answer：Alright, so I have this function ( f(x) = sqrt{x^2 + 3x + 3} + sqrt{x^2 - 3x + 3} ) and I need to find its minimum value for all real numbers ( x ). Hmm, okay, let me try to figure this out step by step.First, I remember that when dealing with functions that have square roots, especially sums of square roots, it can sometimes be helpful to complete the square inside each square root. That might simplify things or reveal some geometric interpretation. Let me try that.Looking at the first term under the square root: ( x^2 + 3x + 3 ). To complete the square, I take the coefficient of ( x ), which is 3, divide it by 2 to get ( frac{3}{2} ), and then square it to get ( left(frac{3}{2}right)^2 = frac{9}{4} ). So, I can rewrite ( x^2 + 3x + 3 ) as:[x^2 + 3x + frac{9}{4} + frac{3}{4} = left(x + frac{3}{2}right)^2 + frac{3}{4}]Similarly, for the second term ( x^2 - 3x + 3 ), I'll complete the square as well. The coefficient of ( x ) is -3, so dividing by 2 gives ( -frac{3}{2} ), and squaring that gives ( frac{9}{4} ). So,[x^2 - 3x + frac{9}{4} + frac{3}{4} = left(x - frac{3}{2}right)^2 + frac{3}{4}]So, substituting these back into the original function, I get:[f(x) = sqrt{left(x + frac{3}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2} + sqrt{left(x - frac{3}{2}right)^2 + left(frac{sqrt{3}}{2}right)^2}]Wait, that looks interesting. Each square root now resembles the distance formula between two points in a plane. Specifically, if I think of each square root as the distance from a point ( (x, 0) ) on the x-axis to another point in the plane. Let me write that down.Let’s define two points:- Point A: ( left(-frac{3}{2}, frac{sqrt{3}}{2}right) )- Point B: ( left(frac{3}{2}, -frac{sqrt{3}}{2}right) )Then, the function ( f(x) ) can be interpreted as the sum of the distances from a point ( P(x, 0) ) on the x-axis to points A and B. So,[f(x) = PA + PB]Where ( PA ) is the distance from ( P ) to ( A ), and ( PB ) is the distance from ( P ) to ( B ).Now, I recall that in geometry, the minimal sum of distances from a moving point to two fixed points is achieved when the moving point lies on the line segment connecting those two fixed points. However, in this case, the moving point ( P ) is restricted to the x-axis. So, the minimal value of ( PA + PB ) would be the distance between points A and B themselves, but only if the line segment AB intersects the x-axis. If not, the minimal value would be the shortest path that reflects off the x-axis.Wait, actually, I think this is related to the reflection principle in optimization problems. If I reflect one of the points across the x-axis, then the minimal path from A to B via the x-axis is the straight line distance between A and the reflection of B, or vice versa.Let me try reflecting point B across the x-axis. The reflection of ( B left(frac{3}{2}, -frac{sqrt{3}}{2}right) ) across the x-axis would be ( B' left(frac{3}{2}, frac{sqrt{3}}{2}right) ).Now, the minimal distance from A to B via the x-axis is the straight line distance from A to B'. So, let me compute that distance.First, compute the distance between A ( left(-frac{3}{2}, frac{sqrt{3}}{2}right) ) and B' ( left(frac{3}{2}, frac{sqrt{3}}{2}right) ).The distance formula is:[AB' = sqrt{left(frac{3}{2} - left(-frac{3}{2}right)right)^2 + left(frac{sqrt{3}}{2} - frac{sqrt{3}}{2}right)^2}]Simplify the differences:[xtext{-difference} = frac{3}{2} + frac{3}{2} = 3][ytext{-difference} = 0]So,[AB' = sqrt{3^2 + 0^2} = sqrt{9} = 3]Wait, that gives me 3, but earlier, when I thought about the minimal value, I thought it might be ( 2sqrt{3} ). Hmm, maybe I made a mistake in reflecting the point or interpreting the problem.Let me double-check. Maybe I should reflect point A instead of B. Let's try reflecting point A across the x-axis. The reflection of ( A left(-frac{3}{2}, frac{sqrt{3}}{2}right) ) would be ( A' left(-frac{3}{2}, -frac{sqrt{3}}{2}right) ).Now, the distance between ( A' ) and ( B ) is:[A'B = sqrt{left(frac{3}{2} - left(-frac{3}{2}right)right)^2 + left(-frac{sqrt{3}}{2} - left(-frac{sqrt{3}}{2}right)right)^2}]Simplify the differences:[xtext{-difference} = frac{3}{2} + frac{3}{2} = 3][ytext{-difference} = 0]So,[A'B = sqrt{3^2 + 0^2} = sqrt{9} = 3]Same result. Hmm, so the minimal distance via reflection is 3. But when I plug in ( x = 0 ) into the original function, let's see what value I get.Compute ( f(0) ):[f(0) = sqrt{0 + 0 + 3} + sqrt{0 + 0 + 3} = sqrt{3} + sqrt{3} = 2sqrt{3} approx 3.464]Wait, that's actually larger than 3. So, if the minimal distance via reflection is 3, but ( f(0) = 2sqrt{3} approx 3.464 ), which is larger, then my reflection approach might not be giving the correct minimal value.Perhaps I need to think differently. Maybe the minimal value is indeed ( 2sqrt{3} ), achieved at ( x = 0 ). Let me check another point, say ( x = frac{3}{2} ).Compute ( fleft(frac{3}{2}right) ):First term: ( sqrt{left(frac{3}{2}right)^2 + 3left(frac{3}{2}right) + 3} = sqrt{frac{9}{4} + frac{9}{2} + 3} = sqrt{frac{9}{4} + frac{18}{4} + frac{12}{4}} = sqrt{frac{39}{4}} = frac{sqrt{39}}{2} approx 3.122 )Second term: ( sqrt{left(frac{3}{2}right)^2 - 3left(frac{3}{2}right) + 3} = sqrt{frac{9}{4} - frac{9}{2} + 3} = sqrt{frac{9}{4} - frac{18}{4} + frac{12}{4}} = sqrt{frac{3}{4}} = frac{sqrt{3}}{2} approx 0.866 )So, ( fleft(frac{3}{2}right) approx 3.122 + 0.866 approx 3.988 ), which is larger than ( 2sqrt{3} ).Wait, so maybe ( x = 0 ) is indeed the point where the function attains its minimal value. Let me try another point, say ( x = 1 ).Compute ( f(1) ):First term: ( sqrt{1 + 3 + 3} = sqrt{7} approx 2.645 )Second term: ( sqrt{1 - 3 + 3} = sqrt{1} = 1 )So, ( f(1) approx 2.645 + 1 = 3.645 ), which is still larger than ( 2sqrt{3} approx 3.464 ).Hmm, so it seems like ( x = 0 ) gives the smallest value so far. Let me try ( x = -frac{3}{2} ).Compute ( fleft(-frac{3}{2}right) ):First term: ( sqrt{left(-frac{3}{2}right)^2 + 3left(-frac{3}{2}right) + 3} = sqrt{frac{9}{4} - frac{9}{2} + 3} = sqrt{frac{9}{4} - frac{18}{4} + frac{12}{4}} = sqrt{frac{3}{4}} = frac{sqrt{3}}{2} approx 0.866 )Second term: ( sqrt{left(-frac{3}{2}right)^2 - 3left(-frac{3}{2}right) + 3} = sqrt{frac{9}{4} + frac{9}{2} + 3} = sqrt{frac{9}{4} + frac{18}{4} + frac{12}{4}} = sqrt{frac{39}{4}} = frac{sqrt{39}}{2} approx 3.122 )So, ( fleft(-frac{3}{2}right) approx 0.866 + 3.122 approx 3.988 ), same as before.Okay, so ( x = 0 ) gives ( 2sqrt{3} approx 3.464 ), which is smaller than the other points I've tried. Maybe that's the minimum. But I need to confirm this more rigorously.Let me consider taking the derivative of ( f(x) ) and setting it to zero to find critical points. That should give me the exact minimum.So, ( f(x) = sqrt{x^2 + 3x + 3} + sqrt{x^2 - 3x + 3} )Let me denote:[f(x) = sqrt{g(x)} + sqrt{h(x)}]where ( g(x) = x^2 + 3x + 3 ) and ( h(x) = x^2 - 3x + 3 )Then, the derivative ( f'(x) ) is:[f'(x) = frac{g'(x)}{2sqrt{g(x)}} + frac{h'(x)}{2sqrt{h(x)}}]Compute ( g'(x) ) and ( h'(x) ):[g'(x) = 2x + 3][h'(x) = 2x - 3]So,[f'(x) = frac{2x + 3}{2sqrt{x^2 + 3x + 3}} + frac{2x - 3}{2sqrt{x^2 - 3x + 3}}]To find critical points, set ( f'(x) = 0 ):[frac{2x + 3}{2sqrt{x^2 + 3x + 3}} + frac{2x - 3}{2sqrt{x^2 - 3x + 3}} = 0]Multiply both sides by 2 to eliminate denominators:[frac{2x + 3}{sqrt{x^2 + 3x + 3}} + frac{2x - 3}{sqrt{x^2 - 3x + 3}} = 0]Let me denote ( A = sqrt{x^2 + 3x + 3} ) and ( B = sqrt{x^2 - 3x + 3} ), so the equation becomes:[frac{2x + 3}{A} + frac{2x - 3}{B} = 0]Multiply both sides by ( AB ):[(2x + 3)B + (2x - 3)A = 0]Expand this:[(2x + 3)sqrt{x^2 - 3x + 3} + (2x - 3)sqrt{x^2 + 3x + 3} = 0]This looks complicated. Maybe I can square both sides to eliminate the square roots, but I have to be careful because squaring can introduce extraneous solutions.Let me rearrange the equation:[(2x + 3)sqrt{x^2 - 3x + 3} = - (2x - 3)sqrt{x^2 + 3x + 3}]Now, square both sides:[(2x + 3)^2 (x^2 - 3x + 3) = (2x - 3)^2 (x^2 + 3x + 3)]Let me compute both sides.First, compute ( (2x + 3)^2 ):[(2x + 3)^2 = 4x^2 + 12x + 9]Multiply this by ( x^2 - 3x + 3 ):[(4x^2 + 12x + 9)(x^2 - 3x + 3)]Let me expand this:First, multiply ( 4x^2 ) by each term in ( x^2 - 3x + 3 ):[4x^2 cdot x^2 = 4x^4][4x^2 cdot (-3x) = -12x^3][4x^2 cdot 3 = 12x^2]Next, multiply ( 12x ) by each term:[12x cdot x^2 = 12x^3][12x cdot (-3x) = -36x^2][12x cdot 3 = 36x]Then, multiply ( 9 ) by each term:[9 cdot x^2 = 9x^2][9 cdot (-3x) = -27x][9 cdot 3 = 27]Now, add all these terms together:[4x^4 - 12x^3 + 12x^2 + 12x^3 - 36x^2 + 36x + 9x^2 - 27x + 27]Combine like terms:- ( x^4 ): ( 4x^4 )- ( x^3 ): ( -12x^3 + 12x^3 = 0 )- ( x^2 ): ( 12x^2 - 36x^2 + 9x^2 = -15x^2 )- ( x ): ( 36x - 27x = 9x )- Constants: ( 27 )So, the left side becomes:[4x^4 - 15x^2 + 9x + 27]Now, compute the right side: ( (2x - 3)^2 (x^2 + 3x + 3) )First, compute ( (2x - 3)^2 ):[(2x - 3)^2 = 4x^2 - 12x + 9]Multiply this by ( x^2 + 3x + 3 ):[(4x^2 - 12x + 9)(x^2 + 3x + 3)]Again, expand term by term:First, multiply ( 4x^2 ) by each term in ( x^2 + 3x + 3 ):[4x^2 cdot x^2 = 4x^4][4x^2 cdot 3x = 12x^3][4x^2 cdot 3 = 12x^2]Next, multiply ( -12x ) by each term:[-12x cdot x^2 = -12x^3][-12x cdot 3x = -36x^2][-12x cdot 3 = -36x]Then, multiply ( 9 ) by each term:[9 cdot x^2 = 9x^2][9 cdot 3x = 27x][9 cdot 3 = 27]Now, add all these terms together:[4x^4 + 12x^3 + 12x^2 - 12x^3 - 36x^2 - 36x + 9x^2 + 27x + 27]Combine like terms:- ( x^4 ): ( 4x^4 )- ( x^3 ): ( 12x^3 - 12x^3 = 0 )- ( x^2 ): ( 12x^2 - 36x^2 + 9x^2 = -15x^2 )- ( x ): ( -36x + 27x = -9x )- Constants: ( 27 )So, the right side becomes:[4x^4 - 15x^2 - 9x + 27]Now, set the left side equal to the right side:[4x^4 - 15x^2 + 9x + 27 = 4x^4 - 15x^2 - 9x + 27]Subtract ( 4x^4 - 15x^2 + 27 ) from both sides:[9x = -9x]Which simplifies to:[18x = 0 implies x = 0]So, the critical point is at ( x = 0 ). Now, I need to verify if this is indeed a minimum.Let me check the second derivative or analyze the behavior around ( x = 0 ). Alternatively, since I already computed ( f(0) = 2sqrt{3} ) and other points give higher values, it's likely that ( x = 0 ) is the point of minimum.Therefore, the minimal value of ( f(x) ) is ( 2sqrt{3} ).**Final Answer**The minimum value of the function is boxed{2sqrt{3}}.

answer：Okay, so I have this problem about mutually exclusive events, and I need to figure out which pairs of events are mutually exclusive. There are four pairs given, and I need to check each one. Let me start by recalling what mutually exclusive means. If two events are mutually exclusive, it means they can't happen at the same time. In other words, there's no overlap between them; their intersection is empty. So, if one happens, the other can't happen, and vice versa.Alright, let's go through each pair one by one.**1. A person shoots once, "hits the 7th ring" and "hits the 8th ring".**Hmm, okay. So, the person is shooting once. If they hit the 7th ring, that means they scored a 7. If they hit the 8th ring, they scored an 8. Since they're only shooting once, can they hit both the 7th and 8th rings at the same time? I don't think so. Each shot can only land in one ring, right? So, hitting the 7th ring and hitting the 8th ring are two distinct outcomes. They can't both happen on a single shot. Therefore, these two events are mutually exclusive.**2. Person A and person B each shoot once, "at least one person hits the target" and "A hits the target, but B does not".**Alright, let's parse this. The first event is "at least one person hits the target." That means either A hits, or B hits, or both hit. The second event is specifically "A hits the target, but B does not." So, is there an overlap here? Let's see.If "A hits the target, but B does not" happens, does that mean "at least one person hits the target" also happens? Yes, because A did hit the target. So, the second event is actually a subset of the first event. They can occur together. Therefore, these two events are not mutually exclusive because there's an overlap. The second event is included within the first one.**3. Drawing 2 balls from a bag containing 2 red balls and 2 black balls, "at least one black ball" and "both are red balls".**Okay, so we have a bag with 2 red and 2 black balls. We're drawing two balls. The first event is "at least one black ball," which means either one or both balls are black. The second event is "both are red balls," which is the opposite—meaning neither is black. So, can these two events happen at the same time? If I draw two balls and both are red, then I haven't drawn any black balls. So, "both are red" means "at least one black" didn't happen. Conversely, if I have at least one black ball, then I can't have both red. So, these two events can't occur together. Therefore, they are mutually exclusive.**4. Drawing 2 balls from a bag containing 2 red balls and 2 black balls, "no black balls" and "exactly one red ball".**Alright, let's break this down. The first event is "no black balls," which means both balls are red. The second event is "exactly one red ball," which means one red and one black. So, can these two happen at the same time? If I have no black balls, that means both are red, which is two reds. On the other hand, exactly one red ball would mean one red and one black. These are completely different outcomes. You can't have both no black balls and exactly one red ball in the same draw. So, these two events are mutually exclusive.Wait, hold on. Let me double-check. If I draw two balls, "no black balls" implies both are red. "Exactly one red ball" implies one red and one black. So, yes, they can't happen together. So, they are mutually exclusive.So, summarizing:1. Mutually exclusive.2. Not mutually exclusive.3. Mutually exclusive.4. Mutually exclusive.Therefore, the mutually exclusive pairs are 1, 3, and 4.**Final Answer**The mutually exclusive events are boxed{1, 3, 4}.

answer：Alright, so I've got this problem here: Given a positive integer ( n ), let ( f(n) ) be the smallest positive integer such that the sum of the first ( f(n) ) positive integers is divisible by ( n ). I need to prove that ( f(n) = 2n - 1 ) if and only if ( n ) is a power of 2. Hmm, okay. Let me try to unpack this step by step.First, let's make sure I understand the problem correctly. For any positive integer ( n ), I need to find the smallest ( f(n) ) such that the sum ( 1 + 2 + 3 + dots + f(n) ) is divisible by ( n ). And I have to show that this smallest ( f(n) ) is equal to ( 2n - 1 ) exactly when ( n ) is a power of 2. So, if ( n ) is like 2, 4, 8, 16, etc., then ( f(n) = 2n - 1 ). For other numbers, ( f(n) ) should be something else.Alright, let's start by recalling the formula for the sum of the first ( m ) positive integers. That's ( frac{m(m + 1)}{2} ). So, we're looking for the smallest ( m ) such that ( frac{m(m + 1)}{2} ) is divisible by ( n ). In other words, ( n ) divides ( frac{m(m + 1)}{2} ). So, ( frac{m(m + 1)}{2} ) must be a multiple of ( n ).So, mathematically, we can write this as:[n mid frac{m(m + 1)}{2}]Which implies that:[frac{m(m + 1)}{2} = kn]for some integer ( k ).But we need the smallest such ( m ), which is ( f(n) ). So, ( f(n) ) is the minimal ( m ) such that ( kn = frac{m(m + 1)}{2} ).Now, the problem states that ( f(n) = 2n - 1 ) if and only if ( n ) is a power of 2. So, I need to prove both directions:1. If ( n ) is a power of 2, then ( f(n) = 2n - 1 ).2. If ( f(n) = 2n - 1 ), then ( n ) must be a power of 2.Let me tackle the first part first: If ( n ) is a power of 2, then ( f(n) = 2n - 1 ).Suppose ( n = 2^k ) for some integer ( k geq 1 ). Then, we need to show that the smallest ( m ) such that ( frac{m(m + 1)}{2} ) is divisible by ( 2^k ) is ( m = 2n - 1 = 2^{k+1} - 1 ).Let me compute ( frac{m(m + 1)}{2} ) when ( m = 2^{k+1} - 1 ):[frac{(2^{k+1} - 1)(2^{k+1})}{2} = frac{(2^{k+1} - 1)2^{k+1}}{2} = (2^{k+1} - 1)2^{k}]So, ( frac{m(m + 1)}{2} = (2^{k+1} - 1)2^{k} ). Since ( n = 2^k ), this expression is clearly divisible by ( n ), because ( 2^k ) is a factor.Now, I need to make sure that this is indeed the smallest such ( m ). That is, for any ( m < 2^{k+1} - 1 ), ( frac{m(m + 1)}{2} ) is not divisible by ( 2^k ).Let me think about this. For ( m < 2^{k+1} - 1 ), we can write ( m = 2^{k+1} - 1 - t ) where ( t geq 1 ). Then, ( m + 1 = 2^{k+1} - t ).So, the product ( m(m + 1) = (2^{k+1} - 1 - t)(2^{k+1} - t) ). Hmm, not sure if that's helpful.Alternatively, maybe I can think about the factors of ( m ) and ( m + 1 ). Since ( m ) and ( m + 1 ) are consecutive integers, they are coprime. So, the factors of ( 2^k ) must divide either ( m ) or ( m + 1 ), but not both.So, for ( frac{m(m + 1)}{2} ) to be divisible by ( 2^k ), either ( m ) or ( m + 1 ) must be divisible by ( 2^{k+1} ), because when we divide by 2, we need to have at least ( 2^k ) remaining.Wait, let me think again. The sum is ( frac{m(m + 1)}{2} ). So, for this to be divisible by ( 2^k ), the numerator ( m(m + 1) ) must be divisible by ( 2^{k+1} ), since we're dividing by 2.Therefore, ( m(m + 1) ) must be divisible by ( 2^{k+1} ). Since ( m ) and ( m + 1 ) are consecutive, one of them is even, and the other is odd. So, the even one must be divisible by ( 2^{k+1} ).So, either ( m ) is divisible by ( 2^{k+1} ) or ( m + 1 ) is divisible by ( 2^{k+1} ). But since ( m ) and ( m + 1 ) are consecutive, only one of them can be even, so only one can be divisible by ( 2^{k+1} ).Therefore, the smallest ( m ) such that either ( m ) or ( m + 1 ) is divisible by ( 2^{k+1} ) is ( m = 2^{k+1} - 1 ). Because ( m + 1 = 2^{k+1} ), which is divisible by ( 2^{k+1} ), so ( m = 2^{k+1} - 1 ).Therefore, ( f(n) = 2^{k+1} - 1 = 2n - 1 ), since ( n = 2^k ).Okay, that seems to check out for the first direction. Now, the second direction: If ( f(n) = 2n - 1 ), then ( n ) must be a power of 2.So, suppose that ( f(n) = 2n - 1 ). Then, this means that the smallest ( m ) such that ( frac{m(m + 1)}{2} ) is divisible by ( n ) is ( m = 2n - 1 ).We need to show that ( n ) must be a power of 2.Let me think about this. Suppose ( n ) is not a power of 2. Then, ( n ) has some odd prime factor, say ( p ). Let me see if I can find a smaller ( m ) such that ( frac{m(m + 1)}{2} ) is divisible by ( n ), which would contradict the minimality of ( f(n) = 2n - 1 ).So, if ( n ) is not a power of 2, it can be written as ( n = 2^k cdot m ), where ( m ) is an odd integer greater than 1.Now, let's consider the sum ( frac{r(r + 1)}{2} ). For this to be divisible by ( n = 2^k cdot m ), ( r(r + 1) ) must be divisible by ( 2^{k+1} cdot m ).Since ( r ) and ( r + 1 ) are consecutive integers, they are coprime. Therefore, the factors of ( 2^{k+1} ) must divide one of them, and the factors of ( m ) must divide the other.So, we can set up a system where either:1. ( r equiv 0 mod 2^{k+1} ) and ( r + 1 equiv 0 mod m ), or2. ( r equiv 0 mod m ) and ( r + 1 equiv 0 mod 2^{k+1} ).In either case, we can use the Chinese Remainder Theorem to find a solution ( r ) that satisfies both congruences. Since ( m ) is odd and ( 2^{k+1} ) is a power of 2, they are coprime. Therefore, there exists a unique solution modulo ( 2^{k+1} cdot m ).Therefore, there exists some ( r ) less than ( 2^{k+1} cdot m ) such that ( r(r + 1)/2 ) is divisible by ( n = 2^k cdot m ). But ( 2^{k+1} cdot m = 2n ). So, ( r ) can be as small as ( 2n - 1 ) or smaller.Wait, but ( f(n) = 2n - 1 ) is supposed to be the minimal such ( r ). So, if ( n ) is not a power of 2, then ( r ) can be smaller than ( 2n - 1 ), which would contradict the minimality of ( f(n) = 2n - 1 ).Therefore, ( n ) must be a power of 2.Wait, let me make sure I didn't skip any steps here. So, if ( n ) is not a power of 2, then it has an odd prime factor, which allows us to find a smaller ( r ) such that ( r(r + 1)/2 ) is divisible by ( n ). Therefore, ( f(n) ) cannot be ( 2n - 1 ) in that case. Hence, ( f(n) = 2n - 1 ) only if ( n ) is a power of 2.I think that makes sense. Let me try to think of an example to test this.Take ( n = 4 ), which is a power of 2. Then, ( f(n) = 2*4 - 1 = 7 ). Let's compute the sum up to 7: ( 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 ). And 28 divided by 4 is 7, which is an integer. Now, is there a smaller ( m ) such that the sum is divisible by 4? Let's check:- ( m = 1 ): sum = 1, not divisible by 4.- ( m = 2 ): sum = 3, not divisible by 4.- ( m = 3 ): sum = 6, not divisible by 4.- ( m = 4 ): sum = 10, not divisible by 4.- ( m = 5 ): sum = 15, not divisible by 4.- ( m = 6 ): sum = 21, not divisible by 4.- ( m = 7 ): sum = 28, which is divisible by 4.So, indeed, the smallest ( m ) is 7, which is ( 2n - 1 ).Now, let's take ( n = 6 ), which is not a power of 2. Let's compute ( f(n) ). We need the smallest ( m ) such that ( frac{m(m + 1)}{2} ) is divisible by 6.Let's compute the sums:- ( m = 1 ): 1, not divisible by 6.- ( m = 2 ): 3, not divisible by 6.- ( m = 3 ): 6, which is divisible by 6. So, ( f(6) = 3 ), which is less than ( 2*6 - 1 = 11 ). Therefore, ( f(n) neq 2n - 1 ) when ( n ) is not a power of 2.Another example: ( n = 3 ), which is not a power of 2. Let's find ( f(3) ).- ( m = 1 ): 1, not divisible by 3.- ( m = 2 ): 3, which is divisible by 3. So, ( f(3) = 2 ), which is less than ( 2*3 - 1 = 5 ).Wait, but 3 is a power of 3, not 2, so it's not a power of 2. So, ( f(n) ) is indeed smaller than ( 2n - 1 ).Another example: ( n = 5 ), which is not a power of 2.- ( m = 1 ): 1, not divisible by 5.- ( m = 2 ): 3, not divisible by 5.- ( m = 3 ): 6, not divisible by 5.- ( m = 4 ): 10, which is divisible by 5. So, ( f(5) = 4 ), which is less than ( 2*5 - 1 = 9 ).So, again, ( f(n) ) is smaller than ( 2n - 1 ) when ( n ) is not a power of 2.Wait, but let me check ( n = 1 ). ( n = 1 ) is ( 2^0 ), so it's a power of 2. Then, ( f(1) = 2*1 - 1 = 1 ). The sum up to 1 is 1, which is divisible by 1. That works.What about ( n = 2 )? ( f(2) = 2*2 - 1 = 3 ). The sum up to 3 is 6, which is divisible by 2. And indeed, there's no smaller ( m ) than 3 for which the sum is divisible by 2:- ( m = 1 ): 1, not divisible by 2.- ( m = 2 ): 3, not divisible by 2.- ( m = 3 ): 6, which is divisible by 2.So, that works too.Wait, but hold on. For ( n = 2 ), the sum up to 2 is 3, which is not divisible by 2, but the sum up to 3 is 6, which is divisible by 2. So, ( f(2) = 3 ), which is ( 2*2 - 1 ). So, that's consistent.Another test: ( n = 8 ), which is ( 2^3 ). Then, ( f(n) = 2*8 - 1 = 15 ). Let's compute the sum up to 15: ( frac{15*16}{2} = 120 ). 120 divided by 8 is 15, which is an integer. Now, is there a smaller ( m ) such that the sum is divisible by 8?Let's check:- ( m = 1 ): 1, not divisible by 8.- ( m = 2 ): 3, not divisible by 8.- ( m = 3 ): 6, not divisible by 8.- ( m = 4 ): 10, not divisible by 8.- ( m = 5 ): 15, not divisible by 8.- ( m = 6 ): 21, not divisible by 8.- ( m = 7 ): 28, not divisible by 8.- ( m = 8 ): 36, not divisible by 8.- ( m = 9 ): 45, not divisible by 8.- ( m = 10 ): 55, not divisible by 8.- ( m = 11 ): 66, not divisible by 8.- ( m = 12 ): 78, not divisible by 8.- ( m = 13 ): 91, not divisible by 8.- ( m = 14 ): 105, not divisible by 8.- ( m = 15 ): 120, which is divisible by 8.So, indeed, the smallest ( m ) is 15, which is ( 2n - 1 ).Now, let's think about why this works for powers of 2. It seems that when ( n ) is a power of 2, the minimal ( m ) is forced to be ( 2n - 1 ) because of the way the factors of 2 are distributed in the product ( m(m + 1) ). Since ( m ) and ( m + 1 ) are consecutive, one is even and the other is odd. So, to get enough factors of 2 in the product, the even one has to be a multiple of ( 2^{k+1} ), which leads to ( m = 2^{k+1} - 1 ).On the other hand, when ( n ) is not a power of 2, it has some odd prime factors. This allows us to find a smaller ( m ) where the product ( m(m + 1) ) can be divisible by ( n ) without needing to go all the way up to ( 2n - 1 ). This is because the odd prime factors can be accommodated in the odd part of the product, allowing the even part to not need as many factors of 2.So, putting it all together, the key idea is that for powers of 2, the minimal ( m ) is constrained by the number of factors of 2 needed, which leads to ( m = 2n - 1 ). For other numbers, the presence of odd prime factors allows for a smaller ( m ) to satisfy the divisibility condition.I think I've got a good grasp on this now. The proof hinges on the properties of consecutive integers and their factors, especially regarding powers of 2 and the Chinese Remainder Theorem for handling composite numbers with odd prime factors.

answer：Alright, so I have this problem here with two curves, C₁ and C₂. I need to find their Cartesian equations and then, when α is π/4, find the equation of the circle with segment MN as its diameter where M and N are the intersection points of C₁ and C₂.Starting with part (1): finding the Cartesian equations.First, let's look at curve C₁. It's given parametrically with x and y in terms of t and α. The parametric equations are:x = 2 + t cos αy = 1 + t sin αSo, t is the parameter, and α is between 0 and π. I need to eliminate the parameter t to get the Cartesian equation.Hmm, okay. So, from the first equation, I can solve for t:x = 2 + t cos αSubtract 2: x - 2 = t cos αSo, t = (x - 2)/cos αSimilarly, from the second equation:y = 1 + t sin αSubtract 1: y - 1 = t sin αSo, t = (y - 1)/sin αSince both expressions equal t, I can set them equal to each other:(x - 2)/cos α = (y - 1)/sin αCross-multiplying:(x - 2) sin α = (y - 1) cos αLet me rearrange this:y - 1 = [(sin α)/cos α] (x - 2)Which simplifies to:y - 1 = tan α (x - 2)So, that's the equation of a straight line. So, curve C₁ is a straight line with slope tan α passing through the point (2, 1). That makes sense because parametric equations like this usually represent lines.But wait, what if cos α is zero? That would be when α = π/2. In that case, the original parametric equations become:x = 2 + t * 0 = 2y = 1 + t * 1 = 1 + tSo, when α = π/2, the curve is a vertical line at x = 2. So, I should note that case separately.So, summarizing, the Cartesian equation for C₁ is:If α ≠ π/2, then y = tan α (x - 2) + 1If α = π/2, then x = 2Alright, that's done for C₁.Now, moving on to curve C₂. It's given in polar coordinates:ρ² = 2 / (1 + cos² θ)I need to convert this into Cartesian coordinates. Remember that in polar coordinates, ρ² = x² + y², and cos θ = x / ρ. So, let's substitute these into the equation.Starting with:ρ² = 2 / (1 + cos² θ)Substitute ρ² = x² + y² and cos θ = x / ρ:x² + y² = 2 / [1 + (x² / (x² + y²))]Let me simplify the denominator:1 + (x² / (x² + y²)) = [ (x² + y²) + x² ] / (x² + y² ) = (2x² + y²) / (x² + y²)So, the equation becomes:x² + y² = 2 / [ (2x² + y²) / (x² + y²) ) ] = 2 * (x² + y²) / (2x² + y²)So, multiply both sides by (2x² + y²):(x² + y²)(2x² + y²) = 2(x² + y²)Assuming x² + y² ≠ 0, we can divide both sides by (x² + y²):2x² + y² = 2So, the Cartesian equation is 2x² + y² = 2. Let me write that as:2x² + y² = 2Alternatively, dividing both sides by 2:x² + (y²)/2 = 1So, that's an ellipse centered at the origin, with semi-major axis along the y-axis, since the denominator under y² is larger.Wait, let me check my steps again to make sure I didn't make a mistake.Starting from ρ² = 2 / (1 + cos² θ)Expressed in Cartesian:x² + y² = 2 / [1 + (x² / (x² + y²))]Simplify denominator:1 + x² / (x² + y²) = (x² + y² + x²) / (x² + y²) = (2x² + y²) / (x² + y²)So, plug back in:x² + y² = 2 / [ (2x² + y²) / (x² + y²) ) ] = 2 * (x² + y²) / (2x² + y²)Multiply both sides by (2x² + y²):(x² + y²)(2x² + y²) = 2(x² + y²)Assuming x² + y² ≠ 0, divide both sides:2x² + y² = 2Yes, that seems correct.So, curve C₂ is an ellipse with equation 2x² + y² = 2.Alright, part (1) is done.Moving on to part (2): When α = π/4, curves C₁ and C₂ intersect at points M and N. Find the Cartesian equation of the circle with segment MN as its diameter.First, let's substitute α = π/4 into the equation of C₁.From part (1), when α ≠ π/2, the equation is y = tan α (x - 2) + 1.Since α = π/4, tan(π/4) = 1.So, the equation becomes:y = 1*(x - 2) + 1 = x - 2 + 1 = x - 1So, C₁ is the line y = x - 1.Now, we need to find the points of intersection between this line and the ellipse C₂: 2x² + y² = 2.So, substitute y = x - 1 into the ellipse equation:2x² + (x - 1)^2 = 2Let me expand (x - 1)^2:(x - 1)^2 = x² - 2x + 1So, plug that in:2x² + x² - 2x + 1 = 2Combine like terms:3x² - 2x + 1 = 2Subtract 2 from both sides:3x² - 2x - 1 = 0So, quadratic equation: 3x² - 2x - 1 = 0Let me solve for x using quadratic formula.x = [2 ± sqrt( ( -2 )² - 4*3*(-1) ) ] / (2*3)Compute discriminant:D = 4 + 12 = 16So, sqrt(D) = 4Thus,x = [2 ± 4]/6So, two solutions:x = (2 + 4)/6 = 6/6 = 1x = (2 - 4)/6 = (-2)/6 = -1/3So, x = 1 and x = -1/3.Now, find corresponding y values using y = x - 1.For x = 1:y = 1 - 1 = 0So, point M is (1, 0)For x = -1/3:y = (-1/3) - 1 = -4/3So, point N is (-1/3, -4/3)So, points M(1, 0) and N(-1/3, -4/3)Now, need to find the equation of the circle with MN as diameter.To find the equation of a circle given two endpoints of its diameter, we can use the midpoint as the center and half the distance between M and N as the radius.First, find the midpoint (center) of MN.Midpoint formula: ((x₁ + x₂)/2, (y₁ + y₂)/2)So,x_center = (1 + (-1/3))/2 = (2/3)/2 = 1/3y_center = (0 + (-4/3))/2 = (-4/3)/2 = -2/3So, center is (1/3, -2/3)Now, find the radius. The radius is half the distance between M and N.First, compute distance between M(1, 0) and N(-1/3, -4/3).Distance formula: sqrt[(x₂ - x₁)^2 + (y₂ - y₁)^2]Compute differences:Δx = (-1/3 - 1) = (-4/3)Δy = (-4/3 - 0) = (-4/3)So,Distance MN = sqrt[ (-4/3)^2 + (-4/3)^2 ] = sqrt[ 16/9 + 16/9 ] = sqrt[32/9] = (4√2)/3So, radius is half of that: (2√2)/3Therefore, the equation of the circle is:(x - 1/3)^2 + (y + 2/3)^2 = (2√2 / 3)^2Compute the right-hand side:(2√2 / 3)^2 = (4*2)/9 = 8/9So, equation is:(x - 1/3)^2 + (y + 2/3)^2 = 8/9Let me just double-check my calculations.Midpoint: (1 + (-1/3))/2 = (2/3)/2 = 1/3. Correct.(0 + (-4/3))/2 = (-4/3)/2 = -2/3. Correct.Distance: sqrt[ (-4/3)^2 + (-4/3)^2 ] = sqrt[32/9] = 4√2 / 3. Correct.Radius: 2√2 / 3. Squared is 8/9. Correct.So, the equation is as above.Therefore, the Cartesian equation of the circle is (x - 1/3)^2 + (y + 2/3)^2 = 8/9.**Final Answer**The Cartesian equation of the circle is boxed{left(x - frac{1}{3}right)^2 + left(y + frac{2}{3}right)^2 = frac{8}{9}}.

answer：Okay, so I have this problem where I need to find the smallest natural number ( n ) such that there exist real numbers ( x_1, x_2, ldots, x_n ) satisfying three conditions. Let me try to break this down step by step.First, the conditions are:1. Each ( x_i ) is between ( frac{1}{2} ) and ( 2 ).2. The sum of all ( x_i ) is at least ( frac{7n}{6} ).3. The sum of the reciprocals of all ( x_i ) is at least ( frac{4n}{3} ).Hmm, okay. So, each ( x_i ) is constrained between ( frac{1}{2} ) and ( 2 ). That means each ( x_i ) can't be too small or too large. The second condition is about the sum of these ( x_i )s being relatively large, specifically ( frac{7}{6} ) times ( n ). The third condition is about the sum of their reciprocals being even larger, ( frac{4}{3} ) times ( n ).I think I need to find the smallest ( n ) where such numbers ( x_i ) can exist. Maybe I should start by looking at the inequalities and see if I can find a relationship between them.Let me consider the reciprocals. Since each ( x_i ) is between ( frac{1}{2} ) and ( 2 ), their reciprocals will be between ( frac{1}{2} ) and ( 2 ) as well. Wait, no. If ( x_i ) is ( frac{1}{2} ), then ( frac{1}{x_i} ) is ( 2 ), and if ( x_i ) is ( 2 ), then ( frac{1}{x_i} ) is ( frac{1}{2} ). So, the reciprocals are also in the same interval ( [frac{1}{2}, 2] ).But how does that help me? Maybe I can use some inequality here, like the Cauchy-Schwarz inequality or AM-HM inequality.Let me recall the AM-HM inequality: For positive real numbers, the arithmetic mean is always greater than or equal to the harmonic mean. So, for our case:[frac{x_1 + x_2 + ldots + x_n}{n} geq frac{n}{frac{1}{x_1} + frac{1}{x_2} + ldots + frac{1}{x_n}}]Plugging in the given conditions:[frac{frac{7n}{6}}{n} geq frac{n}{frac{4n}{3}}]Simplifying:[frac{7}{6} geq frac{n}{frac{4n}{3}} = frac{3}{4}]Wait, that's true because ( frac{7}{6} ) is approximately ( 1.1667 ) and ( frac{3}{4} ) is ( 0.75 ). So, this inequality holds, but it doesn't give me any new information because it's just confirming that the AM is greater than the HM, which is always true.Maybe I need a different approach. Let me think about how to maximize the sum of reciprocals given the constraints on the ( x_i )s.If I want the sum of reciprocals to be as large as possible, I should have as many ( x_i )s as possible equal to ( frac{1}{2} ), because their reciprocals would be ( 2 ), which is the maximum. Similarly, if I want the sum of ( x_i )s to be as large as possible, I should have as many ( x_i )s as possible equal to ( 2 ).But in this problem, we have both sums needing to be large. So, there's a trade-off here. If I set some ( x_i )s to ( 2 ), their reciprocals will be ( frac{1}{2} ), which is the minimum. Conversely, if I set some ( x_i )s to ( frac{1}{2} ), their reciprocals will be ( 2 ), which is the maximum.So, maybe I need a balance between the number of ( x_i )s set to ( 2 ) and ( frac{1}{2} ). Let me denote ( p ) as the number of ( x_i )s set to ( frac{1}{2} ) and ( q ) as the number set to ( 2 ). Then, ( p + q = n ).The sum of ( x_i )s would be ( frac{p}{2} + 2q ), and the sum of reciprocals would be ( 2p + frac{q}{2} ).Given the conditions:1. ( frac{p}{2} + 2q geq frac{7n}{6} )2. ( 2p + frac{q}{2} geq frac{4n}{3} )Since ( p + q = n ), I can express ( q ) as ( n - p ). Let me substitute ( q ) in both inequalities.First inequality:[frac{p}{2} + 2(n - p) geq frac{7n}{6}]Simplify:[frac{p}{2} + 2n - 2p geq frac{7n}{6}][2n - frac{3p}{2} geq frac{7n}{6}]Multiply both sides by 6 to eliminate denominators:[12n - 9p geq 7n][12n - 7n geq 9p][5n geq 9p][p leq frac{5n}{9}]Second inequality:[2p + frac{n - p}{2} geq frac{4n}{3}]Simplify:[2p + frac{n}{2} - frac{p}{2} geq frac{4n}{3}][frac{3p}{2} + frac{n}{2} geq frac{4n}{3}]Multiply both sides by 6:[9p + 3n geq 8n][9p geq 5n][p geq frac{5n}{9}]So from the first inequality, ( p leq frac{5n}{9} ), and from the second, ( p geq frac{5n}{9} ). Therefore, ( p = frac{5n}{9} ).Since ( p ) must be an integer (as it's the number of terms), ( frac{5n}{9} ) must be an integer. Therefore, ( n ) must be a multiple of 9. The smallest natural number ( n ) that satisfies this is ( n = 9 ).Let me verify this. If ( n = 9 ), then ( p = frac{5 times 9}{9} = 5 ), so ( p = 5 ) and ( q = 4 ).Sum of ( x_i )s: ( frac{5}{2} + 2 times 4 = frac{5}{2} + 8 = frac{21}{2} = 10.5 ). Since ( frac{7n}{6} = frac{63}{6} = 10.5 ), this satisfies the second condition.Sum of reciprocals: ( 2 times 5 + frac{4}{2} = 10 + 2 = 12 ). Since ( frac{4n}{3} = frac{36}{3} = 12 ), this satisfies the third condition.So, ( n = 9 ) works. Is there a smaller ( n ) that could work? Let's check ( n = 8 ).If ( n = 8 ), then ( p = frac{5 times 8}{9} approx 4.444 ), which is not an integer. So, ( p ) can't be 4.444; it has to be either 4 or 5.If ( p = 4 ), then ( q = 4 ). Sum of ( x_i )s: ( frac{4}{2} + 2 times 4 = 2 + 8 = 10 ). ( frac{7n}{6} = frac{56}{6} approx 9.333 ). So, 10 is greater than 9.333, which is good.Sum of reciprocals: ( 2 times 4 + frac{4}{2} = 8 + 2 = 10 ). ( frac{4n}{3} = frac{32}{3} approx 10.666 ). So, 10 is less than 10.666, which doesn't satisfy the third condition.If ( p = 5 ), then ( q = 3 ). Sum of ( x_i )s: ( frac{5}{2} + 2 times 3 = 2.5 + 6 = 8.5 ). ( frac{7n}{6} = frac{56}{6} approx 9.333 ). So, 8.5 is less than 9.333, which doesn't satisfy the second condition.Therefore, ( n = 8 ) doesn't work. Similarly, checking ( n = 7 ):( p = frac{5 times 7}{9} approx 3.888 ). Not an integer. Trying ( p = 3 ) and ( p = 4 ).If ( p = 3 ), ( q = 4 ). Sum of ( x_i )s: ( 1.5 + 8 = 9.5 ). ( frac{7 times 7}{6} approx 8.166 ). Good.Sum of reciprocals: ( 6 + 2 = 8 ). ( frac{4 times 7}{3} approx 9.333 ). Not enough.If ( p = 4 ), ( q = 3 ). Sum of ( x_i )s: ( 2 + 6 = 8 ). ( frac{49}{6} approx 8.166 ). Not enough.So, ( n = 7 ) doesn't work either. It seems like ( n = 9 ) is indeed the smallest number where both conditions can be satisfied with integer counts of ( p ) and ( q ).I think this makes sense because when ( n = 9 ), the required number of ( p ) is exactly 5, which is an integer, and the sums work out perfectly. For smaller ( n ), the required ( p ) isn't an integer, and adjusting ( p ) up or down causes one of the conditions to fail.Therefore, the smallest natural number ( n ) is 9.

answer：First, I recognize that the user has provided specific ages: they are 45 years old and their son is 15 years old.To find the ratio of their age to their son's age, I need to divide their age by their son's age.So, I calculate 45 divided by 15, which equals 3.This means that the ratio of their age to their son's age is 3 to 1.Therefore, the final ratio is 3:1.