💡Alright, so I have this problem where I need to find all functions ( f: mathbf{R}^{+} rightarrow mathbf{R}^{+} ) that satisfy two conditions. The first condition is a functional equation involving three variables, and the second condition is that the function is strictly increasing for inputs greater than or equal to 1. Let me try to unpack this step by step.First, let me write down the functional equation again to make sure I have it right:For all positive real numbers ( x, y, z ),[f(x y z) + f(x) + f(y) + f(z) = f(sqrt{x y}) cdot f(sqrt{y z}) cdot f(sqrt{z x}).]And the second condition is that for all ( 1 leqslant x < y ), ( f(x) < f(y) ). So, the function is strictly increasing on the interval ([1, infty)).Hmm, functional equations can be tricky, but maybe I can find some specific values or substitutions that simplify the equation. Let me try plugging in some specific values for ( x, y, z ) to see if I can get more information about ( f ).Let's start by setting ( x = y = z = 1 ). That should simplify things a bit.Substituting ( x = y = z = 1 ) into the equation:[f(1 cdot 1 cdot 1) + f(1) + f(1) + f(1) = f(sqrt{1 cdot 1}) cdot f(sqrt{1 cdot 1}) cdot f(sqrt{1 cdot 1}).]Simplifying:[f(1) + f(1) + f(1) + f(1) = f(1) cdot f(1) cdot f(1).]So,[4f(1) = [f(1)]^3.]This is a cubic equation in ( f(1) ). Let me solve for ( f(1) ).Let ( a = f(1) ). Then,[4a = a^3 implies a^3 - 4a = 0 implies a(a^2 - 4) = 0.]So, ( a = 0 ) or ( a = 2 ) or ( a = -2 ). But since ( f ) maps positive reals to positive reals, ( f(1) ) must be positive. Therefore, ( f(1) = 2 ).Okay, so that's a good start. We know that ( f(1) = 2 ).Now, maybe I can try setting two variables equal and see what happens. Let me set ( x = y = z = t ) for some positive real number ( t ).Substituting ( x = y = z = t ):[f(t cdot t cdot t) + f(t) + f(t) + f(t) = f(sqrt{t cdot t}) cdot f(sqrt{t cdot t}) cdot f(sqrt{t cdot t}).]Simplifying:[f(t^3) + 3f(t) = [f(t)]^3.]So,[f(t^3) = [f(t)]^3 - 3f(t).]Hmm, that's an interesting relation. It relates the value of the function at ( t^3 ) to the value at ( t ).Maybe I can explore this further. Let me denote ( f(t) = g(t) ) for simplicity, so the equation becomes:[g(t^3) = [g(t)]^3 - 3g(t).]This looks similar to some trigonometric identities or maybe some exponential functions, but I'm not sure yet.Alternatively, perhaps I can consider the function ( g(t) = f(t) ) and see if it satisfies a particular functional form.Wait, another thought: if I set ( z = 1 ), maybe that will simplify the original equation.Let me set ( z = 1 ). Then the equation becomes:[f(x y cdot 1) + f(x) + f(y) + f(1) = f(sqrt{x y}) cdot f(sqrt{y cdot 1}) cdot f(sqrt{1 cdot x}).]Simplifying:[f(x y) + f(x) + f(y) + 2 = f(sqrt{x y}) cdot f(sqrt{y}) cdot f(sqrt{x}).]Hmm, that's still a bit complicated, but maybe I can make another substitution. Let me set ( x = y = t ), so that ( z = 1 ).Substituting ( x = y = t ) and ( z = 1 ):[f(t cdot t cdot 1) + f(t) + f(t) + f(1) = f(sqrt{t cdot t}) cdot f(sqrt{t cdot 1}) cdot f(sqrt{1 cdot t}).]Simplifying:[f(t^2) + 2f(t) + 2 = f(t) cdot f(sqrt{t}) cdot f(sqrt{t}).]Which becomes:[f(t^2) + 2f(t) + 2 = [f(sqrt{t})]^2 cdot f(t).]Wait, that seems a bit messy. Maybe I can denote ( s = sqrt{t} ), so ( t = s^2 ). Then, the equation becomes:[f(s^4) + 2f(s^2) + 2 = [f(s)]^2 cdot f(s^2).]Hmm, not sure if that helps directly, but perhaps I can relate it to the earlier equation ( f(t^3) = [f(t)]^3 - 3f(t) ).Alternatively, maybe I should consider the possibility that ( f ) is a power function. Let me assume that ( f(t) = t^k + t^{-k} ) for some exponent ( k ). This form is symmetric and often appears in functional equations involving products and reciprocals.Let me test this assumption. Suppose ( f(t) = t^k + t^{-k} ). Then, let's compute both sides of the original equation.First, compute the left-hand side (LHS):[f(x y z) + f(x) + f(y) + f(z) = (x y z)^k + (x y z)^{-k} + x^k + x^{-k} + y^k + y^{-k} + z^k + z^{-k}.]Now, compute the right-hand side (RHS):[f(sqrt{x y}) cdot f(sqrt{y z}) cdot f(sqrt{z x}).]First, compute each ( f ) term:[f(sqrt{x y}) = (sqrt{x y})^k + (sqrt{x y})^{-k} = (x y)^{k/2} + (x y)^{-k/2},][f(sqrt{y z}) = (y z)^{k/2} + (y z)^{-k/2},][f(sqrt{z x}) = (z x)^{k/2} + (z x)^{-k/2}.]Now, multiply these three expressions together:[[(x y)^{k/2} + (x y)^{-k/2}] cdot [(y z)^{k/2} + (y z)^{-k/2}] cdot [(z x)^{k/2} + (z x)^{-k/2}].]This looks complicated, but maybe it simplifies nicely. Let me denote ( a = x^{k/2} ), ( b = y^{k/2} ), and ( c = z^{k/2} ). Then, the terms become:[(a b + frac{1}{a b}) cdot (b c + frac{1}{b c}) cdot (c a + frac{1}{c a}).]Multiplying these out:First, multiply the first two terms:[(a b + frac{1}{a b})(b c + frac{1}{b c}) = a b cdot b c + a b cdot frac{1}{b c} + frac{1}{a b} cdot b c + frac{1}{a b} cdot frac{1}{b c}.]Simplify each term:[= a b^2 c + frac{a}{c} + frac{c}{a} + frac{1}{a b^2 c}.]Now, multiply this result by the third term ( (c a + frac{1}{c a}) ):[(a b^2 c + frac{a}{c} + frac{c}{a} + frac{1}{a b^2 c}) cdot (c a + frac{1}{c a}).]This will result in a lot of terms, but let me see if any patterns emerge.Alternatively, maybe there's a smarter way to multiply these terms. Notice that each term is of the form ( u + 1/u ), so their product might have a pattern.Recall that ( (u + 1/u)(v + 1/v) = u v + u/v + v/u + 1/(u v) ). Similarly, multiplying three such terms would involve all combinations of multiplying the terms in each bracket.But perhaps instead of expanding everything, I can see if the product simplifies to something similar to the LHS.Wait, let me think differently. If I set ( k = 1 ), does the equation hold?Let me test ( k = 1 ). Then, ( f(t) = t + 1/t ).Compute LHS:[f(x y z) + f(x) + f(y) + f(z) = (x y z) + frac{1}{x y z} + x + frac{1}{x} + y + frac{1}{y} + z + frac{1}{z}.]Compute RHS:[f(sqrt{x y}) cdot f(sqrt{y z}) cdot f(sqrt{z x}).]First, compute each ( f ):[f(sqrt{x y}) = sqrt{x y} + frac{1}{sqrt{x y}},][f(sqrt{y z}) = sqrt{y z} + frac{1}{sqrt{y z}},][f(sqrt{z x}) = sqrt{z x} + frac{1}{sqrt{z x}}.]Now, multiply these three:[(sqrt{x y} + frac{1}{sqrt{x y}})(sqrt{y z} + frac{1}{sqrt{y z}})(sqrt{z x} + frac{1}{sqrt{z x}}).]This seems complicated, but let's see if it equals the LHS.Alternatively, maybe I can choose specific values for ( x, y, z ) to test if ( f(t) = t + 1/t ) satisfies the equation.Let me set ( x = y = z = 1 ). Then, LHS:[f(1) + f(1) + f(1) + f(1) = 4f(1) = 4 times 2 = 8.]RHS:[f(1) cdot f(1) cdot f(1) = 2 times 2 times 2 = 8.]Okay, that works.What about ( x = y = z = t )? Then, LHS:[f(t^3) + 3f(t) = (t^3 + 1/t^3) + 3(t + 1/t).]RHS:[f(t) cdot f(t) cdot f(t) = (t + 1/t)^3.]Let me compute both sides.LHS:[t^3 + 1/t^3 + 3t + 3/t.]RHS:[(t + 1/t)^3 = t^3 + 3t^2 cdot (1/t) + 3t cdot (1/t)^2 + (1/t)^3 = t^3 + 3t + 3/t + 1/t^3.]So, LHS equals RHS. That's good.Another test: let me set ( x = 2 ), ( y = 1 ), ( z = 1 ).Compute LHS:[f(2 times 1 times 1) + f(2) + f(1) + f(1) = f(2) + f(2) + 2 + 2 = 2f(2) + 4.]Compute RHS:[f(sqrt{2 times 1}) cdot f(sqrt{1 times 1}) cdot f(sqrt{1 times 2}) = f(sqrt{2}) cdot f(1) cdot f(sqrt{2}).]Which is:[[f(sqrt{2})]^2 cdot 2.]So, LHS = ( 2f(2) + 4 ), RHS = ( 2[f(sqrt{2})]^2 ).Let me compute both sides with ( f(t) = t + 1/t ).LHS:[2(2 + 1/2) + 4 = 2(2.5) + 4 = 5 + 4 = 9.]RHS:[2[(sqrt{2} + 1/sqrt{2})]^2 = 2[(sqrt{2} + sqrt{2}/2)^2] = 2[( (2sqrt{2} + sqrt{2}) / 2 )^2] = 2[(3sqrt{2}/2)^2] = 2[(9 times 2)/4] = 2[18/4] = 2[4.5] = 9.]So, LHS equals RHS again. That's promising.Okay, so it seems that ( f(t) = t + 1/t ) satisfies the functional equation. But the problem says "find all functions," so I need to check if there are other possible functions or if this is the only one.Wait, earlier I assumed ( f(t) = t^k + t^{-k} ). Let me see if this form works for any ( k ).Suppose ( f(t) = t^k + t^{-k} ). Let's see if this satisfies the functional equation for any ( k ).Compute LHS:[f(x y z) + f(x) + f(y) + f(z) = (x y z)^k + (x y z)^{-k} + x^k + x^{-k} + y^k + y^{-k} + z^k + z^{-k}.]Compute RHS:[f(sqrt{x y}) cdot f(sqrt{y z}) cdot f(sqrt{z x}).]Each ( f ) term is:[f(sqrt{x y}) = (x y)^{k/2} + (x y)^{-k/2},][f(sqrt{y z}) = (y z)^{k/2} + (y z)^{-k/2},][f(sqrt{z x}) = (z x)^{k/2} + (z x)^{-k/2}.]Multiplying these together:[[(x y)^{k/2} + (x y)^{-k/2}] cdot [(y z)^{k/2} + (y z)^{-k/2}] cdot [(z x)^{k/2} + (z x)^{-k/2}].]Let me denote ( a = x^{k/2} ), ( b = y^{k/2} ), ( c = z^{k/2} ). Then, the terms become:[(a b + 1/(a b)) cdot (b c + 1/(b c)) cdot (c a + 1/(c a)).]Multiplying these out:First, multiply the first two terms:[(a b + 1/(a b))(b c + 1/(b c)) = a b cdot b c + a b cdot 1/(b c) + 1/(a b) cdot b c + 1/(a b) cdot 1/(b c).]Simplify:[= a b^2 c + a / c + c / a + 1/(a b^2 c).]Now, multiply this by the third term ( (c a + 1/(c a)) ):[(a b^2 c + a / c + c / a + 1/(a b^2 c)) cdot (c a + 1/(c a)).]This will result in several terms. Let me compute each product:1. ( a b^2 c cdot c a = a^2 b^2 c^2 )2. ( a b^2 c cdot 1/(c a) = b^2 )3. ( a / c cdot c a = a^2 )4. ( a / c cdot 1/(c a) = 1 / c^2 )5. ( c / a cdot c a = c^2 )6. ( c / a cdot 1/(c a) = 1 / a^2 )7. ( 1/(a b^2 c) cdot c a = 1 / b^2 )8. ( 1/(a b^2 c) cdot 1/(c a) = 1/(a^2 b^2 c^2) )So, combining all these terms:[a^2 b^2 c^2 + b^2 + a^2 + 1/c^2 + c^2 + 1/a^2 + 1/b^2 + 1/(a^2 b^2 c^2).]Now, recall that ( a = x^{k/2} ), ( b = y^{k/2} ), ( c = z^{k/2} ). Therefore:[a^2 = x^k, quad b^2 = y^k, quad c^2 = z^k,]and[1/a^2 = x^{-k}, quad 1/b^2 = y^{-k}, quad 1/c^2 = z^{-k},]and[a^2 b^2 c^2 = (x y z)^k, quad 1/(a^2 b^2 c^2) = (x y z)^{-k}.]So, substituting back, the RHS becomes:[(x y z)^k + y^k + x^k + z^{-k} + z^k + x^{-k} + y^{-k} + (x y z)^{-k}.]Comparing this to the LHS:[(x y z)^k + (x y z)^{-k} + x^k + x^{-k} + y^k + y^{-k} + z^k + z^{-k}.]They are the same! So, for any ( k ), the function ( f(t) = t^k + t^{-k} ) satisfies the functional equation.Now, I need to check the second condition: ( f ) is strictly increasing for ( x geq 1 ).Let me compute the derivative of ( f(t) = t^k + t^{-k} ) for ( t > 0 ).The derivative is:[f'(t) = k t^{k - 1} - k t^{-k - 1}.]Simplify:[f'(t) = k left( t^{k - 1} - t^{-k - 1} right).]For ( t geq 1 ), we need ( f'(t) > 0 ) to ensure ( f ) is strictly increasing.Let me analyze ( f'(t) ):[f'(t) = k left( t^{k - 1} - t^{-k - 1} right).]Since ( k > 0 ) (because if ( k = 0 ), ( f(t) = 2 ), which is constant and doesn't satisfy the strictly increasing condition), we can factor out ( t^{-k - 1} ):[f'(t) = k t^{-k - 1} left( t^{2k} - 1 right).]For ( t geq 1 ), ( t^{2k} geq 1 ), so ( t^{2k} - 1 geq 0 ). Therefore, ( f'(t) geq 0 ).Moreover, for ( t > 1 ), ( t^{2k} - 1 > 0 ), so ( f'(t) > 0 ). Thus, ( f(t) ) is strictly increasing for ( t > 1 ).Wait, but what about ( t = 1 )? At ( t = 1 ), ( f'(1) = k (1^{k - 1} - 1^{-k - 1}) = k (1 - 1) = 0 ). So, the derivative is zero at ( t = 1 ), but the function is still strictly increasing for ( t > 1 ).Therefore, ( f(t) = t^k + t^{-k} ) satisfies both conditions for any ( k > 0 ).Is there any other function that satisfies the conditions? Let me think.Suppose there exists another function ( f ) that is not of the form ( t^k + t^{-k} ). But given the functional equation's structure and the fact that we derived ( f(t^3) = [f(t)]^3 - 3f(t) ), which is a specific recurrence relation, it's likely that the only solutions are of the form ( t^k + t^{-k} ).Moreover, the strictly increasing condition for ( t geq 1 ) pins down the exponent ( k ) to be positive, as we saw earlier.Therefore, the only functions that satisfy both conditions are ( f(t) = t^k + t^{-k} ) for some ( k > 0 ).I think that's the solution. Let me just recap:1. By setting ( x = y = z = 1 ), we found ( f(1) = 2 ).2. Assuming ( f(t) = t^k + t^{-k} ), we verified that it satisfies the functional equation for any ( k ).3. Checked that the derivative is positive for ( t > 1 ) when ( k > 0 ), ensuring the function is strictly increasing.4. Concluded that these are the only solutions.So, the functions are ( f(x) = x^k + x^{-k} ) for some positive real number ( k ).

💡Alright, so I've got this problem here about 100 people sitting around a round table. Each person is either a Knight, a Liar, or an Eccentric. Knights always tell the truth, Liars always lie, and Eccentrics have this interesting behavior: they tell the truth if a Liar is to their left, lie if a Knight is to their left, and can say anything if another Eccentric is to their left. Everyone at the table says, "A Liar is sitting to my right." I need to figure out how many Liars there are in total, considering all possible scenarios.Okay, let's break this down. First, since it's a round table with 100 people, the arrangement is circular, meaning the first person is to the right of the last person. Each person makes a statement about the person to their right. So, if I can figure out the pattern or sequence of Knights, Liars, and Eccentrics, I can determine the number of Liars.Let's consider the possibilities one by one.**1. All Knights:**If everyone is a Knight, then everyone tells the truth. So, if everyone says, "A Liar is sitting to my right," that would mean everyone to the right is a Liar. But if everyone is a Knight, that's a contradiction because Knights can't lie. So, this scenario is impossible.**2. All Liars:**If everyone is a Liar, then everyone lies. So, when they say, "A Liar is sitting to my right," the truth is that there is no Liar to their right. But if everyone is a Liar, then there is a Liar to their right, which means their statement is true. But Liars can't tell the truth, so this is also a contradiction. Therefore, all Liars is impossible.**3. All Eccentrics:**If everyone is an Eccentric, their behavior depends on who is to their left. Since everyone is an Eccentric, the person to their left is also an Eccentric. The problem states that if an Eccentric has another Eccentric to their left, they can say anything. So, in this case, they could either lie or tell the truth. If they say, "A Liar is sitting to my right," it doesn't necessarily mean anything specific because they can say anything. So, this scenario is possible, but it doesn't give us any information about the number of Liars because there are none.**4. A Mix of Knights and Liars:**Let's consider alternating Knights and Liars. If we have a Knight, then a Liar, then a Knight, and so on. Knights tell the truth, so if a Knight says, "A Liar is sitting to my right," then indeed, there is a Liar to their right. Liars lie, so when a Liar says, "A Liar is sitting to my right," the truth is that there is not a Liar to their right, which would mean there is a Knight to their right. This fits the alternating pattern. So, in this case, we have 50 Knights and 50 Liars.**5. A Mix Including Eccentrics:**Now, let's consider if there are Eccentrics in the mix. If an Eccentric has a Liar to their left, they tell the truth. If they have a Knight to their left, they lie. If they have an Eccentric to their left, they can say anything. Suppose we have a sequence like Knight, Liar, Eccentric, Knight, Liar, Eccentric, and so on. Let's see if this works. The Knight says, "A Liar is sitting to my right," which is true because there is a Liar to their right. The Liar says, "A Liar is sitting to my right," which is a lie, so there is not a Liar to their right, meaning there is an Eccentric to their right. The Eccentric, having a Liar to their left, tells the truth. So, if the Eccentric says, "A Liar is sitting to my right," then there must be a Liar to their right. But in our sequence, after the Eccentric comes a Knight, which is not a Liar. This creates a contradiction because the Eccentric should be telling the truth but is actually lying.Alternatively, if the sequence is Knight, Eccentric, Liar, Knight, Eccentric, Liar, and so on. The Knight says, "A Liar is sitting to my right," which is true because there is an Eccentric to their right, but the Knight is supposed to tell the truth. Wait, no, the Knight is talking about the person to their right, which is an Eccentric, not a Liar. So, the Knight's statement would be false, which contradicts the fact that Knights always tell the truth.This suggests that introducing Eccentrics into the mix complicates the pattern and may not lead to a consistent solution. It seems that the only consistent patterns are either all Eccentrics or an alternating sequence of Knights and Liars.**6. All Eccentrics:**As previously considered, if all are Eccentrics, they can say anything because they have another Eccentric to their left. So, they can say, "A Liar is sitting to my right," regardless of the truth. This means there could be 0 Liars in this scenario.**7. Alternating Knights and Liars:**As established earlier, an alternating sequence of Knights and Liars results in 50 Knights and 50 Liars. Each Knight truthfully states that there is a Liar to their right, and each Liar falsely claims that there is a Liar to their right, which is actually a Knight.**8. Other Combinations:**What if we have more Knights than Liars or vice versa? For example, two Knights followed by one Liar, and so on. Let's test this. If we have Knight, Knight, Liar, Knight, Knight, Liar, etc. The first Knight says, "A Liar is sitting to my right," which is true because there is a Knight to their right, but wait, the Knight should be telling the truth, but the person to their right is a Knight, not a Liar. This is a contradiction because the Knight's statement would be false.Similarly, if we have more Liars than Knights, say Liar, Liar, Knight, Liar, Liar, Knight, etc. The first Liar says, "A Liar is sitting to my right," which is a lie, meaning there is not a Liar to their right, but in reality, there is a Liar to their right. This is also a contradiction.Therefore, any combination other than all Eccentrics or an alternating sequence of Knights and Liars leads to contradictions.**9. Conclusion:**After analyzing all possible scenarios, the only consistent solutions are:- All 100 people are Eccentrics, resulting in 0 Liars.- An alternating sequence of Knights and Liars, resulting in 50 Liars.These are the only possible configurations that satisfy the given conditions without leading to contradictions.

💡Alright, so I have this problem about a sequence. It says the first term is ( a_1 = 1 ), and there's a recursive relation: ( (a_{n+1} - 1)a_n + a_{n+1} = 0 ) for all positive integers ( n ). I need to find the general term formula for ( {a_n} ) and then, using that, find the sum of the first ( n ) terms of another sequence ( {c_n} ) where ( c_n = frac{3^n}{a_n} ).Okay, starting with part (1). The recursive relation is a bit complicated, so maybe I can rearrange it to make it simpler. Let me write it down:( (a_{n+1} - 1)a_n + a_{n+1} = 0 )Hmm, let's distribute the ( a_n ):( a_{n+1}a_n - a_n + a_{n+1} = 0 )Now, let's collect like terms. The terms with ( a_{n+1} ) are ( a_{n+1}a_n ) and ( a_{n+1} ). So factor ( a_{n+1} ) out:( a_{n+1}(a_n + 1) - a_n = 0 )Wait, that's not quite right. Let me try again. So, from the original equation:( (a_{n+1} - 1)a_n + a_{n+1} = 0 )Let me expand it:( a_{n+1}a_n - a_n + a_{n+1} = 0 )Now, group the ( a_{n+1} ) terms:( a_{n+1}(a_n + 1) - a_n = 0 )So, moving the ( a_n ) to the other side:( a_{n+1}(a_n + 1) = a_n )Then, solving for ( a_{n+1} ):( a_{n+1} = frac{a_n}{a_n + 1} )Okay, that looks better. So, the recursive formula is ( a_{n+1} = frac{a_n}{a_n + 1} ). Hmm, this seems like a nonlinear recurrence relation because ( a_{n+1} ) is expressed in terms of ( a_n ) in a fractional form.I wonder if I can transform this into something more manageable. Maybe by taking reciprocals? Let's try that. Let me define ( b_n = frac{1}{a_n} ). Then, ( a_n = frac{1}{b_n} ).Substituting into the recursive formula:( a_{n+1} = frac{frac{1}{b_n}}{frac{1}{b_n} + 1} )Simplify the denominator:( frac{1}{b_n} + 1 = frac{1 + b_n}{b_n} )So, ( a_{n+1} = frac{frac{1}{b_n}}{frac{1 + b_n}{b_n}} = frac{1}{1 + b_n} )But ( a_{n+1} = frac{1}{b_{n+1}} ), so:( frac{1}{b_{n+1}} = frac{1}{1 + b_n} )Taking reciprocals on both sides:( b_{n+1} = 1 + b_n )Ah! That's much simpler. So, ( b_{n+1} = b_n + 1 ). That means the sequence ( {b_n} ) is an arithmetic sequence with common difference 1.Given that ( a_1 = 1 ), so ( b_1 = frac{1}{a_1} = 1 ). Therefore, the sequence ( {b_n} ) starts at 1 and increases by 1 each time. So, the general term for ( b_n ) is:( b_n = b_1 + (n - 1) times 1 = 1 + (n - 1) = n )Therefore, ( b_n = n ). Since ( b_n = frac{1}{a_n} ), we have:( frac{1}{a_n} = n ) => ( a_n = frac{1}{n} )So, the general term for ( {a_n} ) is ( a_n = frac{1}{n} ). That seems straightforward once I took reciprocals.Moving on to part (2). We have ( c_n = frac{3^n}{a_n} ). Since ( a_n = frac{1}{n} ), substituting that in:( c_n = frac{3^n}{frac{1}{n}} = n times 3^n )So, ( c_n = n cdot 3^n ). Now, we need to find the sum of the first ( n ) terms of this sequence, denoted as ( S_n ).So, ( S_n = c_1 + c_2 + c_3 + ldots + c_n = 1 cdot 3^1 + 2 cdot 3^2 + 3 cdot 3^3 + ldots + n cdot 3^n )This looks like a standard sum involving ( k cdot r^k ). I remember there's a formula for the sum ( sum_{k=1}^{n} k r^k ). Let me recall it.The formula is:( sum_{k=1}^{n} k r^k = frac{r(1 - (n + 1) r^n + n r^{n + 1})}{(1 - r)^2} )But let me verify this formula or derive it if I can't remember exactly.Alternatively, I can use the method of generating functions or recurrence relations to find the sum.Let me try the method of multiplying the sum by the common ratio and subtracting.Let me denote:( S = sum_{k=1}^{n} k cdot 3^k )Then, multiply both sides by 3:( 3S = sum_{k=1}^{n} k cdot 3^{k + 1} = sum_{k=1}^{n} k cdot 3^{k} cdot 3 = sum_{k=1}^{n} k cdot 3^{k + 1} )Now, write out the terms:( S = 1 cdot 3^1 + 2 cdot 3^2 + 3 cdot 3^3 + ldots + n cdot 3^n )( 3S = 1 cdot 3^2 + 2 cdot 3^3 + 3 cdot 3^4 + ldots + n cdot 3^{n + 1} )Now, subtract ( S ) from ( 3S ):( 3S - S = 2S = (1 cdot 3^2 + 2 cdot 3^3 + 3 cdot 3^4 + ldots + n cdot 3^{n + 1}) - (1 cdot 3^1 + 2 cdot 3^2 + 3 cdot 3^3 + ldots + n cdot 3^n) )Let me write out the terms:- The first term of ( 3S ) is ( 1 cdot 3^2 ), and the first term of ( S ) is ( 1 cdot 3^1 ). So, subtracting, we get ( 1 cdot 3^2 - 1 cdot 3^1 = 3^2 - 3^1 = 9 - 3 = 6 ).- The second term of ( 3S ) is ( 2 cdot 3^3 ), and the second term of ( S ) is ( 2 cdot 3^2 ). Subtracting: ( 2 cdot 3^3 - 2 cdot 3^2 = 2(27 - 9) = 2 cdot 18 = 36 ).- The third term of ( 3S ) is ( 3 cdot 3^4 ), and the third term of ( S ) is ( 3 cdot 3^3 ). Subtracting: ( 3 cdot 81 - 3 cdot 27 = 3(81 - 27) = 3 cdot 54 = 162 ).- And so on, until the last term.But wait, actually, when subtracting, most of the terms will cancel out except for the first and the last. Let me see:Looking at the subtraction:( 2S = (1 cdot 3^2 + 2 cdot 3^3 + 3 cdot 3^4 + ldots + n cdot 3^{n + 1}) - (1 cdot 3^1 + 2 cdot 3^2 + 3 cdot 3^3 + ldots + n cdot 3^n) )Let me align the terms:- The ( 1 cdot 3^2 ) term in ( 3S ) subtracts the ( 1 cdot 3^1 ) term in ( S ).- The ( 2 cdot 3^3 ) term in ( 3S ) subtracts the ( 2 cdot 3^2 ) term in ( S ).- The ( 3 cdot 3^4 ) term in ( 3S ) subtracts the ( 3 cdot 3^3 ) term in ( S ).- This pattern continues until the ( (n-1) cdot 3^n ) term in ( 3S ) subtracts the ( (n-1) cdot 3^{n-1} ) term in ( S ).- Finally, the last term in ( 3S ) is ( n cdot 3^{n + 1} ), and there's no corresponding term in ( S ) to subtract.So, after subtraction, the terms that remain are:- The first term: ( 1 cdot 3^2 - 1 cdot 3^1 = 3^2 - 3^1 = 9 - 3 = 6 )- The second term: ( 2 cdot 3^3 - 2 cdot 3^2 = 2(3^3 - 3^2) = 2(27 - 9) = 36 )- The third term: ( 3 cdot 3^4 - 3 cdot 3^3 = 3(81 - 27) = 162 )- ...- The ( (n-1) )-th term: ( (n-1) cdot 3^n - (n-1) cdot 3^{n-1} = (n-1)(3^n - 3^{n-1}) )- And the last term: ( n cdot 3^{n + 1} )Wait, actually, I think I made a mistake here. When subtracting, the terms don't just cancel out in the way I thought. Let me try a different approach.Let me write ( 3S - S = 2S ). So,( 2S = (1 cdot 3^2 + 2 cdot 3^3 + 3 cdot 3^4 + ldots + n cdot 3^{n + 1}) - (1 cdot 3^1 + 2 cdot 3^2 + 3 cdot 3^3 + ldots + n cdot 3^n) )Let me shift the index in the first sum to align the terms:Let me denote ( k' = k + 1 ) in the first sum, so when ( k = 1 ), ( k' = 2 ), and when ( k = n ), ( k' = n + 1 ). So, the first sum becomes:( sum_{k'=2}^{n + 1} (k' - 1) cdot 3^{k'} )But ( k' ) is just a dummy variable, so we can write it as:( sum_{k=2}^{n + 1} (k - 1) cdot 3^{k} )Now, subtracting the second sum:( 2S = sum_{k=2}^{n + 1} (k - 1) cdot 3^{k} - sum_{k=1}^{n} k cdot 3^{k} )Let me write out both sums:First sum: ( sum_{k=2}^{n + 1} (k - 1) cdot 3^{k} = sum_{k=2}^{n} (k - 1) cdot 3^{k} + (n) cdot 3^{n + 1} )Second sum: ( sum_{k=1}^{n} k cdot 3^{k} = 1 cdot 3^1 + sum_{k=2}^{n} k cdot 3^{k} )So, subtracting:( 2S = left( sum_{k=2}^{n} (k - 1) cdot 3^{k} + n cdot 3^{n + 1} right) - left( 1 cdot 3^1 + sum_{k=2}^{n} k cdot 3^{k} right) )Simplify term by term:- The ( sum_{k=2}^{n} (k - 1) cdot 3^{k} ) and ( - sum_{k=2}^{n} k cdot 3^{k} ) terms can be combined:( sum_{k=2}^{n} (k - 1 - k) cdot 3^{k} = sum_{k=2}^{n} (-1) cdot 3^{k} = - sum_{k=2}^{n} 3^{k} )- The remaining terms are ( n cdot 3^{n + 1} - 1 cdot 3^1 )So, putting it all together:( 2S = - sum_{k=2}^{n} 3^{k} + n cdot 3^{n + 1} - 3 )Now, let's compute ( sum_{k=2}^{n} 3^{k} ). This is a geometric series from ( k=2 ) to ( k=n ) with first term ( 3^2 = 9 ) and ratio 3.The sum of a geometric series ( sum_{k=0}^{m} ar^k = a frac{r^{m + 1} - 1}{r - 1} ). So, for our case:( sum_{k=2}^{n} 3^{k} = 3^2 cdot frac{3^{n - 1} - 1}{3 - 1} = 9 cdot frac{3^{n - 1} - 1}{2} = frac{9 cdot 3^{n - 1} - 9}{2} = frac{3^{n + 1} - 9}{2} )Wait, let me verify that:Wait, ( sum_{k=2}^{n} 3^k = sum_{k=0}^{n} 3^k - 3^0 - 3^1 = frac{3^{n + 1} - 1}{3 - 1} - 1 - 3 = frac{3^{n + 1} - 1}{2} - 4 = frac{3^{n + 1} - 1 - 8}{2} = frac{3^{n + 1} - 9}{2} ). Yes, that's correct.So, ( sum_{k=2}^{n} 3^k = frac{3^{n + 1} - 9}{2} )Therefore, substituting back into the equation for ( 2S ):( 2S = - left( frac{3^{n + 1} - 9}{2} right) + n cdot 3^{n + 1} - 3 )Simplify term by term:- ( - frac{3^{n + 1} - 9}{2} = - frac{3^{n + 1}}{2} + frac{9}{2} )- ( n cdot 3^{n + 1} ) remains as is- ( -3 ) remains as isSo, combining all terms:( 2S = - frac{3^{n + 1}}{2} + frac{9}{2} + n cdot 3^{n + 1} - 3 )Combine like terms:First, combine the ( 3^{n + 1} ) terms:( (- frac{1}{2} + n) cdot 3^{n + 1} )Then, combine the constants:( frac{9}{2} - 3 = frac{9}{2} - frac{6}{2} = frac{3}{2} )So, overall:( 2S = left( n - frac{1}{2} right) 3^{n + 1} + frac{3}{2} )Now, solve for ( S ):( S = frac{1}{2} left( left( n - frac{1}{2} right) 3^{n + 1} + frac{3}{2} right) )Simplify:( S = frac{n - frac{1}{2}}{2} cdot 3^{n + 1} + frac{3}{4} )Which is:( S = frac{2n - 1}{4} cdot 3^{n + 1} + frac{3}{4} )So, the sum ( S_n ) is:( S_n = frac{(2n - 1) cdot 3^{n + 1} + 3}{4} )Alternatively, factoring out the 3:( S_n = frac{(2n - 1) cdot 3^{n + 1} + 3}{4} )I think that's the final expression for ( S_n ).Let me check with a small ( n ) to see if this formula works. Let's take ( n = 1 ):- ( c_1 = 1 cdot 3^1 = 3 )- So, ( S_1 = 3 )- Plugging into the formula: ( frac{(2 cdot 1 - 1) cdot 3^{2} + 3}{4} = frac{(1) cdot 9 + 3}{4} = frac{12}{4} = 3 ). Correct.Now, ( n = 2 ):- ( c_1 = 3 ), ( c_2 = 2 cdot 9 = 18 )- ( S_2 = 3 + 18 = 21 )- Formula: ( frac{(4 - 1) cdot 3^{3} + 3}{4} = frac{3 cdot 27 + 3}{4} = frac{81 + 3}{4} = frac{84}{4} = 21 ). Correct.Another check, ( n = 3 ):- ( c_1 = 3 ), ( c_2 = 18 ), ( c_3 = 3 cdot 27 = 81 )- ( S_3 = 3 + 18 + 81 = 102 )- Formula: ( frac{(6 - 1) cdot 3^{4} + 3}{4} = frac{5 cdot 81 + 3}{4} = frac{405 + 3}{4} = frac{408}{4} = 102 ). Correct.Seems like the formula works for these small cases. So, I think I did it right.**Final Answer**(1) The general term is ( boxed{dfrac{1}{n}} ).(2) The sum of the first ( n ) terms is ( boxed{dfrac{(2n - 1) cdot 3^{n + 1} + 3}{4}} ).

💡To determine the coordinates of point ( D ), we start by using the midpoint formula. Given that ( M(6,6) ) is the midpoint of ( overline{CD} ) and point ( C ) has coordinates ( (2,10) ), we can set up the equations for the midpoint.Let the coordinates of point ( D ) be ( (x, y) ). According to the midpoint formula:[left( frac{x + 2}{2}, frac{y + 10}{2} right) = (6, 6)]This gives us two separate equations:[frac{x + 2}{2} = 6 quad text{and} quad frac{y + 10}{2} = 6]Solving for ( x ):[frac{x + 2}{2} = 6 x + 2 = 12 x = 10]Solving for ( y ):[frac{y + 10}{2} = 6 y + 10 = 12 y = 2]Therefore, the coordinates of point ( D ) are ( (10, 2) ). The sum of the coordinates of point ( D ) is:[10 + 2 = 12]

💡Okay, so I have this problem here: Given that ( b ) is an even multiple of 5171, find the greatest common divisor (GCD) of ( 4b^2 + 35b + 72 ) and ( 3b + 8 ). Hmm, let's see how to approach this.First, I remember that when finding the GCD of two numbers, especially polynomials, the Euclidean algorithm is a useful tool. The Euclidean algorithm involves dividing the larger number by the smaller one and then taking the remainder, repeating this process until the remainder is zero. The last non-zero remainder is the GCD.So, applying that here, I need to find ( gcd(4b^2 + 35b + 72, 3b + 8) ). Let me denote ( f(b) = 4b^2 + 35b + 72 ) and ( g(b) = 3b + 8 ). The goal is to compute ( gcd(f(b), g(b)) ).To use the Euclidean algorithm, I should divide ( f(b) ) by ( g(b) ) and find the remainder. Let me set that up.Dividing ( 4b^2 + 35b + 72 ) by ( 3b + 8 ). Hmm, how does polynomial division work? I think it's similar to numerical division. Let me try.First term: ( 4b^2 ) divided by ( 3b ) gives ( frac{4}{3}b ). So, I multiply ( 3b + 8 ) by ( frac{4}{3}b ) to get ( 4b^2 + frac{32}{3}b ). Then, subtract this from ( f(b) ):( (4b^2 + 35b + 72) - (4b^2 + frac{32}{3}b) = 0b^2 + left(35b - frac{32}{3}bright) + 72 ).Simplify the coefficients:( 35b - frac{32}{3}b = frac{105}{3}b - frac{32}{3}b = frac{73}{3}b ).So, the remainder after this step is ( frac{73}{3}b + 72 ). Hmm, that's still a bit messy with fractions. Maybe I should have multiplied through by 3 to eliminate denominators earlier on. Let me try that.Alternatively, perhaps I can express ( f(b) ) as a multiple of ( g(b) ) plus some remainder. Let me think.Wait, maybe instead of polynomial division, I can express ( f(b) ) as ( q(b) cdot g(b) + r ), where ( q(b) ) is the quotient and ( r ) is the remainder. Since ( g(b) ) is linear, the remainder should be a constant.Let me set ( f(b) = q(b) cdot (3b + 8) + r ). I need to find ( q(b) ) and ( r ).Assuming ( q(b) ) is linear, let me write ( q(b) = ab + c ). Then,( f(b) = (ab + c)(3b + 8) + r = 3ab^2 + (8a + 3c)b + (8c + r) ).Comparing coefficients with ( f(b) = 4b^2 + 35b + 72 ):- Coefficient of ( b^2 ): ( 3a = 4 ) ⇒ ( a = frac{4}{3} )- Coefficient of ( b ): ( 8a + 3c = 35 ) - Substitute ( a = frac{4}{3} ): ( 8*(4/3) + 3c = 35 ) ⇒ ( 32/3 + 3c = 35 ) - Subtract ( 32/3 ): ( 3c = 35 - 32/3 = (105 - 32)/3 = 73/3 ) ⇒ ( c = 73/9 )- Constant term: ( 8c + r = 72 ) - Substitute ( c = 73/9 ): ( 8*(73/9) + r = 72 ) ⇒ ( 584/9 + r = 72 ) - Convert 72 to ninths: ( 72 = 648/9 ) - So, ( r = 648/9 - 584/9 = 64/9 )Hmm, so the remainder is ( 64/9 ). That seems a bit complicated because of the fractions. Maybe I made a mistake in assuming ( q(b) ) is linear? Wait, no, since ( g(b) ) is linear, the remainder should indeed be a constant, but perhaps I should have scaled the equation to eliminate denominators.Alternatively, maybe I can use the Remainder Theorem. The Remainder Theorem states that the remainder of dividing a polynomial ( f(b) ) by ( (b - k) ) is ( f(k) ). But here, we're dividing by ( 3b + 8 ), which is ( 3(b + 8/3) ). So, the root is ( b = -8/3 ).So, if I plug ( b = -8/3 ) into ( f(b) ), I should get the remainder.Let's compute ( f(-8/3) ):( f(-8/3) = 4*(-8/3)^2 + 35*(-8/3) + 72 )First, compute each term:- ( 4*(64/9) = 256/9 )- ( 35*(-8/3) = -280/3 )- 72 is just 72.Convert all to ninths:- 256/9- -280/3 = -840/9- 72 = 648/9So, adding them up:( 256/9 - 840/9 + 648/9 = (256 - 840 + 648)/9 = (256 + 648 - 840)/9 )Compute numerator:256 + 648 = 904904 - 840 = 64So, ( f(-8/3) = 64/9 ). So, the remainder is indeed 64/9.Therefore, ( f(b) = (3b + 8)*q(b) + 64/9 ). So, the remainder is 64/9.But since we're dealing with integers (since ( b ) is an integer multiple of 5171, which is an integer), the GCD should be an integer. So, perhaps I need to adjust for the denominator.Wait, actually, in the Euclidean algorithm, when dealing with polynomials, the GCD is defined up to a constant factor, but since we're dealing with integer polynomials, the GCD should be an integer. So, maybe I should scale appropriately.Alternatively, perhaps I should consider that the GCD of ( f(b) ) and ( g(b) ) is the same as the GCD of ( g(b) ) and the remainder when ( f(b) ) is divided by ( g(b) ). So, in this case, ( gcd(f(b), g(b)) = gcd(g(b), 64/9) ).But 64/9 is not an integer, which complicates things. Maybe I should have multiplied through by 9 to clear the denominators. Let me try that.If I multiply ( f(b) ) by 9, I get ( 36b^2 + 315b + 648 ). Then, dividing this by ( 3b + 8 ), the remainder would be 64, as 9*(64/9) = 64.So, ( 9f(b) = (3b + 8)*q(b) + 64 ). Therefore, ( gcd(9f(b), 3b + 8) = gcd(3b + 8, 64) ).But since ( gcd(f(b), 3b + 8) ) divides both ( f(b) ) and ( 3b + 8 ), and 9 is co-prime with 3b + 8 (since 3b + 8 is linear and 9 is a constant), the GCD should be the same as ( gcd(3b + 8, 64) ).So, now, the problem reduces to finding ( gcd(3b + 8, 64) ).Given that ( b ) is an even multiple of 5171, let's denote ( b = 2k*5171 ) where ( k ) is an integer. So, ( b ) is even.Therefore, ( 3b + 8 = 3*(2k*5171) + 8 = 6k*5171 + 8 ). Let's compute this modulo 64 to find the GCD.First, note that 6k*5171 modulo 64 and 8 modulo 64.Compute 5171 modulo 64:Divide 5171 by 64:64*80 = 5120, so 5171 - 5120 = 51. So, 5171 ≡ 51 mod 64.Therefore, 6k*5171 ≡ 6k*51 mod 64.Compute 6*51 = 306. 306 mod 64: 64*4=256, 306-256=50. So, 6*51 ≡ 50 mod 64.Thus, 6k*5171 ≡ 50k mod 64.Therefore, ( 3b + 8 ≡ 50k + 8 ) mod 64.So, ( 3b + 8 ≡ 50k + 8 ) mod 64.We need to find ( gcd(50k + 8, 64) ).Note that 50 and 64 have a GCD of 2, since 50 = 2*25 and 64 = 2^6.So, ( gcd(50k + 8, 64) ) will depend on the value of ( k ).But wait, since ( b ) is an even multiple of 5171, ( k ) can be any integer, but we need to find the GCD in general. However, the problem doesn't specify a particular ( b ), just that it's an even multiple. So, perhaps we need to find the GCD in terms of ( k ), but since ( k ) can vary, the GCD could vary as well. But the problem asks for the GCD given that ( b ) is an even multiple, so perhaps we need to find the maximum possible GCD or the GCD that holds for all such ( b ).Wait, actually, the GCD should be the same for all such ( b ), because the problem is asking for the GCD given that ( b ) is an even multiple, not for a specific ( b ). So, perhaps we need to find the GCD that divides all possible ( 3b + 8 ) where ( b ) is an even multiple of 5171.Alternatively, perhaps we can find the GCD by considering the possible divisors of 64 and seeing which ones divide ( 3b + 8 ).Since ( gcd(3b + 8, 64) ) must divide 64, the possible values are 1, 2, 4, 8, 16, 32, 64.We need to find the largest such divisor that divides ( 3b + 8 ) for all even multiples ( b ) of 5171.Wait, no, actually, the GCD is specific to each ( b ), but since ( b ) is given as an even multiple, perhaps the GCD is fixed regardless of ( k ). Let me check.Let me compute ( 3b + 8 ) modulo 2, 4, 8, etc.First, since ( b ) is even, ( b = 2m ). So, ( 3b + 8 = 6m + 8 ). Both 6m and 8 are even, so ( 3b + 8 ) is even. Thus, the GCD is at least 2.Next, check modulo 4:( 3b + 8 = 6m + 8 ). Since ( m ) is an integer, 6m is congruent to 2m mod 4. 8 is 0 mod 4. So, ( 3b + 8 ≡ 2m ) mod 4.But ( m ) can be any integer, so ( 2m ) can be 0 or 2 mod 4. Therefore, ( 3b + 8 ) can be 0 or 2 mod 4. So, it's not necessarily divisible by 4. Therefore, the GCD cannot be 4.Similarly, check modulo 8:( 3b + 8 = 6m + 8 ). 6m mod 8 can be 0, 2, 4, or 6 depending on ( m ). 8 is 0 mod 8. So, ( 3b + 8 ≡ 6m ) mod 8.But ( m ) can be any integer, so ( 6m ) mod 8 can be 0, 2, 4, or 6. Therefore, ( 3b + 8 ) can be 0, 2, 4, or 6 mod 8. So, it's not necessarily divisible by 8. Therefore, the GCD cannot be 8.Similarly, higher powers of 2 (16, 32, 64) won't necessarily divide ( 3b + 8 ) because ( m ) can vary.Therefore, the GCD is 2.Wait, but let me double-check. Since ( b ) is an even multiple of 5171, which is 5171 itself. Let me compute ( 3b + 8 ) when ( b = 2*5171 ).Compute ( 3*(2*5171) + 8 = 6*5171 + 8 ). Let's compute 6*5171:5171 * 6: 5000*6=30000, 171*6=1026, so total is 30000 + 1026 = 31026.Then, 31026 + 8 = 31034.Now, compute ( gcd(31034, 64) ).Compute 31034 ÷ 64:64*484 = 31016 (since 64*400=25600, 64*80=5120, 64*4=256; 25600+5120=30720, 30720+256=30976; wait, that's 64*484=30976. Then, 31034 - 30976 = 58.So, 31034 = 64*484 + 58.Now, compute ( gcd(64, 58) ).64 ÷ 58 = 1 with remainder 6.58 ÷ 6 = 9 with remainder 4.6 ÷ 4 = 1 with remainder 2.4 ÷ 2 = 2 with remainder 0.So, GCD is 2.Therefore, in this case, the GCD is 2.But wait, is this always the case? Let me try another multiple.Let ( b = 4*5171 ). Then, ( 3b + 8 = 12*5171 + 8 ).Compute 12*5171:5171*10=51710, 5171*2=10342, so total is 51710 + 10342 = 62052.Add 8: 62052 + 8 = 62060.Now, compute ( gcd(62060, 64) ).62060 ÷ 64: 64*970 = 62080, which is larger than 62060. So, 64*969 = 62080 - 64 = 62016.62060 - 62016 = 44.So, ( gcd(64, 44) ).64 ÷ 44 = 1 with remainder 20.44 ÷ 20 = 2 with remainder 4.20 ÷ 4 = 5 with remainder 0.So, GCD is 4.Wait, that's different. So, in this case, the GCD is 4.Hmm, that contradicts my earlier conclusion. So, perhaps the GCD isn't fixed at 2, but depends on ( b ). But the problem states that ( b ) is an even multiple of 5171, so ( b = 2k*5171 ). So, the GCD could vary depending on ( k ).But the problem asks for the GCD given that ( b ) is an even multiple of 5171. It doesn't specify a particular ( b ), so perhaps we need to find the GCD in terms of ( b ), but since ( b ) is variable, the GCD could be 2 or higher depending on ( k ).Wait, but in the first case, when ( k=1 ), GCD was 2, and when ( k=2 ), GCD was 4. So, it's not fixed. But the problem is asking for the GCD, so perhaps it's the maximum possible GCD? Or maybe I made a mistake in my approach.Wait, let's go back to the Euclidean algorithm. Earlier, I tried to compute ( gcd(f(b), g(b)) ) by dividing ( f(b) ) by ( g(b) ) and getting a remainder of 64/9, but that led to complications. Then, I scaled up by 9 to get a remainder of 64, leading to ( gcd(g(b), 64) ).But when I tested with specific values, I got different GCDs. So, perhaps the GCD is 2, but in some cases, it can be higher. However, since the problem doesn't specify a particular ( b ), just that it's an even multiple, perhaps the answer is 2, as the minimal GCD, but that doesn't make sense because GCDs can vary.Wait, actually, the GCD is the same for all such ( b ) because the problem is asking for the GCD in terms of ( b ), but ( b ) is given as an even multiple, so perhaps the GCD is fixed. Let me think again.Wait, perhaps I made a mistake in the initial step. Let me try to perform the Euclidean algorithm correctly.Given ( f(b) = 4b^2 + 35b + 72 ) and ( g(b) = 3b + 8 ).Compute ( f(b) ) divided by ( g(b) ). Let me express ( f(b) = q(b)g(b) + r ), where ( q(b) ) is the quotient and ( r ) is the remainder.As I did earlier, ( q(b) = frac{4}{3}b + frac{73}{9} ), and the remainder is ( frac{64}{9} ). But since we're dealing with integer polynomials, perhaps I should consider scaling.Alternatively, perhaps I can use the fact that ( gcd(f(b), g(b)) = gcd(g(b), f(b) mod g(b)) ).So, ( f(b) mod g(b) ) is the remainder when ( f(b) ) is divided by ( g(b) ), which we found to be ( frac{64}{9} ). But since we're dealing with integers, perhaps I should multiply through by 9 to make it an integer.So, ( 9f(b) = (3b + 8)(12b + 73) + 64 ). Therefore, ( gcd(9f(b), 3b + 8) = gcd(3b + 8, 64) ).But since ( gcd(f(b), 3b + 8) ) divides both ( f(b) ) and ( 3b + 8 ), and 9 is co-prime with ( 3b + 8 ) (since ( 3b + 8 ) is linear and 9 is a constant), the GCD should be the same as ( gcd(3b + 8, 64) ).Therefore, ( gcd(f(b), g(b)) = gcd(3b + 8, 64) ).Now, since ( b ) is an even multiple of 5171, let ( b = 2k*5171 ). Then, ( 3b + 8 = 6k*5171 + 8 ).We need to find ( gcd(6k*5171 + 8, 64) ).Let me compute ( 6k*5171 ) modulo 64.First, compute 5171 modulo 64:As before, 5171 ÷ 64 = 80*64 = 5120, remainder 51. So, 5171 ≡ 51 mod 64.Thus, ( 6k*5171 ≡ 6k*51 mod 64 ).Compute 6*51 = 306. 306 ÷ 64 = 4*64=256, remainder 50. So, 306 ≡ 50 mod 64.Therefore, ( 6k*5171 ≡ 50k mod 64 ).Thus, ( 3b + 8 ≡ 50k + 8 mod 64 ).So, ( gcd(50k + 8, 64) ).Now, 50 and 64 have a GCD of 2, as 50 = 2*25 and 64 = 2^6.Therefore, ( gcd(50k + 8, 64) = gcd(2*(25k + 4), 64) = 2*gcd(25k + 4, 32) ).So, now, we need to find ( gcd(25k + 4, 32) ).Since 25 and 32 are co-prime (25 is 5^2, 32 is 2^5), the GCD depends on ( k ).But since ( k ) can be any integer, ( 25k + 4 ) can take various values modulo 32.Let me compute ( 25k + 4 ) mod 32.25 mod 32 is 25, so ( 25k + 4 ≡ 25k + 4 mod 32 ).We need to find the GCD of ( 25k + 4 ) and 32.Since 25 and 32 are co-prime, ( 25k ) cycles through all residues modulo 32 as ( k ) varies.Therefore, ( 25k + 4 ) can take any value modulo 32, depending on ( k ). Therefore, the GCD ( gcd(25k + 4, 32) ) can be 1, 2, 4, 8, 16, or 32, depending on ( k ).But since ( k ) is arbitrary, the GCD could be as large as 32 if ( 25k + 4 ) is a multiple of 32.However, the problem states that ( b ) is an even multiple of 5171, so ( k ) can be any integer. Therefore, the GCD could vary depending on ( k ).But the problem is asking for the GCD given that ( b ) is an even multiple of 5171, not for a specific ( b ). So, perhaps the answer is the greatest common divisor that divides all possible ( 3b + 8 ) for such ( b ). In other words, the GCD must divide all possible ( 3b + 8 ), so it must divide their differences.Let me consider two different values of ( b ): ( b_1 = 2*5171 ) and ( b_2 = 4*5171 ).Compute ( 3b_1 + 8 = 3*2*5171 + 8 = 6*5171 + 8 ).Compute ( 3b_2 + 8 = 3*4*5171 + 8 = 12*5171 + 8 ).The difference between these two is ( (12*5171 + 8) - (6*5171 + 8) = 6*5171 ).So, the GCD must divide ( 6*5171 ).But 5171 is a prime number? Wait, let me check.Wait, 5171: Let's see if it's divisible by small primes.5171 ÷ 7 = 738.714... Not integer.5171 ÷ 11 = 470.09... Not integer.5171 ÷ 13 = 397.769... Not integer.5171 ÷ 17 = 304.176... Not integer.5171 ÷ 19 = 272.157... Not integer.5171 ÷ 23 = 224.826... Not integer.5171 ÷ 29 = 178.31... Not integer.5171 ÷ 31 = 166.806... Not integer.5171 ÷ 37 = 140.027... Not integer.5171 ÷ 43 = 120.255... Not integer.5171 ÷ 47 = 110.021... Not integer.5171 ÷ 53 = 97.566... Not integer.So, it seems 5171 is a prime number. Therefore, ( 6*5171 ) factors are 2, 3, and 5171.But since the GCD must divide both ( 3b + 8 ) and ( 6*5171 ), and ( 3b + 8 ) is even (as ( b ) is even), the GCD must be at least 2.But earlier, when ( k=1 ), the GCD was 2, and when ( k=2 ), it was 4. So, the GCD can be 2 or higher, but it must divide 6*5171.However, since 5171 is prime and doesn't divide 3b + 8 (because 3b + 8 = 6k*5171 + 8, which is 8 mod 5171, and 8 ≠ 0 mod 5171), the GCD cannot include 5171.Similarly, 3 doesn't divide ( 3b + 8 ) because ( 3b + 8 ≡ 0 + 8 ≡ 8 mod 3 ), which is 2 mod 3, so 3 doesn't divide it.Therefore, the GCD must be a power of 2 that divides 64 and also divides ( 3b + 8 ).But since ( 3b + 8 ) can be congruent to different values modulo higher powers of 2 depending on ( k ), the only guaranteed divisor is 2.Wait, but when ( k=2 ), the GCD was 4. So, is 4 a possible GCD? But since the problem doesn't specify a particular ( b ), just that ( b ) is an even multiple, the GCD could be 2 or higher. However, the problem is asking for the GCD, not the possible GCDs. So, perhaps the answer is 2, as it's the minimal possible GCD, but that doesn't make sense because GCDs can vary.Wait, actually, the GCD is specific to each ( b ), but the problem is asking for the GCD given that ( b ) is an even multiple of 5171. It doesn't specify a particular ( b ), so perhaps the answer is 2, as it's the only common divisor for all such ( b ).Wait, but when ( k=1 ), GCD is 2, when ( k=2 ), GCD is 4, so 2 is the only common divisor for all possible ( b ). Therefore, the GCD is 2.Alternatively, perhaps the answer is 2 because the GCD must divide 2, as shown earlier.Wait, let me think differently. Since ( gcd(3b + 8, 64) ) must divide 64, and ( 3b + 8 ) is even, the GCD is at least 2. But can it be higher?If ( 3b + 8 ) is divisible by 4, then GCD is at least 4. But as we saw, when ( k=1 ), ( 3b + 8 = 31034 ), which is 31034 ÷ 2 = 15517, which is odd, so 31034 is 2 mod 4, so GCD is 2.When ( k=2 ), ( 3b + 8 = 62060 ), which is 62060 ÷ 4 = 15515, which is odd, so 62060 is 4 mod 8, so GCD is 4.Wait, but 62060 ÷ 4 = 15515, which is odd, so 62060 is 4 mod 8, so GCD is 4.Wait, but 62060 ÷ 8 = 7757.5, which is not integer, so 62060 is 4 mod 8, so GCD is 4.Similarly, if ( k=3 ), ( b=6*5171 ), then ( 3b + 8 = 18*5171 + 8 ). Compute 18*5171:5171*10=51710, 5171*8=41368, so 51710 + 41368 = 93078. Add 8: 93086.Compute ( 93086 ÷ 8 = 11635.75 ), so 93086 ≡ 6 mod 8. Therefore, ( gcd(93086, 64) ).Compute ( 93086 ÷ 64 = 1454*64 = 93056, remainder 30.So, ( gcd(64, 30) = 2 ).So, when ( k=3 ), GCD is 2.Similarly, for ( k=4 ), ( b=8*5171 ), ( 3b + 8 = 24*5171 + 8 ).24*5171 = 5171*20 + 5171*4 = 103420 + 20684 = 124104. Add 8: 124112.Compute ( 124112 ÷ 64 = 1939*64 = 124096, remainder 16.So, ( gcd(64, 16) = 16 ).Wait, so when ( k=4 ), GCD is 16.Hmm, so the GCD varies depending on ( k ). Therefore, the GCD isn't fixed, but the problem is asking for the GCD given that ( b ) is an even multiple of 5171. It doesn't specify a particular ( b ), so perhaps the answer is the greatest common divisor that divides all possible ( 3b + 8 ), which would be 2, as it's the only common divisor for all such ( b ).But wait, when ( k=4 ), the GCD was 16, which is higher than 2. So, perhaps the answer is 2, as it's the minimal GCD, but that doesn't make sense because the GCD is specific to each ( b ).Wait, perhaps I'm overcomplicating this. Let me go back to the Euclidean algorithm.We have ( gcd(4b^2 + 35b + 72, 3b + 8) = gcd(3b + 8, 64) ).Since ( b ) is an even multiple of 5171, let ( b = 2k*5171 ). Then, ( 3b + 8 = 6k*5171 + 8 ).We need to find ( gcd(6k*5171 + 8, 64) ).As we saw earlier, this GCD can be 2, 4, 8, 16, 32, or 64, depending on ( k ).However, the problem is asking for the GCD given that ( b ) is an even multiple of 5171, not for a specific ( b ). Therefore, the answer must be the greatest common divisor that is common to all possible ( 3b + 8 ) for such ( b ).In other words, the GCD must divide all possible ( 3b + 8 ), so it must divide their differences.Let me consider two consecutive values of ( b ): ( b_1 = 2*5171 ) and ( b_2 = 4*5171 ).Compute ( 3b_2 + 8 - (3b_1 + 8) = 3*(4*5171) + 8 - (3*(2*5171) + 8) = 12*5171 + 8 - 6*5171 - 8 = 6*5171 ).So, the difference is ( 6*5171 ).Therefore, the GCD must divide ( 6*5171 ).But 5171 is prime, so the factors are 1, 2, 3, 6, 5171, 2*5171, 3*5171, 6*5171.But since ( 3b + 8 ) is even, the GCD must be even, so possible factors are 2, 6, 2*5171, 6*5171.But ( 3b + 8 ) modulo 3: ( 3b + 8 ≡ 0 + 8 ≡ 2 mod 3 ), so 3 doesn't divide ( 3b + 8 ). Therefore, the GCD cannot be 6 or 6*5171.Similarly, ( 3b + 8 ) modulo 5171: ( 3b + 8 ≡ 0 + 8 ≡ 8 mod 5171 ), so 5171 doesn't divide ( 3b + 8 ). Therefore, the GCD cannot be 2*5171 or 6*5171.Thus, the only possible common divisor is 2.Therefore, the greatest common divisor is 2.

💡Alright, so I came across this interesting geometry problem: "Show that for every convex polygon whose area is less than or equal to 1, there exists a parallelogram with area 2 containing the polygon." At first glance, it seems a bit abstract, but I think I can break it down.First, let me recall what a convex polygon is. A convex polygon is one where all its interior angles are less than 180 degrees, and every line segment between two vertices stays inside the polygon. So, no indentations or anything; it's a straightforward shape. The area being less than or equal to 1 is a key constraint here.Now, the goal is to show that no matter what convex polygon you have with area ≤1, you can always find a parallelogram with area exactly 2 that can contain it. Hmm, okay. So, a parallelogram is a four-sided figure with opposite sides parallel. The area of a parallelogram is base times height. So, if we can somehow relate the base and height of the parallelogram to the dimensions of the convex polygon, maybe we can find such a parallelogram.I think a good starting point is to consider the minimal parallelogram that can contain the convex polygon. Maybe if I can find the smallest possible parallelogram that can contain any convex polygon of area ≤1, then show that this minimal parallelogram has an area of at most 2. That would solve the problem.But how do I find such a minimal parallelogram? I remember something about affine transformations and how they can transform any convex polygon into another. Maybe that's a stretch, but perhaps there's a way to use affine geometry here.Wait, another thought: maybe I can use the concept of the area ratio. Since the convex polygon has area ≤1, and the parallelogram has area 2, which is double. So, perhaps the parallelogram can be constructed in such a way that it's twice the area of the polygon, ensuring that it can contain it.But I need to be more precise. Let me think about specific cases. Suppose the convex polygon is a triangle with area 1. Can I find a parallelogram with area 2 that contains it? Well, a triangle can be part of a parallelogram if you duplicate it and attach it to itself. So, if I have a triangle, I can create a parallelogram by reflecting it over one of its sides, effectively doubling its area. So, in this case, the parallelogram would have area 2, and it would contain the original triangle.That seems to work for a triangle. What about a quadrilateral? If the convex polygon is a quadrilateral with area ≤1, can I find a parallelogram of area 2 containing it? Maybe by extending the sides or something. Hmm, not sure.Wait, perhaps I need to think about the affine properties. In affine geometry, any convex polygon can be transformed into any other convex polygon of the same number of sides. So, maybe I can transform the given polygon into a standard shape, like a square or a rectangle, and then construct a parallelogram around it.But I'm not entirely sure how to apply that here. Maybe another approach: consider the diameter of the polygon. The diameter is the maximum distance between any two points in the polygon. If I can bound the diameter somehow, I can construct a parallelogram around it.Alternatively, think about the width of the polygon in different directions. For a convex polygon, the width in a particular direction is the distance between the two supporting lines perpendicular to that direction. If I can find two directions where the widths are manageable, I can construct a parallelogram.Wait, I think I remember something called the "minimum area bounding parallelogram." There's an algorithm or a theorem that says that for any convex polygon, there exists a bounding parallelogram with area at most twice the area of the polygon. Is that right?If that's the case, then since our polygon has area ≤1, the bounding parallelogram would have area ≤2. That would directly answer the question. But I need to verify if such a theorem exists.Let me try to recall. I think it's related to the work of mathematicians like Kurt Mahler, who studied lattice points and convex bodies. Or maybe it's something from computational geometry about minimal enclosing shapes.Alternatively, maybe it's a consequence of the Brunn-Minkowski theorem or some other inequality in geometry. Hmm.Wait, another idea: using the concept of the affine perimeter or something like that. But I'm not sure.Alternatively, think about the John ellipsoid, which is the maximal volume ellipsoid contained within a convex body. But we're dealing with parallelograms, not ellipsoids, so maybe that's not directly applicable.Wait, let's think more simply. Suppose I take the convex polygon and find its minimal bounding rectangle. The area of the minimal bounding rectangle is at most twice the area of the polygon. Is that a known result? I think so. For example, in computational geometry, it's known that the area of the smallest enclosing rectangle for a convex polygon is at most twice the area of the polygon.If that's the case, then since a rectangle is a special case of a parallelogram, we can say that the minimal bounding rectangle has area ≤2, and hence, there exists a parallelogram (the rectangle) with area 2 containing the polygon.But wait, is the minimal bounding rectangle always a parallelogram? Yes, because a rectangle is a parallelogram with right angles. So, if we can find a rectangle with area 2 containing the polygon, then we're done.But does the minimal bounding rectangle have area at most twice the area of the polygon? I think that's a known result. Let me try to recall the proof.I think the idea is that for any convex polygon, you can rotate it such that one side is aligned with the x-axis, and then compute the width and height. The area of the bounding rectangle would be width times height. By rotating the polygon, you can minimize the area of the bounding rectangle.But how does that relate to the area of the polygon?Wait, maybe it's more about the fact that the area of the polygon can be related to the area of the bounding rectangle through some integral or geometric argument.Alternatively, think about the fact that the area of the polygon can be expressed as the integral over all directions of its width in that direction, multiplied by some factor. But I'm not sure.Wait, another approach: use the fact that for any convex polygon, there exists a parallelogram such that the polygon is contained within the parallelogram, and the area of the parallelogram is at most twice the area of the polygon.This seems similar to what I was thinking earlier. Maybe this is a standard result in convex geometry.Alternatively, think about the fact that any convex polygon can be enclosed in a parallelogram whose area is at most twice the area of the polygon. If that's the case, then since our polygon has area ≤1, the parallelogram would have area ≤2.But I need to find a way to construct such a parallelogram or prove that it exists.Wait, perhaps using the concept of the affine diameter. The affine diameter is the maximum distance between two parallel supporting lines of the polygon. If I can find two pairs of parallel lines, each pair supporting the polygon, such that the product of the distances between the lines is at most 2.But I'm not sure.Alternatively, think about the fact that any convex polygon can be enclosed in a parallelogram by taking two pairs of parallel lines that touch the polygon. The area of the parallelogram would then be the product of the distances between these lines.If I can show that these distances can be chosen such that their product is at most 2, given that the area of the polygon is ≤1, then we're done.But how?Wait, maybe use the fact that the area of the polygon can be expressed in terms of the integral of its support function over the unit circle. But that might be too advanced.Alternatively, think about the fact that for any convex polygon, the area is equal to half the integral over all angles of the square of the support function. Not sure.Wait, perhaps a simpler approach: consider that any convex polygon can be enclosed in a parallelogram by taking two pairs of parallel lines, each pair touching the polygon at least at one point. The area of the parallelogram is then the product of the distances between these lines.If I can relate these distances to the area of the polygon, maybe I can bound the area of the parallelogram.Alternatively, think about the fact that the area of the polygon can be expressed as the sum of the areas of triangles formed with a fixed point inside the polygon. But not sure.Wait, another idea: use the fact that any convex polygon can be partitioned into triangles, and each triangle can be enclosed in a parallelogram of area twice the triangle's area. Then, combining these, the total area of the enclosing parallelogram would be twice the area of the polygon.But I'm not sure if that works because the parallelograms for each triangle might overlap or not align properly.Wait, maybe think about the dual problem: instead of enclosing the polygon, think about the polygon enclosing something. But not sure.Wait, going back to the minimal bounding rectangle idea. If I can show that the minimal bounding rectangle has area at most 2, then since a rectangle is a parallelogram, we're done.I think the minimal bounding rectangle for a convex polygon has area at most twice the area of the polygon. Is that a theorem? Let me try to recall.Yes, I think it's a known result in computational geometry. The minimal area bounding rectangle for a convex polygon has area at most twice the area of the polygon. So, if our polygon has area ≤1, the minimal bounding rectangle has area ≤2.Therefore, such a rectangle exists, and since a rectangle is a parallelogram, we're done.But wait, is the minimal bounding rectangle always a parallelogram? Yes, because a rectangle is a special case of a parallelogram with right angles.So, if the minimal bounding rectangle has area ≤2, then there exists a parallelogram (the rectangle) with area 2 containing the polygon.But I should verify this result. Let me think about how the minimal bounding rectangle is found.For a convex polygon, the minimal area bounding rectangle can be found by rotating the polygon and computing the area of the bounding rectangle at each angle. The minimal area occurs when one side of the rectangle is aligned with an edge of the polygon or when a vertex touches a side of the rectangle.But regardless of the orientation, the area of the minimal bounding rectangle is at most twice the area of the polygon. I think this is a result from the rotating calipers method.Yes, the rotating calipers technique can be used to find the minimal bounding rectangle for a convex polygon, and it's known that the area of this rectangle is at most twice the area of the polygon.Therefore, since our polygon has area ≤1, the minimal bounding rectangle has area ≤2, and thus, there exists a parallelogram (the rectangle) with area 2 containing the polygon.So, putting it all together, for any convex polygon with area ≤1, the minimal bounding rectangle (which is a parallelogram) has area ≤2, hence proving the statement.I think that makes sense. I should probably look up the exact theorem to confirm, but based on my reasoning, this seems to hold.