answer：To determine the product of three different positive integers that sum to 7, I start by listing all possible sets of three distinct positive integers that add up to 7.First, I consider the smallest possible integers. The set (1, 2, 4) sums to 7, and their product is 1 × 2 × 4 = 8.Next, I check if there are any other sets. The set (1, 3, 3) also sums to 7, but since the integers must be different, this set is not valid.After reviewing all possibilities, the only valid set is (1, 2, 4), and their product is 8.

answer：Okay, so I need to figure out by what percentage Zyka's expenses exceeded his initially planned expenses. Let me try to break this down step by step.First, let's understand the problem. Zyka decided to buy a krumblik, which I assume is a specific item. But in the store, they also sold kryambliks. So, he bought a kryamblik instead and received coupons worth 50% of the cost of the kryamblik. Then, he used these coupons to pay 20% of the cost of the krumblik. After that, he paid the remaining amount to buy the krumblik. We need to find out by what percentage his total expenses exceeded his initial plan of buying only the krumblik.Hmm, okay. Let me assign some variables to make this clearer. Let's say the cost of the kryamblik is K, and the cost of the krumblik is R. So, Zyka bought a kryamblik for K. Then, he received coupons worth 50% of K, which is 0.5K. He used these coupons to pay 20% of the cost of the krumblik. So, 20% of R is 0.2R. Wait, so he used the coupons to pay 0.2R. But the coupons are worth 0.5K. So, does that mean 0.5K equals 0.2R? Or is it that he used the coupons to cover 20% of R, meaning he used 0.2R worth of coupons? Hmm, the wording says he used the coupons to pay 20% of the cost of the krumblik. So, he used 0.2R from the coupons. But the coupons themselves are worth 0.5K. So, is 0.2R equal to 0.5K? Or is he using the coupons to cover 0.2R, regardless of their value?Wait, maybe it's the latter. He used the coupons to pay 20% of the krumblik's cost, which is 0.2R. So, he used 0.2R worth of coupons, which were worth 0.5K. So, 0.2R = 0.5K. Therefore, R = (0.5 / 0.2)K = 2.5K. So, the krumblik costs 2.5 times the kryamblik.Wait, that seems like a big difference. Let me double-check. If he used coupons worth 0.5K to pay 20% of R, then 0.5K = 0.2R. So, R = (0.5 / 0.2)K = 2.5K. Yeah, that seems right.So, the krumblik is 2.5 times more expensive than the kryamblik. Okay, so if he had just bought the krumblik, he would have spent R = 2.5K. But instead, he bought the kryamblik for K, got coupons worth 0.5K, used 0.2R (which is 0.5K) to pay part of the krumblik, and then paid the remaining amount.Wait, so let's see. He bought the kryamblik for K. Then, he used the coupons (0.5K) to pay 20% of R, which is 0.2R. Since 0.2R = 0.5K, then R = 2.5K as before.Now, after using the coupons, he still needs to pay the remaining 80% of R, which is 0.8R. So, how much did he pay in total?He paid K for the kryamblik, and then 0.8R for the krumblik. So, total expenses are K + 0.8R.But since R = 2.5K, substituting that in, total expenses = K + 0.8*(2.5K) = K + 2K = 3K.Wait, so he spent 3K in total. His initial plan was to buy only the krumblik, which would have cost R = 2.5K. So, his expenses exceeded by 3K - 2.5K = 0.5K.Therefore, the percentage increase is (0.5K / 2.5K) * 100% = (0.5 / 2.5) * 100% = 0.2 * 100% = 20%.Wait, so his expenses exceeded by 20%? Hmm, let me check my steps again.1. Cost of kryamblik: K2. Coupons received: 0.5K3. Coupons used to pay 20% of krumblik: 0.2R4. So, 0.5K = 0.2R => R = 2.5K5. Total cost: K (kryamblik) + (R - 0.2R) = K + 0.8R6. Substitute R: K + 0.8*(2.5K) = K + 2K = 3K7. Initial plan: R = 2.5K8. Excess: 3K - 2.5K = 0.5K9. Percentage: (0.5K / 2.5K) * 100% = 20%Yes, that seems consistent. So, his expenses exceeded by 20%.But wait, let me think again. Is the krumblik more expensive than the kryamblik? Because he bought the kryamblik first, which might have been cheaper. So, if the krumblik is 2.5K, and the kryamblik is K, then yes, the krumblik is more expensive.Alternatively, maybe I misread the problem. Let me read it again."Zyaka decided to buy a krumblik. In the store, they also sold kryambliks. Zyka bought a kryamblik and received coupons worth 50% of the cost of the kryamblik. He used these coupons to pay 20% of the cost of the krumblik. By paying the remaining amount, he also bought the krumblik. By what percentage did Zyka's expenses for buying both the kryamblik and the krumblik exceed his initially planned expenses for buying only the krumblik?"So, he initially planned to buy only the krumblik, which would have cost R. Instead, he bought the kryamblik (cost K), got coupons worth 0.5K, used them to pay 0.2R, and then paid the rest (0.8R) to buy the krumblik. So, total expenses: K + 0.8R.But since 0.5K = 0.2R, R = 2.5K.So, total expenses: K + 0.8*(2.5K) = K + 2K = 3K.Planned expenses: R = 2.5K.Excess: 3K - 2.5K = 0.5K.Percentage: (0.5K / 2.5K) * 100% = 20%.Yes, that seems correct. So, the answer is 20%.Wait, but let me think if there's another way to interpret the problem. Maybe the coupons are used towards the krumblik, but the krumblik is bought in addition to the kryamblik. So, he bought both, but initially, he only planned to buy the krumblik. So, his expenses are K + R, but he used coupons to reduce the cost of the krumblik by 0.2R, so he only paid 0.8R. So, total expenses: K + 0.8R.But since the coupons are worth 0.5K, which he used to pay 0.2R, so 0.5K = 0.2R => R = 2.5K.Therefore, total expenses: K + 0.8*(2.5K) = K + 2K = 3K.Planned expenses: R = 2.5K.Excess: 3K - 2.5K = 0.5K.Percentage: (0.5K / 2.5K) * 100% = 20%.Yes, same result. So, I think 20% is correct.Alternatively, maybe the coupons are applied to the krumblik, so he paid 80% of R, but the coupons are worth 50% of K, which is 0.5K. So, 0.5K = 0.2R => R = 2.5K.So, total expenses: K (kryamblik) + 0.8R (krumblik) = K + 0.8*(2.5K) = 3K.Planned: R = 2.5K.Excess: 0.5K, which is 20% of 2.5K.Yes, that's consistent.I think I've verified it multiple ways, so I'm confident the answer is 20%.**Final Answer**The percentage by which Zyka's expenses exceeded his planned expenses is boxed{20%}.

answer：Okay, so I have this trigonometry problem here: Given that cos α = -99/101 and α is between 180 degrees and 270 degrees, I need to find the value of cot α. Hmm, let me think about how to approach this.First, I remember that cotangent is the reciprocal of tangent, so cot α = 1/tan α. But I also know that tan α is sin α over cos α, so cot α would be cos α over sin α. That might be useful because I know cos α already, which is -99/101. So if I can find sin α, I can then find cot α.But wait, I need to figure out what sin α is. Since I know cos α and the quadrant where α is, I can use the Pythagorean identity to find sin α. The identity is sin² α + cos² α = 1. So, sin² α = 1 - cos² α. Let me plug in the value of cos α.So, cos α is -99/101. Squaring that, I get (-99/101)². Let me calculate that: 99 squared is 9801, and 101 squared is 10201. So, cos² α is 9801/10201. Then, 1 - cos² α is 1 - 9801/10201. Let me compute that.1 is equal to 10201/10201, so subtracting 9801/10201 gives (10201 - 9801)/10201. Let me do the subtraction: 10201 minus 9801 is 400. So, sin² α is 400/10201. Therefore, sin α is the square root of 400/10201, which is 20/101. But wait, is it positive or negative?Since α is between 180 degrees and 270 degrees, that places it in the third quadrant. In the third quadrant, both sine and cosine are negative. So, sin α should be negative. Therefore, sin α is -20/101.Alright, so now I have sin α = -20/101 and cos α = -99/101. So, cot α is cos α over sin α, which is (-99/101) divided by (-20/101). Let me compute that.Dividing two fractions is the same as multiplying by the reciprocal, so (-99/101) * (101/-20). The 101s cancel out, so we have (-99)/(-20). Two negatives make a positive, so that simplifies to 99/20.Wait, that seems straightforward. Let me double-check my steps to make sure I didn't make a mistake.1. I started with cos α = -99/101 and the angle in the third quadrant.2. I used the Pythagorean identity to find sin α. Calculated sin² α as 1 - (99/101)², which is 1 - 9801/10201 = 400/10201.3. Took the square root to get sin α = ±20/101. Since it's in the third quadrant, sin α is negative, so -20/101.4. Then, cot α is cos α / sin α, which is (-99/101)/(-20/101) = 99/20.Hmm, seems correct. Let me think if there's another way to approach this, maybe using the unit circle or something else.Alternatively, I can think about constructing a right triangle where the adjacent side is -99 and the hypotenuse is 101. But since we're in the third quadrant, both adjacent and opposite sides are negative. So, the adjacent side is -99, hypotenuse is 101, and the opposite side would be -20, as we found earlier.Therefore, cotangent, which is adjacent over opposite, would be (-99)/(-20) = 99/20. Yep, same result.Just to make sure, let me compute the numerical values to see if it makes sense.cos α = -99/101 ≈ -0.9802. Since it's in the third quadrant, the angle is between 180 and 270 degrees. So, the reference angle would be arccos(99/101). Let me compute that.arccos(99/101) ≈ arccos(0.9802) ≈ 11.3 degrees. So, the actual angle α would be 180 + 11.3 ≈ 191.3 degrees. Wait, but 191.3 is still in the third quadrant, right? Wait, no, 180 to 270 is the third quadrant, so 191.3 is indeed in the third quadrant.So, sin α would be negative, as we found, and cot α is positive because both cos and sin are negative, making their ratio positive. So, 99/20 is approximately 4.95, which is a positive number, which makes sense because in the third quadrant, tangent is positive, so cotangent is also positive.Therefore, I think my answer is correct.**Final Answer**The value of (cot alpha) is boxed{dfrac{99}{20}}.

answer：Okay, so I have this cubic equation:[ax^3 + (a + 2b) x^2 + (b - 3a) x + (8 - a) = 0]And I know that two of its roots are -2 and 3. I need to find the third root. Hmm, let me think about how to approach this.First, I remember that for a cubic equation of the form (px^3 + qx^2 + rx + s = 0), the sum of the roots is (-q/p), the sum of the products of the roots two at a time is (r/p), and the product of the roots is (-s/p). Maybe I can use Vieta's formulas here.But wait, in this equation, the coefficients are expressed in terms of a and b. So, I might need to find the values of a and b first before I can find the third root. That makes sense because the coefficients depend on a and b, which aren't given.Given that -2 and 3 are roots, I can plug these values into the equation to get two equations involving a and b. Then, I can solve those equations to find a and b.Let me start by plugging in x = -2 into the equation:[a(-2)^3 + (a + 2b)(-2)^2 + (b - 3a)(-2) + (8 - a) = 0]Calculating each term:First term: (a(-8) = -8a)Second term: ((a + 2b)(4) = 4a + 8b)Third term: ((b - 3a)(-2) = -2b + 6a)Fourth term: (8 - a)Adding all these together:-8a + 4a + 8b - 2b + 6a + 8 - a = 0Let me combine like terms:For a: (-8a + 4a + 6a - a) = ( (-8 + 4 + 6 -1 )a ) = (1a) = aFor b: (8b - 2b) = 6bConstants: 8So, the equation becomes:a + 6b + 8 = 0Let me write that as equation (1):a + 6b = -8Now, let's plug in x = 3 into the original equation:[a(3)^3 + (a + 2b)(3)^2 + (b - 3a)(3) + (8 - a) = 0]Calculating each term:First term: (a(27) = 27a)Second term: ((a + 2b)(9) = 9a + 18b)Third term: ((b - 3a)(3) = 3b - 9a)Fourth term: (8 - a)Adding all these together:27a + 9a + 18b + 3b - 9a + 8 - a = 0Combine like terms:For a: (27a + 9a - 9a - a) = (26a)For b: (18b + 3b) = 21bConstants: 8So, the equation becomes:26a + 21b + 8 = 0Let me write that as equation (2):26a + 21b = -8Now, I have two equations:1) a + 6b = -82) 26a + 21b = -8I need to solve this system of equations for a and b.Let me use the substitution method. From equation (1), I can express a in terms of b:a = -8 - 6bNow, substitute this into equation (2):26(-8 - 6b) + 21b = -8Let me compute 26*(-8 -6b):26*(-8) = -20826*(-6b) = -156bSo, equation becomes:-208 -156b + 21b = -8Combine like terms:-208 -135b = -8Now, add 208 to both sides:-135b = -8 + 208-135b = 200Divide both sides by -135:b = 200 / (-135) = -40/27Wait, let me check my calculations again because 200 divided by 135 simplifies.200 and 135 are both divisible by 5:200 ÷ 5 = 40135 ÷ 5 = 27So, yes, b = -40/27Now, substitute b back into equation (1) to find a:a + 6*(-40/27) = -8Compute 6*(-40/27):6*(-40)/27 = (-240)/27 = (-80)/9So,a - 80/9 = -8Add 80/9 to both sides:a = -8 + 80/9Convert -8 to ninths:-8 = -72/9So,a = (-72/9) + (80/9) = (8/9)So, a = 8/9 and b = -40/27Alright, now that I have a and b, I can write the original equation with these values.Let me substitute a = 8/9 and b = -40/27 into the original equation:First, compute each coefficient:1. Leading coefficient: a = 8/92. x^2 coefficient: a + 2b = (8/9) + 2*(-40/27) = (8/9) - (80/27)Convert 8/9 to 24/27:24/27 - 80/27 = (-56)/273. x coefficient: b - 3a = (-40/27) - 3*(8/9) = (-40/27) - (24/9) = (-40/27) - (8/3)Convert 8/3 to 72/27:(-40/27) - (72/27) = (-112)/274. Constant term: 8 - a = 8 - (8/9) = (72/9 - 8/9) = 64/9So, the equation becomes:(8/9)x^3 + (-56/27)x^2 + (-112/27)x + 64/9 = 0Hmm, that's a bit messy with fractions. Maybe I can multiply through by 27 to eliminate denominators.Multiply each term by 27:27*(8/9)x^3 + 27*(-56/27)x^2 + 27*(-112/27)x + 27*(64/9) = 0Simplify each term:27*(8/9) = 3*8 = 2427*(-56/27) = -5627*(-112/27) = -11227*(64/9) = 3*64 = 192So, the equation becomes:24x^3 -56x^2 -112x + 192 = 0Hmm, that's a bit simpler. Let me see if I can factor this equation.But wait, I know two of the roots are -2 and 3, so (x + 2) and (x - 3) are factors. Let me perform polynomial division or use synthetic division to factor them out.Alternatively, since I know two roots, I can write the cubic as:24(x + 2)(x - 3)(x - c) = 0Where c is the third root.Let me expand this and compare coefficients to find c.First, multiply (x + 2)(x - 3):(x + 2)(x - 3) = x^2 - 3x + 2x -6 = x^2 - x -6Now, multiply by (x - c):(x^2 - x -6)(x - c) = x^3 - c x^2 - x^2 + c x -6x + 6cCombine like terms:x^3 + (-c -1)x^2 + (c -6)x + 6cNow, multiply by 24:24x^3 + 24*(-c -1)x^2 + 24*(c -6)x + 24*6c = 0Which simplifies to:24x^3 -24(c + 1)x^2 + 24(c -6)x + 144c = 0Compare this with our earlier equation:24x^3 -56x^2 -112x + 192 = 0So, equate coefficients:1. Coefficient of x^3: 24 = 24 (matches)2. Coefficient of x^2: -24(c + 1) = -56So,-24(c + 1) = -56Divide both sides by -8:3(c + 1) = 7So,3c + 3 = 73c = 4c = 4/3Wait, let me check the other coefficients to make sure.3. Coefficient of x: 24(c -6) = -112So,24(c -6) = -112Divide both sides by 8:3(c -6) = -143c - 18 = -143c = 4c = 4/3Same result.4. Constant term: 144c = 192So,144c = 192c = 192 / 144 = 4/3Consistent again.So, c = 4/3 is the third root.Alternatively, I could have used Vieta's formula directly once I had a and b.Let me try that method too to confirm.Given the original equation:ax^3 + (a + 2b)x^2 + (b - 3a)x + (8 - a) = 0With roots r1 = -2, r2 = 3, and r3 = c.From Vieta's:Sum of roots: r1 + r2 + r3 = -(a + 2b)/aSo,-2 + 3 + c = -(a + 2b)/aSimplify left side:1 + c = -(a + 2b)/aMultiply both sides by a:a(1 + c) = -(a + 2b)But we know a = 8/9 and b = -40/27.Let me compute -(a + 2b):= -(8/9 + 2*(-40/27)) = -(8/9 - 80/27)Convert 8/9 to 24/27:= -(24/27 - 80/27) = -(-56/27) = 56/27So,a(1 + c) = 56/27But a = 8/9, so:(8/9)(1 + c) = 56/27Multiply both sides by 27 to eliminate denominators:27*(8/9)(1 + c) = 56Simplify 27*(8/9) = 3*8 = 24So,24(1 + c) = 56Divide both sides by 24:1 + c = 56/24 = 14/6 = 7/3So,c = 7/3 - 1 = 4/3Same result. So, the third root is 4/3.Alternatively, using the product of roots:r1*r2*r3 = -(8 - a)/aSo,(-2)*3*c = -(8 - a)/aCompute left side:-6c = -(8 - a)/aMultiply both sides by -1:6c = (8 - a)/aSo,6c = (8 - a)/aCompute (8 - a)/a:a = 8/9, so 8 - a = 8 - 8/9 = 72/9 - 8/9 = 64/9Thus,(8 - a)/a = (64/9)/(8/9) = 64/9 * 9/8 = 8So,6c = 8c = 8/6 = 4/3Consistent again.So, regardless of the method, the third root is 4/3.I think that's solid. So, the third root is 4/3.**Final Answer**The third root is boxed{dfrac{4}{3}}.

answer：Alright, so I have this functional equation to solve: ( f(xf(y) + x) = xy + f(x) ) for all real numbers ( x ) and ( y ). The goal is to find the number of possible values of ( f(2) ), which is ( n ), and the sum of all possible values of ( f(2) ), which is ( s ). Then, I need to compute ( n times s ).First, functional equations can sometimes be tricky, but they often have standard techniques to approach them. I remember that plugging in specific values for ( x ) or ( y ) can help simplify the equation and find properties of the function ( f ).Let me start by plugging in ( x = 0 ). That often simplifies things because multiplying by zero can eliminate variables.So, if ( x = 0 ), the equation becomes:[ f(0 cdot f(y) + 0) = 0 cdot y + f(0) ]Simplifying both sides:Left side: ( f(0) )Right side: ( 0 + f(0) = f(0) )So, ( f(0) = f(0) ). Hmm, that doesn't give me any new information. It's just an identity, so I need to try another value.How about ( x = 1 )? Let's see what happens.Plugging ( x = 1 ) into the equation:[ f(1 cdot f(y) + 1) = 1 cdot y + f(1) ]Simplify:Left side: ( f(f(y) + 1) )Right side: ( y + f(1) )So, ( f(f(y) + 1) = y + f(1) ). Interesting. This seems like it might be useful. Maybe I can denote ( c = f(1) ) to make it simpler. So, ( f(f(y) + 1) = y + c ).This suggests that ( f ) is invertible because if ( f ) composed with something gives a linear function, which is invertible, then ( f ) itself must be invertible. So, ( f ) is bijective.Wait, is that necessarily true? Let me think. If ( f ) is injective or surjective, then it's bijective. Since ( f(f(y) + 1) = y + c ), which is a linear function, it's both injective and surjective. So, ( f ) must be bijective because the composition is bijective. Therefore, ( f ) is a bijection.Knowing that ( f ) is bijective is helpful because it means ( f ) has an inverse function. Maybe I can use this property later on.Next, let me try plugging in ( y = 0 ). That might also simplify the equation.So, ( y = 0 ):[ f(xf(0) + x) = x cdot 0 + f(x) ]Simplify:Left side: ( f(xf(0) + x) )Right side: ( 0 + f(x) = f(x) )So, ( f(xf(0) + x) = f(x) ).Since ( f ) is injective (because it's bijective), the arguments must be equal. Therefore:[ x f(0) + x = x ]Simplify:Factor out ( x ):[ x (f(0) + 1) = x ]Assuming ( x neq 0 ), we can divide both sides by ( x ):[ f(0) + 1 = 1 ]So, ( f(0) = 0 ).Great, so now I know that ( f(0) = 0 ). That's useful information.Going back to the equation when ( x = 1 ):[ f(f(y) + 1) = y + c ]Where ( c = f(1) ).Since ( f ) is bijective, I can consider its inverse function. Let me denote ( f^{-1} ) as the inverse of ( f ). Then, applying ( f^{-1} ) to both sides:[ f(y) + 1 = f^{-1}(y + c) ]So, ( f^{-1}(y + c) = f(y) + 1 ).Hmm, not sure if that's immediately helpful, but let's keep that in mind.Another approach is to try to find the form of ( f ). Maybe it's linear? Let me assume that ( f ) is a linear function. Let's suppose ( f(x) = kx + d ), where ( k ) and ( d ) are constants. Let's test this assumption.Plugging ( f(x) = kx + d ) into the original equation:Left side: ( f(xf(y) + x) = k(xf(y) + x) + d = kx f(y) + kx + d )Right side: ( xy + f(x) = xy + kx + d )So, equate the two:[ kx f(y) + kx + d = xy + kx + d ]Simplify both sides by subtracting ( kx + d ):[ kx f(y) = xy ]So, ( kx f(y) = xy ) for all ( x, y ).Since this must hold for all ( x ) and ( y ), let's consider ( x neq 0 ). Then, we can divide both sides by ( x ):[ k f(y) = y ]So, ( f(y) = frac{y}{k} ).But we assumed ( f(y) = ky + d ). So, setting these equal:[ ky + d = frac{y}{k} ]For this to hold for all ( y ), the coefficients of ( y ) and the constants must match.Coefficient of ( y ):[ k = frac{1}{k} ]So, ( k^2 = 1 ) which implies ( k = 1 ) or ( k = -1 ).Constant term:[ d = 0 ]So, ( d = 0 ).Therefore, the possible linear functions are ( f(y) = y ) or ( f(y) = -y ).Let me check both possibilities in the original equation.First, ( f(y) = y ):Left side: ( f(xf(y) + x) = f(x y + x) = x y + x )Right side: ( x y + f(x) = x y + x )So, both sides are equal. That works.Second, ( f(y) = -y ):Left side: ( f(xf(y) + x) = f(-x y + x) = -(-x y + x) = x y - x )Right side: ( x y + f(x) = x y + (-x) = x y - x )So, both sides are equal. That also works.So, both linear functions satisfy the equation. Therefore, ( f ) could be either ( f(x) = x ) or ( f(x) = -x ).Wait, but earlier, when I plugged in ( x = 1 ), I had ( f(f(y) + 1) = y + c ), where ( c = f(1) ). Let me check if both functions satisfy this.For ( f(x) = x ), ( c = f(1) = 1 ). Then, ( f(f(y) + 1) = f(y + 1) = y + 1 ), which is equal to ( y + c = y + 1 ). So, that works.For ( f(x) = -x ), ( c = f(1) = -1 ). Then, ( f(f(y) + 1) = f(-y + 1) = -(-y + 1) = y - 1 ). On the other hand, ( y + c = y - 1 ). So, that also works.So, both functions satisfy the equation. Therefore, there are two possible functions: ( f(x) = x ) and ( f(x) = -x ).Therefore, the possible values of ( f(2) ) are ( f(2) = 2 ) and ( f(2) = -2 ). So, the number of possible values ( n = 2 ), and the sum ( s = 2 + (-2) = 0 ). Therefore, ( n times s = 2 times 0 = 0 ).Wait, but before I conclude, let me make sure that these are the only possible solutions. I assumed that ( f ) is linear, but is that necessarily the case?Functional equations can sometimes have non-linear solutions, but in this case, since we were able to derive that ( f ) is bijective and we found linear solutions, maybe those are the only solutions.Alternatively, let's try to find more properties of ( f ) without assuming linearity.We know that ( f(0) = 0 ). Let's try to find ( f ) in terms of itself.From the original equation:[ f(xf(y) + x) = xy + f(x) ]Let me factor out ( x ) in the argument of ( f ):[ f(x(f(y) + 1)) = xy + f(x) ]Since ( f ) is bijective, I can set ( x = 1 ) to get:[ f(f(y) + 1) = y + f(1) ]Which we already did earlier. Let me denote ( c = f(1) ), so:[ f(f(y) + 1) = y + c ]Now, let me consider substituting ( y ) with ( f^{-1}(z) ), since ( f ) is bijective, every ( z ) can be written as ( f(y) ) for some ( y ).So, let ( y = f^{-1}(z) ). Then:[ f(z + 1) = f^{-1}(z) + c ]So, ( f(z + 1) = f^{-1}(z) + c ).Hmm, not sure if that helps directly, but let's try to find ( f(1) ). Let me plug ( y = 1 ) into the original equation.So, ( y = 1 ):[ f(xf(1) + x) = x cdot 1 + f(x) ]Simplify:Left side: ( f(x(f(1) + 1)) )Right side: ( x + f(x) )So, ( f(x(f(1) + 1)) = x + f(x) ).Let me denote ( d = f(1) + 1 ). So, ( f(d x) = x + f(x) ).Hmm, this seems like another functional equation. Let me see if I can use this.If ( d = 0 ), then ( f(0) = x + f(x) ) for all ( x ). But ( f(0) = 0 ), so ( 0 = x + f(x) ), which would imply ( f(x) = -x ). That's one of our solutions.If ( d neq 0 ), then perhaps we can express ( f ) in terms of itself. Let me write the equation as:[ f(d x) = x + f(x) ]Let me try to find ( f ) in terms of ( f(d x) ). Maybe I can iterate this equation.Suppose I replace ( x ) with ( d x ):[ f(d (d x)) = d x + f(d x) ]Which is:[ f(d^2 x) = d x + f(d x) ]But from the original equation, ( f(d x) = x + f(x) ). So, substitute that in:[ f(d^2 x) = d x + x + f(x) = (d + 1) x + f(x) ]Similarly, if I replace ( x ) with ( d^2 x ) in the original equation:[ f(d (d^2 x)) = d^2 x + f(d^2 x) ]Which is:[ f(d^3 x) = d^2 x + f(d^2 x) ]But from above, ( f(d^2 x) = (d + 1) x + f(x) ). So:[ f(d^3 x) = d^2 x + (d + 1) x + f(x) = (d^2 + d + 1) x + f(x) ]Hmm, I see a pattern here. Let me see if I can generalize this.Suppose I have ( f(d^n x) = left( sum_{k=0}^{n-1} d^k right) x + f(x) ). Let me test this for ( n = 1, 2, 3 ).For ( n = 1 ):Left side: ( f(d x) )Right side: ( sum_{k=0}^{0} d^k x + f(x) = x + f(x) )Which matches the original equation.For ( n = 2 ):Left side: ( f(d^2 x) )Right side: ( sum_{k=0}^{1} d^k x + f(x) = (1 + d) x + f(x) )Which matches what we had earlier.For ( n = 3 ):Left side: ( f(d^3 x) )Right side: ( sum_{k=0}^{2} d^k x + f(x) = (1 + d + d^2) x + f(x) )Which also matches.So, by induction, I can assume that for any positive integer ( n ):[ f(d^n x) = left( sum_{k=0}^{n-1} d^k right) x + f(x) ]Now, if ( d neq 1 ), the sum ( sum_{k=0}^{n-1} d^k ) is a geometric series, which equals ( frac{d^n - 1}{d - 1} ).So, substituting that in:[ f(d^n x) = frac{d^n - 1}{d - 1} x + f(x) ]But let's think about the behavior as ( n ) grows. If ( |d| > 1 ), then ( d^n ) grows without bound, which might cause issues unless ( f ) is linear. Similarly, if ( |d| < 1 ), ( d^n ) approaches zero, which might also impose restrictions on ( f ).Alternatively, let's consider the case when ( d = 1 ). If ( d = 1 ), then the sum ( sum_{k=0}^{n-1} 1^k = n ). So, ( f(1^n x) = n x + f(x) ), which implies ( f(x) = n x + f(x) ). Subtracting ( f(x) ) from both sides gives ( 0 = n x ) for all ( x ), which is only possible if ( n = 0 ). But ( n ) is a positive integer, so this is a contradiction. Therefore, ( d ) cannot be 1.Thus, ( d neq 1 ), so we can proceed with the geometric series.Now, let's see if we can find ( f(x) ) in terms of itself. Let me rearrange the equation:[ f(d^n x) - f(x) = frac{d^n - 1}{d - 1} x ]If I let ( n ) approach infinity, depending on the value of ( d ), the left side might have a limit, but the right side will either go to infinity or zero.Case 1: ( |d| > 1 ). Then, ( d^n ) goes to infinity as ( n ) increases. So, ( f(d^n x) - f(x) ) would go to infinity if ( f ) behaves nicely, but the right side ( frac{d^n - 1}{d - 1} x ) also goes to infinity. However, without knowing the behavior of ( f ) at infinity, it's hard to conclude.Case 2: ( |d| < 1 ). Then, ( d^n ) approaches zero as ( n ) increases. So, ( f(d^n x) ) approaches ( f(0) = 0 ). Therefore, the left side ( f(d^n x) - f(x) ) approaches ( -f(x) ). The right side ( frac{d^n - 1}{d - 1} x ) approaches ( frac{-1}{d - 1} x ).So, equating the limits:[ -f(x) = frac{-1}{d - 1} x ]Multiply both sides by -1:[ f(x) = frac{1}{d - 1} x ]So, ( f(x) ) is linear in this case. Let me denote ( k = frac{1}{d - 1} ), so ( f(x) = k x ).Wait, earlier, we assumed ( f ) is linear and found that ( f(x) = x ) or ( f(x) = -x ). So, perhaps this is consistent.Let me check if ( f(x) = k x ) satisfies the original equation.Plugging ( f(x) = k x ) into the original equation:Left side: ( f(x f(y) + x) = f(x k y + x) = k(x k y + x) = k^2 x y + k x )Right side: ( x y + f(x) = x y + k x )So, equate the two:[ k^2 x y + k x = x y + k x ]Subtract ( k x ) from both sides:[ k^2 x y = x y ]Since this must hold for all ( x, y ), we can divide both sides by ( x y ) (assuming ( x, y neq 0 )):[ k^2 = 1 ]Thus, ( k = 1 ) or ( k = -1 ), which are the solutions we found earlier.So, in the case ( |d| < 1 ), we get that ( f ) must be linear with ( k = 1 ) or ( k = -1 ).But wait, earlier, we had ( d = f(1) + 1 ). Let me see what ( d ) is for each case.If ( f(x) = x ), then ( f(1) = 1 ), so ( d = 1 + 1 = 2 ). But ( |d| = 2 > 1 ), which is in the first case, not the second. Hmm, that's a problem because we assumed ( |d| < 1 ) to get ( f ) linear, but in reality, ( d = 2 ) for ( f(x) = x ).Similarly, if ( f(x) = -x ), then ( f(1) = -1 ), so ( d = -1 + 1 = 0 ). But ( d = 0 ) is a special case because earlier, when we set ( d = f(1) + 1 ), if ( d = 0 ), then ( f(0) = x + f(x) ) for all ( x ), which implies ( f(x) = -x ). So, that's consistent.Wait, so when ( d = 0 ), we get ( f(x) = -x ). When ( d = 2 ), which is ( |d| > 1 ), we might not get a contradiction because ( f(x) = x ) is a solution. So, maybe both cases are covered.But in the case ( |d| > 1 ), we didn't get a contradiction, but in the case ( |d| < 1 ), we forced ( f ) to be linear, which gave us the solutions. So, perhaps the only solutions are ( f(x) = x ) and ( f(x) = -x ).Therefore, going back, the only possible functions are ( f(x) = x ) and ( f(x) = -x ), which gives ( f(2) = 2 ) or ( f(2) = -2 ).Hence, the number of possible values ( n = 2 ), and the sum ( s = 2 + (-2) = 0 ). Therefore, ( n times s = 2 times 0 = 0 ).But wait, just to be thorough, let me check if there are any other possible functions besides the linear ones. Suppose ( f ) is not linear. Is that possible?Given that ( f ) is bijective, and we have the equation ( f(xf(y) + x) = xy + f(x) ), it's not obvious how a non-linear function could satisfy this for all ( x ) and ( y ). The structure of the equation seems to enforce a linear relationship, especially since we can express ( f ) in terms of itself with scaling factors, leading us to the linear solutions.Therefore, it's reasonable to conclude that the only solutions are ( f(x) = x ) and ( f(x) = -x ), giving ( f(2) = 2 ) and ( f(2) = -2 ).So, the number of possible values ( n = 2 ), the sum ( s = 0 ), so ( n times s = 0 ).**Final Answer**boxed{0}

answer：To solve the first problem, I need to find the other rational number when the product of two rational numbers is -2 and one of them is 1/7. I'll set up the equation (1/7) * x = -2 and solve for x by multiplying both sides by 7 to isolate x, resulting in x = -14.For the second problem, I need to determine the third rational number when the sum of three rational numbers is -5, and two of the numbers are 1 and -4. I'll set up the equation 1 + (-4) + y = -5, simplify it to -3 + y = -5, and then solve for y by adding 3 to both sides, which gives y = -2.