these factors, let's see their sums:1, 1, 2, 1001: sum is 1 + 1 + 2 + 1001 = 1005, way too high.1, 1, 7, 286: sum is 1 + 1 + 7 + 286 = 295, too high.1, 1, 11, 182: sum is 1 + 1 + 11 + 182 = 195, too high.1, 1, 13, 154: sum is 1 + 1 + 13 + 154 = 169, too high.1, 1, 14, 143: sum is 1 + 1 + 14 + 143 = 159, too high.1, 1, 22, 91: sum is 1 + 1 + 22 + 91 = 115, too high.1, 1, 26, 77: sum is 1 + 1 + 26 + 77 = 105, too high.So, none of these combinations work because their sums are way over 40.What if I have three 1s? Then, the product would be 1 × 1 × 1 × a = 2002, so a = 2002. Then, the four numbers would be 1, 1, 1, 2002. Sum is 1 + 1 + 1 + 2002 = 2005, which is way too high.So, that's not an option.Therefore, the only valid combinations are the two I found earlier: 2, 7, 11, 13 and 1, 14, 11, 13.Wait, let me make sure I haven't missed any other combinations. Maybe there's a way to have more than two composite numbers without using 1. Let me think.If I have four numbers, all greater than 1, but not necessarily prime. So, I need to factor 2002 into four factors, each greater than 1. Let's see.Starting with the prime factors: 2, 7, 11, 13. That's one set.Alternatively, can I combine some factors differently? For example, 2 × 7 = 14, so 14, 11, 13, and 1. But that's the same as before.Alternatively, 2 × 11 = 22, so 22, 7, 13, 1. Sum is 43, too high.Similarly, 2 × 13 = 26, so 26, 7, 11, 1. Sum is 45, too high.7 × 11 = 77, so 77, 2, 13, 1. Sum is 93, too high.7 × 13 = 91, so 91, 2, 11, 1. Sum is 105, too high.11 × 13 = 143, so 143, 2, 7, 1. Sum is 153, too high.So, all other combinations either result in a sum over 40 or are duplicates of the two valid sets I found.Therefore, the only possible sets are 2, 7, 11, 13 and 1, 14, 11, 13.I think that's it. I can't think of any other combinations that would satisfy both the product and the sum condition.

answer：Okay, so I have this problem here: "The range of the function ( f(x) = sqrt{x} ) is (　　)" with options A to D. I need to figure out which one is correct. Let me think about what the range of a function means. The range is all the possible output values (y-values) the function can produce. So, for ( f(x) = sqrt{x} ), I need to determine what values ( f(x) ) can take on.First, I remember that the square root function is only defined for non-negative numbers because you can't take the square root of a negative number in the set of real numbers. So, the domain of ( f(x) = sqrt{x} ) is all ( x ) such that ( x geq 0 ). That means ( x ) can be zero or any positive number.Now, considering the outputs, when ( x = 0 ), ( f(0) = sqrt{0} = 0 ). So, the smallest value ( f(x) ) can take is 0. What happens as ( x ) increases? If I plug in larger and larger values of ( x ), say ( x = 1 ), ( x = 4 ), ( x = 9 ), etc., the square root of these numbers gives me 1, 2, 3, and so on. It seems like as ( x ) increases without bound, ( sqrt{x} ) also increases without bound. So, the function can take on any non-negative value starting from 0 up to infinity.Let me check the options given:A: ( [0, +infty) ) B: ( (-infty, 0] ) C: ( (0, +infty) ) D: ( [1, +infty) )Option A says the range is all real numbers from 0 to infinity, including 0. That makes sense because when ( x = 0 ), ( f(x) = 0 ), and as ( x ) increases, ( f(x) ) can be any positive number up to infinity.Option B is all real numbers from negative infinity to 0. But since the square root function only outputs non-negative numbers, this can't be right.Option C is all positive real numbers excluding 0. But wait, when ( x = 0 ), ( f(x) = 0 ), so 0 should be included in the range. Therefore, option C is incorrect because it doesn't include 0.Option D is from 1 to infinity. That would mean the smallest value ( f(x) ) can take is 1, but we know that when ( x = 0 ), ( f(x) = 0 ), which is less than 1. So, option D is also incorrect.Therefore, the correct answer should be option A: ( [0, +infty) ).But just to make sure I haven't missed anything, let me think about the graph of ( f(x) = sqrt{x} ). It starts at the point (0,0) and as ( x ) increases, the graph rises gradually to the right. There's no upper limit to how high ( f(x) ) can go because as ( x ) approaches infinity, ( sqrt{x} ) also approaches infinity. So, the range should indeed cover all non-negative real numbers starting from 0.I don't think there's any restriction on the outputs other than they have to be non-negative. So, 0 is included, and every positive number is attainable by some ( x ) value. That confirms that option A is correct.Another way to think about it is considering the inverse of the function. If I solve ( y = sqrt{x} ) for ( x ), I get ( x = y^2 ). Since ( x ) must be greater than or equal to 0, ( y^2 ) is always non-negative. Therefore, ( y ) must also be non-negative because squaring any real number gives a non-negative result. This again tells me that the range of ( f(x) ) is ( [0, +infty) ).I also remember that for functions involving even roots (like square roots), the outputs are always non-negative. Odd roots, like cube roots, can take on negative values as well, but square roots specifically are confined to non-negative outputs. So, this reinforces that the range starts at 0 and goes up without bound.Is there any chance that 0 isn't included? Well, when ( x = 0 ), ( f(x) = 0 ), so it's definitely included. If it wasn't, the range would be ( (0, +infty) ), but since 0 is attainable, it's included in the range.So, after considering all these points, I'm confident that the range of ( f(x) = sqrt{x} ) is ( [0, +infty) ), which is option A.