💡First, identify the amount Ian paid to each person:- **Colin:** 20- **Helen:** Twice as much as Colin, so 20 * 2 = 40- **Benedict:** Half as much as Helen, so 40 / 2 = 20Next, calculate the total amount paid to all three:20 (Colin) + 40 (Helen) + 20 (Benedict) = 80Finally, add the remaining 20 to the total debts to find the total amount Ian won:80 (total debts) + 20 (remaining) = 100

💡Alright, so I've got this problem here where Petya writes down a bunch of fractions based on a natural number ( n ). The fractions are ( frac{0}{n}, frac{1}{n-1}, frac{2}{n-2}, ldots, frac{n-1}{1} ). The problem says that ( n ) is divisible by some natural number ( d ), and I need to prove that among these fractions, there's one that equals ( d-1 ).Hmm, okay. Let's break this down. First, since ( n ) is divisible by ( d ), that means ( n = d times k ) for some integer ( k ). So, ( n ) is just ( d ) multiplied by another natural number ( k ).Now, looking at the fractions, they start from ( frac{0}{n} ) and go up to ( frac{n-1}{1} ). Each fraction has a numerator that increases by 1 and a denominator that decreases by 1. So, the general form of each fraction is ( frac{i}{n - i} ) where ( i ) ranges from 0 to ( n-1 ).I need to find a specific fraction in this list that equals ( d - 1 ). Let's think about what ( d - 1 ) represents. Since ( d ) divides ( n ), ( d ) is a factor of ( n ), and ( d - 1 ) is just one less than that factor.Maybe I can express ( d - 1 ) as a fraction ( frac{i}{n - i} ) for some ( i ). Let's set up the equation:[frac{i}{n - i} = d - 1]If I solve for ( i ), I can find the specific numerator that gives me the fraction equal to ( d - 1 ). Let's do that:Multiply both sides by ( n - i ):[i = (d - 1)(n - i)]Expand the right side:[i = (d - 1)n - (d - 1)i]Bring all terms with ( i ) to one side:[i + (d - 1)i = (d - 1)n]Factor out ( i ):[i(1 + d - 1) = (d - 1)n]Simplify inside the parentheses:[i(d) = (d - 1)n]Now, solve for ( i ):[i = frac{(d - 1)n}{d}]Since ( n = d times k ), substitute that in:[i = frac{(d - 1)d times k}{d}]The ( d ) in the numerator and denominator cancels out:[i = (d - 1)k]So, ( i = k(d - 1) ). Now, since ( k ) and ( d ) are natural numbers, ( i ) is also a natural number. Moreover, ( i ) must be less than ( n ) because the numerators in the fractions go up to ( n - 1 ).Let's verify that ( i = k(d - 1) ) is indeed less than ( n ):Since ( n = d times k ), and ( d ) is at least 1, ( k(d - 1) ) will be less than ( d times k ) as long as ( d - 1 < d ), which is always true for natural numbers ( d geq 1 ).Therefore, ( i = k(d - 1) ) is a valid numerator in the list of fractions, and the corresponding fraction is:[frac{i}{n - i} = frac{k(d - 1)}{d times k - k(d - 1)} = frac{k(d - 1)}{k} = d - 1]So, this fraction simplifies to ( d - 1 ), which is exactly what we needed to prove.Wait, let me double-check that step where I simplified the denominator:The denominator is ( n - i = d times k - k(d - 1) ). Let's compute that:[d times k - k(d - 1) = dk - dk + k = k]Yes, that's correct. So, the denominator simplifies to ( k ), and the numerator is ( k(d - 1) ), so the fraction becomes ( frac{k(d - 1)}{k} = d - 1 ).Alright, that makes sense. So, by choosing ( i = k(d - 1) ), we get a fraction equal to ( d - 1 ). Since ( i ) is within the range of numerators (from 0 to ( n - 1 )), this fraction is indeed among the ones Petya wrote on the board.I think that covers it. I've found the specific fraction that equals ( d - 1 ) by expressing ( n ) in terms of ( d ) and solving for the appropriate numerator. It all checks out.

💡Alright, so I have this problem about a sine function, f(x) = A sin(ωx + φ). The problem says that the graph intersects the x-axis, and the distance between two adjacent intersection points is π/2. Also, one of the lowest points on the graph is M(2π/3, -2). I need to find the analytical expression of the function and its monotonically increasing interval for part 1, and then find the range of f(x) when x is between π/12 and π/2 for part 2.Okay, let's start with part 1. The function is f(x) = A sin(ωx + φ). I know that A is the amplitude, ω affects the period, and φ is the phase shift. The graph intersects the x-axis, so those are the points where f(x) = 0. The distance between two adjacent zeros is π/2. Hmm, so the period of the sine function is related to the distance between zeros.Wait, for a sine function, the distance between two adjacent zeros is half the period, right? Because the sine function crosses the x-axis every half period. So if the distance between two adjacent zeros is π/2, then the period T should be 2*(π/2) = π. So T = π.The period of a sine function is given by T = 2π / ω. So if T = π, then ω = 2π / T = 2π / π = 2. So ω is 2. Got that.Next, the amplitude A. The problem mentions a lowest point at M(2π/3, -2). Since the sine function oscillates between -A and A, the lowest point is -A. So if the y-coordinate is -2, that means A = 2. So A is 2.Now, we have f(x) = 2 sin(2x + φ). We need to find φ. We know that at x = 2π/3, the function reaches its minimum value, which is -2. So let's plug that into the function:2 sin(2*(2π/3) + φ) = -2Simplify inside the sine:2*(2π/3) = 4π/3So, 2 sin(4π/3 + φ) = -2Divide both sides by 2:sin(4π/3 + φ) = -1We know that sin(θ) = -1 when θ = 3π/2 + 2πk, where k is any integer. So,4π/3 + φ = 3π/2 + 2πkSolve for φ:φ = 3π/2 - 4π/3 + 2πkLet's compute 3π/2 - 4π/3. To subtract these, they need a common denominator. The common denominator is 6.3π/2 = 9π/64π/3 = 8π/6So, 9π/6 - 8π/6 = π/6Therefore, φ = π/6 + 2πkBut the problem states that 0 < φ < π/2. So let's see what k can be.If k = 0, φ = π/6, which is within (0, π/2). If k = 1, φ = π/6 + 2π, which is way larger than π/2. Similarly, negative k would make φ negative, which is not allowed. So φ = π/6.Therefore, the function is f(x) = 2 sin(2x + π/6).Now, the first part is done. The analytical expression is f(x) = 2 sin(2x + π/6).Next, find the monotonically increasing interval. For a sine function, it's increasing where its derivative is positive. Let's compute the derivative:f'(x) = 2 * cos(2x + π/6) * 2 = 4 cos(2x + π/6)Wait, actually, the derivative of sin(u) is cos(u) * u', so:f'(x) = 2 * cos(2x + π/6) * 2 = 4 cos(2x + π/6)So f'(x) = 4 cos(2x + π/6). The function is increasing when f'(x) > 0, so when cos(2x + π/6) > 0.The cosine function is positive in the intervals (-π/2 + 2πk, π/2 + 2πk) for any integer k.So, we need to solve:-π/2 + 2πk < 2x + π/6 < π/2 + 2πkSubtract π/6 from all parts:-π/2 - π/6 + 2πk < 2x < π/2 - π/6 + 2πkCompute -π/2 - π/6:Convert to sixths: -3π/6 - π/6 = -4π/6 = -2π/3Similarly, π/2 - π/6 = 3π/6 - π/6 = 2π/6 = π/3So:-2π/3 + 2πk < 2x < π/3 + 2πkDivide all parts by 2:-π/3 + πk < x < π/6 + πkSo, the function is increasing on intervals (-π/3 + πk, π/6 + πk) for all integers k.But since the function is periodic with period π, these intervals repeat every π units.So, in terms of x, the function is increasing on intervals [kπ - π/3, kπ + π/6] for all integers k.Wait, but in the original problem, the function is defined for all real numbers, so we can express the increasing intervals as [kπ - π/3, kπ + π/6], where k is any integer.Alright, that's part 1 done.Now, part 2: Find the range of f(x) when x is between π/12 and π/2.So, x ∈ [π/12, π/2]. Let's find f(x) = 2 sin(2x + π/6) over this interval.First, let's find the range of the argument inside the sine function, which is 2x + π/6.When x = π/12:2*(π/12) + π/6 = π/6 + π/6 = π/3When x = π/2:2*(π/2) + π/6 = π + π/6 = 7π/6So, the argument 2x + π/6 varies from π/3 to 7π/6 as x goes from π/12 to π/2.So, we need to find the range of sin(θ) where θ ∈ [π/3, 7π/6].Let me recall the sine function's behavior. From π/3 to 7π/6.π/3 is 60 degrees, where sin(π/3) = √3/2 ≈ 0.866.Then, as θ increases to π/2 (90 degrees), sin(θ) increases to 1.Then, from π/2 to π (180 degrees), sin(θ) decreases back to 0.From π to 3π/2 (270 degrees), sin(θ) decreases to -1.From 3π/2 to 2π, it increases back to 0.But in our case, θ goes up to 7π/6, which is π + π/6, so 210 degrees. At 7π/6, sin(7π/6) = -1/2.So, let's plot the sine curve from π/3 to 7π/6.At π/3: sin(π/3) = √3/2 ≈ 0.866At π/2: sin(π/2) = 1At π: sin(π) = 0At 3π/2: sin(3π/2) = -1At 7π/6: sin(7π/6) = -1/2So, the maximum value of sin(θ) in this interval is 1, achieved at θ = π/2.The minimum value is -1, achieved at θ = 3π/2.But wait, our interval goes up to 7π/6, which is beyond 3π/2? Wait, no, 7π/6 is less than 3π/2 (which is 4.712), since 7π/6 ≈ 3.665.Wait, 3π/2 is 4.712, and 7π/6 is approximately 3.665, so 7π/6 is before 3π/2.So, in the interval [π/3, 7π/6], the sine function starts at √3/2, goes up to 1 at π/2, then decreases to 0 at π, then continues decreasing to -1/2 at 7π/6.Wait, but does it reach -1 in this interval? No, because 3π/2 is outside the interval. So the minimum value is -1/2.Wait, but hold on, let me think again.Wait, 7π/6 is π + π/6, which is 210 degrees. The sine of 210 degrees is -1/2.So, in the interval [π/3, 7π/6], the sine function starts at √3/2, goes up to 1, comes back down to 0 at π, and then goes down to -1/2 at 7π/6.Therefore, the maximum value is 1, and the minimum value is -1/2.But wait, hold on, is that correct? Because between π and 7π/6, the sine function is decreasing from 0 to -1/2, but it doesn't reach -1 in this interval.So, the range of sin(θ) for θ ∈ [π/3, 7π/6] is [-1/2, 1].Therefore, f(x) = 2 sin(θ) will have a range of 2*(-1/2) to 2*1, which is [-1, 2].Wait, but let me double-check.At θ = π/3: sin(π/3) = √3/2 ≈ 0.866, so f(x) = 2*(√3/2) = √3 ≈ 1.732At θ = π/2: sin(π/2) = 1, so f(x) = 2*1 = 2At θ = π: sin(π) = 0, so f(x) = 0At θ = 7π/6: sin(7π/6) = -1/2, so f(x) = 2*(-1/2) = -1So, the function f(x) starts at approximately 1.732, goes up to 2, then down to 0, and then down to -1.So, the maximum value is 2, and the minimum value is -1.Therefore, the range of f(x) over x ∈ [π/12, π/2] is [-1, 2].Wait, but hold on, is that correct? Because when θ is between π/3 and 7π/6, sin(θ) is between -1/2 and 1, so multiplying by 2, f(x) is between -1 and 2.Yes, that seems correct.But let me visualize the graph to make sure.From x = π/12 to x = π/2, the argument 2x + π/6 goes from π/3 to 7π/6.So, the sine wave starts at sin(π/3) = √3/2, goes up to sin(π/2) = 1, then down to sin(π) = 0, then further down to sin(7π/6) = -1/2.So, the highest point is 1, and the lowest is -1/2, but since we multiply by 2, the highest is 2 and the lowest is -1.Therefore, the range is [-1, 2].Wait, but hold on, is there any point where f(x) is less than -1 in this interval?No, because the lowest point in the interval is at 7π/6, which is -1/2, so f(x) is -1 there.So, the function doesn't go below -1 in this interval.Therefore, the range is from -1 to 2.So, to summarize:1. The function is f(x) = 2 sin(2x + π/6), and it's increasing on intervals [kπ - π/3, kπ + π/6] for all integers k.2. The range of f(x) when x is between π/12 and π/2 is [-1, 2].I think that's it.**Final Answer**1. The function is boxed{2sinleft(2x + dfrac{pi}{6}right)} and its monotonically increasing intervals are boxed{left[kpi - dfrac{pi}{3}, kpi + dfrac{pi}{6}right]} for all integers (k).2. The range of (f(x)) when (x in left[ dfrac{pi}{12}, dfrac{pi}{2} right]) is boxed{[-1, 2]}.

💡Alright, so I have this functional equation to solve: ( f(xy) = max { f(x + y), f(x)f(y) } ) for all real numbers ( x ) and ( y ). Hmm, functional equations can be tricky, but let me try to break this down step by step.First off, I need to find all functions ( f: mathbb{R} rightarrow mathbb{R} ) that satisfy this condition. The equation involves both multiplication and addition of ( x ) and ( y ), and it's set equal to the maximum of two expressions: one involving the function at the sum ( x + y ) and the other involving the product of the function values at ( x ) and ( y ).Maybe I should start by considering some specific values for ( x ) and ( y ) to see if I can get some properties of ( f ). Let's try plugging in ( x = 0 ). If ( x = 0 ), then the equation becomes:( f(0 cdot y) = max { f(0 + y), f(0)f(y) } )Simplifying, that's:( f(0) = max { f(y), f(0)f(y) } )Hmm, interesting. So for any ( y ), ( f(0) ) is the maximum of ( f(y) ) and ( f(0)f(y) ). Let's think about what this implies.If ( f(0) ) is greater than or equal to ( f(y) ) for all ( y ), then ( f(0) ) is an upper bound for ( f ). Alternatively, if ( f(0)f(y) ) is greater than or equal to ( f(y) ), that would mean ( f(0) geq 1 ) because ( f(0)f(y) geq f(y) ) implies ( f(0) geq 1 ) if ( f(y) ) is positive, or ( f(0) leq 1 ) if ( f(y) ) is negative. But since ( f(0) ) is the maximum, it must be that ( f(0) geq f(y) ) for all ( y ).Wait, so if ( f(0) ) is the maximum of ( f(y) ) and ( f(0)f(y) ), then ( f(0) geq f(y) ) for all ( y ). That suggests that ( f(0) ) is the maximum value of ( f ) on the entire real line.Okay, so ( f(0) ) is the maximum value of ( f ). Let's denote ( M = f(0) ). So ( M geq f(y) ) for all ( y ).Now, let's consider another special case: ( x = 1 ) and ( y = 1 ). Plugging into the equation:( f(1 cdot 1) = max { f(1 + 1), f(1)f(1) } )Simplifying:( f(1) = max { f(2), [f(1)]^2 } )So, ( f(1) ) is the maximum of ( f(2) ) and ( [f(1)]^2 ). Since ( M ) is the maximum value of ( f ), we know ( f(1) leq M ) and ( f(2) leq M ).If ( f(1) = M ), then ( [f(1)]^2 = M^2 ). But since ( M ) is the maximum, ( M^2 leq M ) only if ( M leq 1 ). Because if ( M > 1 ), then ( M^2 > M ), which would contradict ( M ) being the maximum. So, if ( f(1) = M ), then ( M leq 1 ).Alternatively, if ( f(1) < M ), then ( [f(1)]^2 leq f(1) ) because ( f(1) leq 1 ) (since ( M leq 1 )). Therefore, ( f(1) ) is the maximum of ( f(2) ) and something less than or equal to ( f(1) ), so ( f(1) geq f(2) ).This suggests that ( f ) might be decreasing or at least non-increasing as we move away from 0, but I need more information.Let's try another case: ( x = y = 0 ). Plugging into the equation:( f(0 cdot 0) = max { f(0 + 0), f(0)f(0) } )Simplifying:( f(0) = max { f(0), [f(0)]^2 } )Since ( f(0) = M ), this becomes:( M = max { M, M^2 } )Which implies that ( M geq M^2 ). So, ( M^2 leq M ), which again gives ( M leq 1 ) because ( M ) is non-negative (since it's the maximum value of ( f )).So, ( M leq 1 ).Now, let's consider ( x = -1 ) and ( y = -1 ). Then:( f((-1)(-1)) = max { f(-1 + (-1)), f(-1)f(-1) } )Simplifying:( f(1) = max { f(-2), [f(-1)]^2 } )Hmm, interesting. So ( f(1) ) is the maximum of ( f(-2) ) and ( [f(-1)]^2 ). Since ( M ) is the maximum, ( f(-2) leq M ) and ( [f(-1)]^2 leq M^2 leq M ) because ( M leq 1 ).So, ( f(1) leq M ), and ( f(1) ) is the maximum of two values each less than or equal to ( M ). Therefore, ( f(1) leq M ), which we already knew.But this doesn't give me much new information. Maybe I need to consider other specific values or perhaps try to find a general form.Another approach is to consider the case where ( x + y = xy ). Let's see if that's possible.Suppose ( x + y = xy ). Then, the functional equation becomes:( f(xy) = max { f(xy), f(x)f(y) } )Which simplifies to:( f(xy) = max { f(xy), f(x)f(y) } )This implies that ( f(xy) geq f(x)f(y) ).But from the original equation, ( f(xy) ) is equal to the maximum of ( f(x + y) ) and ( f(x)f(y) ). So, if ( x + y = xy ), then ( f(xy) geq f(x)f(y) ).But since ( f(xy) ) is the maximum, it must be that ( f(xy) geq f(x + y) ) as well. But in this case, ( x + y = xy ), so ( f(xy) geq f(xy) ), which is trivial.Not sure if that helps. Maybe another approach.Let me consider the case where ( y = 1 ). Then, the equation becomes:( f(x cdot 1) = max { f(x + 1), f(x)f(1) } )Simplifying:( f(x) = max { f(x + 1), f(x)f(1) } )So, for any ( x ), ( f(x) ) is the maximum of ( f(x + 1) ) and ( f(x)f(1) ).This suggests that ( f(x) geq f(x + 1) ) and ( f(x) geq f(x)f(1) ).From ( f(x) geq f(x + 1) ), we can see that ( f ) is non-increasing as we move to the right on the real line.From ( f(x) geq f(x)f(1) ), we have two cases:1. If ( f(x) neq 0 ), then we can divide both sides by ( f(x) ), yielding ( 1 geq f(1) ).2. If ( f(x) = 0 ), then the inequality holds trivially.So, in the case where ( f(x) neq 0 ), we get ( f(1) leq 1 ).But we already knew ( M leq 1 ), and ( f(1) leq M ), so this doesn't add new information.Wait, but if ( f(x) geq f(x + 1) ) for all ( x ), then ( f ) is non-increasing. So, as ( x ) increases, ( f(x) ) doesn't increase.Similarly, if we consider ( y = -1 ), let's see what happens.Plugging ( y = -1 ) into the original equation:( f(-x) = max { f(x - 1), f(x)f(-1) } )Hmm, not sure if that helps directly. Maybe I need to consider more properties.Another thought: suppose ( f ) is a constant function. Let's test that.Assume ( f(x) = c ) for all ( x ). Then, the equation becomes:( c = max { c, c cdot c } )Which simplifies to:( c = max { c, c^2 } )So, ( c geq c^2 ). Which implies ( c^2 leq c ), so ( c leq 1 ) since ( c geq 0 ) (because if ( c ) were negative, ( c^2 ) would be positive and greater than ( c ), contradicting ( c geq c^2 )).Therefore, constant functions ( f(x) = c ) where ( 0 leq c leq 1 ) satisfy the equation.But are there any non-constant solutions?Suppose ( f ) is not constant. Then, there exist ( x ) and ( y ) such that ( f(x) neq f(y) ).But from earlier, we saw that ( f(0) = M ) is the maximum value, and ( f ) is non-increasing as ( x ) increases.Wait, but if ( f ) is non-increasing, then as ( x ) approaches positive infinity, ( f(x) ) approaches some limit, possibly zero.But let's think about ( x = 2 ) and ( y = 0.5 ). Then, ( xy = 1 ), and ( x + y = 2.5 ).So, the equation becomes:( f(1) = max { f(2.5), f(2)f(0.5) } )Since ( f ) is non-increasing, ( f(2.5) leq f(2) leq f(1) ), and ( f(0.5) geq f(1) ) because ( 0.5 < 1 ).Therefore, ( f(2)f(0.5) leq f(1)f(0.5) ). But since ( f(0.5) geq f(1) ), ( f(1)f(0.5) geq [f(1)]^2 ).But from earlier, when ( x = y = 1 ), we had ( f(1) = max { f(2), [f(1)]^2 } ). So, ( f(1) geq [f(1)]^2 ), which implies ( f(1) leq 1 ).Putting this together, ( f(1)f(0.5) geq [f(1)]^2 ), but ( f(1) geq f(2.5) ) and ( f(1) geq f(2)f(0.5) ).This seems a bit convoluted. Maybe I need a different approach.Let me consider the case where ( x = y ). Then, the equation becomes:( f(x^2) = max { f(2x), [f(x)]^2 } )So, for any ( x ), ( f(x^2) ) is the maximum of ( f(2x) ) and ( [f(x)]^2 ).Since ( x^2 ) is always non-negative, and ( f ) is non-increasing, ( f(x^2) leq f(0) = M ).Also, ( f(2x) leq f(0) = M ) because ( 2x ) can be any real number, but ( f ) is non-increasing, so as ( |x| ) increases, ( f(2x) ) decreases.Similarly, ( [f(x)]^2 leq [f(0)]^2 = M^2 leq M ) since ( M leq 1 ).Therefore, both ( f(2x) ) and ( [f(x)]^2 ) are less than or equal to ( M ), so ( f(x^2) leq M ).But ( f(x^2) ) is the maximum of two terms each less than or equal to ( M ), so ( f(x^2) leq M ), which is consistent.Not sure if that helps.Another idea: suppose ( f ) is identically zero. Then, ( f(xy) = 0 ), and ( max { f(x + y), f(x)f(y) } = max { 0, 0 } = 0 ). So, that works.Similarly, if ( f ) is identically ( M ) where ( 0 leq M leq 1 ), as we saw earlier, that works.But are there any other solutions?Suppose ( f ) is not constant. Then, there exists some ( x ) such that ( f(x) < M ).But since ( f ) is non-increasing, for ( x > 0 ), ( f(x) leq f(0) = M ), and for ( x < 0 ), ( f(x) geq f(0) = M ) because as ( x ) decreases, ( f(x) ) increases (since it's non-increasing).Wait, that might not necessarily be true. Non-increasing means that as ( x ) increases, ( f(x) ) decreases or stays the same. So, for ( x < 0 ), as ( x ) becomes more negative, ( f(x) ) could be increasing or decreasing.Hmm, maybe I need to consider negative values more carefully.Let me consider ( x = -1 ) and ( y = -1 ). Then:( f(1) = max { f(-2), [f(-1)]^2 } )Since ( f ) is non-increasing, ( f(-2) geq f(-1) geq f(0) = M ). Wait, that can't be right because ( f(0) = M ) is the maximum.Wait, if ( f ) is non-increasing, then for ( x < y ), ( f(x) geq f(y) ). So, as ( x ) decreases, ( f(x) ) increases.Therefore, ( f(-2) geq f(-1) geq f(0) = M ).But ( f(-2) geq M ) and ( f(-1) geq M ), but ( M ) is the maximum, so ( f(-2) = M ) and ( f(-1) = M ).Wait, that's interesting. So, for all ( x leq 0 ), ( f(x) = M ).Because if ( x ) is negative, then ( f(x) geq f(0) = M ), but ( M ) is the maximum, so ( f(x) = M ) for all ( x leq 0 ).Similarly, for ( x > 0 ), ( f(x) leq M ).So, ( f(x) = M ) for all ( x leq 0 ), and ( f(x) leq M ) for ( x > 0 ).Now, let's consider ( x > 0 ) and ( y > 0 ). Then, ( xy > 0 ) and ( x + y > 0 ).So, the functional equation becomes:( f(xy) = max { f(x + y), f(x)f(y) } )Since ( f(x) leq M ) and ( f(y) leq M ), ( f(x)f(y) leq M^2 leq M ).Also, ( f(x + y) leq M ).So, ( f(xy) ) is the maximum of two terms each less than or equal to ( M ), so ( f(xy) leq M ).But ( f(xy) ) is also equal to this maximum, so ( f(xy) geq f(x + y) ) and ( f(xy) geq f(x)f(y) ).But since ( f(xy) leq M ), we have ( f(xy) = M ) only if either ( f(x + y) = M ) or ( f(x)f(y) = M ).But ( f(x + y) leq M ), so ( f(x + y) = M ) only if ( x + y leq 0 ). But ( x > 0 ) and ( y > 0 ), so ( x + y > 0 ), hence ( f(x + y) < M ).Similarly, ( f(x)f(y) leq M^2 leq M ), so ( f(x)f(y) = M ) only if ( f(x) = f(y) = M ), but ( x > 0 ) and ( y > 0 ), so ( f(x) leq M ) and ( f(y) leq M ), but unless ( f(x) = f(y) = M ), which would require ( x leq 0 ) and ( y leq 0 ), which they are not.Therefore, ( f(xy) ) cannot be equal to ( M ) in this case, so ( f(xy) ) must be less than ( M ).But wait, ( f(xy) ) is the maximum of ( f(x + y) ) and ( f(x)f(y) ), both of which are less than ( M ). So, ( f(xy) ) is less than ( M ).This suggests that for ( x > 0 ) and ( y > 0 ), ( f(xy) ) is determined by the maximum of two values less than ( M ).But how does this help me find ( f )?Maybe I should consider specific forms of ( f ). For example, suppose ( f(x) = M ) for ( x leq 0 ) and ( f(x) = c ) for ( x > 0 ), where ( c leq M ).Then, for ( x > 0 ) and ( y > 0 ), the equation becomes:( f(xy) = max { f(x + y), f(x)f(y) } )Which would be:( c = max { c, c cdot c } )So, ( c = max { c, c^2 } ). This implies ( c geq c^2 ), so ( c leq 1 ).But we already have ( c leq M leq 1 ), so this is consistent.But does this function satisfy the equation for all ( x ) and ( y )?Let's test it.Case 1: ( x leq 0 ) and ( y leq 0 ).Then, ( xy geq 0 ) (since negative times negative is positive), and ( x + y leq 0 ).So, ( f(xy) = c ) if ( xy > 0 ), or ( f(xy) = M ) if ( xy = 0 ).Wait, if ( x leq 0 ) and ( y leq 0 ), then ( xy geq 0 ). So, ( f(xy) = c ) if ( xy > 0 ), and ( f(0) = M ) if ( xy = 0 ).On the other hand, ( f(x + y) = M ) since ( x + y leq 0 ), and ( f(x)f(y) = M cdot M = M^2 ).So, the equation becomes:If ( xy > 0 ):( c = max { M, M^2 } )But ( M geq M^2 ) since ( M leq 1 ), so ( max { M, M^2 } = M ).Thus, ( c = M ).If ( xy = 0 ):( f(0) = M = max { f(x + y), f(x)f(y) } )Since ( x + y leq 0 ), ( f(x + y) = M ), and ( f(x)f(y) = M cdot M = M^2 ).So, ( M = max { M, M^2 } ), which is true.Therefore, in this case, ( c = M ).So, for ( x leq 0 ) and ( y leq 0 ), the equation holds only if ( c = M ).But earlier, we considered ( c leq M ). So, if ( c = M ), then ( f(x) = M ) for all ( x ), which is the constant function.Wait, so if I assume ( f(x) = M ) for ( x leq 0 ) and ( f(x) = c ) for ( x > 0 ), then the only way the equation holds for ( x leq 0 ) and ( y leq 0 ) is if ( c = M ). Therefore, ( f ) must be constant everywhere.So, that suggests that the only solutions are constant functions ( f(x) = c ) where ( 0 leq c leq 1 ).But let me double-check this.Suppose ( f(x) = c ) for all ( x ), with ( 0 leq c leq 1 ). Then, the equation becomes:( c = max { c, c cdot c } )Which simplifies to:( c = max { c, c^2 } )Since ( c geq c^2 ) for ( 0 leq c leq 1 ), this holds true.Therefore, constant functions ( f(x) = c ) with ( 0 leq c leq 1 ) satisfy the equation.Are there any non-constant solutions?From the earlier reasoning, if ( f ) is not constant, then for ( x > 0 ), ( f(x) < M ). But when we considered ( x leq 0 ) and ( y leq 0 ), it forced ( f(x) = M ) for all ( x leq 0 ). Then, for ( x > 0 ) and ( y > 0 ), the equation requires ( f(xy) = max { f(x + y), f(x)f(y) } ). But since ( f(x) leq M ) and ( f(y) leq M ), and ( f(x + y) leq M ), it's possible that ( f(xy) ) could be less than ( M ), but we need to ensure that ( f(xy) ) is indeed the maximum of those two terms.However, if ( f ) is not constant, then there must be some ( x > 0 ) where ( f(x) < M ). Let's suppose ( f(a) = c < M ) for some ( a > 0 ). Then, consider ( x = a ) and ( y = 1 ). The equation becomes:( f(a cdot 1) = max { f(a + 1), f(a)f(1) } )Simplifying:( c = max { f(a + 1), c cdot f(1) } )Since ( f(a + 1) leq c ) (because ( a + 1 > a > 0 ) and ( f ) is non-increasing), and ( c cdot f(1) leq c ) (since ( f(1) leq 1 )), the maximum is ( c ). So, this holds.But does this impose any new conditions? Not really, because both terms are less than or equal to ( c ).Wait, but if ( f(a + 1) leq c ), and ( f(a + 1) leq f(a) ) because ( a + 1 > a ), which is consistent with ( f ) being non-increasing.But I'm not sure if this leads to any contradictions. Maybe I need to consider more specific cases.Suppose ( f ) is not constant, so there exists some ( a > 0 ) such that ( f(a) < M ). Let's consider ( x = a ) and ( y = a ). Then:( f(a^2) = max { f(2a), [f(a)]^2 } )Since ( a^2 > 0 ), ( f(a^2) leq M ). Also, ( f(2a) leq f(a) ) because ( 2a > a ), and ( [f(a)]^2 leq f(a) ) because ( f(a) leq 1 ).Therefore, ( f(a^2) ) is the maximum of two terms each less than or equal to ( f(a) ). So, ( f(a^2) leq f(a) ).But ( f(a^2) ) is also equal to this maximum, so ( f(a^2) geq f(2a) ) and ( f(a^2) geq [f(a)]^2 ).But since ( f(a^2) leq f(a) ), we have ( f(a) geq f(2a) ) and ( f(a) geq [f(a)]^2 ).The second inequality ( f(a) geq [f(a)]^2 ) implies ( f(a) leq 1 ), which we already know.But the first inequality ( f(a) geq f(2a) ) is consistent with ( f ) being non-increasing.So, this doesn't lead to a contradiction.Another approach: suppose ( f ) is continuous. Then, since ( f ) is non-increasing and continuous, it must be constant. But the problem doesn't specify continuity, so I can't assume that.Wait, but without continuity, it's hard to conclude much. Maybe I need to consider the behavior as ( x ) approaches zero.Let me consider ( x ) approaching zero from the positive side. Let ( x = epsilon ) where ( epsilon ) is a small positive number. Then, ( f(epsilon) leq M ).Consider ( y = 1/epsilon ). Then, ( xy = 1 ), and ( x + y = epsilon + 1/epsilon ).So, the equation becomes:( f(1) = max { f(epsilon + 1/epsilon), f(epsilon)f(1/epsilon) } )Since ( epsilon + 1/epsilon geq 2 ) by AM-GM inequality, and ( f ) is non-increasing, ( f(epsilon + 1/epsilon) leq f(2) leq f(1) ).Also, ( f(epsilon) leq M ) and ( f(1/epsilon) leq M ), so ( f(epsilon)f(1/epsilon) leq M^2 leq M ).But ( f(1) leq M ), so ( f(1) = max { f(epsilon + 1/epsilon), f(epsilon)f(1/epsilon) } leq M ).But ( f(1) leq M ), and ( M ) is the maximum, so this holds.But does this give me any information about ( f(1) )?If ( f(1) < M ), then ( f(epsilon + 1/epsilon) leq f(1) ) and ( f(epsilon)f(1/epsilon) leq f(1) ).But as ( epsilon ) approaches zero, ( epsilon + 1/epsilon ) approaches infinity, so ( f(epsilon + 1/epsilon) ) approaches the limit of ( f ) at infinity, which is some value ( L leq f(1) ).Similarly, ( f(epsilon) ) approaches ( f(0) = M ), and ( f(1/epsilon) ) approaches ( L ).So, ( f(epsilon)f(1/epsilon) ) approaches ( M cdot L ).Therefore, ( f(1) = max { L, M cdot L } ).But ( L leq f(1) ), and ( M cdot L leq M cdot f(1) leq f(1) ) because ( M leq 1 ).So, ( f(1) = max { L, M cdot L } leq f(1) ), which is consistent.But unless ( L = f(1) ), this doesn't give me much.Wait, if ( L = f(1) ), then ( f(1) = max { f(1), M cdot f(1) } ).Since ( M leq 1 ), ( M cdot f(1) leq f(1) ), so ( f(1) = f(1) ), which holds.But if ( L < f(1) ), then ( f(1) = max { L, M cdot L } leq f(1) ), which is still consistent.So, this approach doesn't seem to lead to a contradiction unless ( f(1) = M ).Wait, if ( f(1) = M ), then from earlier, ( M leq 1 ).But if ( f(1) = M ), then from the case ( x = y = 1 ), we have:( f(1) = max { f(2), [f(1)]^2 } )Which implies ( M = max { f(2), M^2 } ).Since ( M geq f(2) ) and ( M geq M^2 ), this holds.But if ( f(1) < M ), then ( f(1) = max { L, M cdot L } ), which could be less than ( M ).But I'm not sure if this leads to any contradictions.Another idea: suppose ( f ) is not constant, so there exists some ( a > 0 ) such that ( f(a) < M ). Then, consider ( x = a ) and ( y = 1/a ). Then, ( xy = 1 ), and ( x + y = a + 1/a ).So, the equation becomes:( f(1) = max { f(a + 1/a), f(a)f(1/a) } )Since ( a + 1/a geq 2 ), ( f(a + 1/a) leq f(2) leq f(1) ).Also, ( f(a) < M ) and ( f(1/a) leq M ), so ( f(a)f(1/a) leq M cdot f(a) leq M^2 leq M ).But ( f(1) leq M ), so ( f(1) = max { f(a + 1/a), f(a)f(1/a) } leq M ).But ( f(a + 1/a) leq f(1) ) and ( f(a)f(1/a) leq M cdot f(a) leq M cdot M = M^2 leq M ).So, ( f(1) ) is the maximum of two terms each less than or equal to ( M ), which is consistent.But again, this doesn't seem to lead to a contradiction.Wait, but if ( f(a) < M ), then ( f(a)f(1/a) leq M cdot f(a) < M^2 leq M ). So, ( f(a)f(1/a) < M ).Therefore, ( f(1) = max { f(a + 1/a), f(a)f(1/a) } leq max { f(1), M } = M ).But ( f(1) leq M ), so this holds.But if ( f(1) = M ), then ( f(a + 1/a) leq M ) and ( f(a)f(1/a) < M ), so ( f(1) = M ) is still valid.If ( f(1) < M ), then ( f(1) = max { f(a + 1/a), f(a)f(1/a) } leq M ), which is still consistent.So, this approach also doesn't lead to a contradiction.Given all this, it seems that the only solutions are constant functions ( f(x) = c ) where ( 0 leq c leq 1 ).But let me try one more thing. Suppose ( f ) is not constant, so there exists some ( a > 0 ) such that ( f(a) < M ). Let's consider ( x = a ) and ( y = 0 ). Then:( f(0) = max { f(a), f(a)f(0) } )Simplifying:( M = max { f(a), M cdot f(a) } )Since ( f(a) < M ), ( M cdot f(a) < M^2 leq M ).Therefore, ( max { f(a), M cdot f(a) } = M ).But ( M = max { f(a), M cdot f(a) } ), which implies ( M geq f(a) ) and ( M geq M cdot f(a) ).But ( M geq f(a) ) is already true, and ( M geq M cdot f(a) ) simplifies to ( 1 geq f(a) ), which is also true since ( f(a) leq M leq 1 ).So, this doesn't lead to a contradiction either.Given that I can't find any contradictions assuming ( f ) is non-constant, but all the specific cases seem to be consistent with ( f ) being constant, I'm starting to think that the only solutions are constant functions.Therefore, the functions satisfying the equation are all constant functions ( f(x) = c ) where ( 0 leq c leq 1 ).

💡Alright, so I have this problem about finite commutative rings. It says that a finite commutative ring ( R ) has a multiplicative identity element (which is usually denoted as 1) if and only if the annihilator of ( R ) is zero. The annihilator of ( R ) is defined as the set of all elements ( a ) in ( R ) such that ( aR = 0 ). So, the problem is asking me to prove that ( R ) has a 1 if and only if the only element that annihilates the entire ring is zero itself.Hmm, okay. Let me break this down into two parts: the "if" part and the "only if" part. That is, I need to show that:1. If ( R ) has a multiplicative identity, then the annihilator of ( R ) is zero.2. If the annihilator of ( R ) is zero, then ( R ) has a multiplicative identity.Let me start with the first part because it seems a bit more straightforward.**Part 1: If ( R ) has a multiplicative identity, then the annihilator of ( R ) is zero.**Okay, suppose ( R ) has a multiplicative identity, which we'll call 1. Now, the annihilator of ( R ) is the set ( text{Ann}(R) = { a in R mid aR = 0 } ). So, if ( a ) is in the annihilator, then multiplying ( a ) by any element of ( R ) gives zero. But since 1 is in ( R ), we can consider ( a cdot 1 ). But ( a cdot 1 = a ). So, if ( a ) is in the annihilator, then ( a = 0 ). Therefore, the annihilator can only contain the zero element. That shows that if ( R ) has a multiplicative identity, then ( text{Ann}(R) = 0 ).Okay, that part wasn't too bad. Now, the second part is probably trickier.**Part 2: If the annihilator of ( R ) is zero, then ( R ) has a multiplicative identity.**Alright, so now we're assuming that ( text{Ann}(R) = 0 ). That means the only element that annihilates the entire ring is zero. We need to show that under this condition, ( R ) must have a multiplicative identity.Since ( R ) is a finite commutative ring, I remember that finite commutative rings have some nice properties. For instance, they can be decomposed into local rings, but I'm not sure if that's directly helpful here. Maybe I can think about the structure of ( R ).Another thought: in a finite ring, every element is either a unit or a zero divisor. Since ( R ) is commutative, this might help. But I'm not sure yet.Wait, the annihilator being zero means that there are no non-zero elements that annihilate the entire ring. So, for any non-zero ( a in R ), there exists some ( b in R ) such that ( ab neq 0 ). That seems useful.Maybe I can use some kind of counting argument or consider the endomorphism ring of ( R ). Let me think about that.Consider the ring ( text{End}_R(R) ), which is the ring of ( R )-module endomorphisms of ( R ). Since ( R ) is commutative, ( text{End}_R(R) ) is also a ring, and there's a natural ring homomorphism ( iota: R to text{End}_R(R) ) defined by sending ( a ) to the endomorphism ( x mapsto ax ).If ( iota ) is injective, then ( R ) is isomorphic to a subring of ( text{End}_R(R) ). Since ( R ) is finite, ( text{End}_R(R) ) is also finite. Maybe I can show that ( iota ) is surjective, which would imply that ( R ) is isomorphic to ( text{End}_R(R) ), which has a multiplicative identity (the identity endomorphism). Therefore, ( R ) would have a multiplicative identity.But how do I show that ( iota ) is surjective? Well, since ( R ) is finite, ( text{End}_R(R) ) is also finite, and if ( iota ) is injective, then it's also surjective if the dimensions match. But I'm not sure about that.Wait, maybe I can use the fact that ( text{Ann}(R) = 0 ). Since ( text{Ann}(R) = 0 ), the kernel of ( iota ) is zero, so ( iota ) is injective. So, ( R ) is isomorphic to a subring of ( text{End}_R(R) ). Now, ( text{End}_R(R) ) has a multiplicative identity, which is the identity map on ( R ). If I can show that the image of ( iota ) contains this identity map, then ( R ) itself must have a multiplicative identity.So, how do I show that the identity map is in the image of ( iota )? Maybe by constructing an element ( e in R ) such that ( iota(e) ) is the identity map. That would mean ( e cdot x = x ) for all ( x in R ), so ( e ) is a multiplicative identity.To construct such an ( e ), perhaps I can use some kind of idempotent element or consider the structure of ( R ) as a module over itself.Another approach: Since ( R ) is finite and commutative, it's a product of local rings. If each local ring has an identity, then their product would too. But I need to ensure that each local ring has an identity, which might come from the annihilator condition.Wait, maybe I can use the fact that in a finite ring, the Jacobson radical is zero if and only if the ring is semisimple. But I'm not sure if that's directly applicable here.Alternatively, since ( R ) is finite, it's Artinian, so it's a product of local Artinian rings. If each local ring has an identity, then ( R ) does as well. But I need to tie this back to the annihilator being zero.Hmm, maybe I'm overcomplicating things. Let's go back to the endomorphism ring idea. Since ( iota ) is injective, ( R ) is isomorphic to a subring of ( text{End}_R(R) ). Now, ( text{End}_R(R) ) is a finite ring, so it has a multiplicative identity. If I can show that this identity is in the image of ( iota ), then ( R ) has an identity.To do this, maybe I can use the fact that ( R ) is finite and consider the composition of endomorphisms. Since ( text{End}_R(R) ) is finite, every element has finite order, so maybe I can find an element ( e in R ) such that ( e ) acts as the identity on ( R ).Alternatively, perhaps I can use the fact that ( R ) has no non-zero annihilators to show that the natural map ( R to text{End}_R(R) ) is surjective.Wait, here's another idea. Since ( R ) is finite, it's a vector space over some finite field, right? Because finite rings are modules over finite fields. So, maybe I can use linear algebra techniques.But ( R ) isn't necessarily a field, it's a ring. Hmm.Wait, no, ( R ) is a finite commutative ring, so it's a product of finite local rings. Each local ring has a residue field which is a finite field. Maybe I can use the structure theorem for finite commutative rings.But I'm not sure if that's necessary. Let's try to think more elementarily.Since ( text{Ann}(R) = 0 ), for any ( a in R setminus {0} ), there exists some ( b in R ) such that ( ab neq 0 ). So, no non-zero element annihilates the entire ring.Now, consider the set ( S = { f in text{End}_R(R) mid f text{ is surjective} } ). Since ( R ) is finite, every surjective endomorphism is also injective, so ( S ) consists of automorphisms.But I'm not sure if that helps directly.Wait, another thought: Since ( R ) is finite, the set of endomorphisms ( text{End}_R(R) ) is also finite. Therefore, the image of ( iota ) is a finite subring of ( text{End}_R(R) ). If I can show that this subring contains the identity, then we're done.Alternatively, maybe I can use the fact that in a finite ring, every element is either a unit or a zero divisor. Since ( text{Ann}(R) = 0 ), there are no non-zero zero divisors? Wait, no, that's not necessarily true. The annihilator being zero just means that no single element annihilates the entire ring, but there could still be zero divisors.Wait, actually, in a finite ring, if an element is a zero divisor, then it's in the annihilator of some non-zero element. But since the annihilator of the entire ring is zero, maybe that implies that there are no non-zero zero divisors? Hmm, I'm not sure.Wait, no. For example, consider the ring ( mathbb{Z}/6mathbb{Z} ). It has zero divisors like 2 and 3, but the annihilator of the entire ring is zero because there's no element that annihilates all elements. So, the annihilator being zero doesn't imply that there are no zero divisors.So, scratch that idea.Maybe I can use the fact that ( R ) is finite and commutative to argue about the existence of an identity element. Since ( R ) is finite, it's Artinian, so it's a product of local rings. If each local ring has an identity, then ( R ) does as well. But I need to show that each local ring has an identity, which might come from the annihilator condition.Wait, in a local ring, the annihilator of the maximal ideal is zero? Not necessarily. Hmm.Alternatively, maybe I can use the fact that in a finite ring, the Jacobson radical is nilpotent. If the annihilator is zero, then the Jacobson radical is zero, making the ring semisimple. But I'm not sure.Wait, no, the Jacobson radical being zero implies that the ring is von Neumann regular, but I'm not sure if that directly gives an identity.Hmm, this is getting a bit tangled. Let me try a different approach.Since ( R ) is finite, let's consider the set ( R ) itself. Let me think about the multiplication maps. For each ( a in R ), the map ( mu_a: R to R ) defined by ( mu_a(x) = ax ) is an ( R )-module endomorphism.Now, since ( R ) is finite, the set of endomorphisms ( text{End}_R(R) ) is also finite. The map ( iota: R to text{End}_R(R) ) given by ( a mapsto mu_a ) is a ring homomorphism. If ( iota ) is injective, then ( R ) is isomorphic to a subring of ( text{End}_R(R) ).We know that ( text{Ann}(R) = 0 ), so the kernel of ( iota ) is zero, meaning ( iota ) is injective. Therefore, ( R ) is isomorphic to a subring of ( text{End}_R(R) ).Now, ( text{End}_R(R) ) has a multiplicative identity, which is the identity map on ( R ). If I can show that this identity map is in the image of ( iota ), then ( R ) has a multiplicative identity.So, how can I show that the identity map is in the image of ( iota )? That would mean there exists an element ( e in R ) such that ( mu_e ) is the identity map, i.e., ( e cdot x = x ) for all ( x in R ). Thus, ( e ) would be a multiplicative identity in ( R ).To find such an ( e ), maybe I can use the fact that ( R ) is finite and consider the composition of endomorphisms. Since ( text{End}_R(R) ) is finite, every element has finite order, so perhaps I can find an idempotent element or use some kind of fixed point theorem.Alternatively, since ( text{End}_R(R) ) is finite, it's a finite-dimensional algebra over some finite field. Maybe I can use linear algebra techniques to show that the identity is in the image.Wait, another idea: Since ( iota ) is injective, and ( R ) is finite, the image of ( iota ) is a finite subring of ( text{End}_R(R) ). Now, in a finite ring, every element is either a unit or a zero divisor. But since ( text{Ann}(R) = 0 ), the image of ( iota ) has no non-zero zero divisors? Wait, no, because ( text{End}_R(R) ) could have zero divisors.Hmm, maybe not. Let me think differently.Since ( R ) is finite, the set ( R ) is finite, so the set of endomorphisms ( text{End}_R(R) ) is also finite. Therefore, the image of ( iota ) is a finite subring of ( text{End}_R(R) ). Now, in a finite ring, every element is either a unit or a zero divisor. But since ( text{Ann}(R) = 0 ), the image of ( iota ) has no non-zero zero divisors? Wait, no, because ( text{End}_R(R) ) could have zero divisors.Wait, perhaps I can use the fact that ( text{End}_R(R) ) is a finite ring with identity, and ( iota ) is an injective ring homomorphism. Then, the image of ( iota ) is a subring of ( text{End}_R(R) ) that contains the identity if and only if ( R ) has an identity.But I need to show that the image contains the identity. Maybe I can use some kind of fixed point argument or consider the trace.Alternatively, perhaps I can use the fact that in a finite ring, the Jacobson radical is nilpotent, and if the annihilator is zero, then the Jacobson radical is zero, making the ring semisimple. But I'm not sure if that directly helps.Wait, another approach: Since ( R ) is finite and commutative, it's a product of finite local rings. If each local ring has an identity, then ( R ) does as well. So, maybe I can reduce the problem to the case where ( R ) is local.So, suppose ( R ) is a finite local commutative ring with ( text{Ann}(R) = 0 ). Then, does ( R ) have an identity?In a local ring, the annihilator of the maximal ideal is zero? Not necessarily. Wait, in a local ring, the maximal ideal ( mathfrak{m} ) has the property that ( mathfrak{m} ) consists of all non-units. But if ( text{Ann}(R) = 0 ), then no non-zero element annihilates the entire ring.Wait, in a local ring, if the annihilator of the entire ring is zero, does that imply that the ring has an identity? Maybe.Alternatively, perhaps I can use the fact that in a finite local ring, the maximal ideal is nilpotent. If the annihilator is zero, then the maximal ideal can't be the entire ring, so there must be units.Wait, in a finite local ring, the set of units is non-empty because the ring is finite and local. So, maybe the units generate the ring, and one of them is the identity.Hmm, I'm not sure. Maybe I need to think about specific examples.Consider the ring ( mathbb{Z}/p^nmathbb{Z} ) for a prime ( p ) and integer ( n geq 1 ). This is a finite local ring with maximal ideal ( pmathbb{Z}/p^nmathbb{Z} ). The annihilator of the entire ring is zero because if ( a in mathbb{Z}/p^nmathbb{Z} ) and ( a cdot mathbb{Z}/p^nmathbb{Z} = 0 ), then ( a = 0 ). So, in this case, the ring has an identity, which is 1.Another example: the ring ( mathbb{F}_p ), the finite field with ( p ) elements. It has an identity and its annihilator is zero.What about a product of such rings? For example, ( mathbb{Z}/2mathbb{Z} times mathbb{Z}/3mathbb{Z} ). This ring has an identity, which is ( (1,1) ), and its annihilator is zero because if ( (a,b) ) annihilates the entire ring, then ( a ) must annihilate ( mathbb{Z}/2mathbb{Z} ) and ( b ) must annihilate ( mathbb{Z}/3mathbb{Z} ), so ( a = 0 ) and ( b = 0 ).So, in these examples, when the annihilator is zero, the ring has an identity. That's consistent with the problem statement.But how do I generalize this? Maybe I can use the structure theorem for finite commutative rings, which states that every finite commutative ring is isomorphic to a product of finite local rings. If each local ring has an identity, then their product does as well. So, if I can show that each local ring in the product has an identity, then ( R ) has an identity.But in a finite local ring, does ( text{Ann}(R) = 0 ) imply the existence of an identity? I think so, because in a finite local ring, the set of units is non-empty, and if the annihilator is zero, then the units must generate the entire ring, including the identity.Wait, in a finite local ring, the set of units is ( R setminus mathfrak{m} ), where ( mathfrak{m} ) is the maximal ideal. If ( text{Ann}(R) = 0 ), then ( mathfrak{m} ) cannot be the entire ring, so there must be units. Moreover, the units must act as invertible elements on the ring, so their product must include the identity.Hmm, I'm not entirely sure, but it seems plausible.Alternatively, maybe I can use the fact that in a finite ring, the set of units forms a group under multiplication. If the annihilator is zero, then the group of units acts faithfully on the ring, which might imply the existence of an identity.Wait, another idea: Since ( R ) is finite and ( text{Ann}(R) = 0 ), the multiplication maps ( mu_a ) are injective for all ( a neq 0 ). Because if ( a neq 0 ), then ( aR neq 0 ), so ( mu_a ) is injective. Since ( R ) is finite, injective implies surjective, so ( mu_a ) is bijective for all ( a neq 0 ). Therefore, every non-zero element is a unit.Wait, that's a crucial point! If ( mu_a ) is injective and ( R ) is finite, then ( mu_a ) is also surjective. Therefore, for every ( a neq 0 ), there exists some ( b in R ) such that ( ab = 1 ). So, every non-zero element is a unit.Therefore, ( R ) is a division ring. But since ( R ) is commutative, it's a field. Wait, but ( R ) is a finite commutative ring where every non-zero element is a unit, so it's a finite field, which certainly has a multiplicative identity.Wait, hold on. Is that correct? If every non-zero element is a unit, then ( R ) is a division ring, and since it's commutative, it's a field. But ( R ) is given as a finite commutative ring, so if it's a field, it has an identity.But wait, does ( R ) necessarily have an identity? Because in general, a division ring has an identity, but we're assuming ( R ) is a ring, not necessarily with identity. However, in our case, we've deduced that every non-zero element is a unit, which in the context of rings, usually implies the existence of an identity.Wait, let me think carefully. In a ring where every non-zero element is a unit, does the ring necessarily have a multiplicative identity?Yes, actually. Suppose ( R ) is a ring where every non-zero element is a unit. Let ( a ) be any non-zero element. Then there exists ( b ) such that ( ab = 1 ). So, ( 1 ) exists in ( R ), making ( R ) a division ring with identity. Since ( R ) is commutative, it's a field.Therefore, in our case, since ( text{Ann}(R) = 0 ), every non-zero element is a unit, so ( R ) is a field, hence has an identity.Wait, but in the problem statement, ( R ) is just a finite commutative ring, not necessarily a field. So, does this mean that if ( text{Ann}(R) = 0 ), then ( R ) must be a field? That seems too strong because there are finite commutative rings with identity that are not fields, like ( mathbb{Z}/4mathbb{Z} ). But in ( mathbb{Z}/4mathbb{Z} ), the annihilator is zero because if ( a ) annihilates the entire ring, then ( a ) must be zero. However, ( mathbb{Z}/4mathbb{Z} ) is not a field because 2 is a zero divisor.Wait, hold on. In ( mathbb{Z}/4mathbb{Z} ), the annihilator is zero because if ( a ) annihilates the entire ring, then ( a ) must be zero. However, not every non-zero element is a unit. For example, 2 is not a unit because 2*2=0, which is not 1. So, in this case, ( text{Ann}(R) = 0 ), but ( R ) is not a field because not every non-zero element is a unit.Hmm, so my earlier reasoning that ( text{Ann}(R) = 0 ) implies every non-zero element is a unit must be flawed.Wait, where did I go wrong? I said that if ( text{Ann}(R) = 0 ), then for any ( a neq 0 ), ( mu_a ) is injective, hence surjective because ( R ) is finite. Therefore, ( mu_a ) is bijective, so ( a ) is a unit.But in ( mathbb{Z}/4mathbb{Z} ), ( a = 2 ) is not a unit, but ( mu_2 ) is not injective because ( 2*2 = 0 ). Wait, but ( text{Ann}(R) = 0 ) in ( mathbb{Z}/4mathbb{Z} ) because the only element that annihilates the entire ring is zero. However, ( 2 ) does not annihilate the entire ring because ( 2*1 = 2 neq 0 ). So, ( mu_2 ) is not injective because ( 2*2 = 0 ), but ( 2 ) is not in the annihilator.Wait, so in ( mathbb{Z}/4mathbb{Z} ), ( text{Ann}(R) = 0 ), but ( mu_2 ) is not injective because ( 2*2 = 0 ), which shows that ( mu_2 ) is not injective. But ( 2 ) is not in the annihilator because ( 2*1 = 2 neq 0 ). So, my earlier reasoning that ( text{Ann}(R) = 0 ) implies that ( mu_a ) is injective for all ( a neq 0 ) is incorrect.Wait, no. If ( text{Ann}(R) = 0 ), then for any ( a neq 0 ), there exists some ( b ) such that ( ab neq 0 ). But that doesn't necessarily mean that ( mu_a ) is injective. For ( mu_a ) to be injective, we need that ( a x = 0 ) implies ( x = 0 ). But ( text{Ann}(R) = 0 ) only tells us that there's no ( a neq 0 ) such that ( a x = 0 ) for all ( x ). It doesn't prevent ( a ) from annihilating some specific ( x neq 0 ).So, in ( mathbb{Z}/4mathbb{Z} ), ( 2 ) annihilates ( 2 ), but ( 2 ) doesn't annihilate the entire ring, so ( text{Ann}(R) = 0 ). Therefore, ( mu_2 ) is not injective, but ( text{Ann}(R) = 0 ).So, my earlier conclusion that ( text{Ann}(R) = 0 ) implies every non-zero element is a unit is incorrect. Therefore, I need to find another approach.Maybe I can use the fact that ( R ) is finite and consider the set of idempotent elements. If I can find an idempotent element ( e ) such that ( e ) acts as the identity on ( R ), then ( e ) is the multiplicative identity.Alternatively, perhaps I can use the fact that in a finite ring, the set of units is non-empty and forms a group. If I can show that the group of units acts transitively on the ring or something like that, maybe I can find an identity.Wait, another idea: Since ( R ) is finite, let's consider the sum of all elements in ( R ). Maybe that sum has some special properties. But I'm not sure.Alternatively, perhaps I can use the fact that in a finite ring, the Jacobson radical is nilpotent. If the annihilator is zero, then the Jacobson radical is zero, making the ring semisimple. But I'm not sure how that helps.Wait, in a semisimple ring, every module is a direct sum of simple modules. But I'm not sure if that directly implies the existence of an identity.Hmm, this is getting complicated. Maybe I need to look up some theorems or properties related to finite commutative rings and annihilators.Wait, I recall that in a finite ring, the annihilator of the ring being zero implies that the ring is a division ring. But since it's commutative, it's a field. But earlier, I saw that ( mathbb{Z}/4mathbb{Z} ) has ( text{Ann}(R) = 0 ) but is not a field. So, that must not be correct.Wait, no, in ( mathbb{Z}/4mathbb{Z} ), the annihilator is zero, but it's not a division ring because not every non-zero element is a unit. So, maybe the statement is that in a finite ring, if the annihilator is zero and every non-zero element is a unit, then it's a division ring. But that's a bit circular.Wait, perhaps I need to use the fact that in a finite ring, if the annihilator is zero, then the ring is a division ring. But as we saw, that's not true because ( mathbb{Z}/4mathbb{Z} ) is a counterexample.So, maybe the correct statement is that in a finite ring, if the annihilator is zero and the ring is reduced (i.e., has no nilpotent elements), then it's a division ring. But I'm not sure.Alternatively, perhaps I can use the fact that in a finite ring, the set of units is non-empty and forms a group, and if the annihilator is zero, then the group of units acts transitively on the ring, forcing the existence of an identity.Wait, I'm not making progress here. Let me try to think of another approach.Since ( R ) is finite and commutative, it's a product of finite local rings. Suppose ( R = R_1 times R_2 times dots times R_n ), where each ( R_i ) is a finite local ring. If each ( R_i ) has an identity, then ( R ) does as well. So, it's sufficient to show that each ( R_i ) has an identity.Now, in each local ring ( R_i ), the annihilator ( text{Ann}(R_i) ) is zero because if ( a in R_i ) and ( a R_i = 0 ), then ( a R = 0 ), so ( a = 0 ). Therefore, each ( R_i ) has ( text{Ann}(R_i) = 0 ).So, now I can focus on the case where ( R ) is a finite local commutative ring with ( text{Ann}(R) = 0 ). Does such a ring have an identity?In a finite local ring, the maximal ideal ( mathfrak{m} ) is nilpotent. If ( text{Ann}(R) = 0 ), then ( mathfrak{m} ) cannot be the entire ring, so there must be units in ( R ). Moreover, the units are ( R setminus mathfrak{m} ).Now, in a finite local ring, the set of units forms a group under multiplication. If I can show that this group acts transitively on the ring or something like that, maybe I can find an identity.Alternatively, perhaps I can use the fact that in a finite local ring, the residue field ( R/mathfrak{m} ) is a finite field, which has an identity. Maybe lifting this identity to ( R ) gives the desired result.Wait, another idea: Since ( R ) is a finite local ring with ( text{Ann}(R) = 0 ), the maximal ideal ( mathfrak{m} ) is nilpotent, say ( mathfrak{m}^k = 0 ) for some ( k ). Now, consider the element ( 1 + m ) where ( m in mathfrak{m} ). Since ( mathfrak{m} ) is nilpotent, ( 1 + m ) is a unit. Therefore, the set of units is non-empty and forms a group.But how does this help me find an identity?Wait, perhaps I can use the fact that in a finite ring, the sum of all units is either zero or a unit. But I'm not sure.Alternatively, maybe I can use the fact that in a finite local ring, the identity can be expressed as a linear combination of idempotents or something like that.Wait, I'm stuck here. Maybe I need to look for a different approach altogether.Let me think about the original problem again. It says that ( R ) is a finite commutative ring with ( text{Ann}(R) = 0 ) if and only if it has a multiplicative identity.I've shown the "if" part, that if ( R ) has an identity, then ( text{Ann}(R) = 0 ). Now, for the "only if" part, I need to show that if ( text{Ann}(R) = 0 ), then ( R ) has an identity.Another idea: Since ( R ) is finite, it's a vector space over some finite field. Let me assume that ( R ) has characteristic ( p ), so it's a vector space over ( mathbb{F}_p ). Then, ( text{End}_R(R) ) is isomorphic to ( text{End}_{mathbb{F}_p}(R) ), which is a matrix ring over ( mathbb{F}_p ). Since ( R ) is finite-dimensional over ( mathbb{F}_p ), ( text{End}_{mathbb{F}_p}(R) ) is finite-dimensional as well.But I'm not sure if this helps. Maybe I can use the fact that ( text{End}_{mathbb{F}_p}(R) ) has an identity, and if ( iota ) is surjective, then ( R ) has an identity.Wait, but I don't know if ( iota ) is surjective. I only know it's injective.Hmm, perhaps I can use the fact that ( text{End}_{mathbb{F}_p}(R) ) is a finite-dimensional algebra over ( mathbb{F}_p ), and ( R ) is a subalgebra. Then, by some theorem, maybe ( R ) has an identity.Wait, I recall that in finite-dimensional algebras over a field, every subalgebra with trivial annihilator has an identity. Maybe that's the case here.Alternatively, perhaps I can use the fact that in a finite-dimensional algebra over a field, if the annihilator is zero, then the algebra has an identity. Is that a theorem?I think it is. Let me recall. In a finite-dimensional algebra over a field, if the left annihilator of the algebra is zero, then the algebra has an identity. Similarly for the right annihilator. Since ( R ) is commutative, left and right annihilators coincide.Therefore, in our case, since ( R ) is a finite-dimensional algebra over ( mathbb{F}_p ) with ( text{Ann}(R) = 0 ), it must have an identity.Therefore, ( R ) has a multiplicative identity.Okay, that seems to work. So, putting it all together:1. If ( R ) has an identity, then ( text{Ann}(R) = 0 ).2. If ( text{Ann}(R) = 0 ), then ( R ) is a finite-dimensional algebra over ( mathbb{F}_p ) with trivial annihilator, hence it has an identity.Therefore, the two statements are equivalent.I think that's the solution. It took a while, but breaking it down into parts and considering the structure of finite commutative rings helped me see the connection between the annihilator and the existence of an identity.

💡Okay, let's see. Fred has 10 red balloons, Sam has 46, and Dan has 16. I need to find out how many red balloons they have altogether. Hmm, so I guess I should add up all their balloons. First, Fred has 10. That's straightforward. Then Sam has 46. Okay, so 10 plus 46. Let me do that in my head. 10 plus 40 is 50, and then plus 6 is 56. So Fred and Sam together have 56 balloons. Now, Dan has 16 balloons. I need to add that to the 56. So 56 plus 16. Hmm, 56 plus 10 is 66, and then plus 6 more is 72. So altogether, Fred, Sam, and Dan have 72 red balloons. Wait, does the cost of the balloons matter here? The problem mentioned that the balloons cost 10 dollars, but I don't think that affects the total number of balloons. I think it's just extra information. So I can ignore that part for this question. Just to double-check, let me add them again. Fred: 10, Sam: 46, Dan: 16. 10 plus 46 is 56, plus 16 is 72. Yeah, that seems right. I don't think I made any mistakes there. So, the total number of red balloons they have is 72.