🧠:[<think>Alright, let me try to solve this problem step by step. So, we have triangle ABC with sides AB = 48 cm and AC = 30 cm. There's also a median CP = 18 cm. We need to find the perimeter and the area of triangle ABC. Hmm, okay.First, let me recall that a median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. In this case, CP is the median from vertex C to the midpoint P of side AB. Since AB is 48 cm, the midpoint P divides AB into two segments each of length 24 cm. So, AP = PB = 24 cm.Now, the problem is that we know two sides of the triangle (AB and AC) and the length of the median CP. But we need to find the perimeter, which requires knowing the third side BC. Once we have BC, the perimeter will just be AB + BC + AC. So, the key here is to find BC. Then, once we have all three sides, we can also compute the area using Heron's formula, perhaps.But how do we find BC? Since we have a median involved, I think there's a formula that relates the length of a median to the sides of the triangle. Let me recall. There's the Apollonius theorem, which states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. Wait, let me write that formula properly.Apollonius theorem: For a triangle ABC, with median CP (where P is the midpoint of AB), then ( AC^2 + BC^2 = 2CP^2 + 2AP^2 ). Let me check that. Yes, that seems right. Since AP is half of AB, which is 24 cm. So plugging in the known values, we can solve for BC.Let me write down the formula:( AC^2 + BC^2 = 2CP^2 + 2AP^2 )Substituting the known values:AC = 30 cm, CP = 18 cm, AP = 24 cm.So,( 30^2 + BC^2 = 2*(18)^2 + 2*(24)^2 )Calculating each term:30^2 = 90018^2 = 324, so 2*324 = 64824^2 = 576, so 2*576 = 1152Therefore,900 + BC^2 = 648 + 1152Adding 648 and 1152 gives 1800.So,900 + BC^2 = 1800Subtract 900 from both sides:BC^2 = 900Therefore, BC = sqrt(900) = 30 cm.Wait, that's interesting. So BC is 30 cm? Let me verify that again because that seems a bit surprising.If BC is 30 cm, then all sides would be 48, 30, and 30? Wait, no. Wait, AB is 48 cm, AC is 30 cm, and BC is 30 cm? So triangle ABC would have sides 48, 30, 30. Hmm, is that possible?Wait, let me check my calculation again to make sure I didn't make a mistake. Let's go through each step.Starting with Apollonius' theorem:( AC^2 + BC^2 = 2CP^2 + 2AP^2 )Plugging in the numbers:30^2 + BC^2 = 2*(18)^2 + 2*(24)^2Compute left side: 900 + BC^2Right side: 2*(324) + 2*(576) = 648 + 1152 = 1800So 900 + BC^2 = 1800 => BC^2 = 900 => BC = 30So that's correct. Therefore, BC is indeed 30 cm. Hmm. So the triangle has sides 48, 30, 30. Let's check if such a triangle is possible.The triangle inequality states that the sum of the lengths of any two sides must be greater than the third side. So, 30 + 30 > 48? 60 > 48, which is true. Then, 30 + 48 > 30? 78 > 30, which is true. And 30 + 48 > 30, same thing. So yes, the triangle is valid.Wait, but this would make triangle ABC an isosceles triangle with AC = BC = 30 cm. But the median CP is given as 18 cm. Let me check if in such a triangle, the median length makes sense.Alternatively, maybe I can compute the coordinates to verify. Let me try that approach.Let's place the triangle on a coordinate system. Let me set point A at (0, 0), and point B at (48, 0), so that AB is along the x-axis. Then, the midpoint P of AB is at (24, 0). Point C is somewhere in the plane, and we know AC = 30 cm, and CP = 18 cm.So coordinates:A = (0, 0)B = (48, 0)P = (24, 0)Let coordinates of C be (x, y). Then:AC = 30 cm, so the distance from A to C is sqrt(x^2 + y^2) = 30 => x^2 + y^2 = 900.CP = 18 cm, so the distance from C to P is sqrt((x - 24)^2 + y^2) = 18 => (x - 24)^2 + y^2 = 324.We can set up these two equations:1) x^2 + y^2 = 9002) (x - 24)^2 + y^2 = 324Subtracting equation 2 from equation 1:x^2 + y^2 - [(x - 24)^2 + y^2] = 900 - 324Simplify left side:x^2 - (x^2 - 48x + 576) = 576Which becomes:x^2 - x^2 + 48x - 576 = 576So:48x - 576 = 576Add 576 to both sides:48x = 1152Divide by 48:x = 24So x = 24. Plugging back into equation 1:24^2 + y^2 = 90024^2 = 576, so:576 + y^2 = 900Thus, y^2 = 324Therefore, y = 18 or y = -18. Since we're talking about a triangle in a plane, we can take y as positive, so y = 18.Therefore, the coordinates of C are (24, 18). Then, the coordinates of B are (48, 0). So BC can be calculated as the distance between (24, 18) and (48, 0).Compute BC:sqrt((48 - 24)^2 + (0 - 18)^2) = sqrt(24^2 + (-18)^2) = sqrt(576 + 324) = sqrt(900) = 30. So that's correct. Therefore, BC is indeed 30 cm.Wait a second, so point C is at (24, 18). Therefore, the coordinates are:A(0,0), B(48,0), C(24,18). Let's confirm that AC is 30.Distance from A(0,0) to C(24,18):sqrt((24)^2 + 18^2) = sqrt(576 + 324) = sqrt(900) = 30. Correct.Similarly, CP is the distance from C(24,18) to P(24,0):sqrt((24 - 24)^2 + (0 - 18)^2) = sqrt(0 + 324) = 18. Correct.Therefore, all given lengths are satisfied. So BC is indeed 30 cm. So perimeter is AB + BC + AC = 48 + 30 + 30 = 108 cm.Then, the area. Since we have coordinates, perhaps it's easiest to compute the area using coordinates. The coordinates are A(0,0), B(48,0), C(24,18). We can use the shoelace formula.Shoelace formula for area:Area = |(Ax(By - Cy) + Bx(Cy - Ay) + Cx(Ay - By))/2|Plugging in:Ax = 0, Ay = 0Bx = 48, By = 0Cx = 24, Cy = 18So:Area = |(0*(0 - 18) + 48*(18 - 0) + 24*(0 - 0))/2|Simplify:= |0 + 48*18 + 0| / 2= |864| / 2 = 432 cm²Alternatively, since AB is the base along the x-axis from (0,0) to (48,0), the length of the base is 48 cm. The height is the y-coordinate of point C, which is 18 cm. So area is (base * height)/2 = (48 * 18)/2 = 432 cm². Same result.Alternatively, since the triangle is isosceles with sides AC = BC = 30 cm and base AB = 48 cm, we can compute the area by splitting the triangle into two right triangles. The height h from point C to AB splits AB into two 24 cm segments. So using Pythagoras in one of the right triangles:h^2 + 24^2 = 30^2h^2 = 900 - 576 = 324h = 18 cm, which matches the coordinates. So area is (48 * 18)/2 = 432 cm².Therefore, the perimeter is 108 cm and the area is 432 cm².But wait, just to make sure there isn't another possible configuration. For example, could point C be located below the x-axis? If so, would that change anything? Let's check.If C were at (24, -18), the distance AC would still be 30, CP would still be 18, but the area would still be the same since the height is absolute value. So the area is positive regardless. So either way, the area is 432 cm². So that's consistent.Alternatively, could there be another triangle where BC is different? Wait, according to the calculations using Apollonius' theorem and the coordinate method, BC is uniquely determined as 30 cm. So there is only one possible triangle here. Therefore, the answer is unique.Therefore, the perimeter is 48 + 30 + 30 = 108 cm, and the area is 432 cm².Wait, but let me double-check using Heron's formula as another method. Heron's formula states that the area of a triangle with sides a, b, c is sqrt[s(s - a)(s - b)(s - c)], where s is the semi-perimeter.So, the sides are 48, 30, 30. The semi-perimeter s = (48 + 30 + 30)/2 = 108/2 = 54.Area = sqrt[54*(54 - 48)*(54 - 30)*(54 - 30)] = sqrt[54*6*24*24]Compute inside the sqrt:54*6 = 32424*24 = 576So 324*576 = Let's compute that.First, 324 * 576.324 * 500 = 162,000324 * 76 = 24,624Total = 162,000 + 24,624 = 186,624So sqrt[186,624] = 432. Because 432^2 = (400 + 32)^2 = 400^2 + 2*400*32 + 32^2 = 160,000 + 25,600 + 1,024 = 186,624. Correct.Therefore, Heron's formula gives the same area, 432 cm². So that checks out.So, all methods confirm that the perimeter is 108 cm and the area is 432 cm².Therefore, the final answers are perimeter 108 cm and area 432 cm².Final AnswerThe perimeter of triangle (ABC) is boxed{108} cm and its area is boxed{432} cm².</think>Given triangle (ABC) with ( |AB| = 48 , text{cm} ), ( |AC| = 30 , text{cm} ), and the median ( |CP| = 18 , text{cm} ) (where (P) is the midpoint of (AB)), we need to find the perimeter and area of triangle (ABC).First, using Apollonius' theorem for the median (CP):[AC^2 + BC^2 = 2CP^2 + 2AP^2]Substituting the known values (AC = 30 , text{cm}), (CP = 18 , text{cm}), and (AP = 24 , text{cm}):[30^2 + BC^2 = 2 cdot 18^2 + 2 cdot 24^2][900 + BC^2 = 648 + 1152][BC^2 = 900][BC = 30 , text{cm}]The perimeter of triangle (ABC) is:[AB + BC + AC = 48 + 30 + 30 = 108 , text{cm}]To find the area, we use the coordinates of the points. Placing (A) at ((0, 0)), (B) at ((48, 0)), and midpoint (P) at ((24, 0)), we determine the coordinates of (C). Using the distances (AC = 30 , text{cm}) and (CP = 18 , text{cm}), we find (C) at ((24, 18)).The area can be calculated using the base (AB = 48 , text{cm}) and height (18 , text{cm}) (the y-coordinate of (C)):[text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 48 times 18 = 432 , text{cm}^2]Verification using Heron's formula with sides (48 , text{cm}), (30 , text{cm}), and (30 , text{cm}) confirms the area:[s = frac{48 + 30 + 30}{2} = 54][text{Area} = sqrt{s(s - a)(s - b)(s - c)} = sqrt{54 times 6 times 24 times 24} = sqrt{186624} = 432 , text{cm}^2]Thus, the perimeter of triangle (ABC) is (boxed{108}) cm and its area is (boxed{432}) cm².,<think>Okay, let's see. I have this triangle ABC where AB is 48 cm, AC is 30 cm, and the median CP is 18 cm. I need to find the perimeter and the area of triangle ABC. Hmm, medians... I remember a median connects a vertex to the midpoint of the opposite side. So CP is the median from C to AB, meaning P is the midpoint of AB. Since AB is 48 cm, then AP and PB should each be 24 cm, right?Alright, so first, maybe I can draw a diagram in my mind. Triangle ABC with AB = 48, AC = 30, and CP = 18. I need to find BC to get the perimeter. Once I have all three sides, the perimeter is just the sum. The area might be trickier. Maybe I can use Heron's formula if I know all three sides? But first, I need to find BC.Since CP is a median, perhaps I can use Apollonius's theorem. I remember that theorem relates the lengths of the sides of a triangle to the length of a median. The theorem states that the sum of the squares of any two sides of the triangle is equal to twice the square of the median to the third side plus twice the square of the half third side. Let me write that formula down:AB² + AC² = 2AP² + 2CP²Wait, is that correct? Let me check. Apollonius's theorem says that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the third side's half. So actually, maybe it's:AB² + AC² = 2AP² + 2CP²But AP is half of AB, right? Since P is the midpoint. So AP is 24 cm. So substituting the known values:AB is 48, so AB² is 48² = 2304AC is 30, so AC² is 30² = 900AP is 24, so AP² is 24² = 576CP is 18, so CP² is 18² = 324So plugging into the theorem:2304 + 900 = 2*576 + 2*324Left side: 2304 + 900 = 3204Right side: 2*576 = 1152; 2*324 = 648; 1152 + 648 = 1800Wait, 3204 ≠ 1800. That's a problem. That means either I applied the theorem incorrectly or I mixed up the sides.Wait, maybe I mixed up which sides the theorem refers to. Let me check Apollonius's theorem again. The theorem states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. So if the median is to side AB, then the formula would be AC² + BC² = 2CP² + 2(AP²). Wait, no. Let me think.Wait, Apollonius's theorem for median CP (which is to side AB) would be:AC² + BC² = 2CP² + 2AP²Wait, that might be the correct formula. Because the median is to AB, so the sides adjacent to the median are AC and BC. So the formula would be:AC² + BC² = 2CP² + 2AP²So then, we can plug in the known values:30² + BC² = 2*(18²) + 2*(24²)Compute each term:30² = 900BC² is unknown.18² = 324, so 2*324 = 64824² = 576, so 2*576 = 1152Therefore:900 + BC² = 648 + 1152Calculate the right side: 648 + 1152 = 1800So:900 + BC² = 1800Subtract 900 from both sides:BC² = 900Therefore, BC = sqrt(900) = 30 cmWait, so BC is 30 cm? That seems a bit surprising, but okay. Let me check if this makes sense.So the sides would be AB = 48 cm, AC = 30 cm, and BC = 30 cm. Then the perimeter is 48 + 30 + 30 = 108 cm. Wait, but then two sides are equal (AC and BC), so the triangle would be isoceles with AB as the base. But in that case, the median CP to AB would also be the altitude. Let's check if that holds.If ABC is isoceles with AC = BC = 30 cm, then the altitude from C to AB would split AB into two 24 cm segments. Then the length of the altitude can be found via Pythagoras:In triangle APC, which is a right triangle (since the altitude is perpendicular to AB), AC is the hypotenuse, AP is 24 cm, so the altitude h is sqrt(AC² - AP²) = sqrt(900 - 576) = sqrt(324) = 18 cm.Wait, that's exactly the length of the median CP given. So that checks out! So in this case, the median is indeed equal to the altitude because the triangle is isoceles. So BC is 30 cm. Therefore, the perimeter is 48 + 30 + 30 = 108 cm.Okay, that seems good. Now, the area. Since it's an isoceles triangle with base AB = 48 cm and height CP = 18 cm, the area is (base * height)/2 = (48 * 18)/2 = (864)/2 = 432 cm².Alternatively, using Heron's formula: perimeter is 108, so semi-perimeter s = 54. Then area is sqrt[s(s - AB)(s - AC)(s - BC)] = sqrt[54*(54 - 48)*(54 - 30)*(54 - 30)] = sqrt[54*6*24*24]. Compute that:54*6 = 324; 24*24 = 576; so sqrt[324*576] = sqrt[324]*sqrt[576] = 18*24 = 432. Same result. So that's consistent.Wait, so all checks out. Therefore, the perimeter is 108 cm and the area is 432 cm².But let me just step back and make sure I didn't miss anything. So the key was applying Apollonius's theorem correctly. Initially, I thought there was a contradiction when I tried plugging in the numbers, but that's because I had the formula wrong. Once I corrected the formula to AC² + BC² = 2CP² + 2AP², everything worked out. Then verifying that the triangle is isoceles and that the median equals the altitude made sense. So I think this is correct.Final AnswerThe perimeter of triangle (ABC) is boxed{108} cm and its area is boxed{432} cm².</think>Given triangle (ABC) with ( |AB| = 48 , text{cm} ), ( |AC| = 30 , text{cm} ), and the median ( |CP| = 18 , text{cm} ), we need to find the perimeter and area of triangle (ABC).1. Identify the midpoint (P) of (AB): - Since (P) is the midpoint, ( |AP| = |PB| = frac{48}{2} = 24 , text{cm} ).2. Apply Apollonius's theorem: - Apollonius's theorem states that in any triangle, the sum of the squares of two sides is equal to twice the square of the median to the third side plus twice the square of the half third side. For median (CP): [ AC^2 + BC^2 = 2CP^2 + 2AP^2 ] - Substituting the known values: [ 30^2 + BC^2 = 2 times 18^2 + 2 times 24^2 ] - Calculate each term: [ 900 + BC^2 = 2 times 324 + 2 times 576 ] [ 900 + BC^2 = 648 + 1152 ] [ 900 + BC^2 = 1800 ] - Solving for (BC^2): [ BC^2 = 1800 - 900 = 900 ] [ BC = sqrt{900} = 30 , text{cm} ]3. Calculate the perimeter: - The sides of the triangle are (AB = 48 , text{cm}), (AC = 30 , text{cm}), and (BC = 30 , text{cm}). [ text{Perimeter} = 48 + 30 + 30 = 108 , text{cm} ]4. Calculate the area: - Since (ABC) is isosceles with (AC = BC = 30 , text{cm}) and median (CP = 18 , text{cm}) (which is also the altitude), the area can be calculated as: [ text{Area} = frac{1}{2} times text{base} times text{height} = frac{1}{2} times 48 times 18 = 432 , text{cm}^2 ]Thus, the perimeter of triangle (ABC) is (boxed{108}) cm and its area is (boxed{432}) cm².]

🔑:3. We are given triangle A P C. Its sides have lengths |A P|=24 mathrm{~cm}, |P C|=18 mathrm{~cm}, and |C A|=30 mathrm{~cm}.1 mathrm{POINT}The sum of the areas of the squares constructed on the two shorter sides is18^{2}+24^{2}=324+576=9001 mathrm{POINT}The area of the square on its longest side is 30^{2}=900.1 mathrm{POINT}Since 18^{2}+24^{2}=30^{2}, by the converse of the Pythagorean theorem, we conclude that triangle A P C is a right triangle with a right angle at vertex P.2 POINTSTriangles triangle A P C and triangle B P C are right triangles and share the leg overline{C P}. Given that |A P|=|B P|, it follows that |A C|=|B C|=30 mathrm{~cm}, i.e., triangle A B C is isosceles with base overline{A B}.The perimeter of triangle A B C is O=48+2 cdot 30=108 mathrm{~cm}.1 mathrm{POINT}Since overline{C P} perp overline{A B}, the median overline{C P} is also the altitude from vertex C.The area of triangle A B C is P_{A B C}=frac{|A B| cdot|C P|}{2}=frac{48 cdot 18}{2}=432 mathrm{~cm}^{2}. qquadTOTAL 10 POINTS

🧠:[<think>Okay, so I have this problem here. Let me read it again to make sure I get all the details. It says that given a+b+c is not equal to zero, and the ratios (a+b)/c, (b+c)/a, and (a+c)/b are all equal to p. Then, the line y = px + p does not pass through one of the quadrants, and I need to figure out which quadrant that is. The options are first, second, third, or fourth.Alright, let's start by understanding what the problem is asking. First, there's a condition involving a, b, c, and p. Since all three fractions equal p, we can set up equations from that. Then, once we find p, we need to analyze the line y = px + p to see which quadrant it doesn't pass through. The answer choices are the four quadrants, so I need to figure out which one is excluded.First step: dealing with the ratios. Since (a+b)/c = p, (b+c)/a = p, and (a+c)/b = p, these give us three equations:1. (a + b)/c = p2. (b + c)/a = p3. (a + c)/b = pHmm. Since all three equal p, maybe I can set them equal to each other. Let me write equations 1 and 2 equal to each other: (a + b)/c = (b + c)/a. Cross-multiplying gives a(a + b) = c(b + c). Similarly, set equations 2 and 3 equal: (b + c)/a = (a + c)/b. Cross-multiplying gives b(b + c) = a(a + c). And then equations 1 and 3: (a + b)/c = (a + c)/b. Cross-multiplying gives b(a + b) = c(a + c).This seems a bit complicated with three variables. Maybe there's a better way. Let me think. Since all three ratios equal p, maybe I can express a + b, b + c, and a + c in terms of p and the denominators. So:a + b = pcb + c = paa + c = pbSo now I have three equations:1. a + b = pc2. b + c = pa3. a + c = pbHmm. Maybe adding all three equations together? Let's try that. Adding the left sides: (a + b) + (b + c) + (a + c) = 2a + 2b + 2c. Adding the right sides: pc + pa + pb = p(a + b + c). So:2(a + b + c) = p(a + b + c)But the problem states that a + b + c ≠ 0, so we can divide both sides by (a + b + c):2 = pSo p is 2. Wait, that seems straightforward. So p equals 2. Let me check if that makes sense.If p is 2, then from equation 1: a + b = 2cEquation 2: b + c = 2aEquation 3: a + c = 2bLet me see if such a, b, c exist. Let's assume a, b, c are real numbers (they probably are, since we're dealing with quadrants). Let's try to solve these equations.From equation 1: a + b = 2c. Let's express a = 2c - b.Substitute a into equation 2: b + c = 2a = 2(2c - b) = 4c - 2bSo, b + c = 4c - 2bBringing like terms together: b + 2b = 4c - c => 3b = 3c => b = cSimilarly, substitute a = 2c - b. If b = c, then a = 2c - c = c. So a = c, b = c, so a = b = c. Then, substituting into equation 3: a + c = 2b => c + c = 2c => 2c = 2c. That's true. So all variables equal to each other. So a = b = c. Then, a + b + c = 3a ≠ 0, which is okay because a can't be zero (since denominators are a, b, c; so a, b, c can't be zero). So if a = b = c ≠ 0, then the equations hold, and p = 2. So p is indeed 2.Therefore, the line is y = 2x + 2.Now, need to figure out which quadrant this line doesn't pass through. Let's recall the quadrants:- First quadrant: x > 0, y > 0- Second quadrant: x < 0, y > 0- Third quadrant: x < 0, y < 0- Fourth quadrant: x > 0, y < 0So the line y = 2x + 2. Let's analyze its slope and intercept. The slope is 2, which is positive, and the y-intercept is 2, also positive. A line with positive slope and positive y-intercept typically crosses the first, second, and third quadrants? Wait, no. Let me think.Wait, when you have a positive slope and positive y-intercept, the line goes from the second quadrant (since when x is very negative, y is still positive but going downward) to the first quadrant and into the fourth? Wait, actually, let me sketch it mentally.When x is 0, y is 2. So the y-intercept is (0,2). Then, as x increases, y increases (slope positive), so it goes through the first quadrant. As x decreases (negative values), y decreases. So when x is negative, say x approaches negative infinity, y approaches negative infinity as well because slope is positive. Wait, but the slope is positive. So when x is negative, y = 2x + 2. If x is negative, 2x is negative, so y is 2x + 2. So for large negative x, y is negative. For x = -1, y = 0. So the line crosses the y-axis at (0,2), goes up to the right through the first quadrant, and goes down to the left, crossing the x-axis at (-1,0). So then, from the second quadrant (left of y-axis) to the first and fourth? Wait, when x is negative but y is positive, that's the second quadrant. But once x is negative enough that y becomes negative, then it's in the third quadrant.Wait, let's check when x is negative:If x is between -1 and 0, then 2x is between -2 and 0, so y = 2x + 2 is between 0 and 2. So between x = -1 and 0, y is positive. So that part is in the second quadrant. For x < -1, y becomes negative. So when x < -1, y is negative. So in that case, the line is passing through the third quadrant (x negative, y negative). Then, crossing the x-axis at (-1,0), going into the fourth quadrant? Wait, but when x is positive, y is positive, so first quadrant. So the line passes through first, second, and third quadrants? Wait, but when x is positive, y = 2x + 2 is positive (first quadrant). When x is between -1 and 0, y is positive (second quadrant). When x is less than -1, y is negative (third quadrant). So it never goes into the fourth quadrant, which is x positive, y negative. Because when x is positive, y is positive. So the line y = 2x + 2 doesn't pass through the fourth quadrant.Therefore, the answer should be (D) fourth quadrant.But let me double-check. Maybe I made a mistake. Let's see.Take a point in the fourth quadrant, say (1, -1). Let's check if y = 2x + 2 passes through here. Plug x=1: y = 2*1 + 2 = 4. So the point (1,4) is on the line, which is in the first quadrant. To get a point in the fourth quadrant, we need y negative when x is positive. So let's solve for x when y is negative. y = 2x + 2 < 0 => 2x < -2 => x < -1. But then x would be negative, so that's the third quadrant. So there's no x positive where y is negative. Therefore, the line does not pass through the fourth quadrant. So answer is D.But just to be thorough, let's check intercepts. The x-intercept is when y=0: 0 = 2x + 2 => x = -1. So the line crosses the x-axis at (-1, 0). The y-intercept is (0, 2). So the line goes from (-1,0) upwards to the left into the second quadrant and upwards to the right into the first quadrant. So it never goes into the fourth quadrant. Therefore, the correct answer is D.Final Answerboxed{D}</think>Given (a + b + c neq 0) and the ratios (frac{a+b}{c} = frac{b+c}{a} = frac{a+c}{b} = p), we start by setting up the equations:1. (a + b = pc)2. (b + c = pa)3. (a + c = pb)Adding all three equations, we get:[(a + b) + (b + c) + (a + c) = pc + pa + pb]Simplifying both sides, we have:[2(a + b + c) = p(a + b + c)]Since (a + b + c neq 0), we can divide both sides by (a + b + c) to find:[2 = p]Thus, (p = 2). The line given is (y = 2x + 2).To determine which quadrant the line does not pass through, we analyze the line (y = 2x + 2). The slope is positive (2) and the y-intercept is positive (2). The line crosses the y-axis at ((0, 2)) and the x-axis at ((-1, 0)). - For positive (x), (y) is positive (first quadrant).- For (x) between (-1) and (0), (y) is positive (second quadrant).- For (x < -1), (y) is negative (third quadrant).Since the line does not have any points where (x) is positive and (y) is negative, it does not pass through the fourth quadrant.The answer is (boxed{D}).,<think>Okay, so I have this problem here where I need to find out which quadrant the line y = px + p doesn't pass through. The options are the first, second, third, or fourth quadrant. The value of p is given by these equations: (a+b)/c = (b+c)/a = (a+c)/b = p, and it's also mentioned that a + b + c ≠ 0. Hmm, okay, let's break this down step by step.First, let's understand the given equations. All three ratios equal to p. So, (a + b)/c = p, (b + c)/a = p, and (a + c)/b = p. Since all of them equal p, they must be equal to each other as well. That means (a + b)/c = (b + c)/a, and similarly for the others. Maybe I can set these equal to each other and solve for a, b, c?But before that, maybe there's a better approach. Since all three expressions equal p, perhaps I can set up equations and solve for a, b, c in terms of p. Let's try that.From the first equation: (a + b)/c = p ⇒ a + b = p*c.Second equation: (b + c)/a = p ⇒ b + c = p*a.Third equation: (a + c)/b = p ⇒ a + c = p*b.So now we have three equations:1. a + b = p*c2. b + c = p*a3. a + c = p*bHmm, three equations with three variables. Maybe I can solve this system. Let's see.Let me try to express each variable in terms of another. For example, from equation 1: a = p*c - b. Then plug this into equation 3. So substituting a into equation 3: (p*c - b) + c = p*b. Let's simplify that:p*c - b + c = p*b ⇒ (p*c + c) - b = p*b ⇒ c*(p + 1) = p*b + b ⇒ c*(p + 1) = b*(p + 1).Wait, that's interesting. So c*(p + 1) = b*(p + 1). If p + 1 ≠ 0, then we can divide both sides by (p + 1), so c = b. But if p + 1 = 0, then p = -1, and then we have to check if that's possible.So first case: p + 1 ≠ 0 ⇒ c = b.Similarly, maybe we can find relations between a and c, or a and b.Alternatively, let's substitute a = p*c - b into equation 2: b + c = p*a ⇒ b + c = p*(p*c - b). Let's expand that:b + c = p^2*c - p*b.Bring all terms to one side: b + c - p^2*c + p*b = 0 ⇒ b*(1 + p) + c*(1 - p^2) = 0.But from the previous result, if p + 1 ≠ 0, then c = b. So substituting c = b into this equation:b*(1 + p) + b*(1 - p^2) = 0 ⇒ b[(1 + p) + (1 - p^2)] = 0.Simplify inside the brackets: 1 + p + 1 - p^2 = 2 + p - p^2. So:b*(2 + p - p^2) = 0.Since a + b + c ≠ 0, and if we had c = b, then a = p*c - b = p*b - b = b*(p - 1). So a + b + c = b*(p - 1) + b + b = b*(p - 1 + 1 + 1) = b*(p + 1). Since a + b + c ≠ 0, then either b ≠ 0 and p + 1 ≠ 0. But we already assumed p + 1 ≠ 0, so this is okay.So, back to the equation: b*(2 + p - p^2) = 0. Since a + b + c ≠ 0 and b is part of that sum, b cannot be zero. Therefore, 2 + p - p^2 = 0 ⇒ -p^2 + p + 2 = 0 ⇒ p^2 - p - 2 = 0.Solving this quadratic equation: p = [1 ± sqrt(1 + 8)]/2 = [1 ± 3]/2. So p = (1 + 3)/2 = 2 or p = (1 - 3)/2 = -1.Wait, p = 2 or p = -1. But earlier, we considered the case where p + 1 ≠ 0, which would exclude p = -1. So if p = -1, we need to check that case separately. So let's check both possibilities.Case 1: p ≠ -1. Then we have c = b, and from the equation above, p = 2 or p = -1, but since p ≠ -1, then p = 2. So p = 2.Then, c = b, and from a = p*c - b = 2b - b = b. So a = b = c. So in this case, all variables are equal. Then, a + b + c = 3a ≠ 0, which is fine.So, when p = 2, a = b = c. Let me check if this satisfies the original equations.Take (a + b)/c = (a + a)/a = 2a/a = 2 = p. Similarly, (b + c)/a = (a + a)/a = 2 = p, same with the third equation. So that works. So p = 2 is valid.Case 2: p = -1. Then we need to check if this is possible. Let's go back to the original equations.If p = -1, then:(a + b)/c = -1 ⇒ a + b = -cSimilarly, (b + c)/a = -1 ⇒ b + c = -aAnd (a + c)/b = -1 ⇒ a + c = -bSo we have:1. a + b = -c2. b + c = -a3. a + c = -bLet me add all three equations:(a + b) + (b + c) + (a + c) = -c - a - bSimplify left side: 2a + 2b + 2c = - (a + b + c)Bring all terms to left side: 2a + 2b + 2c + a + b + c = 0 ⇒ 3a + 3b + 3c = 0 ⇒ 3(a + b + c) = 0 ⇒ a + b + c = 0.But the problem states that a + b + c ≠ 0. Therefore, p = -1 is not possible because it leads to a contradiction. Hence, p cannot be -1, so the only valid solution is p = 2.Therefore, p = 2. So the line is y = 2x + 2. Now, we need to determine which quadrant this line does not pass through.To find out which quadrants the line passes through, let's analyze the line y = 2x + 2.First, let's find the x-intercept and y-intercept.The y-intercept is when x = 0: y = 2*0 + 2 = 2. So the point is (0, 2), which is on the positive y-axis, between the first and second quadrants.The x-intercept is when y = 0: 0 = 2x + 2 ⇒ 2x = -2 ⇒ x = -1. So the point is (-1, 0), which is on the negative x-axis, between the second and third quadrants.Now, plotting this line, it goes through (0, 2) and (-1, 0). Let's see the direction. The slope is 2, which is positive, so the line rises from left to right.Starting from the x-intercept (-1, 0), as x increases, y increases. Moving to the right from (-1, 0), when x becomes positive, y becomes 2x + 2, which is positive, so it goes into the first quadrant. When moving to the left from (-1, 0), as x decreases (more negative), y = 2x + 2 becomes more negative (since 2x is negative and adding 2 might not be enough). Let's check for x = -2: y = 2*(-2) + 2 = -4 + 2 = -2. So the point (-2, -2) is in the third quadrant. So the line passes through the third quadrant as well.But wait, does it pass through the fourth quadrant? The fourth quadrant has positive x and negative y. Let's check if there's any x > 0 where y is negative. For x > 0, y = 2x + 2. Since x is positive, 2x is positive, and adding 2 makes it even more positive. So y is positive for all x > 0. Therefore, in the fourth quadrant (positive x, negative y), the line does not pass through because y is always positive there. Similarly, in the first quadrant (positive x, positive y), it does pass. In the second quadrant (negative x, positive y), at x = -0.5, for example, y = 2*(-0.5) + 2 = -1 + 2 = 1, so point (-0.5, 1) is in the second quadrant. In the third quadrant (negative x, negative y), as we saw with x = -2, y = -2, so that's in the third quadrant. The fourth quadrant, as established, has no points from this line because for x > 0, y is positive. So the line does not pass through the fourth quadrant.Therefore, the answer should be (D) fourth quadrant.But wait, let me confirm again. The line y = 2x + 2. Let's check the quadrants:First quadrant: x > 0, y > 0. For example, x=1, y=4. Yes, passes through.Second quadrant: x < 0, y > 0. For example, x=-0.5, y=1. Yes, passes through.Third quadrant: x < 0, y < 0. For example, x=-2, y=-2. Yes, passes through.Fourth quadrant: x > 0, y < 0. But for any x > 0, y = 2x + 2 is positive. Therefore, never enters the fourth quadrant. So the line does not pass through the fourth quadrant. So answer is D.Wait, but the options are:(A) first(B) second(C) third(D) fourthYes, so D is the correct answer. Therefore, the line does not pass through the fourth quadrant.Final Answerboxed{D}</think>Given the equations (frac{a+b}{c} = frac{b+c}{a} = frac{a+c}{b} = p) and (a + b + c neq 0), we need to determine which quadrant the line (y = px + p) does not pass through.First, we convert the given ratios into equations:1. (a + b = pc)2. (b + c = pa)3. (a + c = pb)By substituting and solving these equations, we find that (p = 2) (since (p = -1) leads to a contradiction with (a + b + c neq 0)).The line is then (y = 2x + 2). To determine the quadrants it passes through:- The y-intercept is ((0, 2)), which is between the first and second quadrants.- The x-intercept is ((-1, 0)), which is between the second and third quadrants.- The slope is positive, so the line rises from left to right.Checking the quadrants:1. First quadrant: For (x > 0), (y = 2x + 2) is positive (e.g., (x = 1), (y = 4)).2. Second quadrant: For (x < 0), (y = 2x + 2) can be positive (e.g., (x = -0.5), (y = 1)).3. Third quadrant: For (x < 0), (y = 2x + 2) can be negative (e.g., (x = -2), (y = -2)).4. Fourth quadrant: For (x > 0), (y = 2x + 2) is always positive, so it never enters the fourth quadrant.Thus, the line does not pass through the fourth quadrant.(boxed{D})]

🔑:6. D.From the given information, we havebegin{array}{l}left{begin{array}{l}a+b=c p, a+c=a p, c+a=b pend{array}right. Rightarrowleft{begin{array}{l}2(a+b+c)=p(a+b+c) a+b+c neq 0end{array}right. Rightarrow p=2 .end{array}Therefore, the line y=p x+p cannot pass through the fourth quadrant.

🧠:[<think>Okay, let's see. The problem says that the sum of the squares of 50 consecutive odd integers is 300,850. I need to find the largest odd integer whose square is the last term in this sum. Hmm. Alright, let's break this down step by step.First, let's recall that consecutive odd integers can be represented in a sequence. Since they are consecutive odds, each term is 2 more than the previous one. For example, if the first odd integer is n, then the next one is n + 2, then n + 4, and so on. Since there are 50 terms, the last term would be n + 2*(50 - 1) = n + 98. So, the sequence is n, n+2, n+4, ..., n+98.Our goal is to find the value of n + 98, which is the largest odd integer in this sequence, such that the sum of the squares of these 50 terms equals 300,850.So, the sum we need to compute is the sum from k=0 to 49 of (n + 2k)^2, and that should equal 300,850. Let's write that out mathematically:Sum_{k=0}^{49} (n + 2k)^2 = 300,850I need to expand this sum and solve for n. Once I find n, I can add 98 to it to get the largest term.Let me recall the formula for the sum of squares of an arithmetic sequence. The general formula for the sum of squares of terms in an arithmetic progression is a bit complicated, but maybe I can expand the expression (n + 2k)^2 and then sum term by term.Expanding (n + 2k)^2 gives n^2 + 4nk + 4k^2. Therefore, the sum becomes:Sum_{k=0}^{49} [n^2 + 4nk + 4k^2] = 300,850This can be separated into three separate sums:Sum_{k=0}^{49} n^2 + 4n Sum_{k=0}^{49} k + 4 Sum_{k=0}^{49} k^2 = 300,850Let's compute each of these sums individually.First, Sum_{k=0}^{49} n^2. Since n is constant with respect to k (it doesn't depend on k), this is just 50n^2.Second, 4n Sum_{k=0}^{49} k. The sum of k from 0 to 49 is (49)(49 + 1)/2 = 49*50/2 = 1225. So, this term becomes 4n * 1225 = 4900n.Third, 4 Sum_{k=0}^{49} k^2. The sum of k^2 from 0 to m is given by m(m + 1)(2m + 1)/6. Here, m = 49. So, plugging that in:Sum_{k=0}^{49} k^2 = (49)(50)(99)/6. Let me compute that step by step.First, 49*50 = 2450. Then, 2450*99. Let's compute 2450*100 = 245,000, subtract 2450 to get 245,000 - 2,450 = 242,550. Then, divide by 6: 242,550 / 6 = 40,425. So, the sum of k^2 from 0 to 49 is 40,425. Therefore, 4 times that is 4*40,425 = 161,700.Putting all these together, the entire sum is:50n^2 + 4900n + 161,700 = 300,850Now, let's subtract 300,850 from both sides to set the equation to zero:50n^2 + 4900n + 161,700 - 300,850 = 0Compute 161,700 - 300,850 = -139,150. So, the equation becomes:50n^2 + 4900n - 139,150 = 0To simplify this equation, let's divide all terms by 50:n^2 + 98n - 2783 = 0So, n^2 + 98n - 2783 = 0Now, this is a quadratic equation in n. Let's attempt to solve it using the quadratic formula. The quadratic formula states that for an equation ax^2 + bx + c = 0, the solutions are:x = [-b ± sqrt(b^2 - 4ac)] / (2a)In our case, a = 1, b = 98, c = -2783. Plugging these into the formula:n = [-98 ± sqrt(98^2 - 4*1*(-2783))]/(2*1)First, compute the discriminant:D = 98^2 - 4*1*(-2783) = 9604 + 4*2783Compute 4*2783:2783*2 = 55665566*2 = 11,132So, D = 9604 + 11,132 = 20,736Then sqrt(D) = sqrt(20,736). Let me see, 144^2 = 20,736 (since 140^2 = 19,600, 141^2 = 19,881, 142^2=20,164, 143^2=20,449, 144^2=20,736). So sqrt(20,736) = 144.Therefore, the solutions are:n = [-98 ± 144]/2So, two possibilities:n = (-98 + 144)/2 = (46)/2 = 23n = (-98 - 144)/2 = (-242)/2 = -121Since we are dealing with integers, both 23 and -121 are integers. However, we need to check if these solutions make sense in the context. The problem mentions "50 consecutive odd integers," but it doesn't specify whether they are positive or negative. However, the sum of their squares is 300,850, which is a positive number. If we take n = -121, then the terms would be negative odd integers, but their squares would still be positive. So, both solutions could potentially work.Wait, but let's check both possibilities.First, let's check n = 23. Then the sequence is from 23 to 23 + 98 = 121. The largest term is 121.If n = -121, then the sequence is from -121 to -121 + 98 = -23. The largest term here is -23. But since -23 is larger than -121, but in terms of magnitude, 121 is larger. But the problem asks for the largest odd integer whose square is the last term. If the last term is -23, then its square is 529, but the largest odd integer would actually be 121, but wait, wait. Wait, if n is -121, then the sequence is -121, -119, ..., -23. So the last term is -23. But -23 is greater than -121. So in terms of actual integer size, -23 is larger, but if we consider the magnitude, 121 is larger. However, since the problem says "the largest odd integer," in terms of numerical value, not absolute value, then -23 is larger than -121. But if the problem is referring to the largest in absolute terms, then it's 121. Hmm, this is a bit ambiguous.But let's check the sum for both n = 23 and n = -121. Because maybe one of them doesn't actually give the sum 300,850. Let's verify.First, let's check n = 23. The sum is 50n^2 + 4900n + 161,700. Let's compute that:50*(23)^2 + 4900*(23) + 161,700First, 23^2 = 529. 50*529 = 26,4504900*23 = let's compute 49*23 = 1,127, then add two zeros: 112,700Then, add 161,700.So total sum: 26,450 + 112,700 + 161,700 = 26,450 + 112,700 = 139,150; 139,150 + 161,700 = 300,850. That's correct.Now check n = -121. Let's compute the same expression:50*(-121)^2 + 4900*(-121) + 161,700First, (-121)^2 = 14,641. 50*14,641 = 732,0504900*(-121) = -4900*121. Let's compute 4900*100 = 490,000; 4900*20 = 98,000; 4900*1 = 4,900. So total 490,000 + 98,000 + 4,900 = 592,900. So 4900*(-121) = -592,900Then, adding 161,700.So total sum: 732,050 - 592,900 + 161,700 = (732,050 - 592,900) + 161,700 = 139,150 + 161,700 = 300,850. So that also checks out.Therefore, both n = 23 and n = -121 give the correct sum. Therefore, there are two sequences: one starting at 23 and ending at 121, and another starting at -121 and ending at -23.But the problem asks for "the largest odd integer whose square is the last term of this sum." So, the last term in the first sequence is 121, whose square is 14,641. In the second sequence, the last term is -23, whose square is 529. Therefore, between 121 and -23, the largest odd integer is 121. Because even though -23 is an integer, 121 is larger in value. Since the problem says "the largest odd integer," without specifying positivity, but generally, when asked for the largest integer, it refers to the greatest value, which would be 121. So, the answer should be 121.Wait, but let me make sure. If the sequence is from -121 to -23, the last term is -23, which is greater than -121, but still negative. So if we are to take the largest odd integer in the sequence, it's -23, but -23 is less than 121. So depending on interpretation, but mathematically, 121 is larger than -23. Therefore, 121 is the correct answer.Therefore, the largest odd integer is 121.Final AnswerThe largest odd integer is boxed{121}.</think>Given the sum of the squares of 50 consecutive odd integers is 300,850, we need to find the largest odd integer whose square is the last term in this sum.1. Represent the sequence of 50 consecutive odd integers starting from ( n ) as ( n, n+2, n+4, ldots, n+98 ).2. The sum of the squares of these integers is given by: [ sum_{k=0}^{49} (n + 2k)^2 = 300,850 ]3. Expand and separate the sum: [ sum_{k=0}^{49} (n^2 + 4nk + 4k^2) = 50n^2 + 4n sum_{k=0}^{49} k + 4 sum_{k=0}^{49} k^2 ]4. Calculate each part: - ( sum_{k=0}^{49} n^2 = 50n^2 ) - ( sum_{k=0}^{49} k = frac{49 cdot 50}{2} = 1225 ), so ( 4n sum_{k=0}^{49} k = 4900n ) - ( sum_{k=0}^{49} k^2 = frac{49 cdot 50 cdot 99}{6} = 40425 ), so ( 4 sum_{k=0}^{49} k^2 = 161700 )5. Combine these results: [ 50n^2 + 4900n + 161700 = 300850 ]6. Simplify the equation: [ 50n^2 + 4900n - 139150 = 0 implies n^2 + 98n - 2783 = 0 ]7. Solve the quadratic equation using the quadratic formula: [ n = frac{-98 pm sqrt{98^2 + 4 cdot 2783}}{2} = frac{-98 pm 144}{2} ]8. The solutions are ( n = 23 ) and ( n = -121 ).9. Determine the largest odd integer: - For ( n = 23 ), the sequence ends at ( 23 + 98 = 121 ). - For ( n = -121 ), the sequence ends at ( -121 + 98 = -23 ).Since 121 is the larger value, the largest odd integer is (boxed{121}).,<think>Okay, so I have this problem where the sum of the squares of 50 consecutive odd integers is 300,850. I need to find the largest odd integer whose square is the last term in this sum. Hmm, let's break this down step by step.First, let's recall that consecutive odd integers can be represented in terms of a variable. If I let the first odd integer be, say, n, then the next ones would be n + 2, n + 4, and so on. Since they are consecutive odd numbers, each subsequent number is 2 more than the previous one. But since there are 50 terms, the last term would be n + 2*(50-1) = n + 98. Wait, 50 terms starting from n: the first term is n, the second is n + 2, ..., the 50th term is n + 2*(50 - 1) = n + 98. So the largest integer is n + 98, and I need to find this value.But the problem states that the sum of their squares is 300,850. So I need to set up an equation where the sum from k=0 to 49 of (n + 2k)^2 equals 300,850. Let me write that out:Sum_{k=0}^{49} (n + 2k)^2 = 300,850Expanding this sum might be a bit tedious, but maybe there's a formula for the sum of squares of an arithmetic sequence. Let me recall. The general formula for the sum of squares of an arithmetic progression: if we have terms a, a + d, a + 2d, ..., a + (m-1)d, then the sum of their squares is m*a^2 + 2*a*d*Sum_{k=0}^{m-1} k + d^2*Sum_{k=0}^{m-1} k^2.Alternatively, maybe expand (n + 2k)^2 and sum term by term.Yes, let's try expanding (n + 2k)^2 first:(n + 2k)^2 = n^2 + 4nk + 4k^2So the sum becomes Sum_{k=0}^{49} [n^2 + 4nk + 4k^2] = 50n^2 + 4n*Sum_{k=0}^{49} k + 4*Sum_{k=0}^{49} k^2Let's compute each part separately.First term: 50n^2Second term: 4n times the sum of k from 0 to 49. The sum of k from 0 to m-1 is (m-1)*m/2. So here, m = 50, so the sum is 49*50/2 = 1225. Therefore, the second term is 4n*1225 = 4900nThird term: 4 times the sum of k^2 from 0 to 49. The formula for the sum of squares from 0 to m-1 is (m-1)*m*(2m-1)/6. So here, m = 50, so the sum is 49*50*99/6. Let me compute that.49*50 = 2450; 2450*99 = 2450*(100 - 1) = 245000 - 2450 = 242550. Then divide by 6: 242550 / 6 = 40425. So the third term is 4*40425 = 161,700Putting it all together:50n^2 + 4900n + 161,700 = 300,850Now, subtract 300,850 from both sides to set the equation to zero:50n^2 + 4900n + 161,700 - 300,850 = 0Compute 161,700 - 300,850 = -139,150So:50n^2 + 4900n - 139,150 = 0Let's simplify this equation by dividing all terms by 50 to make the numbers smaller:n^2 + 98n - 2783 = 0Wait, 50n^2 /50 = n^2, 4900n /50 = 98n, and -139,150 /50 = -2783. So yes, the equation becomes:n^2 + 98n - 2783 = 0Now, we need to solve this quadratic equation for n. Let's use the quadratic formula. For an equation ax^2 + bx + c = 0, the solutions are:n = [-b ± sqrt(b^2 - 4ac)] / (2a)Here, a = 1, b = 98, c = -2783. So compute discriminant first:Discriminant D = b^2 - 4ac = 98^2 - 4*1*(-2783)Compute 98^2: 98*98. Let's compute 100*100 = 10,000, subtract 2*100*2 + 2^2 = 400 + 4 = 404, so 100 - 2 = 98, so 98^2 = 9604Then, -4*1*(-2783) = +4*2783 = 11,132So D = 9604 + 11,132 = 20,736sqrt(20,736). Let me see. 144^2 = 20,736, since 140^2 = 19,600, 140^2 + 8*140 + 16 = 19,600 + 1,120 + 16 = 20,736. So sqrt(D) = 144Therefore, n = [-98 ± 144]/2We have two solutions:n = (-98 + 144)/2 = 46/2 = 23n = (-98 - 144)/2 = (-242)/2 = -121Since we are dealing with integers, both 23 and -121 are integers. However, the problem says "the largest odd integer whose square is the last term of this sum". So we need to check which of these starting terms gives a last term that is the largest.Wait, but odd integers can be positive or negative. If the first term is 23, then the last term is 23 + 98 = 121. If the first term is -121, then the last term is -121 + 98 = -23. Now, comparing 121 and -23, 121 is definitely larger. So regardless of the starting term, the sequence starting at 23 will have the largest last term (121), whereas starting at -121 will have the last term as -23, which is smaller.Therefore, the answer should be 121. But let me verify this.Wait, let's check the sum if n = 23. Let's compute the sum of squares from 23 to 121 (50 terms) and see if it's 300,850.Alternatively, let's check our quadratic equation:If n = 23, then the sum is 50*(23)^2 + 4900*(23) + 161,700Compute each term:50*(529) = 26,4504900*23: Let's compute 49*23 = 1,127, then multiply by 100: 112,700Third term is 161,700Adding them up: 26,450 + 112,700 = 139,150; 139,150 + 161,700 = 300,850. Perfect, that's correct.If n = -121, compute the sum:50*(-121)^2 + 4900*(-121) + 161,700First term: 50*(14,641) = 732,050Second term: 4900*(-121) = -592,900Third term: 161,700Total: 732,050 - 592,900 = 139,150; 139,150 + 161,700 = 300,850. So that also checks out.Therefore, both sequences starting at 23 and -121 will have the sum of squares equal to 300,850. But the problem asks for the largest odd integer whose square is the last term. The last term in the first sequence is 23 + 98 = 121, and in the second sequence is -121 + 98 = -23. Since 121 is larger than -23, the answer is 121.But wait, just to be thorough, is there another possible sequence? For example, maybe starting with a different integer where the last term is even larger? But given that the quadratic equation only has two solutions, those are the only possible starting points. Since the quadratic equation was derived directly from the sum, there are only two possible sequences, one positive and one negative. Hence, 121 is the correct answer.Wait another thought: The quadratic equation gives starting points n = 23 and n = -121. The problem is about 50 consecutive odd integers. Since odd numbers can be both positive and negative, but in this case, the sum is positive (300,850). So both sequences sum to the same positive value, but their last terms are 121 and -23. Since the question asks for the largest odd integer, which is 121. So I think that's solid.But just to make sure, maybe I can test with smaller numbers. Let's say, if we have 2 consecutive odd integers whose squares sum to, say, (1^2 + 3^2) = 1 + 9 = 10. Let's see what the formula gives.If n = 1, the sum is 1 + 9 = 10. The quadratic equation would be 2n^2 + 4n*Sum(k=0 to 1) k + 4*Sum(k=0 to1)k^2. Wait, wait, in the previous problem, for 50 terms, but for 2 terms:Sum_{k=0}^{1} (n + 2k)^2 = (n)^2 + (n + 2)^2 = n^2 + n^2 +4n +4 = 2n^2 +4n +4. If this sum is 10, then 2n^2 +4n +4 =10 => 2n^2 +4n -6=0 => n^2 +2n -3=0. Solutions: n = [-2 ± sqrt(4 +12)]/2 = [-2 ± sqrt(16)]/2 = [-2 ±4]/2. So n= (2)/2=1 or n=(-6)/2=-3. Then the last terms are 1 +2=3 or -3 +2= -1. So the largest is 3. Which is correct, as the first example sums 1 and 3. So same logic, works here. Therefore, the method is correct.Therefore, returning to the original problem, the answer is 121.Final AnswerThe largest odd integer is boxed{121}.</think>Given the problem of finding the largest odd integer whose square is the last term in the sum of the squares of 50 consecutive odd integers totaling 300,850, we start by representing the integers. Let the first odd integer be ( n ). The 50 consecutive odd integers are ( n, n+2, n+4, ldots, n+98 ). The sum of their squares is given by:[sum_{k=0}^{49} (n + 2k)^2 = 300,850]Expanding and summing the squares, we get:[sum_{k=0}^{49} (n^2 + 4nk + 4k^2) = 50n^2 + 4n sum_{k=0}^{49} k + 4 sum_{k=0}^{49} k^2]Calculating each term separately:1. The first term is ( 50n^2 ).2. The second term involves the sum of the first 49 integers: ( sum_{k=0}^{49} k = frac{49 cdot 50}{2} = 1225 ), so the second term is ( 4n cdot 1225 = 4900n ).3. The third term involves the sum of the squares of the first 49 integers: ( sum_{k=0}^{49} k^2 = frac{49 cdot 50 cdot 99}{6} = 40425 ), so the third term is ( 4 cdot 40425 = 161,700 ).Combining these, the equation becomes:[50n^2 + 4900n + 161,700 = 300,850]Simplifying:[50n^2 + 4900n - 139,150 = 0]Dividing by 50:[n^2 + 98n - 2783 = 0]Solving this quadratic equation using the quadratic formula ( n = frac{-b pm sqrt{b^2 - 4ac}}{2a} ):[n = frac{-98 pm sqrt{98^2 + 4 cdot 2783}}{2} = frac{-98 pm 144}{2}]This gives two solutions: ( n = 23 ) and ( n = -121 ).The last term for ( n = 23 ) is ( 23 + 98 = 121 ), and for ( n = -121 ) it is ( -121 + 98 = -23 ). The largest odd integer is 121.Thus, the largest odd integer whose square is the last term in the sum is (boxed{121}).]

🔑:1. Answer: 121Solution. Let the integers be X+2 X+4 cdots X+100. Then(X+2)^{2}+(X+4)^{2}+cdots+(X+100)^{2}=300850Let y=X+51 and regrouping the terms, we obtainleft[(y-49)^{2}+(y+49)^{2}right]+left[(y-47)^{2}+(y+47)^{2}right]+cdots+left[(y-1)^{2}+(y+1)^{2}right]=300850which simplifies to50 y^{2}+2left(1^{2}+3^{2}+5^{2}+7^{2}+cdots+49^{2}right)=300850Using the fact that 1^{2}+3^{2}+5^{2}+cdots+(2 n-1)^{2}=frac{4}{3} n^{3}-frac{1}{3} n, we obtain y=72. Hence X=21, so that the required number is 121 .

🧠:[<think>Okay, so I need to find the smallest number that has a digit sum of 62, and it has to use at least three different digits. Hmm, let's start by understanding what the problem is asking. First, the digit sum of a number is the sum of all its digits. For example, the digit sum of 123 is 1+2+3=6. The problem wants the smallest number possible with a digit sum of 62, but there's a catch: the number must include at least three different digits. That means I can't just use all 9s and maybe an 8 or something; I need to have at least three distinct digits in the number.When they say the smallest number, I think they mean the numerical value, not the number of digits. So, smaller numbers have fewer digits and, for numbers with the same number of digits, the one that is smaller when compared digit by digit from left to right. For example, 100 is smaller than 101, even though they have the same number of digits.To get the smallest number possible with a given digit sum, the usual strategy is to make the number have as few digits as possible (since fewer digits mean a smaller number) and arrange the digits in ascending order (so that smaller digits are on the left). However, since we need at least three different digits, we might need to adjust this strategy.Let's first consider the case without the three different digits requirement. The standard approach is to use as many 9s as possible, since 9 is the largest single-digit number, which allows us to reach the digit sum with the fewest digits. For a digit sum of 62, let's divide 62 by 9 to see how many 9s we need:62 ÷ 9 = 6 with a remainder of 8. So that would be 6 nines and an 8. Therefore, the number would be 8999999 (since we want the smaller digits on the left to make the number as small as possible). Wait, hold on: if we have six 9s and an 8, arranging them in ascending order would put the 8 first, followed by six 9s. So the number is 8999999. That has a digit sum of 8 + 6*9 = 8 + 54 = 62. That seems right. But in this case, there are only two different digits: 8 and 9. So that violates the requirement of having at least three different digits. Therefore, we need to adjust this. So, how do we get a third different digit into the number without making the number too large? Let's think. The minimal number of digits is 7 (since 6*9=54, 54+8=62). If we have to introduce a third digit, we need to replace one of the existing digits with a different digit. However, replacing a 9 with a smaller digit will require increasing another digit to compensate, which might make the number larger. Alternatively, replacing the 8 with a smaller digit and distributing the remaining sum accordingly. Let's explore both possibilities.First, starting from the original 8999999. If we need a third digit, we can replace one of the 9s with another digit. Let's say we take one 9 and turn it into an 8. Then, we have two 8s and five 9s. But that's still only two different digits. Hmm, no. So maybe replace a 9 with a 7. Then, the digit sum would decrease by 2 (since 9 becomes 7). To compensate, we need to add 2 to another digit. So, for example, replace one 9 with a 7 and increase the 8 to a 10. But wait, digits can't be more than 9. So that approach might not work.Alternatively, when we replace a 9 with a lower digit, we need to distribute the difference to other digits. Let's think. If we reduce a 9 to an 8, that's a decrease of 1. Then we need to add 1 somewhere else. But since we already have an 8 in the original number, replacing a 9 with an 8 would still result in two different digits (8 and 9). So that's not helpful.Wait, maybe we can replace the 8 with a 7 and then adjust the remaining digits. Let's see. If we have 7 instead of 8, then the total digit sum would decrease by 1. So we need to add 1 somewhere else. But we can only add to other digits. For instance, changing the first digit from 8 to 7 and then adding 1 to another digit. The next digits are all 9s, so adding 1 to a 9 would make it 10, which is invalid. So that doesn't work either.Alternatively, perhaps we need to have more digits. If we can't achieve three different digits in seven digits, maybe we need to add an extra digit. Let's see. If we go to eight digits, then instead of six 9s and an 8, we could have seven 9s and a digit that is 62 - 7*9 = 62 - 63 = -1, which is impossible. Wait, that's not helpful.Wait, perhaps splitting the 8 into two digits. If instead of having an 8 followed by six 9s, we split the 8 into a 7 and a 1. Then the number would be 7 followed by a 1 and six 9s. That would be 71999999. But the digit sum here is 7 + 1 + 6*9 = 7 + 1 + 54 = 62. Now, we have three different digits: 7, 1, and 9. But the number is 71999999, which is larger than 7999999, which is 7999999. Wait, no. Wait, 71999999 is an 8-digit number, while 7999999 is a 7-digit number. Therefore, 7999999 is smaller. But 7999999 has digits 7, 9. Only two different digits. So, 71999999 is longer and bigger. Not good.Alternatively, maybe split the 8 into a 7 and a 1, but in the middle of the number. Wait, but the number should be as small as possible, so leading digits should be as small as possible. So putting a 1 in the first digit would make the number smaller, but if we split the 8 into a 1 and a 7, putting the 1 first, then we have 1 followed by 7 and six 9s: 17999999. That's an 8-digit number with digits 1, 7, 9. However, 17999999 is an 8-digit number, which is larger than the original 7-digit 8999999. So even though we now have three different digits, the number is longer and hence larger. So that's worse. Alternatively, maybe we can find another way to split the digits. Let's think. Let's start from the minimal number of digits, which is seven. To have three different digits, we need to replace some digits. Let's see. If we take the original 8999999 (digits: 8,9). If we replace one 9 with a 7 and another 9 with a 8. Wait, that would be 8, then 8,7, and the rest 9s. So the number would be 8879999. The digit sum here is 8 +8 +7 +5*9= 8+8+7+45=68, which is too much. Wait, that's more than 62. Wait, that can't be. Wait, original digit sum was 8 +6*9=62. If we replace two 9s with 8 and 7, then total change is -1 -2 = -3. So the new digit sum would be 62 -3=59, which is less than required. So that's not helpful.Alternatively, let's think differently. Let's suppose we need to have three different digits. Let's try to use the digits 9, 8, and 7. Let's see how many of each we need. Let's say we have x 9s, y 8s, z 7s. Then:9x +8y +7z =62and x + y + z >=1 (but the total digits would be x + y + z, which we want to minimize). Also, since we need at least three different digits, z should be at least 1.But this might get complex. Maybe trial and error.Alternatively, perhaps using digits 9, 8, and another digit. Let's try to use 9, 8, and 7.Suppose we replace one 9 with a 7. Then, the total sum would decrease by 2. To compensate, we need to add 2 to another digit. But digits can't exceed 9. So, for example, if we have the original number 8999999 (sum 62). Replace one 9 with a 7 (sum decreases by 2). Then, we need to add 2 to another digit. The original 8 could become a 10, but that's invalid. Alternatively, split the 2 over two digits. For example, turn one 9 into a 7, and turn another 9 into a 9+1=10, which is invalid. Hmm.Alternatively, maybe replace two 9s with two 8s. Then, sum decreases by 2 (since 9 to 8 is -1 each). Then, the total sum would be 62 -2 =60. Not enough. Alternatively, replace three 9s with 8s, then sum decreases by 3, leading to 62 -3=59. Still not.Alternatively, replace one 9 with an 8 and one 9 with a 7. Then total decrease is 1 +2=3. So sum becomes 62 -3=59. Still not.Alternatively, maybe replace the 8 with a 7, and add a 1. So, instead of 8 followed by six 9s, we have 7,1 followed by five 9s. Then, the number is 7199999. Digit sum is 7 +1 +5*9=7 +1 +45=53. Wait, that's way too low. So not helpful.Alternatively, replace the 8 with a 7 and a 2. So 7299999. Then digit sum is 7 +2 +5*9=7 +2 +45=54. Still too low.Alternatively, replace the 8 with a 7 and a 3. 7399999. Sum is 7+3+45=55. Not enough. Hmm. This approach isn't working because replacing the 8 with two digits would require distributing the 8 into two digits summing to 8, but unless we add more digits, which would increase the total number of digits. But increasing digits would make the number longer and hence larger, which we don't want. So maybe this is not the way.Wait, perhaps instead of starting with 8 followed by six 9s, which is two digits, we need to rearrange the digits to include a third digit. Let's try.Suppose we have 7 digits. To have three different digits, maybe we can have one 8, five 9s, and one 7. Let's check the sum: 8 +5*9 +7=8 +45 +7=60. That's 60, which is 2 less than needed. So we need to add 2 more. How? Since digits can't exceed 9, maybe increase two of the existing digits. For example, turn two 9s into 10s? No, that's invalid.Alternatively, maybe instead of 8,5,7,1, perhaps start with a higher digit. Wait, maybe 9,9,9,9,9,9,8 is the original. If we replace two of the 9s with an 8 and a 7, then sum is 62 -2=60. Then, to get back to 62, we need to add 2. But how?Alternatively, maybe use three different digits, but more of them. Let's try:Suppose we use 9, 8, and 7. Let’s denote the number of 9s as a, 8s as b, and 7s as c. Then:9a +8b +7c =62And a + b + c should be as small as possible.Also, we need at least one of each: a≥1, b≥1, c≥1.Let’s try to maximize the number of 9s first.Let’s set a=6. Then 9*6=54. Remaining sum=62-54=8. So we need to make 8 with 8 and 7, using at least one of each. So 8=8*1 +7*0, but we need at least one 7. 8-7=1, which can't be made with 8s. So not possible.a=5. 9*5=45. Remaining sum=62-45=17. We need to make 17 with 8s and 7s, at least one of each.Let’s try 8*1 +7*1=15. 17-15=2. Not possible. 8*2 +7*1=23. Too much. 8*1 +7*2=22. Too much. So not possible.a=4. 9*4=36. Remaining sum=62-36=26. Need 26 with 8s and 7s, at least one each.Let's see. 8*3 +7*1=24 +7=31. Too much. 8*2 +7*2=16 +14=30. Still too much. 8*1 +7*3=8 +21=29. Still too much. 8*0 +7*4=28. Still too much. Hmm, not possible.a=3. 9*3=27. Remaining sum=62-27=35. Need 35 with 8s and 7s. 8*4 +7*1=32 +7=39. Too much. 8*3 +7*2=24 +14=38. Still too much. 8*2 +7*3=16 +21=37. Still too much. 8*1 +7*4=8 +28=36. Still too much. 8*0 +7*5=35. Exactly. But we need at least one 8. So that's invalid.a=2. 9*2=18. Remaining sum=62-18=44. Need 44 with 8s and 7s. 8*5 +7*2=40 +14=54. Too much. 8*4 +7*4=32 +28=60. Still too much. This seems not helpful.a=1. 9*1=9. Remaining sum=62-9=53. 53 with 8s and 7s. 8*6 +7*1=48 +7=55. Close. 8*5 +7*3=40 +21=61. No. Not helpful.a=0. Not possible since we need three different digits.So using 9s, 8s, and 7s doesn't seem possible. Maybe we need a different combination of digits.What if we use digits 9, 8, and something else, like 5? Let's try.But this is getting too arbitrary. Maybe another approach.Since we need three different digits, let's try to minimize the number of digits. Let's assume we need at least three different digits. Let's consider the minimal number of digits, which is seven. So, we have seven digits. The maximum digit sum with seven digits is 7*9=63. We need 62, which is 1 less. So the standard minimal number is 8999999 (8 followed by six 9s). But that uses only two digits. So we need to modify this to include a third digit.We need to reduce one of the digits by 1 to get from 63 to 62. But in the standard case, that 8 is already 8. Wait, no. If the maximum digit sum is 63 with seven 9s, then subtracting 1 gives 62, which would be 8999999 (since we subtract 1 from the first digit, turning a 9 into an 8). But that's only two digits.Therefore, to get three different digits, perhaps we can subtract 1 from a different digit instead. For example, instead of subtracting 1 from the first digit (making it 8), subtract 1 from the second digit, making it 8, and keep the first digit as 9. So, the number would be 9899999. That has digits 9,8,9,9,9,9,9. So still two different digits. Not helpful.Alternatively, subtract 1 from two different digits. Let's say subtract 1 from the first and second digits. Then we have 8 and 8, followed by five 9s. 8899999. Still two different digits.Alternatively, subtract 1 from the first digit and subtract 1 from the last digit. So, first digit is 8, then five 9s, and the last digit is 8. So 8999998. Still two different digits. Not helpful.Alternatively, subtract 1 from two different digits. For example, subtract 1 from the first digit (making it 8) and subtract 1 from the third digit (making it 8). So the number would be 8989999. That's digits 8,9,8,9,9,9,9. Still two different digits. Hmm.Alternatively, subtract 1 from one of the digits and subtract 2 from another. For instance, subtract 1 from the first digit (8) and subtract 2 from the second digit (7). Then, the number would be 8799999. Now, digits are 8,7,9,9,9,9,9. So three different digits: 8,7,9. The digit sum is 8 +7 +5*9 =8 +7 +45=60. Wait, but we need 62. So that's 2 less. Hmm, this approach is not working.Wait, this is getting confusing. Let me recap. If we have seven digits, maximum sum 63. To get 62, we need to subtract 1. The standard way is to have one 8 and six 9s. To get three different digits, we need to subtract more than 1 in total? But that would decrease the digit sum further. Wait, no. Wait, actually, if we start from 63 (all 9s), subtract 1 to get 62 (one 8 and six 9s). But we need three different digits. So, perhaps instead of just subtracting 1, we need to subtract in such a way that we create a third digit. For example, subtract 1 from one 9 to make it 8, and subtract another 1 from a different 9 to make it 7, but then add 1 back somewhere else. Wait, but you can't add; you can only subtract. Since the total digit sum is 63, subtract a total of 1 to get 62. So, you can only subtract 1 once. So, you can't create a third digit by subtracting in two places because that would subtract 2 total, which would lead to 61. Therefore, maybe impossible to have three different digits in seven digits. So, we need to use more digits.Therefore, maybe eight digits. Let's see. If we use eight digits, the maximum digit sum is 8*9=72. We need 62, which is 10 less. So, we need to subtract 10. But how?To get three different digits, we need to have at least three distinct digits. Let's try distributing the subtracted 10 across the digits. The standard approach is to have as many 9s as possible. So, subtract 1 from ten different digits, turning them into 8s. But we only have eight digits. Wait, 8 digits. To subtract 10, we need to reduce some digits by more than 1. For example, subtract 1 from two digits (making them 8) and subtract 2 from one digit (making it 7). Total subtracted: 1+1+2=4. Not enough. Wait, perhaps:To subtract 10 from 72, we need to decrease some digits. Let's do this in a way that introduces a third digit. For example, turn one 9 into a 9 - x, and another 9 into a 9 - y, such that x + y =10. But since each digit can only go down to 0, the maximum we can subtract from a single digit is 9 (turning it into 0). But we need to have at least three different digits, so maybe:Let's decrease one 9 to 8 (subtract 1), another 9 to 7 (subtract 2), and another 9 to 6 (subtract 3). Total subtracted:1+2+3=6. Then, we still need to subtract 4 more. Hmm, maybe subtract 1 from four more digits. So, total subtracted:6+4=10. So, starting from eight 9s, subtract 1,2,3 from three digits and subtract 1 from four others. The digits would be 8,7,6, and four 8s (from subtracting 1 from four 9s). Wait, that's confusing.Alternatively, let's think of it as:Start with eight 9s: 99999999 (sum 72). Need sum 62, so subtract 10. To create three different digits, we need at least three digits reduced by different amounts. For example:- Turn one 9 into a 7 (subtract 2)- Turn one 9 into an 8 (subtract 1)- Turn one 9 into a 6 (subtract 3)- Then, total subtracted:2+1+3=6. Still need to subtract 4 more. Subtract 1 from four more digits: turning four 9s into 8s.So total subtracted:2+1+3+4=10. So the digits would be:7,8,6,8,8,8,8,9. Wait, but that would be 7,8,6, and five 8s. Wait, no, we started with eight digits:Original:9,9,9,9,9,9,9,9After subtracting:First digit:9-2=7Second digit:9-1=8Third digit:9-3=6Fourth to seventh digits:9-1=8 (four times)Eighth digit:9 (unchanged)So the digits are:7,8,6,8,8,8,8,9Arrange them in ascending order for the smallest number:6,7,8,8,8,8,8,9. So the number is 67888889. Let's check the digit sum:6+7+8*5 +9=6+7+40+9=62. Yes, that works. And the digits used are 6,7,8,9. Wait, four different digits. But the problem only requires at least three different digits, so this is acceptable. However, maybe we can do better with fewer different digits?Wait, if we use three different digits. Let's try:Subtract 3 from one digit (making it 6), subtract 2 from another (making it 7), and subtract 5 from another (making it 4). Total subtracted:3+2+5=10. Then digits would be 6,7,4, and the rest 9s. But this introduces three different digits:4,6,7,9. Still four digits. Hmm.Alternatively, subtract 1, 2, and 7 from three digits. But subtracting 7 from a digit would make it 2, which is a new digit. So 9-7=2, 9-2=7, 9-1=8. Then digits:2,7,8, and the rest 9s. Sum:2+7+8+5*9=2+7+8+45=62. So the number arranged is 2 followed by 7,8, and five 9s: 27899999. But this is an 8-digit number. Let's see if this is smaller than the previous 67888889. Comparing the first digit:2 vs 6. 2 is smaller, so 27899999 is smaller. But wait, does this have three different digits? 2,7,8,9. That's four different digits. The problem requires at least three, so this is okay.But maybe we can get a smaller number by using three different digits instead of four. Let's see.Suppose we subtract 1 from one digit (8), subtract 3 from another (6), and subtract 6 from another (3). Total subtracted:1+3+6=10. Then digits:8,6,3, and five 9s. Arranged:3,6,8,9,9,9,9,9. The number is 36899999. The digit sum:3+6+8+5*9=3+6+8+45=62. This is also an 8-digit number. Comparing to 27899999, 278 is smaller than 368, so 27899999 is smaller.Alternatively, subtract 4 and 6 from two digits. 9-4=5, 9-6=3, and subtract 0 from the rest. Wait, total subtracted:4+6=10. Then digits:5,3, and six 9s. Sum:5+3+6*9=8+54=62. The number arranged is 3,5,9,9,9,9,9,9: 35999999. This is an 8-digit number. Compare to 27899999: 278 vs 35. 35 is smaller, but wait, 35 is 3 followed by 5, which is 35, whereas 278 is 2 followed by 7, so 2 is smaller. Therefore, 27899999 is smaller than 35999999.So far, the smallest number I've found with three different digits is 27899999. But wait, is there a way to have a smaller number?Wait, let's try subtracting 1,1, and 8. 9-1=8, 9-1=8, 9-8=1. Then digits:8,8,1, and five 9s. Sum:8+8+1+5*9=17+45=62. The number arranged is 1,8,8,9,9,9,9,9: 18899999. This has digits 1,8,9. Three different digits. The number is 18899999. That's smaller than 27899999 because it starts with 1 instead of 2. Okay, so 18899999. Let's check the digit sum:1+8+8+6*9=1+8+8+54=71. Wait, that's way over. Wait, no, wait. Wait, original digits are 1,8,8, and five 9s. So 1 +8 +8 +5*9=1+8+8+45=62. Correct. So the number is 1 followed by 8,8,9,9,9,9,9: 18899999. That's an 8-digit number. That's better. Is there a way to make it even smaller? Let's see. If we can have the first digit as 1, then the next digits as small as possible. Let's try to use 1,9, and another digit. Let's see.Suppose we have a 1, then we need the remaining seven digits to sum to 62-1=61. Let's try to use two more different digits. For example, 9s and 8s. Let's see how many 9s and 8s we need to sum to 61 over seven digits. Let’s let x be the number of 9s and y be the number of 8s. Then:9x +8y =61x + y =7From the second equation: x=7 - y. Substitute into the first equation:9(7 - y) +8y =6163 -9y +8y =6163 - y =61-y= -2 → y=2Therefore, x=7 -2=5. So five 9s and two 8s. Therefore, the digits after the 1 would be two 8s and five 9s. Arranged in ascending order: 1,8,8,9,9,9,9,9 → 18899999. Which is the same as before. So this is the minimal number with a leading 1 and three different digits.But wait, could we have a smaller number with a leading 1 and different digits? Let's see. Maybe using a digit lower than 8. Let's try 7. Let's see.We need the remaining seven digits to sum to 61. Suppose we use 9s, 8s, and 7s. Let's see. Let’s assume we have x 9s, y 8s, z 7s. Then:9x +8y +7z=61x + y + z=7We need to find non-negative integers x,y,z satisfying these equations. Let's try:If z=1:Then 9x +8y=61 -7=54x + y=7 -1=6So x=6 - ySubstitute:9(6 - y) +8y=5454 -9y +8y=5454 - y=54→ y=0Thus, x=6, y=0, z=1. So digits would be six 9s and one 7. But in addition to the leading 1. So total digits:1,9,9,9,9,9,9,7. Arranged:1,7,9,9,9,9,9,9. That is 17999999. Let's check digit sum:1 +7 +6*9=1 +7 +54=62. Correct. So the number is 17999999. Which is smaller than 18899999 because 17 is smaller than 18. So this is better.Now, digits used are 1,7,9. Three different digits. So this satisfies the requirement. Can we do even better?Let's try z=2:9x +8y=61 -14=47x + y=7 -2=5x=5 - ySo 9(5 - y) +8y=4745 -9y +8y=4745 - y=47→ -y=2 → y=-2. Not possible.z=3:9x +8y=61 -21=40x + y=7 -3=4x=4 - y9(4 - y) +8y=4036 -9y +8y=4036 - y=40→ -y=4 → y=-4. Not possible.Similarly, higher z will lead to more negative y. So z=1 is the only possibility here. Therefore, 17999999 is the number.Can we use a 6 instead of 7? Let's try.If we have a 1, a 6, and the rest 9s and 8s. So remaining seven digits need to sum to 61. Let’s try including a 6.Wait, if we have a 1 and a 6, then the remaining six digits need to sum to 61 -6=55. Wait, no. Wait, total digits after the leading 1 are seven. If one of them is 6, then the remaining six digits must sum to 61 -6=55. So:Let x be the number of 9s, y the number of 8s. Then:9x +8y=55x + y=6So x=6 - y9(6 - y) +8y=5554 -9y +8y=5554 - y=55→ -y=1 → y=-1. Not possible.So can't use a 6 here. How about a 5?Similarly, if we have a 1 and a 5 in the remaining seven digits, sum becomes 61 -5=56 for the remaining six digits.9x +8y=56x + y=6x=6 - y9(6 - y) +8y=5654 -9y +8y=5654 - y=56→ -y=2 → y=-2. Not possible.Same with 4:Sum remaining:61 -4=579x +8y=57x + y=6x=6 - y9(6 - y) +8y=5754 -9y +8y=5754 - y=57→ -y=3 → y=-3. Not possible.3:Sum remaining:61 -3=589x +8y=58x + y=6x=6 - y9(6 - y) +8y=5854 -9y +8y=5854 - y=58→ -y=4 → y=-4. Not possible.2:Sum remaining:61 -2=599x +8y=59x + y=6x=6 - y9(6 - y) +8y=5954 -9y +8y=5954 - y=59→ -y=5 → y=-5. Not possible.1:Sum remaining:61 -1=609x +8y=60x + y=6x=6 - y9(6 - y) +8y=6054 -9y +8y=6054 - y=60→ -y=6 → y=-6. Not possible.0:Sum remaining:61 -0=619x +8y=61x + y=7x=7 - y9(7 - y) +8y=6163 -9y +8y=6163 - y=61→ y=2. Then x=5. So digits would be 9*5 +8*2=45 +16=61. So this is possible. So, if we don't include a lower digit besides 1,9,8. So digits would be 1, five 9s, two 8s. Arranged:1,8,8,9,9,9,9,9 → 18899999. Which we already considered. But we saw that 17999999 is smaller.So 17999999 is better. Let's see if we can get even lower by using more than three different digits. Wait, the problem allows three or more, so maybe using four different digits could result in a smaller number.Let's try. If we can introduce a 1, a 5, a 9, and other digits. Wait, but this is getting too vague. Alternatively, think of splitting the 7 into lower digits.Wait, in 17999999, we have a 1, a 7, and six 9s. The 7 is the second digit. If we can split that 7 into smaller digits, maybe getting a lower number. For example, instead of 7, use 6 and 1. Then, the digits would be 1,6,1, and five 9s. But this would make the number 11699999. Let's check the digit sum:1+1+6 +5*9=8 +45=53. No, that's too low. Not helpful.Alternatively, split the 7 into 5 and 2. Then digits:1,5,2, and five 9s. Number:12599999. Sum:1+2+5 +5*9=8 +45=53. Still too low.Alternatively, split the 7 into 3 and 4. Number:13499999. Sum:1+3+4 +5*9=8 +45=53. Still no good.Alternatively, maybe split the 7 into two 3s and a 1. But that would require more digits. Wait, we only have seven digits after the leading 1. If we split the 7 into smaller digits, but that would require more digits, which we don't have. So this approach might not work.Alternatively, let's try replacing one of the 9s in 17999999 with an 8 and a 1. For example, instead of 17999999, replace a 9 with 8 and 1. Then the number becomes 179999819. Wait, that's nine digits, which is longer. So the number would be larger. So no good.Alternatively, maybe replace two 9s with two 8s and a 0. But 0 would make the number have leading zeros, which isn't allowed. For example, 17889999. But that's 1,7,8,8,9,9,9,9. Digit sum:1+7+8+8+4*9=1+7+8+8+36=60. Not enough.Alternatively, replace a 9 with a 7 and a 2. Then digits:1,7,7,2, and five 9s. Number:12779999. Digit sum:1+2+7+7 +4*9=17 +36=53. Not enough.This isn't working. So 17999999 seems to be the minimal number so far. But wait, let's check another approach. Let's see if we can get a seven-digit number with three different digits.Earlier, we saw that a seven-digit number with three different digits might not be possible because we can't subtract enough from the original 8999999 without reducing the digit sum too much. But let's verify.Original number:8999999 (sum 62). If we need three different digits, we have to replace at least one of the digits with a different digit. Let's try replacing one 9 with a 7. Then, sum becomes 8 +5*9 +7 +1*9=8 +45 +7 +9=69. Wait, no. Wait, original sum is 8 +6*9=62. If we replace one 9 with a 7, the new sum is 8 +5*9 +7=8 +45 +7=60. That's two less. To get back to 62, we need to add 2. Since we can't increase a digit beyond 9, perhaps split another 9 into two digits. For example, replace a 9 with a 9 +2=11, but that's invalid. Alternatively, split the 7 into a 7 and add 2 to another digit. Wait, no. This seems not helpful.Alternatively, replace two 9s with 8 and 7. Then sum is 8 +4*9 +8 +7=8 +36 +8 +7=59. Still too low.Alternatively, replace the 8 with a 9 and replace two 9s with lower digits. For example, 9 followed by five 9s and two 8s. Wait, sum is 9 +5*9 +2*8=9 +45 +16=70. Too high.Alternatively, maybe 9,9,9,9,9,9, 62 -54=8. So 8999999 again. Still two digits.Alternatively, maybe 9,9,9,9,9,8,8. Sum is 5*9 +2*8=45 +16=61. Not enough. Add 1 to a digit: 9,9,9,9,9,9,7. Sum is 62. But that's two digits again.Alternatively, 9,9,9,9,8,8,9. Sum is 62. Still two digits.Thus, in seven digits, it's impossible to have three different digits. Therefore, the minimal number of digits required is eight. So, the smallest number is an eight-digit number. So far, we have 17999999. Let's check if that's the smallest.Is there a way to make a number smaller than 17999999 with three different digits and digit sum 62?Let's try starting with 1, then the next digits as small as possible. For example, 11 followed by digits. But the remaining six digits need to sum to 62 -1 -1=60. With six digits summing to 60. The maximum digit sum for six digits is 6*9=54. 60 exceeds that. So impossible.Alternatively, start with 1, then next digit 2. Then remaining five digits need to sum to 62 -1 -2=59. Five digits summing to 59. Max sum is 45. Still impossible.Similarly, starting with 1,2,3: sum 6. Remaining digits need to sum to 62-6=56 with four digits. Still impossible.So, leading digits can't be too small. Hence, the smallest possible starting digit is 1, followed by a 7, as in 17999999. Let's see if there's a way to get a smaller number by having the second digit smaller than 7. For example, 1,6,... Let's check.If we have 1,6, then the remaining six digits need to sum to 62 -1 -6=55. With six digits, the maximum sum is 54. So impossible.How about 1,7: remaining six digits sum to 62 -1 -7=54. That's possible with six 9s: 999999. So the number is 1,7,9,9,9,9,9,9:17999999. That works.If we try 1,7,8: sum so far is 1+7+8=16. Remaining five digits need to sum to 62 -16=46. 46 with five digits. Max is 45. Not enough.Alternatively, 1,7,7: sum 15. Remaining five digits need to sum to 47. Max 45. Still no.Alternatively, 1,5,... Sum 1+5=6. Remaining six digits need to sum to 56. Max 54. No.So, seems that 17999999 is the minimal number with three different digits. Let's check another possibility: starting with 1,8, but with a different third digit.For example, 1,8,7: sum 16. Remaining five digits sum to 46. Max possible with five digits is 45. So no.Alternatively, 1,8,6: sum 15. Remaining five digits need to sum to 47. Still too high.Alternatively, 1,8,5: sum 14. Remaining five digits need to sum to 48. Still over.Alternatively, 1,8,8: sum 17. Remaining five digits need to sum to 45. Which is five 9s. So the number is 1,8,8,9,9,9,9,9:18899999. Which we already considered. It's larger than 17999999.Alternatively, starting with 1,9: sum 10. Remaining six digits need to sum to 52. Six digits summing to 52: Let's use as many 9s as possible. 52 ÷9=5 with remainder 7. So five 9s and one 7. Therefore, digits:1,9,7,9,9,9,9,9. Arrange them:1,7,9,9,9,9,9,9:17999999. Same as before.Therefore, 17999999 is the smallest possible number with three different digits and digit sum 62. Wait, but let me check if there's a way to make the number start with a smaller digit than 1. But digits can't start with 0. So 1 is the smallest possible first digit. Therefore, 17999999 is the smallest possible number. However, let's verify once more.Wait, 17999999 uses digits 1,7,9. Three different digits. The digit sum is 1+7+6*9=8+54=62. Correct. The number is 17,999,999. Is there a smaller eight-digit number with three different digits and sum 62? Let's think.Suppose we have a number starting with 1, then the next digits are 6,9,9,9,9,9,9. Sum is 1+6+6*9=7+54=61. Not enough. Need 1 more. So, replace a 9 with a 10. Not possible. Alternatively, replace a 9 with an 8 and add 1 to another. So 1,6,8,9,9,9,9,9. Sum:1+6+8+5*9=15+45=60. Still not.Alternatively, 1,7,9,9,9,9,9,9. Which is 17999999. Sum 62. Correct.Alternatively, what if we use digits 1,8,9 arranged differently. For example, 1,1,8,9,9,9,9,9. Sum:1+1+8+5*9=10+45=55. Not enough. Alternatively, 1,2,9,9,9,9,9,9. Sum:1+2+6*9=3+54=57. Not enough. So, indeed, 17999999 seems to be the smallest possible. But wait, let's check another combination. Suppose we use digits 1,8,9, and another digit. For example, 1,7,8,9. Wait, but that would be four different digits. Let's see.Suppose we have the number 1,7,8,9,9,9,9,9. That's 17899999. Sum:1+7+8+5*9=16+45=61. Not enough. If we replace a 9 with a 10, invalid. Alternatively, replace a 9 with an 8 and add 1 to another. 1,7,8,8,9,9,9,10. Not valid.Alternatively, 1,7,9,9,9,9,9,9 is 17999999. Which is correct.Alternatively, if we use a different arrangement. Let's think outside the box. What if we use 19999999. Sum:1+8*9=1+72=73. Too much. Subtract 11. How? Replace two 9s with 8 and 7. 1,9,9,9,9,9,8,7. Arrange:1,7,8,9,9,9,9,9. Digit sum:1+7+8+5*9=16+45=61. Still need 1 more. Not helpful.Alternatively, 1,9,9,9,9,9,9,8. Sum:1+6*9+8=1+54+8=63. Subtract 1: replace a 9 with an 8. 1,8,9,9,9,9,9,8. Sum:1+2*8+5*9=1+16+45=62. Number is 18899999. Which is larger than 17999999.So, after exploring these possibilities, it seems that 17999999 is the smallest number that meets the criteria: digit sum of 62, uses at least three different digits (1,7,9), and is the smallest numerical value.But wait, before finalizing, let me check another possibility: can we have a number like 2699999? Wait, that's seven digits. Digit sum:2+6+5*9=8+45=53. No. Not enough.Alternatively, 3599999. Digit sum:3+5+5*9=8+45=53. Still no.Alternatively, 4499999. Digit sum:4+4+5*9=8+45=53. Still low.Alternatively, 5599999. 5+5+5*9=10+45=55. Still no.Wait, seems that all seven-digit numbers with three different digits can't reach 62. Therefore, the minimal number is indeed an eight-digit number, and the smallest one is 17999999.But wait, I just thought of something. What if we use a 9, 8, and a 5 in the number? For example, 5999998. Let's check digit sum:5+7*9+8=5+63+8=76. Too much.Alternatively, 5899999. Digit sum:5+8+5*9=13+45=58. Not enough.Alternatively, 5699999. Sum:5+6+5*9=11+45=56. Nope.Alternatively, 8399999. Digit sum:8+3+5*9=11+45=56. Not enough.Alternatively, 7499999. Sum:7+4+5*9=11+45=56. Still low.Alternatively, 9299999. Sum:9+2+5*9=11+45=56. No.Hmm, none of these seven-digit numbers work. So eight-digit is necessary. Therefore, I think 17999999 is the answer. But let me check once more for any oversight.Wait, let's verify the digit sum of 17999999:1 (first digit) +7 (second) +9+9+9+9+9+9 (last six digits).Calculating:1 +7 =8Six 9s: 6*9=54Total:8 +54=62. Correct.Digits used:1,7,9. Three different digits. Check.Is there a smaller eight-digit number with three different digits summing to 62?If we can find a number starting with 1, followed by a digit less than 7, but as we saw earlier, that's impossible because the remaining digits can't compensate for the sum. For example, 1,6,... requires the remaining six digits to sum to 55, which is impossible. Similarly for lower digits.Thus, 17999999 is indeed the smallest possible number.Final AnswerThe smallest number is boxed{17999999}.</think>To find the smallest number with a digit sum of 62 using at least three different digits, we start by considering the usual strategy of using as many 9s as possible. However, this approach only uses two different digits (8 and 9), so we need to adjust it to include a third digit.1. Initial Approach with Two Digits: - The digit sum of 62 can be achieved with the number 8999999 (one 8 and six 9s), but this uses only two different digits.2. Introducing a Third Digit: - We need to modify the number to include a third digit. By distributing the necessary sum reduction across the digits, we can introduce a third digit while maintaining the digit sum of 62.3. Exploring Eight-Digit Numbers: - We start with eight digits, as seven digits with three different digits are not feasible. The maximum digit sum for eight digits is 72 (eight 9s). We need to reduce this sum by 10 to reach 62. - By replacing one 9 with a 7 and another 9 with an 8, we get the digits 1, 7, and 9. This results in the number 17999999.4. Verification: - The number 17999999 has digits 1, 7, and 9, which are three different digits. - The digit sum is calculated as (1 + 7 + 6 times 9 = 1 + 7 + 54 = 62), which meets the requirement. - Arranging the digits in ascending order gives the smallest possible number, which is 17999999.After verifying all possibilities and ensuring the number is the smallest possible with the required digit sum and three different digits, the final answer is:[boxed{17999999}],<think>Okay, let's see. I need to find the smallest number that has a digit sum of 62, and it has to use at least three different digits. Hmm, how do I approach this?First, I remember that to get the smallest number possible, you want the number to have the least number of digits, and the digits should be arranged in ascending order. But wait, no, actually, the smallest number would have the digits arranged in ascending order from left to right, right? So like, the smaller digits should be on the left. Wait, no, actually, to make the number as small as possible, you want the higher place values (like the leftmost digits) to be as small as possible. That way, the number doesn't get too big just because the first digit is large.So, if we want the smallest number with a digit sum of 62, we need to use the largest digits possible towards the end to minimize the number of digits. Wait, but if we use larger digits, we can get a higher digit sum with fewer digits. So maybe that's conflicting with the idea of using different digits?Wait, let me think again. The digit sum is fixed at 62, so we need digits that add up to 62. To make the number as small as possible, we need the number of digits to be as few as possible because a number with fewer digits is smaller than one with more digits. But at the same time, we need to arrange the digits in ascending order (smallest digits first) to keep the number small.But there's a catch here: if we use larger digits, we can reach the digit sum faster (with fewer digits), which would result in a smaller number. But if we use too many large digits, maybe we can't use three different digits. Hmm.Wait, the problem says that at least three different digits must be used. So the number must have at least three distinct digits. So, I can't just use all 9s and a 5, for example, because that's two digits. I need at least three different digits.So, first, let's figure out the minimal number of digits required for a digit sum of 62. If we use as many 9s as possible, which is the largest digit, then we can get the minimal number of digits. Let's see: 62 divided by 9 is 6 with a remainder of 8. So, 6 nines would give 54, and then we have 8 left. So that would be a 7-digit number: 8 followed by six 9s: 8999999. But this uses only two digits: 8 and 9. So that's two different digits, which doesn't satisfy the requirement of three different digits. So we need to adjust.So, how can we introduce a third digit? Let's think. Instead of having a single 8 and six 9s, maybe we can split the 8 into smaller digits. For example, if we use a 7 and a 1 instead of an 8. Then the digits would be 7, 1, and nine 9s. Wait, but that would increase the number of digits. Wait, original was 7 digits (8 and six 9s). If we split 8 into 7 and 1, then we have 1, 7, and six 9s. That's 1 + 7 + 6*9 = 1 + 7 + 54 = 62. So that's three different digits: 1, 7, 9. The number would be 1 followed by 7 and then six 9s. But arranging them in ascending order would mean the digits go from smallest to largest. So the number would be 1, 7, 9, 9, 9, 9, 9, 9. But that's 8 digits. Wait, but the original number was 7 digits: 8 followed by six 9s. So splitting 8 into 1 and 7 increases the number of digits by one, making the number longer. Therefore, the number becomes 17799999 vs. 8999999. But 17799999 is 8 digits, and 8999999 is 7 digits, so 8999999 is smaller. But 8999999 only uses two digits, so it's invalid. So we need to find a way to have three different digits without increasing the number of digits too much.Alternatively, perhaps we can replace two of the 9s with other digits. Let's see. Let's take the original 8999999, which is 7 digits. To introduce a third digit, maybe replace one of the 9s with a different digit. Let's see: if we subtract 1 from one of the 9s to make it 8, then we have two 8s and five 9s, but that's still only two different digits. If we subtract 2 from a 9 to make it 7, then we have 8, 7, and five 9s. So that's three different digits. The digit sum would be 8 + 7 + 5*9 = 8 + 7 + 45 = 60. Wait, but we need 62. So that's two less. Hmm.Wait, maybe that approach isn't correct. Let's start over.We need a digit sum of 62 with at least three different digits. Let's try to use as many 9s as possible but adjust to get the third digit.Suppose we have six 9s: 6*9 = 54. Then we need 62 - 54 = 8 more. So if we have an 8, that gives 8 + 54 = 62, but that's two digits. So we need to split the 8 into two digits that sum to 8, but also introduce a third digit. For example, 7 and 1. Then we have digits 7, 1, and six 9s. The total digits would be 1, 7, 9, 9, 9, 9, 9, 9. That's 8 digits. To arrange this as the smallest number, we need to sort the digits in ascending order: 1, 7, 9, 9, 9, 9, 9, 9. So the number is 17799999. But this is 8 digits. Alternatively, if we split the 8 into different digits, like 6 and 2, we would have digits 6, 2, and six 9s. Then the sorted number would be 2, 6, 9, 9, 9, 9, 9, 9, which is 26999999. That's still 8 digits, but starting with 2, which is larger than 1. So 17799999 is smaller.Alternatively, splitting 8 into 5 and 3: 5, 3, six 9s. Sorted: 3,5,9,9,9,9,9,9: 35999999. Still larger.So the smallest one here is 17799999. But is there a way to have three different digits with fewer digits? For example, 7 digits? Let's check.If we have seven digits. Let's try to have three different digits. Let's say we use five 9s, one 8, and one 5. Then the sum would be 5*9 + 8 +5 = 45 +8 +5=58. Not enough.Wait, need 62. So 62 divided by 9 is 6 with remainder 8, so six 9s and an 8. But that's two digits. To get three digits, we need to replace some 9s and the 8 with other digits such that the total sum remains 62 but we have three different digits.Let me try replacing one of the 9s with a lower digit. Let's say instead of six 9s and one 8 (total 62), we can replace one 9 with an 8 and one 8 with a 7. Wait, that would be five 9s, two 8s, and a 7? Wait, maybe this is getting confusing. Let me think step by step.Original: six 9s (54) + one 8 = 62. Now, if we want to introduce a third digit, we need to replace some digits. Let's say we replace one 9 with a 9 - x and the 8 with 8 + x, such that the total sum remains the same. Wait, but that won't change the sum. Wait, maybe instead, we need to decrease one digit and increase another. Wait, maybe replace a 9 with a lower digit and compensate by increasing another digit. But since 9 is the highest digit, we can't increase beyond 9. Hmm.Alternatively, maybe split the 8 into two digits. For example, instead of 8 and six 9s, use 7 and 1 and five 9s. Wait, that's total digits: 1,7, five 9s. Which is 7 digits? Wait, 1 +7 +5*9=1+7+45=53. Wait, that's not 62. Wait, no, that's wrong. Wait, original sum was 8 + 6*9 = 8 +54=62. If we split the 8 into 7 and1, then we have 7 +1 +6*9=7+1+54=62. But that's 1,7, six 9s: total digits 8. So number is 17799999. But this is 8 digits. The original was 7 digits. So this is longer. So is there a way to have three different digits with 7 digits?Suppose we have seven digits. Let’s use as many 9s as possible. Let’s say five 9s: 5*9=45. Then we need 62-45=17 more. So we have two digits left. If we use 9 and 8, but that's two digits. Wait, 17 split into two digits. The maximum digit is 9, so 9 and 8 sum to 17. So digits would be five 9s, one 9, one 8. Wait, that's six 9s and one 8: which is two digits. So again, two digits. To get three digits, maybe split the 17 into three digits. Wait, but we have seven digits total. If five 9s, then two remaining digits. To split 17 into two digits but with three different digits. Wait, impossible. Because two digits can only be two different digits at most. So to get three different digits in seven digits, we need to have at least three different digits in the seven digits.So, perhaps use five 9s, and two digits that sum to 17. For example, 9 and 8, but that's two digits. To have three different digits, we need to replace one of the 9s and the 8 with another digit.Wait, let's try this: original digits (six 9s and one 8). To get three different digits, replace one 9 with an 8 and replace the original 8 with a 7. Then we have five 9s, two 8s, and one 7. Wait, but that's three different digits. The sum would be 5*9 + 2*8 +7=45+16+7=68, which is too much. So that's over.Alternatively, maybe subtract 1 from one of the 9s and add 1 to the 8. So instead of six 9s and one 8 (sum 62), we have five 9s, one 8, and one 9-1=8, and 8+1=9. Wait, that doesn't change anything. Hmm.Alternatively, maybe decrease a 9 by 1 to make it 8, and then decrease the original 8 by 1 to make it 7, so that we have five 9s, one 8, one 7, and the original 8 is now a 7. Wait, the sum would be 5*9 +8 +7=45 +15=60. That's 2 less than 62. So we need to compensate by adding 2 somewhere. Maybe increase two of the digits by 1. But we can't go over 9. So this seems complicated.Alternatively, perhaps instead of six 9s and one 8, we can have five 9s, one 9, one 8, and replace one of the 9s with a different digit. Wait, this isn't making sense.Let me approach this differently. Let's consider that we need at least three different digits. Let's use three different digits: 9, 8, and another digit. Let's say we use as many 9s as possible, then 8s, and then a smaller digit.Suppose we have x 9s, y 8s, and z of another digit, say a.So the equation would be: 9x + 8y + a*z = 62.And we need x + y + z to be as small as possible, since fewer digits make a smaller number. Also, we need at least three different digits, so z >=1, and a <8 or something.Wait, but since we want the smallest number, we need the digits to be as small as possible in the higher places. So perhaps we need to have the smaller digits towards the left. So, maybe we need to have the smallest digit first, then the next, etc.Alternatively, if we use three different digits, perhaps the digits could be 9, 8, and 7, but maybe a smaller digit would allow for a smaller number.Wait, but if we use a smaller digit, like 1, we might need more digits, but the number could be smaller. For example, 17799999 vs. a number with digits 9,8,7 but arranged to be smaller.Wait, maybe there's a way to have three different digits with the same number of digits as 8999999 (7 digits) but with three different digits. Let's check.If we have 7 digits, with three different digits, sum 62.Let’s denote the digits as a, b, c, where a < b < c. Let’s say we have x of a, y of b, z of c. Then x + y + z =7, and a x + b y + c z =62.We need to find the minimal number (which would be the digits arranged in ascending order). So to minimize the number, we need the leftmost digit to be as small as possible.So let's assume the smallest digit is 1. Then, we need to have 1 in the leftmost position, followed by other digits.Let’s try:Suppose we have one 1, and then six digits of 9 and 8.So sum is 1 + 6*9 = 1 +54=55. Not enough. 1 +5*9 +8=1 +45 +8=54. Still not enough. 1 +4*9 +2*8=1 +36 +16=53. Still not.Wait, maybe 1 + 6*9 + something. Wait, but 1 +6*9=55. To reach 62, we need 7 more. So, if we have one 1, five 9s, and one 8. Sum is 1 +5*9 +8=1 +45 +8=54. Still 8 short. Hmm.Alternatively, two 1s and five 9s: 2 +45=47. Not close.Alternatively, maybe 1, 8, and 9. Let's see.Suppose we have one 1, one 8, and five 9s: sum is 1 +8 +45=54. Still 8 short. Need 8 more. So maybe replace two of the 9s with two 8s. So 1, three 8s, and four 9s. Sum is 1 +24 +36=61. Still one short.Alternatively, 1, four 8s, and three 9s: 1 +32 +27=60. Still two short.Alternatively, 1, five 8s, and two 9s: 1 +40 +18=59. Still three short.This approach isn't working. Maybe starting with 1 is not feasible because we need a higher digit sum.Let me try starting with a higher digit. Let's say the smallest digit is 2.Then, with digits 2, 8, 9.Suppose one 2, one 8, five 9s: sum=2 +8 +45=55. Not enough.Two 2s, one 8, four 9s: 4 +8 +36=48. Still low.Three 2s, one 8, three 9s: 6 +8 +27=41. Nope.Alternatively, maybe 2, 9, and another digit.Wait, maybe this is getting too convoluted. Let's try a different approach.If we have to use three different digits, then maybe the digits could be 9, 8, and 7. Let's see.Suppose we use as many 9s as possible, then 8s, and then 7s.Let’s compute how many 9s we can have. 62 divided by 9 is 6 with remainder 8. So six 9s would give 54, remainder 8. So if we replace one 9 with an 8 + the remainder becomes 8 +8=16? Wait, not sure.Wait, original six 9s: sum 54. Need 8 more. If we replace one 9 with a 9 +8=17, but digits can't exceed 9. So instead, maybe replace one 9 with a 9 - x and add x to the remainder.Wait, if we reduce one 9 by 1 to make it 8, then the total sum becomes 54 -1 =53, and the remainder needed is 62 -53=9. Then we can add a 9. So that would be five 9s, one 8, and one 9: which is six 9s and one 8. Still two digits. Not helpful.Alternatively, reduce two 9s by 1 each, making them 8s. Then total sum is 54 -2=52. Remainder needed is 62 -52=10. Add a 10? Not possible. So split 10 into two digits, 9 and 1. So then we have five 9s, two 8s, one 1. But the digits would be 1,8,9. So three different digits. Total digits: five 9s, two 8s, one 1: total digits 5+2+1=8. The number would be 1 followed by two 8s and five 9s: sorted ascendingly. Wait, but arranging the digits in ascending order: 1,8,8,9,9,9,9,9. So the number is 18899999. Which is eight digits, but with three different digits. However, earlier we had 17799999 with eight digits as well. Comparing 17799999 and 18899999, 177 is smaller than 188, so 17799999 is better.But perhaps there's a way to make it with seven digits. Let's check.If we need three different digits in seven digits, with a total sum of 62. Let’s try:Let’s let the digits be a, b, c with a < b < c. Let's try to maximize the number of 9s.Suppose we have five 9s, one 8, and one 5. Then sum is 5*9 +8 +5=45+8+5=58. Not enough.Five 9s, one 7, one 6: 45+7+6=58. Still not.Six 9s and one 8: 62, but only two digits.What if we have four 9s, two 8s, and one 7: 4*9 +2*8 +7=36+16+7=59. Still not.Four 9s, one 8, two 7s: 36+8+14=58.Three 9s, five 8s: 27+40=67. Too much.Three 9s, four 8s, one 7: 27+32+7=66. Still too much.Wait, maybe this trial and error isn't effective. Let me set up equations.Let’s denote the number of 9s as x, 8s as y, and another digit as z. Let's say the third digit is d, where d <8.So total sum: 9x +8y +d*z=62.Total digits: x + y + z =n, which we want to minimize.But also, x,y,z >=1 because we need three different digits.Let’s suppose z=1 (only one instance of the third digit). Then, 9x +8y +d=62, and x + y +1 =n.We want to maximize x and y to minimize n.Let’s try d=7 (the largest possible below 8). Then:9x +8y +7=62 =>9x +8y=55.Looking for integer solutions where x,y >=1.Let’s solve for y: 8y=55 -9x => y=(55 -9x)/8.We need (55 -9x) divisible by 8 and positive.Trying x=1: 55-9=46. 46/8=5.75. Not integer.x=2: 55-18=37. 37/8=4.625. Not integer.x=3: 55-27=28. 28/8=3.5. Not integer.x=4: 55-36=19. 19/8=2.375.x=5: 55-45=10. 10/8=1.25.x=6:55-54=1. 1/8=0.125. Not integer.No solutions with z=1 and d=7.Next, try d=6.9x +8y +6=62 =>9x +8y=56.y=(56-9x)/8.x=1: (56-9)/8=47/8=5.875.x=2: (56-18)/8=38/8=4.75.x=3:56-27=29/8=3.625.x=4:56-36=20/8=2.5.x=5:56-45=11/8=1.375.x=6:56-54=2/8=0.25.No integer solutions.d=5:9x +8y +5=62 =>9x +8y=57.y=(57 -9x)/8.x=1:48/8=6. y=6. So x=1, y=6, z=1. Total digits=1+6+1=8. So the digits are 1 9, 6 8s, 1 5. Arranged in order: 5,8,8,8,8,8,8,9. Wait, but the digits are 5,8,9. Wait, no, actually, the digits would be one 9, six 8s, and one 5. So sorted ascendingly: 5,8,8,8,8,8,8,9. So the number is 58888889. But this has three different digits, sum 5 +6*8 +9=5 +48 +9=62. But this is 8 digits. Comparing to earlier 17799999, which is 17799999 vs 58888889. 17799999 is smaller because it starts with 1.Wait, but 58888889 starts with 5, which is larger. So 17799999 is smaller. So this is worse.d=4:9x +8y +4=62 =>9x +8y=58.y=(58 -9x)/8.x=2:58-18=40/8=5. So x=2, y=5. z=1. Digits: two 9s, five 8s, one 4. Total digits=8. Sorted: 4,8,8,8,8,8,9,9. Number: 488888899. Again, starts with 4, which is higher than 1, so 17799999 is better.d=3:9x +8y +3=62 =>9x +8y=59.y=(59-9x)/8.x=3:59-27=32/8=4. So x=3, y=4, z=1. Digits: three 9s, four 8s, one 3. Sorted: 3,8,8,8,8,9,9,9. Number: 388888999. Again, starts with 3, which is worse than 1.d=2:9x +8y +2=62 =>9x +8y=60.y=(60 -9x)/8.x=4:60 -36=24/8=3. So x=4, y=3, z=1. Digits: four 9s, three 8s, one 2. Sorted: 2,8,8,8,9,9,9,9. Number: 28889999. Starts with 2, worse than 1.d=1:9x +8y +1=62 =>9x +8y=61.y=(61 -9x)/8.x=5:61 -45=16/8=2. So x=5, y=2, z=1. Digits: five 9s, two 8s, one 1. Sorted: 1,8,8,9,9,9,9,9. Number: 18899999. Wait, same as before. So this is 18899999, which is same as if we split the 8 into 1 and7 earlier.Wait, so with d=1, we get 18899999, which is 8 digits. But if we split the original 8 into 7 and1, we get 17799999. Which is better? 1 followed by two 7s vs. 1 followed by two 8s. 177 is smaller than 188, so 17799999 is better.So, even when trying different d's, the minimal number we get is 17799999 with 8 digits. But is there a way to get a smaller number with three different digits but more digits? Probably not, since more digits would make the number larger. Wait, but maybe there's a way with same number of digits but a smaller arrangement.Wait, 17799999 vs. other 8-digit numbers. For example, if we have digits 1, 7, 9. Let's see, can we arrange them in a way that some of the 7s are moved to the right? But since we have to sort the digits in ascending order, the leftmost digit must be the smallest. So 1 first, then 7, then 9s. So 17799999 is indeed the smallest possible arrangement with those digits.But maybe we can find another combination of digits with three different digits that sum to 62 but allows for a smaller number with same or fewer digits.Wait, what if instead of using 1, 7, and 9, we use smaller digits? For example, 1, 6, and 9.Let's try. Suppose we use one 1, one 6, and the rest 9s. Then the total sum would be 1 +6 + (n-2)*9=62. So 7 +9*(n-2)=62. 9*(n-2)=55. 55 is not divisible by 9. 55/9≈6.11. So n-2≈6.11, so n≈8.11. So n=9. So 1,6, and seven 9s. Sum:1+6+63=70. Too much. Not helpful.Alternatively, maybe two 1s, one 6, and six 9s: 2 +6 +54=62. That works. So digits: two 1s, one 6, six 9s. Total digits:9. Sorted: 1,1,6,9,9,9,9,9,9. The number is 116999999. That's 9 digits, which is longer than 17799999 (8 digits), so 17799999 is smaller.Alternatively, using 1, 5, and 9.Suppose one 1, one 5, rest 9s. Sum:1+5 +9*(n-2)=62. 6 +9*(n-2)=62. 9*(n-2)=56. 56/9≈6.22. n-2=7, so n=9. 1,5, seven 9s: sum=1+5+63=69. Too much.Alternatively, two 1s, two 5s, and five 9s: sum=2 +10 +45=57. Not enough.This approach isn't working. Let's try another angle.Perhaps using three different digits where one of them is 9 and another is 8, and the third is a smaller digit, arranged in such a way that the total digits are still minimal.Wait, let's think of how to split the required sum into 9s, 8s, and a third digit.If we start with six 9s (sum 54) and an 8 (sum 62), but we need to replace some digits to get a third digit.For example, replace one 9 and the 8 with two digits: one 7 and one 9. Wait, that just swaps 8 for 7 and keeps the same number of 9s. Wait, sum would be 5*9 +8 +7=45+8+7=60. Not enough.Alternatively, replace two 9s and the 8 with one 7, one 8, and one 9. Wait, confused.Alternatively, subtract 1 from two 9s to make them 8s, and add 2 to the original 8 to make it 10, which is invalid. Not helpful.Alternatively, subtract 1 from a 9 to make it 8, subtract 1 from the original 8 to make it 7, so we have five 9s, one 8, one 7. Sum is 5*9 +8 +7=45+15=60. Still need 2 more. Maybe add 1 to two of the digits. But the 9s can't go higher. So maybe increase the 7 to 9? No, that would just give back the original digits.This seems stuck. Maybe it's not possible to have three different digits with 7 digits. Therefore, the minimal number of digits required is 8, using digits 1,7,9, as in 17799999.Wait, but let's check another possibility. What if we use three 9s, three 8s, and two 7s. Let's calculate the sum: 3*9 +3*8 +2*7=27+24+14=65. Too much.Alternatively, four 9s, two 8s, and two 7s: 36+16+14=66. Still too much.Alternatively, five 9s, one 8, two 6s: 45 +8 +12=65. Still over.Alternatively, five 9s, one 7, two 6s: 45 +7 +12=64. Still over.Hmm. Maybe subtracting digits. Let's try four 9s, three 8s, and one 5: 36 +24 +5=65. Still too much.This is getting frustrating. Maybe 17799999 is indeed the minimal.Wait, but let's think outside the box. What if instead of starting with 1, we start with a higher digit but use fewer digits. Wait, but more digits make the number larger. For example, 17799999 is 8 digits, but if we can have a 7-digit number with three different digits, that would be smaller. But earlier attempts suggest it's not possible. Let me verify again.Suppose 7 digits. Let's try to have three different digits summing to 62.Let’s let’s have x 9s, y 8s, z 7s, where x + y + z =7.Sum:9x +8y +7z=62.Looking for non-negative integers x,y,z >=1.Let’s express as:9x +8y +7z=62 and x+y+z=7.Let’s substitute z=7 -x -y into the equation:9x +8y +7(7 -x -y)=62.Simplify:9x +8y +49 -7x -7y=62.2x + y +49=62.2x + y=13.So y=13 -2x.Since x and y must be at least 1 and z=7 -x -y >=1.So:x >=1,y=13 -2x >=1 =>13 -2x >=1 =>2x <=12 =>x <=6,z=7 -x -y=7 -x -(13 -2x)=7 -x -13 +2x= x -6 >=1 =>x -6 >=1 =>x >=7.But x <=6 from above. Contradiction. So no solution.Therefore, it's impossible to have three different digits in a 7-digit number summing to 62. Thus, the minimal number of digits is 8.Therefore, the smallest number is 17799999.But let's confirm. Let's check another combination for 8 digits.Suppose we have two 1s, one 8, and five 9s. Sum:2 +8 +45=55. Not enough.Three 1s, one 8, four 9s:3 +8 +36=47. Not enough.Alternatively, one 2, one 8, six 9s:2 +8 +54=64. Over.Wait, no. Wait, 8 digits: for example, one 1, one 7, six 9s:1 +7 +54=62. Exactly. That's the 17799999. So this seems to be the minimal.Alternatively, what if we use digits like 1, 6, and 9. For example, one 1, two 6s, five 9s: sum=1 +12 +45=58. Not enough. Three 6s:1 +18 +45=64. Over.Alternatively, one 1, one 5, six 9s:1 +5 +54=60. Close.Alternatively, one 1, two 5s, five 9s:1 +10 +45=56. Not enough.So no, 17799999 is still the only combination that works.Wait, another idea: maybe using three different digits where the third digit is higher than 9? No, digits can only be 0-9.Alternatively, using a 0? Wait, but 0 can't be in the leading position. If we use a 0, we have to place it somewhere else. But using 0 would allow for more flexibility. Let's try.For example, using digits 0, 9, and 8. Let's see.If we have one 0, then the remaining digits need to sum to 62. Let's try:Suppose one 0, six 9s, one 8: sum=0 +54 +8=62. But that uses digits 0,8,9. Three different digits. The number would be 0 followed by six 9s and one 8: but leading zero is not allowed. So we have to arrange the digits without leading zero. So the smallest number would be 8 followed by 0 and six 9s: 8099999. But this is two different digits (8 and 9) plus a zero, but zeros are allowed as non-leading digits. Wait, but does the problem allow zero as one of the three different digits? The problem says "at least three different digits are used". So digits 0,8,9 are three different digits. So the number could be 8099999. Let's check.Number: 8099999. Digit sum:8+0+9+9+9+9+9=8+0+54=62. Yes. Digits used: 0,8,9. Three different digits. The number is 8099999, which is 7 digits. Wait, this is 7 digits, which is shorter than 17799999. So this would be smaller.But hold on, 8099999 vs 17799999. Which is smaller? Let's compare:8099999 is 7 digits: 8 followed by 0 and five 9s.17799999 is 8 digits: 1 followed by 7 and six 9s.Wait, but 8099999 is a 7-digit number, which is smaller than any 8-digit number. So 8099999 is smaller. But does it meet the requirement? It uses digits 0,8,9—three different digits. Yes. So this seems like a valid solution.But why didn't I think of this earlier? Because I assumed that introducing a 0 would require more digits, but here, it actually allows fewer digits.So the key insight was that using a 0 allows us to keep the number of digits to 7, same as the original two-digit solution, but introducing a 0 as the third digit.So, the digits are 8,0,9. The number is 8099999. Let's check digit sum:8+0+9*5=8+0+45=53. Wait, wait, no. Wait, 8099999 has digits: 8,0,9,9,9,9,9. So that's seven digits. The sum is 8 +0 +9*5=8 +0 +45=53. Wait, that's not 62. What?Oh no! Wait, there's a mistake here. If it's seven digits: the digits are one 8, one 0, and five 9s. Sum:8 +0 +5*9=8 +0 +45=53. Not 62. Oops, my mistake. So that's not correct.So that approach doesn't work. Then, how to get a digit sum of 62 with a 0?Let me try again.Suppose we want to include a 0. Let's say we have one 0, and the rest digits sum to 62. So total digits would be n, with one 0 and (n-1) digits summing to 62. To minimize n, we need to maximize the digits. So using as many 9s as possible.So 62 divided by 9 is 6 with remainder 8. So six 9s sum to 54, plus 8 gives 62. So total digits would be six 9s, one 8, one 0: total digits 8. So the number is 0,8,9,9,9,9,9,9. But leading zero is not allowed. So arrange the digits to have the smallest number without leading zero. The smallest number is 80999999. Wait, but that's eight digits: 8,0,9,9,9,9,9,9. Sum:8 +0 +6*9=8 +0 +54=62. Yes. So digits used:0,8,9. Three different digits. The number is 80999999. Which is eight digits. Compared to 17799999, which is also eight digits. Which is smaller?Comparing 80999999 and 17799999:First digit:8 vs 1. 1 is smaller, so 17799999 is smaller.So even though 80999999 uses a 0, the leading digit is 8, which makes it larger than 1... So 17799999 is smaller.But wait, if we can arrange the digits with 0 somewhere else but starting with a smaller digit. For example, if we have a 1, a 0, and the rest 9s. Let's see.Sum would be1 +0 +9*(n-2)=62. So 1 +0 +9*(n-2)=62 =>9*(n-2)=61. 61 isn't divisible by 9. So no solution.If we have two 1s, one 0, and the rest 9s:2 +0 +9*(n-3)=62 =>9*(n-3)=60 =>n-3=60/9≈6.66. Not integer.Three 1s, one 0, rest 9s:3 +0 +9*(n-4)=62 =>9*(n-4)=59. Not divisible.Alternatively, other digits. Suppose we have one 7, one 0, rest 9s. Sum:7 +0 +9*(n-2)=62 =>9*(n-2)=55. Not divisible.One 8, one 0, rest 9s:8 +0 +9*(n-2)=62 =>9*(n-2)=54 =>n-2=6 =>n=8. Which brings us back to 80999999. So again, same as before.Alternatively, two 8s, one 0, rest 9s:16 +0 +9*(n-3)=62 =>9*(n-3)=46. Not divisible.So no luck.Alternatively, include more digits. For example, two 0s, but leading zeros aren't allowed. So maybe one 0, one 8, and six 9s, but that's the same as before.So 80999999 is eight digits and uses three different digits, but it's larger than 17799999.Therefore, 17799999 is still the smallest.Wait, but is there another combination with three different digits in eight digits that sum to 62 and is smaller than 17799999?For example, using 1,6,9.Suppose we have two 1s, one 6, and five 9s: sum=2 +6 +45=53. Not enough.Three 1s, one 6, four 9s:3 +6 +36=45. No.Alternatively, one 1, two 6s, five 9s:1 +12 +45=58. Not enough.Alternatively, one 1, one 7, six 9s:1 +7 +54=62. That's the 17799999.Alternatively, one 2, one 7, six 9s:2 +7 +54=63. Over.Alternatively, one 1, one 8, six 9s:1 +8 +54=63. Over.So no, only 1,7,9 combination gives exactly 62.Another idea: use digits 1,7,9 in eight digits. Alternatively, maybe use more of the smaller digits to allow fewer 9s but keeping the sum. Wait, but more smaller digits would require more digits, which isn't helpful.Wait, let's try using more 7s. For example, two 7s, one 1, and five 9s: sum=14 +1 +45=60. Not enough.Three 7s, one 1, four 9s:21 +1 +36=58. Still no.Alternatively, one 1, three 7s, four 9s:1 +21 +36=58. Not enough.Not helpful.Alternatively, one 1, one 7, and six 9s:1 +7 +54=62. Exactly. Which is the 17799999.So, after exhausting all possibilities, it seems that 17799999 is the smallest number that meets the criteria: digit sum of 62, uses three different digits (1,7,9), and is the smallest possible number.But wait, another thought. What if we use digits like 1, 8, and 9 but arrange them with the 1 first, then 8, then 9s. For example, 18999999. Sum:1+8+7*9=1+8+63=72. Way over. Not good.Alternatively, 1, 8, and two 9s:1+8+18=27. Not relevant.Alternatively, 1 followed by multiple 8s and 9s. But the sum would not reach 62 without more digits.So, no. Conclusion: 17799999 is the answer.Final AnswerThe smallest number is boxed{17799999}.</think>To find the smallest number with a digit sum of 62 using at least three different digits, we need to consider the following steps:1. Maximizing the use of the largest digit (9) to minimize the number of digits. However, using only 9s and another digit would not satisfy the requirement of three different digits.2. Introducing a third digit by splitting the remaining sum into smaller digits while keeping the total number of digits as small as possible.3. Arranging the digits in ascending order to form the smallest possible number.Key steps and reasoning:- The initial approach using six 9s and one 8 (summing to 62) only uses two digits, which is invalid.- Splitting the 8 into smaller digits (1 and 7) introduces a third digit. This results in the digits 1, 7, and nine 9s, but we need to check the total number of digits and their arrangement.- The digits 1, 7, and nine 9s sum to 62 and use three different digits. Arranging these digits in ascending order gives the number 17799999.- Checking other combinations with a third digit (including 0) either resulted in more digits or a larger number when arranged properly.After thorough consideration and verification, the smallest number with a digit sum of 62 using at least three different digits is:[boxed{17799999}]]

🔑:Answer: 17999999.Solution: 62=9 cdot 6+8. That is, the number is at least a seven-digit number. But if it is a seven-digit number, then in its decimal representation there are exactly 6 nines and 1 eight. According to the problem, the number must contain at least three different digits, so it is at least an eight-digit number. The first digit is no less than 1, and the sum of the first and second digits is no less than 8. By splitting 8 into 1+7, we get the answer 17999999.

🧠:[<think>Alright, let's see. I need to find the maximum and minimum values of z given that x, y, z are real numbers with x + y + z = 5 and xy + yz + zx = 3. Hmm, okay. So, the problem gives me two equations involving three variables. Since there are three variables and only two equations, it seems like z can vary within some range, and I need to find that range. First, maybe I can express x and y in terms of z, then substitute into the equations. Let me try that. If I take the first equation, x + y + z = 5, then I can write x + y = 5 - z. Similarly, the second equation is xy + yz + zx = 3. Let me see if I can express xy in terms of z as well. Expanding xy + yz + zx, that's equal to xy + z(x + y). Since I already know that x + y = 5 - z, substitute that in. So, xy + z(5 - z) = 3. Therefore, xy = 3 - z(5 - z) = 3 - 5z + z². So, now I have expressions for both x + y and xy in terms of z. Hmm, this is starting to look like a quadratic equation. If I consider x and y as roots of a quadratic equation, then the sum of the roots is x + y = 5 - z, and the product is xy = 3 - 5z + z². So, the quadratic would be t² - (5 - z)t + (3 - 5z + z²) = 0. For x and y to be real numbers, the discriminant of this quadratic must be non-negative. Right, the discriminant D of a quadratic at² + bt + c is D = b² - 4ac. So, applying that here:D = [-(5 - z)]² - 4 * 1 * (3 - 5z + z²)= (5 - z)² - 4(3 - 5z + z²)Let me compute this step by step. First, expand (5 - z)²:(5 - z)² = 25 - 10z + z²Then compute 4*(3 - 5z + z²):4*(3 - 5z + z²) = 12 - 20z + 4z²Now subtract the second expression from the first:25 - 10z + z² - (12 - 20z + 4z²) = 25 - 10z + z² -12 + 20z -4z² = (25 -12) + (-10z +20z) + (z² -4z²)= 13 + 10z -3z²So, the discriminant D = -3z² +10z +13. For the quadratic to have real roots, we need D ≥ 0. So, set:-3z² +10z +13 ≥ 0Multiply both sides by -1 (which reverses the inequality):3z² -10z -13 ≤ 0Now, solve the quadratic inequality 3z² -10z -13 ≤ 0.First, find the roots of the equation 3z² -10z -13 =0.Using the quadratic formula:z = [10 ± sqrt(100 + 156)] / 6Because the discriminant here is D = (-10)^2 -4*3*(-13) = 100 + 156 = 256. sqrt(256) = 16.Therefore,z = [10 ±16]/6So, the two roots are:z = (10 +16)/6 = 26/6 = 13/3 ≈ 4.333...andz = (10 -16)/6 = (-6)/6 = -1So, the quadratic 3z² -10z -13 is a parabola opening upwards (since the coefficient of z² is positive) and it is ≤0 between its roots. Therefore, the inequality 3z² -10z -13 ≤0 holds when z is between -1 and 13/3.Therefore, the discriminant D is non-negative only when z is in [-1, 13/3]. Therefore, the possible values of z are between -1 and 13/3. Therefore, the minimum value of z is -1 and the maximum is 13/3.Wait, but let me confirm this. Let me check if when z is at these endpoints, x and y are real numbers. For example, when z =13/3, then x + y =5 -13/3 = (15 -13)/3 =2/3. Then, xy =3 -5*(13/3) + (13/3)^2. Let's compute that:xy =3 - (65/3) + (169/9) = convert to ninths: 27/9 -195/9 +169/9 = (27 -195 +169)/9 = (1)/9. So, xy=1/9. Then, x + y =2/3. So, the quadratic equation is t² - (2/3)t +1/9 =0. The discriminant here is (2/3)^2 -4*(1)*(1/9) =4/9 -4/9=0. So, discriminant is zero, so x and y are equal, both (2/3)/2=1/3. So, x=y=1/3. So, that's a valid solution. Similarly, when z=-1, then x + y =5 -(-1)=6. Then, xy=3 -5*(-1) + (-1)^2=3 +5 +1=9. So, x + y=6, xy=9. Then quadratic equation is t² -6t +9=0, which factors as (t -3)^2=0. So, x=y=3. So, that's also valid. Therefore, these are indeed attainable values when x and y are equal. So, the maximum value of z is 13/3 and the minimum is -1. Therefore, the answer should be maximum 13/3 and minimum -1.But let me just check if there are any other possible cases. Wait, but we derived this through substitution and using discriminant conditions. Since x and y have to be real numbers, this is the necessary and sufficient condition. Therefore, these are indeed the extremal values. Therefore, the answer is z can be as small as -1 and as large as 13/3. So, the maximum is 13/3 and the minimum is -1.Final AnswerThe maximum value of ( z ) is (boxed{dfrac{13}{3}}) and the minimum value is (boxed{-1}).</think>Given that ( x, y, z ) are real numbers, and ( x + y + z = 5 ) and ( xy + yz + zx = 3 ), we need to find the maximum and minimum values of ( z ).First, we express ( x + y ) and ( xy ) in terms of ( z ):- From ( x + y + z = 5 ), we get ( x + y = 5 - z ).- From ( xy + yz + zx = 3 ), we substitute ( x + y = 5 - z ) to get ( xy + z(5 - z) = 3 ). Solving for ( xy ), we get ( xy = 3 - 5z + z^2 ).We then consider ( x ) and ( y ) as roots of the quadratic equation ( t^2 - (5 - z)t + (3 - 5z + z^2) = 0 ). For ( x ) and ( y ) to be real, the discriminant of this quadratic must be non-negative:[Delta = (5 - z)^2 - 4(3 - 5z + z^2)]Expanding and simplifying the discriminant:[Delta = 25 - 10z + z^2 - 12 + 20z - 4z^2 = -3z^2 + 10z + 13]We require ( -3z^2 + 10z + 13 geq 0 ). Solving the quadratic inequality ( 3z^2 - 10z - 13 leq 0 ), we find the roots using the quadratic formula:[z = frac{10 pm sqrt{100 + 156}}{6} = frac{10 pm 16}{6}]This gives the roots ( z = frac{26}{6} = frac{13}{3} ) and ( z = frac{-6}{6} = -1 ). The quadratic ( 3z^2 - 10z - 13 ) is non-positive between these roots, so ( z ) must lie between (-1) and (frac{13}{3}).Checking the endpoints:- When ( z = frac{13}{3} ), ( x + y = frac{2}{3} ) and ( xy = frac{1}{9} ), giving real solutions ( x = y = frac{1}{3} ).- When ( z = -1 ), ( x + y = 6 ) and ( xy = 9 ), giving real solutions ( x = y = 3 ).Thus, the maximum value of ( z ) is (boxed{dfrac{13}{3}}) and the minimum value is (boxed{-1}).,<think>Okay, so I need to find the maximum and minimum values of z given that x, y, z are real numbers satisfying x + y + z = 5 and xy + yz + zx = 3. Hmm, let's see. I remember that in problems involving symmetric equations with three variables, sometimes you can reduce it to two variables by using substitution. Let me try that.First, from the first equation, x + y + z = 5, so if I solve for x + y, that gives x + y = 5 - z. Similarly, the second equation is xy + yz + zx = 3. Let me see if I can express xy in terms of z. If I expand xy + yz + zx, it's equal to xy + z(x + y). Since I already know x + y = 5 - z, substitute that in. So the second equation becomes xy + z(5 - z) = 3. Therefore, xy = 3 - z(5 - z) = 3 - 5z + z².Now, I have x + y = 5 - z and xy = z² - 5z + 3. Wait a second, these are the sum and product of two numbers x and y. So, x and y are roots of the quadratic equation t² - (5 - z)t + (z² - 5z + 3) = 0. Since x and y are real numbers, the discriminant of this quadratic must be non-negative. The discriminant D is [ (5 - z) ]² - 4 * 1 * (z² - 5z + 3 ) ≥ 0.Let me compute that discriminant. First, expand (5 - z)²: 25 - 10z + z². Then subtract 4(z² - 5z + 3): so that's 25 -10z + z² -4z² +20z -12. Combine like terms:25 -12 = 13-10z +20z = 10zz² -4z² = -3z²So overall, the discriminant D = -3z² +10z +13 ≥0Therefore, -3z² +10z +13 ≥0Multiply both sides by -1 (which reverses the inequality):3z² -10z -13 ≤0Now, we can solve this quadratic inequality. First, find the roots of 3z² -10z -13 =0.Using the quadratic formula: z = [10 ± sqrt(100 + 156)] /6Because discriminant here is 100 +4*3*13 =100 +156=256. sqrt(256)=16.So z = [10 ±16]/6First root: (10 +16)/6=26/6=13/3≈4.333Second root: (10 -16)/6=(-6)/6=-1Therefore, the quadratic 3z² -10z -13 is a parabola opening upwards (since coefficient of z² is positive). So it is ≤0 between its roots. Hence, the solution to 3z² -10z -13 ≤0 is z ∈ [-1, 13/3].Therefore, z must be between -1 and 13/3. Therefore, the minimum value of z is -1 and the maximum value is 13/3.Wait, but let me check if this is correct. Let's verify.Suppose z = -1. Then x + y =5 - (-1)=6, and xy= (-1)^2 -5*(-1) +3=1 +5 +3=9. So the quadratic equation becomes t² -6t +9=0, which factors as (t -3)^2=0. So x = y =3. Then check if x + y + z=3+3+(-1)=5, which is correct. And xy + yz +zx=9 + (3*(-1) +3*(-1))=9 -3 -3=3. Correct.Similarly, if z=13/3. Then x + y=5 -13/3= (15 -13)/3=2/3. And xy=(13/3)^2 -5*(13/3)+3=169/9 -65/3 +3=169/9 -195/9 +27/9=(169 -195 +27)/9=(1)/9. So quadratic equation is t² - (2/3)t +1/9=0. Multiply through by 9:9t² -6t +1=0. Discriminant is 36 -36=0. So roots t=(6 ±0)/18=1/3. So x=y=1/3. Check x + y + z=1/3 +1/3 +13/3=15/3=5. Correct. And xy + yz +zx=(1/3)(1/3) + (1/3)(13/3) + (1/3)(13/3)=1/9 +13/9 +13/9=27/9=3. Correct. So both endpoints work.Therefore, the minimum value of z is -1 and maximum is 13/3. So that's the answer.Final AnswerThe minimum value of ( z ) is (boxed{-1}) and the maximum value is (boxed{dfrac{13}{3}}).</think>Given that ( x, y, z ) are real numbers satisfying ( x + y + z = 5 ) and ( xy + yz + zx = 3 ), we need to find the maximum and minimum values of ( z ).First, we express ( x + y ) and ( xy ) in terms of ( z ):[x + y = 5 - z][xy = 3 - z(5 - z) = z^2 - 5z + 3]Since ( x ) and ( y ) are roots of the quadratic equation ( t^2 - (5 - z)t + (z^2 - 5z + 3) = 0 ), the discriminant of this quadratic must be non-negative for ( x ) and ( y ) to be real:[(5 - z)^2 - 4(z^2 - 5z + 3) geq 0]Simplifying the discriminant:[25 - 10z + z^2 - 4(z^2 - 5z + 3) geq 0][25 - 10z + z^2 - 4z^2 + 20z - 12 geq 0][-3z^2 + 10z + 13 geq 0]Solving the quadratic inequality ( -3z^2 + 10z + 13 geq 0 ):[3z^2 - 10z - 13 leq 0]Finding the roots of the quadratic equation ( 3z^2 - 10z - 13 = 0 ):[z = frac{10 pm sqrt{256}}{6} = frac{10 pm 16}{6}][z = frac{26}{6} = frac{13}{3} quad text{and} quad z = frac{-6}{6} = -1]The quadratic ( 3z^2 - 10z - 13 ) is a parabola opening upwards, so it is non-positive between its roots ( -1 ) and ( frac{13}{3} ). Therefore, the values of ( z ) must lie in the interval ([-1, frac{13}{3}]).Verification for ( z = -1 ) and ( z = frac{13}{3} ) confirms that these values satisfy the original equations. Thus, the minimum value of ( z ) is (-1) and the maximum value is (frac{13}{3}).[text{Minimum value of } z: boxed{-1}][text{Maximum value of } z: boxed{dfrac{13}{3}}]]

🔑:Solution: because x+y+z=5,therefore y=5-x-z text {. }Substitute equation (1) into x y+y z+z x=3, we getx(5-x-z)+(5-x-z) z+z x=3 text {. }Rearranging gives x^{2}+(z-5) x+left(z^{2}-5 z+3right)=0.because x is a real number,therefore Delta=(z-5)^{2}-4left(z^{2}-5 z+3right) geqslant 0 text {, }which simplifies to 3 z^{2}-10 z-13 leqslant 0. Solving this, we get -1 leqslant z leqslant frac{13}{3}.Therefore, the maximum value of z is frac{13}{3}, and the minimum value is -1.

🧠:[<think>Okay, so I need to find all polynomials P(x) with real coefficients such that for all nonzero real numbers x,P(x) + P(1/x) = [P(x + 1/x) + P(x - 1/x)] / 2.Hmm, let's start by understanding what the equation is saying. For any x ≠ 0, the value of P at x plus P at 1/x is equal to the average of P evaluated at x + 1/x and P evaluated at x - 1/x. Since this has to hold for all nonzero real x, maybe we can substitute specific values of x to get some constraints on the polynomial P.First, let me note that if x is a nonzero real number, then so is 1/x. So the equation relates the values of P at x and 1/x to the values of P at x + 1/x and x - 1/x. Since P is a polynomial, maybe we can look for polynomials that have some symmetry or satisfy certain functional equations.Let me consider some simple polynomials first. Let's try constant polynomials. Suppose P(x) = c, where c is a constant. Then the left-hand side becomes c + c = 2c. The right-hand side becomes [c + c]/2 = 2c/2 = c. Wait, so 2c = c implies c = 0. So the zero polynomial is a solution. But are there non-constant solutions?Next, let's try linear polynomials. Let P(x) = ax + b. Then P(1/x) = a/x + b. So the left-hand side is ax + b + a/x + b = ax + a/x + 2b.Now the right-hand side: we need to compute P(x + 1/x) and P(x - 1/x). Let's compute each:P(x + 1/x) = a(x + 1/x) + b = ax + a/x + b.Similarly, P(x - 1/x) = a(x - 1/x) + b = ax - a/x + b.So the average of these two is [ (ax + a/x + b) + (ax - a/x + b) ] / 2 = [2ax + 2b] / 2 = ax + b.Therefore, the right-hand side is ax + b. Comparing to the left-hand side, which is ax + a/x + 2b. So we have:ax + a/x + 2b = ax + b.Subtracting ax from both sides gives a/x + 2b = b. Therefore, a/x + b = 0 for all nonzero x. But this can only be true if a = 0 and b = 0. Therefore, the only linear solution is the zero polynomial. So no non-zero linear polynomial works.Now let's try quadratic polynomials. Let P(x) = ax² + bx + c. Then P(1/x) = a/x² + b/x + c. So the left-hand side becomes:ax² + bx + c + a/x² + b/x + c = ax² + a/x² + bx + b/x + 2c.Now compute the right-hand side. First, compute P(x + 1/x) and P(x - 1/x):P(x + 1/x) = a(x + 1/x)² + b(x + 1/x) + c.Let's expand this:= a(x² + 2 + 1/x²) + b(x + 1/x) + c= a x² + 2a + a/x² + b x + b/x + c.Similarly, P(x - 1/x) = a(x - 1/x)² + b(x - 1/x) + c.Expanding that:= a(x² - 2 + 1/x²) + b x - b/x + c= a x² - 2a + a/x² + b x - b/x + c.Now take the average of these two:[P(x + 1/x) + P(x - 1/x)] / 2= [ (a x² + 2a + a/x² + b x + b/x + c) + (a x² - 2a + a/x² + b x - b/x + c) ] / 2Combine terms:= [2a x² + 0a + 2a/x² + 2b x + 0b/x + 2c] / 2= (2a x² + 2a/x² + 2b x + 2c)/2= a x² + a/x² + b x + c.So the right-hand side is a x² + a/x² + b x + c.Compare this to the left-hand side, which was ax² + a/x² + bx + b/x + 2c.Set them equal:ax² + a/x² + bx + b/x + 2c = a x² + a/x² + b x + c.Subtract the right-hand side from the left-hand side:(ax² + a/x² + bx + b/x + 2c) - (a x² + a/x² + b x + c) = 0 + 0 + 0 + b/x + c = 0.Therefore, we have b/x + c = 0 for all nonzero x. Again, this must hold for all x ≠ 0. So set coefficients equal to zero:b = 0 and c = 0. Then, with b = 0 and c = 0, the equation is satisfied. Therefore, quadratic polynomials of the form P(x) = a x² are solutions.So non-zero quadratic polynomials can be solutions. So far, we have that constant zero polynomial, and quadratic polynomials of the form a x² are solutions.Now, let's check cubic polynomials. Let P(x) = a x³ + b x² + c x + d. Then P(1/x) = a/x³ + b/x² + c/x + d. The left-hand side becomes:a x³ + b x² + c x + d + a/x³ + b/x² + c/x + d= a x³ + a/x³ + b x² + b/x² + c x + c/x + 2d.Now compute the right-hand side. First compute P(x + 1/x) and P(x - 1/x):Starting with P(x + 1/x) = a(x + 1/x)^3 + b(x + 1/x)^2 + c(x + 1/x) + d.Let's expand each term:(x + 1/x)^3 = x³ + 3x + 3/x + 1/x³,(x + 1/x)^2 = x² + 2 + 1/x².So:P(x + 1/x) = a(x³ + 3x + 3/x + 1/x³) + b(x² + 2 + 1/x²) + c(x + 1/x) + d= a x³ + 3a x + 3a/x + a/x³ + b x² + 2b + b/x² + c x + c/x + d.Similarly, P(x - 1/x) = a(x - 1/x)^3 + b(x - 1/x)^2 + c(x - 1/x) + d.Expanding each term:(x - 1/x)^3 = x³ - 3x + 3/x - 1/x³,(x - 1/x)^2 = x² - 2 + 1/x².So:P(x - 1/x) = a(x³ - 3x + 3/x - 1/x³) + b(x² - 2 + 1/x²) + c(x - 1/x) + d= a x³ - 3a x + 3a/x - a/x³ + b x² - 2b + b/x² + c x - c/x + d.Now average these two expressions:[P(x + 1/x) + P(x - 1/x)] / 2= [ (a x³ + 3a x + 3a/x + a/x³ + b x² + 2b + b/x² + c x + c/x + d) + (a x³ - 3a x + 3a/x - a/x³ + b x² - 2b + b/x² + c x - c/x + d) ] / 2Combine like terms:= [2a x³ + (3a x - 3a x) + (3a/x + 3a/x) + (a/x³ - a/x³) + 2b x² + (2b - 2b) + 2b/x² + (c x + c x) + (c/x - c/x) + 2d] / 2Simplify term by term:- 2a x³- 0 (from 3a x - 3a x)- 6a/x- 0 (from a/x³ - a/x³)- 2b x²- 0 (from 2b - 2b)- 2b/x²- 2c x- 0 (from c/x - c/x)- 2dSo the average becomes:(2a x³ + 6a/x + 2b x² + 2b/x² + 2c x + 2d) / 2= a x³ + 3a/x + b x² + b/x² + c x + d.Now, set the left-hand side equal to the right-hand side:Left-hand side was: a x³ + a/x³ + b x² + b/x² + c x + c/x + 2d.Right-hand side is: a x³ + 3a/x + b x² + b/x² + c x + d.Set them equal:a x³ + a/x³ + b x² + b/x² + c x + c/x + 2d = a x³ + 3a/x + b x² + b/x² + c x + d.Subtracting the right-hand side from the left-hand side:(a x³ - a x³) + (a/x³) + (b x² - b x²) + (b/x² - b/x²) + (c x - c x) + (c/x - 3a/x) + (2d - d) = 0.Simplify:a/x³ + (c - 3a)/x + d = 0.This must hold for all nonzero x. Therefore, each coefficient of the powers of x must be zero:1. Coefficient of 1/x³: a = 0.2. Coefficient of 1/x: c - 3a = 0. But since a = 0, this gives c = 0.3. Constant term: d = 0.Therefore, for cubic polynomials, we must have a = 0, c = 0, d = 0. The remaining term is b x², which we already considered in the quadratic case. So cubic polynomials reduce to quadratic solutions here. Therefore, cubic polynomials only contribute the same quadratic solutions. So no new solutions from cubic.Moving on to quartic polynomials. Let's try P(x) = a x⁴ + b x³ + c x² + d x + e.This might get complicated, but perhaps we can look for patterns. Alternatively, maybe the solutions are only quadratic and constants? Wait, but we saw that constants must be zero. Wait, constants had to be zero. So far, only quadratics of the form a x² are solutions.Wait, maybe all even-powered polynomials? Let's see. Suppose P(x) is a monomial: P(x) = a x^{2n}. Let's test if such a polynomial satisfies the equation.Let P(x) = a x^{2n}. Then P(1/x) = a x^{-2n}. The left-hand side is a x^{2n} + a x^{-2n}.The right-hand side is [P(x + 1/x) + P(x - 1/x)] / 2. Let's compute P(x + 1/x) and P(x - 1/x):P(x + 1/x) = a (x + 1/x)^{2n}, P(x - 1/x) = a (x - 1/x)^{2n}.So the average is [ (x + 1/x)^{2n} + (x - 1/x)^{2n} ] / 2 * a.Therefore, we need to check if x^{2n} + x^{-2n} equals [ (x + 1/x)^{2n} + (x - 1/x)^{2n} ] / 2.Is this identity true?Wait, let's test for small n. Let n = 1: x² + 1/x² vs [ (x + 1/x)^2 + (x - 1/x)^2 ] / 2.Compute the right-hand side:[ (x² + 2 + 1/x²) + (x² - 2 + 1/x²) ] / 2 = (2x² + 2/x²)/2 = x² + 1/x². So yes, holds for n=1.n=2: x⁴ + 1/x⁴ vs [ (x + 1/x)^4 + (x - 1/x)^4 ] /2.Compute (x + 1/x)^4 = x⁴ + 4x² + 6 + 4/x² + 1/x⁴.(x - 1/x)^4 = x⁴ - 4x² + 6 - 4/x² + 1/x⁴.Average them: [2x⁴ + 12 + 2/x⁴]/2 = x⁴ + 6 + 1/x⁴.Compare to x⁴ + 1/x⁴. So unless 6=0, which it isn't, this is not equal. Therefore, for n=2, the identity does not hold. Hence, x⁴ + 1/x⁴ ≠ [ (x +1/x)^4 + (x -1/x)^4 ] /2. Therefore, the monomial P(x) = x^4 does not satisfy the equation.Wait, but that seems contradictory. Wait, let's check my calculation again.(x + 1/x)^4: expand (x + 1/x)^4:= x⁴ + 4x³*(1/x) + 6x²*(1/x²) + 4x*(1/x³) + 1/x⁴= x⁴ + 4x² + 6 + 4/x² + 1/x⁴.Similarly, (x - 1/x)^4 = x⁴ - 4x³*(1/x) + 6x²*(1/x²) - 4x*(1/x³) + 1/x⁴= x⁴ - 4x² + 6 - 4/x² + 1/x⁴.Average of these two:[ (x⁴ + 4x² + 6 + 4/x² + 1/x⁴) + (x⁴ - 4x² + 6 - 4/x² + 1/x⁴) ] / 2= [2x⁴ + 0x² + 12 + 0/x² + 2/x⁴] /2= (2x⁴ + 12 + 2/x⁴)/2= x⁴ + 6 + 1/x⁴.So indeed, compared to x⁴ +1/x⁴, which is different. Therefore, x⁴ + 1/x⁴ is not equal to x⁴ +6 +1/x⁴. Therefore, the identity does not hold for n=2. Therefore, monomials of degree 4 are not solutions.Hmm. So the identity holds for n=1 but not for higher n. So quadratic monomials work, but higher even degrees don't. Therefore, perhaps only quadratic monomials and their linear combinations?Wait, but let's think again. Suppose P(x) is a quadratic polynomial, then it works. What if we have higher-degree polynomials that can be expressed in terms of x²? For example, if P(x) is a polynomial in x², say P(x) = Q(x²). Let's check if such polynomials can satisfy the equation.Let me suppose that P(x) = Q(x²) for some polynomial Q. Then, P(1/x) = Q(1/x²). Let's substitute into the equation:Q(x²) + Q(1/x²) = [ Q( (x +1/x)^2 ) + Q( (x -1/x)^2 ) ] / 2.Compute (x +1/x)^2 = x² + 2 + 1/x², and (x -1/x)^2 = x² - 2 + 1/x².So the equation becomes:Q(x²) + Q(1/x²) = [ Q(x² + 2 + 1/x²) + Q(x² - 2 + 1/x²) ] /2.Let me set y = x² + 1/x². Then note that (x +1/x)^2 = y + 2, and (x -1/x)^2 = y - 2. Wait, no, actually:Wait, if y = x² + 1/x², then (x +1/x)^2 = y + 2, and (x -1/x)^2 = y - 2. So, substituting into the equation:Q(x²) + Q(1/x²) = [ Q(y + 2) + Q(y - 2) ] / 2.But x² and 1/x² are related to y. Since y = x² + 1/x², then x² and 1/x² are roots of the equation t^2 - y t +1=0. So x² and 1/x² are t1 and t2 where t1 + t2 = y and t1 t2 =1.Therefore, the equation becomes:Q(t1) + Q(t2) = [ Q(y + 2) + Q(y -2) ] /2.But y = t1 + t2. So for variables t1 and t2 with t1 t2 =1, Q(t1) + Q(t2) = [ Q(t1 + t2 + 2) + Q(t1 + t2 -2) ] / 2.This seems like a functional equation for Q. Let's see if we can find Q such that this holds.Suppose we let Q(t) be a linear function: Q(t) = a t + b. Then:Left-hand side: Q(t1) + Q(t2) = a(t1 + t2) + 2b = a y + 2b.Right-hand side: [ Q(y + 2) + Q(y -2) ] /2 = [a(y +2) + b + a(y -2) + b]/2 = [2a y + 2b]/2 = a y + b.Therefore, equation becomes a y + 2b = a y + b => 2b = b => b=0. So Q(t) = a t. Therefore, P(x) = a x². Which is the solution we already found.If Q(t) is quadratic, let Q(t) = a t² + b t + c. Then:Left-hand side: Q(t1) + Q(t2) = a(t1² + t2²) + b(t1 + t2) + 2c.Since t1 t2 =1, t1² + t2² = (t1 + t2)^2 - 2 t1 t2 = y² - 2.Therefore, Left-hand side = a(y² -2) + b y + 2c = a y² - 2a + b y + 2c.Right-hand side: [ Q(y + 2) + Q(y -2) ] /2.Compute Q(y +2) = a(y +2)^2 + b(y +2) + c = a(y² +4y +4) + b y + 2b + c.Q(y -2) = a(y -2)^2 + b(y -2) + c = a(y² -4y +4) + b y - 2b + c.Average them:[ a(y² +4y +4) + b y + 2b + c + a(y² -4y +4) + b y - 2b + c ] /2= [ 2a y² + 8a + 2b y + 2c ] /2= a y² +4a + b y + c.Set equal to left-hand side:a y² - 2a + b y + 2c = a y² +4a + b y + c.Subtract right-hand side from left-hand side:(a y² -2a + b y +2c) - (a y² +4a + b y +c) = (-6a + c) = 0.Therefore, -6a + c =0 => c =6a.Therefore, Q(t) can be a quadratic polynomial with c=6a, i.e., Q(t) = a t² + b t +6a.But remember that P(x) = Q(x²) = a x⁴ + b x² +6a. Let's check if such a polynomial satisfies the original equation.Take P(x) = a x⁴ + b x² +6a. Then P(1/x) = a/x⁴ + b/x² +6a.Left-hand side: a x⁴ + b x² +6a + a/x⁴ + b/x² +6a = a x⁴ + a/x⁴ + b x² + b/x² +12a.Right-hand side: [ P(x +1/x) + P(x -1/x) ] /2.Compute P(x +1/x):= a(x +1/x)^4 + b(x +1/x)^2 +6a.As before, (x +1/x)^4 = x⁴ +4x² +6 +4/x² +1/x⁴.So P(x +1/x) = a(x⁴ +4x² +6 +4/x² +1/x⁴) + b(x² +2 +1/x²) +6a.= a x⁴ +4a x² +6a +4a/x² +a/x⁴ +b x² +2b +b/x² +6a.Combine like terms:= a x⁴ + a/x⁴ + (4a + b) x² + (4a + b)/x² + (6a +2b +6a).Similarly, P(x -1/x):= a(x -1/x)^4 + b(x -1/x)^2 +6a.Expand (x -1/x)^4 = x⁴ -4x² +6 -4/x² +1/x⁴.(x -1/x)^2 =x² -2 +1/x².So:= a(x⁴ -4x² +6 -4/x² +1/x⁴) + b(x² -2 +1/x²) +6a.= a x⁴ -4a x² +6a -4a/x² +a/x⁴ +b x² -2b +b/x² +6a.Combine like terms:= a x⁴ + a/x⁴ + (-4a + b) x² + (-4a + b)/x² + (6a -2b +6a).Now average the two:[P(x +1/x) + P(x -1/x)] /2= [ (a x⁴ + a/x⁴ + (4a + b) x² + (4a + b)/x² +12a +2b ) + (a x⁴ + a/x⁴ + (-4a + b) x² + (-4a + b)/x² +12a -2b ) ] /2Combine terms:= [2a x⁴ + 2a/x⁴ + ( (4a + b) + (-4a + b) )x² + ( (4a + b) + (-4a + b) )/x² +24a ] /2Simplify:= [2a x⁴ + 2a/x⁴ + 2b x² + 2b/x² +24a ] /2= a x⁴ + a/x⁴ +b x² +b/x² +12a.Compare to the left-hand side, which was a x⁴ + a/x⁴ +b x² +b/x² +12a. They match! Therefore, for Q(t) =a t² +b t +6a, the polynomial P(x) = Q(x²) satisfies the equation. Therefore, quartic polynomials of the form P(x) =a x⁴ +b x² +6a are solutions.Wait, but earlier when we tried P(x)=x⁴, which would correspond to a=1, b=0, c=6 (since P(x)=x⁴ +6*1= x⁴ +6). But earlier, when we tried P(x)=x⁴, we saw that it didn't satisfy the equation. Wait, but that's because if a=1, b=0, then P(x)=x⁴ +6. Then P(x) + P(1/x) =x⁴ +6 +1/x⁴ +6 =x⁴ +1/x⁴ +12. Then the right-hand side would be [P(x+1/x) + P(x-1/x)] /2. Let's check with P(x)=x⁴ +6.Wait, but P(x)=a x⁴ +b x² +6a. If a=1, b=0, then P(x)=x⁴ +6. Then P(x +1/x) = (x +1/x)^4 +6. As computed earlier, (x +1/x)^4 =x⁴ +4x² +6 +4/x² +1/x⁴. Therefore, P(x +1/x)=x⁴ +4x² +6 +4/x² +1/x⁴ +6 =x⁴ +1/x⁴ +4x² +4/x² +12. Similarly, P(x -1/x)= (x -1/x)^4 +6 =x⁴ -4x² +6 -4/x² +1/x⁴ +6 =x⁴ +1/x⁴ -4x² -4/x² +12. Then average these two: [ (x⁴ +1/x⁴ +4x² +4/x² +12 ) + (x⁴ +1/x⁴ -4x² -4/x² +12 ) ] /2 = [2x⁴ +2/x⁴ +24 ] /2 =x⁴ +1/x⁴ +12. Which matches P(x) + P(1/x) =x⁴ +1/x⁴ +12. Therefore, it works! So actually, the quartic polynomial P(x)=x⁴ +6 does satisfy the equation. Wait, so my previous conclusion that x⁴ doesn't work was because I forgot the constant term. So when Q(t)= t² +6, which gives P(x)=x⁴ +6, then it works.Therefore, quartic polynomials of the form a x⁴ +b x² +6a are solutions. So we can have solutions with higher degrees as long as they follow this structure.Similarly, we can consider Q(t) being higher degree. Let me try Q(t) of degree 3. Let Q(t) =a t³ +b t² +c t +d. Then, substituting into the functional equation:Q(t1) + Q(t2) = [ Q(y +2) + Q(y -2) ] /2,where y = t1 + t2, and t1 t2 =1.Compute Q(t1) + Q(t2):= a(t1³ + t2³) + b(t1² + t2²) + c(t1 + t2) + 2d.Express t1³ + t2³ in terms of y and t1 t2:t1³ + t2³ = (t1 + t2)^3 -3 t1 t2 (t1 + t2) = y³ -3y.Similarly, t1² + t2² = y² -2.Thus,Q(t1) + Q(t2) = a(y³ -3y) + b(y² -2) + c y + 2d.Now compute [ Q(y +2) + Q(y -2) ] /2.First, compute Q(y +2):= a(y +2)^3 + b(y +2)^2 + c(y +2) +d= a(y³ +6y² +12y +8) + b(y² +4y +4) +c y +2c +d.Q(y -2):= a(y -2)^3 + b(y -2)^2 + c(y -2) +d= a(y³ -6y² +12y -8) + b(y² -4y +4) +c y -2c +d.Average:[Q(y +2) + Q(y -2)] /2= [ a(y³ +6y² +12y +8) + b(y² +4y +4) +c y +2c +d + a(y³ -6y² +12y -8) + b(y² -4y +4) +c y -2c +d ] /2Combine like terms:= [2a y³ +24a y + (6a y² -6a y²) + (b y² +b y²) + (4b y -4b y) + (8a -8a) + (4b +4b) + (c y +c y) + (2c -2c) +2d ] /2Wait, step by step:Expand the terms:From Q(y +2):- a y³ +6a y² +12a y +8a- b y² +4b y +4b- c y +2c- dFrom Q(y -2):- a y³ -6a y² +12a y -8a- b y² -4b y +4b- c y -2c- dCombine all terms:a y³ +6a y² +12a y +8a +b y² +4b y +4b +c y +2c +d +a y³ -6a y² +12a y -8a +b y² -4b y +4b +c y -2c +dNow combine:a y³ + a y³ = 2a y³6a y² -6a y² = 012a y +12a y =24a y8a -8a =0b y² +b y² =2b y²4b y -4b y =04b +4b =8bc y +c y =2c y2c -2c =0d +d =2d.Therefore, total expression:2a y³ +24a y +2b y² +8b +2c y +2d.Divide by 2:a y³ +12a y +b y² +4b +c y +d.Set this equal to Q(t1) + Q(t2) which was a(y³ -3y) + b(y² -2) + c y + 2d.Therefore:a y³ -3a y +b y² -2b +c y +2d = a y³ +12a y +b y² +4b +c y +d.Subtract the right-hand side from the left-hand side:(a y³ -3a y +b y² -2b +c y +2d) - (a y³ +12a y +b y² +4b +c y +d)= (0 y³) + (-15a y) + (0 y²) + (-6b) + (0 y) + (d) =0.Therefore:-15a y -6b +d =0 for all y.Since this must hold for all y, each coefficient must be zero:-15a =0 ⇒ a=0-6b =0 ⇒ b=0d=0.Thus, Q(t) reduces to Q(t)=c t. But then Q(t) would be linear, but we considered Q(t) as cubic. However, if a=0, b=0, d=0, then Q(t)=c t. But earlier, we saw that linear Q(t) require c=0. Wait, in the linear case, Q(t)=c t would have to satisfy Q(t1) + Q(t2) = [ Q(y +2) + Q(y -2) ] /2. So Q(t1)+Q(t2)=c(t1 + t2)=c y. The right-hand side would be [c(y +2) +c(y -2)] /2 = [2c y]/2 =c y. Therefore, this holds for any c. Wait, but in our previous analysis when we considered linear Q(t), we required Q(t) to satisfy Q(t1) + Q(t2) = [ Q(y +2) + Q(y -2) ] /2. Which for linear Q(t)=c t +d, we have:Q(t1) + Q(t2)=c(t1 + t2)+2d= c y +2d.Right-hand side: [ Q(y +2) + Q(y -2) ] /2 = [c(y +2) +d +c(y -2) +d]/2 = [2c y +2d]/2 =c y +d.So equation is c y +2d =c y +d ⇒ d=0.Therefore, linear Q(t)=c t are allowed, which correspond to P(x)=c x². Which are the quadratic solutions we already found.But in the cubic Q(t) case, we ended up with only Q(t)=c t (with a=0,b=0,d=0). But since we started with Q(t) cubic, unless c=0, but then Q(t)=0. So in that case, the only cubic Q(t) that would satisfy the equation is Q(t)=0. Therefore, P(x)=0.So, in general, for higher-degree Q(t), unless their coefficients satisfy certain conditions, they don't contribute new solutions. So far, the solutions we have are:1. The zero polynomial.2. Quadratic polynomials of the form a x².3. Quartic polynomials of the form a x⁴ +b x² +6a.But wait, when we considered Q(t) quadratic, leading to P(x)=a x⁴ +b x² +6a, then in that case, 6a is a constant term. So this is a quartic polynomial. But if a=0, then P(x)=b x², which is the quadratic case. So the quartic case includes the quadratic case when a=0. Wait, no. If a=0, then P(x)=b x² +0, which is the quadratic case. So the quartic family includes the quadratic solutions as a subset when a=0. Similarly, if we consider Q(t) of higher degrees, maybe we can get higher-degree polynomials that include lower-degree solutions.Wait, let's check if there are solutions of higher degrees. Let me try Q(t) of degree 4: Q(t) =a t^4 +b t^3 +c t^2 +d t +e.But this is going to get very complicated. Alternatively, maybe all solutions are of the form P(x) = Q(x² +1/x²)? Not sure. Wait, but given the structure of the equation, maybe there is a pattern where we can express P(x) in terms of x² +1/x². However, x² +1/x² is related to y =x +1/x. Wait, (x +1/x)^2 =x² +2 +1/x², so x² +1/x² = y² -2. Therefore, maybe P(x) can be expressed in terms of y² -2. Not sure.Alternatively, notice that in the quadratic case, P(x) =a x², then x² = (x +1/x)^2 - 2 - 2*(x -1/x)^2 / (something). Wait, maybe not.Alternatively, maybe all solutions are linear combinations of (x^{2n} + x^{-2n}) for n≥0, but with coefficients such that the combination equals the average of combinations at x +1/x and x -1/x. But this might be similar to generating functions.Alternatively, given that we have solutions for Q(t) linear and quadratic, maybe there's a pattern where Q(t) can be any polynomial satisfying a certain kind of functional equation. Let's see.The functional equation for Q(t) is:Q(t1) + Q(t2) = [ Q(y +2) + Q(y -2) ] /2,where y = t1 + t2 and t1 t2 =1.This resembles a functional equation where the sum of Q over reciprocal pairs equals the average of Q evaluated at y +2 and y -2.Perhaps this suggests that Q(t) must satisfy a certain kind of symmetry or recursion. For example, when Q(t) is linear or quadratic, we saw that specific forms work. Maybe in general, Q(t) must satisfy Q(t) + Q(1/t) = [ Q(t +1/t +2) + Q(t +1/t -2) ] /2. But since t1 = t and t2=1/t, y = t +1/t. Wait, substituting y = t +1/t, then the equation becomes Q(t) + Q(1/t) = [Q(y +2) + Q(y -2)] /2. Which is similar to the original equation.But this seems recursive. For example, applying this to Q(t), we can express Q(y ±2) in terms of Q(t) and Q(1/t). However, unless Q(t) has a specific form, this recursion might not terminate.Alternatively, suppose Q(t) satisfies Q(t) + Q(1/t) = 2Q( sqrt(t) ) or something, but not sure.Alternatively, let's consider that for the equation Q(t) + Q(1/t) = [Q(y +2) + Q(y -2)] /2, where y = t +1/t, maybe Q(t) must be a linear combination of Chebyshev polynomials or something similar, as they often satisfy such functional equations.Chebyshev polynomials satisfy T_n(t +1/t) = t^n + t^{-n}, for example. But maybe not directly applicable here.Alternatively, suppose we let t = e^{iθ}, then 1/t = e^{-iθ}, and y = t +1/t = 2 cosθ. Then Q(t) + Q(1/t) = Q(e^{iθ}) + Q(e^{-iθ}) = 2 Re Q(e^{iθ}). Similarly, Q(y +2) = Q(2 cosθ +2) and Q(y -2) = Q(2 cosθ -2). But unless Q has a specific form, this might not lead anywhere.Alternatively, consider that if Q(t) is a polynomial such that Q(t) + Q(1/t) is a function of y = t +1/t. Then since Q(t) + Q(1/t) must equal [Q(y +2) + Q(y -2)] /2, perhaps Q(t) + Q(1/t) must be a function that's invariant under certain transformations.Alternatively, let's think of Q(t) + Q(1/t) = S(y), where S(y) = [Q(y +2) + Q(y -2)] /2.So we have a functional equation S(y) = [ Q(y +2) + Q(y -2) ] /2, but S(y) is Q(t) + Q(1/t) where y = t +1/t. This might be difficult to solve in general.But let's note that when Q(t) is a monomial, say Q(t) = t^n, then Q(t) + Q(1/t) = t^n + t^{-n}. Let's see what this would require:For Q(t) = t^n, the left-hand side is t^n + t^{-n}.The right-hand side would be [ (y +2)^n + (y -2)^n ] /2, where y = t +1/t.So we need to check if t^n + t^{-n} = [ (y +2)^n + (y -2)^n ] /2.But when n=1: Left-hand side = t +1/t = y. Right-hand side = [ (y +2) + (y -2) ] /2 = (2y)/2 = y. So holds for n=1.n=2: Left-hand side = t² +1/t² = y² -2. Right-hand side: [ (y +2)^2 + (y -2)^2 ] /2 = [ y² +4y +4 + y² -4y +4 ] /2 = (2y² +8)/2 = y² +4. So set equal to y² -2: y² -2 = y² +4 ⇒ -2=4, which is false. Therefore, n=2 does not work. Wait, but when Q(t)=t², we saw that the quartic polynomial P(x)=x⁴ +6 works. But this seems contradictory. Wait, Q(t)=t² would correspond to P(x)=x⁴, but then Q(t) + Q(1/t)=t² +1/t² = y² -2. Then [ Q(y +2) + Q(y -2) ] /2 = [ (y +2)^2 + (y -2)^2 ] /2 = y² +4. So to have y² -2 = y² +4, which is impossible. Therefore, Q(t)=t² does not satisfy the equation. However, when we set Q(t)=t² +6, then Q(t) + Q(1/t)=t² +1/t² +12 = y² -2 +12 = y² +10. Then [ Q(y +2) + Q(y -2) ] /2 = [ (y +2)^2 +6 + (y -2)^2 +6 ] /2 = [ (y² +4y +4 +6) + (y² -4y +4 +6) ] /2 = [2y² +20]/2 = y² +10. Which matches. So Q(t)=t² +6 works. Therefore, adding a constant term allows the equation to hold.Therefore, for monomials plus constants, it's possible. So maybe in general, for Q(t) = t^n + c_n, where c_n is chosen appropriately, we can have the equation hold. Let's test n=3.Let Q(t)=t^3 +c_3.Then Q(t) + Q(1/t) = t^3 +1/t^3 +2c_3.[ Q(y +2) + Q(y -2) ] /2 = [ (y +2)^3 +c_3 + (y -2)^3 +c_3 ] /2.Compute (y +2)^3 + (y -2)^3 = [y³ +6y² +12y +8] + [y³ -6y² +12y -8] = 2y³ +24y.Therefore, average is (2y³ +24y +2c_3)/2 =y³ +12y +c_3.Set equal to t^3 +1/t^3 +2c_3. But y = t +1/t. So t^3 +1/t^3 = (t +1/t)^3 -3(t +1/t) = y³ -3y. Therefore:Left-hand side = y³ -3y +2c_3.Right-hand side =y³ +12y +c_3.Setting equal: y³ -3y +2c_3 = y³ +12y +c_3 ⇒ -15y +c_3=0. This must hold for all y, which is impossible unless coefficient of y is zero and c_3=0. But then -15y=0 for all y, which is not possible. Therefore, Q(t)=t^3 +c_3 doesn't work. Hence, no cubic solutions of this form.Similarly, for Q(t)=t^4 +c_4.Left-hand side: t^4 +1/t^4 +2c_4.Right-hand side: [ (y +2)^4 + (y -2)^4 ] /2 +c_4.Compute (y +2)^4 + (y -2)^4 = 2y^4 + 24y² + 32. Therefore, average is y^4 +12y² +16.Set equal to t^4 +1/t^4 +2c_4. Note that t^4 +1/t^4 = (t^2 +1/t^2)^2 -2 = (y² -2)^2 -2 = y^4 -4y² +2.Therefore, left-hand side: y^4 -4y² +2 +2c_4.Right-hand side: y^4 +12y² +16 +c_4.Set equal:y^4 -4y² +2 +2c_4 = y^4 +12y² +16 +c_4 ⇒ -16y² -14 +c_4=0.Again, impossible for all y unless coefficients are zero:-16=0, which is not possible. Therefore, quartic monomials plus constants don't work. However, when Q(t)=t^2 +6, which is a quadratic polynomial with a constant term, it works. So perhaps the only way this works is when Q(t) is a quadratic polynomial with a specific constant term.Alternatively, looking back at the quadratic Q(t):Q(t) = a t^2 +b t +6a.Wait, when we had Q(t) quadratic, we found that such a Q(t) satisfies the equation. Therefore, maybe the general solution is linear combinations of such Q(t) for different degrees? Wait, but how?Wait, when we set Q(t) to be quadratic, we could get solutions. If we set Q(t) to be of higher degree, maybe with specific coefficients, we can get solutions. For example, in the quartic case, Q(t) =a t^2 +b t +6a led to P(x)=a x^4 +b x^2 +6a. Wait, but this Q(t) is quadratic, not quartic. So P(x) is quartic because Q(t) is quadratic in t=x². Therefore, maybe all solutions are of the form P(x) = Q(x²) where Q(t) is a quadratic polynomial of the form a t^2 +b t +6a. Then, since Q(t) can be any quadratic polynomial with the constant term fixed as 6a, then the general solution would be P(x) = a x^4 +b x² +6a for real constants a and b.But wait, when a=0, this reduces to P(x)=b x², which is the quadratic solution we found earlier. So this family includes the quadratic solutions as a subset. Therefore, the general solution is P(x) =a x^4 +b x² +6a, where a and b are real constants.But let's verify this. Take P(x) =a x^4 +b x² +6a. Then P(x) + P(1/x) =a x^4 +a/x^4 +b x² +b/x² +12a. Earlier, we saw that the right-hand side [P(x +1/x) + P(x -1/x)] /2 also equals a x^4 +a/x^4 +b x² +b/x² +12a. Therefore, it works. Therefore, quartic polynomials of this form are solutions, including quadratic ones when a=0.But are there higher-degree solutions? For example, if we set Q(t) to be a cubic polynomial, but we saw that leads to only the zero polynomial. So maybe the only solutions are combinations of these quartic (including quadratic) polynomials.Alternatively, perhaps there are solutions of every even degree. Let's try a sextic polynomial. Let Q(t) =a t^3 +b t^2 +c t +d. Then P(x)=Q(x²)=a x^6 +b x^4 +c x² +d.But this seems complicated. Alternatively, let's see if there's a pattern with the previous solutions:Quadratic solution: P(x) =b x².Quartic solution: P(x)=a x^4 +b x² +6a.Suppose we consider a sextic solution by adding another term. Let P(x)=a x^6 +b x^4 +c x² +d. Then compute P(x) + P(1/x) and compare to [P(x +1/x) + P(x -1/x)] /2. This would be quite involved, but maybe if we set a=1 and find the necessary constants.Alternatively, maybe the solutions are generated recursively. For example, if we have a solution P(x) of degree 2n, then maybe another solution of degree 2n+2 can be formed by adding a term involving x^{2n+2} and adjusting lower-degree terms. But without more information, it's hard to see.Alternatively, notice that when we considered Q(t) as a quadratic polynomial, we were able to get quartic solutions. If we consider Q(t) as a cubic polynomial, but found that only the zero polynomial works, maybe the only solutions are combinations of the quartic and quadratic polynomials. But since the quartic family already includes the quadratic ones (by setting a=0), perhaps all solutions are of the form P(x) =a x^4 +b x² +c, where c=6a. Wait, because when we had Q(t)=a t² +b t +6a, then P(x)=Q(x²)=a x^4 +b x² +6a. Therefore, c=6a. So the constant term is fixed as 6a. Hence, the general solution is P(x)=a x^4 +b x² +6a, where a and b are real constants.But let me verify if a linear combination of solutions is a solution. Suppose P1(x) and P2(x) are solutions. Then, for any constants k1 and k2, is k1 P1(x) +k2 P2(x) also a solution?Yes, because the equation is linear. Let's check:Left-hand side: (k1 P1 +k2 P2)(x) + (k1 P1 +k2 P2)(1/x) =k1 (P1(x) +P1(1/x)) +k2 (P2(x) +P2(1/x)).Right-hand side: [ (k1 P1 +k2 P2)(x +1/x) + (k1 P1 +k2 P2)(x -1/x) ] /2 =k1 [P1(x +1/x) +P1(x -1/x)] /2 +k2 [P2(x +1/x) +P2(x -1/x)] /2.Since P1 and P2 satisfy the equation, this equals k1 (P1(x) +P1(1/x)) +k2 (P2(x) +P2(1/x)), which matches the left-hand side. Therefore, linear combinations of solutions are solutions. Therefore, the set of solutions is a vector space.Now, from our previous analysis, we have that the solutions are polynomials of the form a x^4 +b x² +6a. Let's check the dimension. If a and b are arbitrary real numbers, then this is a two-dimensional vector space. However, when a=0, we get P(x)=b x², which is the quadratic solution, and when a≠0, we get quartic polynomials. However, wait, if we consider that 6a is a constant term dependent on a, then for each a, we have a quartic polynomial with a specific constant term. Therefore, the general solution is any linear combination of the quadratic solutions and the quartic solutions. However, in our quartic case, the constant term is tied to the coefficient of x^4. Therefore, these are not separate solutions but part of a family.But actually, when we set Q(t) =a t² +b t +6a, then P(x)=a x^4 +b x² +6a. Here, if a=0, then P(x)=b x², which is quadratic. If a≠0, then it's quartic. Therefore, the general solution is indeed the set of all polynomials of the form a x^4 +b x² +6a for real numbers a and b. This includes quadratic polynomials when a=0.But wait, can we have higher-degree polynomials? For instance, let's suppose there is a solution of degree 6. Let me assume P(x)=a x^6 +b x^4 +c x² +d. Then, compute both sides of the equation.But this would be very tedious. Alternatively, maybe no such solutions exist. Given that when we tried cubic Q(t), only the zero solution worked, and for quadratic Q(t) we got up to quartic P(x). So perhaps the only solutions are quartic and quadratic. To confirm, suppose P(x) is a polynomial of degree n>4. Then, in the equation P(x) +P(1/x) = [P(x +1/x) +P(x -1/x)] /2, the highest-degree term on the left-hand side is degree n, and on the right-hand side, expanding P(x +1/x) would have degree n as well, but when averaged, it might not cancel out. Let's analyze the leading terms.Suppose P(x) =a_n x^n + lower-degree terms.Then P(1/x) =a_n x^{-n} + lower-degree terms.So the left-hand side has leading term a_n x^n and a term a_n x^{-n}.The right-hand side: compute P(x +1/x). The leading term of (x +1/x)^n is x^n. So P(x +1/x) has leading term a_n x^n. Similarly, P(x -1/x) has leading term a_n x^n. Therefore, the average of P(x +1/x) and P(x -1/x) would have leading term a_n x^n. So comparing both sides:Left-hand side has a_n x^n +a_n x^{-n}.Right-hand side has a_n x^n + lower-degree terms.Therefore, equating both sides, we must have a_n x^{-n} present on the left but not on the right. Therefore, this term must cancel out, implying a_n=0. Hence, degree n must be even, and the coefficients for x^{-n} must cancel out. Wait, but how?Wait, if P(x) is of degree n, then P(x +1/x) will have degree n, but when expanded, it will have terms like x^n, x^{n-2}, etc., depending on n. Similarly, P(x -1/x) will also have leading term x^n. But the average of these two would still have leading term x^n. On the left-hand side, the leading term is x^n and x^{-n}. For the equation to hold, the coefficients of x^n must match, which they do (both sides have a_n x^n). But the left-hand side also has a term a_n x^{-n}, which must be canceled by something on the right-hand side. However, the right-hand side, when expanded, will have terms from lower degrees. Therefore, unless a_n=0, the term x^{-n} cannot be canceled. Therefore, to have a_n x^{-n}=0 for all x≠0, we must have a_n=0. Therefore, there are no solutions with degree n>4. Similarly, if n=4, we saw that the term x^{-4} can be canceled if the polynomial has a specific form, which we found earlier. Wait, when P(x)=a x^4 +b x² +6a, then P(1/x)=a x^{-4} +b x^{-2} +6a. So left-hand side P(x) + P(1/x)=a x^4 +a x^{-4} +b x² +b x^{-2} +12a. On the right-hand side, when we computed [P(x +1/x) +P(x -1/x)] /2, we obtained a x^4 +a x^{-4} +b x² +b x^{-2} +12a. Therefore, the x^{-4} terms are present on both sides and cancel out. Hence, the leading coefficient a can be non-zero because the x^{-4} term is canceled by the corresponding term on the right-hand side. But this only works because the specific combination in the quartic polynomial leads to the x^{-4} term being matched on both sides. For higher-degree polynomials, unless similar cancellation occurs, which would require specific coefficients, such terms cannot be canceled. Therefore, the only possible solutions are those with degrees up to 4.Thus, combining all these results, the general solution is the set of all polynomials of the form P(x) =a x⁴ +b x² +6a where a and b are real constants. When a=0, this reduces to P(x)=b x². When a≠0, it's a quartic polynomial. Therefore, all solutions are either quadratic polynomials of the form b x² or quartic polynomials of the form a x⁴ +b x² +6a.But wait, to confirm, let's take a=1 and b=0: P(x)=x⁴ +6. As checked earlier, this works. Take a=1 and b=5: P(x)=x⁴ +5x² +6. Let's verify if this satisfies the equation.Left-hand side: x⁴ +5x² +6 +1/x⁴ +5/x² +6= x⁴ +1/x⁴ +5x² +5/x² +12.Right-hand side: [P(x +1/x) + P(x -1/x)] /2.Compute P(x +1/x):= (x +1/x)^4 +5(x +1/x)^2 +6.Expand:(x +1/x)^4 =x⁴ +4x² +6 +4/x² +1/x⁴.5(x +1/x)^2 =5x² +10 +5/x².So total: x⁴ +4x² +6 +4/x² +1/x⁴ +5x² +10 +5/x² +6= x⁴ +9x² +22 +9/x² +1/x⁴.Similarly, P(x -1/x):= (x -1/x)^4 +5(x -1/x)^2 +6.Expand:(x -1/x)^4 =x⁴ -4x² +6 -4/x² +1/x⁴.5(x -1/x)^2 =5x² -10 +5/x².So total: x⁴ -4x² +6 -4/x² +1/x⁴ +5x² -10 +5/x² +6= x⁴ +x² +2 +1/x⁴ +1/x².Average of P(x +1/x) and P(x -1/x):[ (x⁴ +9x² +22 +9/x² +1/x⁴ ) + (x⁴ +x² +2 +1/x⁴ +1/x²) ] /2= [2x⁴ +10x² +24 +10/x² +2/x⁴ ] /2= x⁴ +5x² +12 +5/x² +1/x⁴.Which is equal to the left-hand side: x⁴ +1/x⁴ +5x² +5/x² +12. Therefore, it works.Therefore, the general solution is indeed P(x) =a x⁴ +b x² +6a for real constants a and b. This includes the quadratic case when a=0. Hence, all such polynomials satisfy the given functional equation.Final AnswerThe polynomials are all quadratic and quartic polynomials of the form boxed{a x^4 + b x^2 + 6a} where (a) and (b) are real constants.</think>To find all polynomials (P(x)) with real coefficients such that for all nonzero real numbers (x),[P(x) + Pleft(frac{1}{x}right) = frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2},]we start by testing simple polynomials and gradually consider higher degrees.1. Constant Polynomials: Only the zero polynomial satisfies the equation.2. Linear Polynomials: Only the zero polynomial satisfies the equation.3. Quadratic Polynomials: Polynomials of the form (P(x) = ax^2) are solutions.4. Cubic Polynomials: Only the zero polynomial satisfies the equation.5. Quartic Polynomials: Polynomials of the form (P(x) = a x^4 + b x^2 + 6a) are solutions. This form includes quadratic polynomials when (a = 0).To verify, we consider the functional equation for polynomials (P(x) = Q(x^2)) where (Q(t)) satisfies a specific functional equation. We find that (Q(t)) must be a quadratic polynomial of the form (Q(t) = a t^2 + b t + 6a), leading to (P(x) = a x^4 + b x^2 + 6a).Higher-degree polynomials are shown to not satisfy the equation due to unmatched terms, confirming that the solutions are limited to quadratic and quartic forms.Thus, the polynomials are all quadratic and quartic polynomials of the form[boxed{a x^4 + b x^2 + 6a}]where (a) and (b) are real constants.,<think>Alright, let me try to work through this problem step by step. The task is to find all polynomials ( P(x) ) with real coefficients such that for all nonzero real numbers ( x ),[P(x) + Pleft(frac{1}{x}right) = frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2}.]First, I need to understand what this equation is saying. It relates the values of the polynomial at ( x ) and ( 1/x ) to the average of its values at ( x + 1/x ) and ( x - 1/x ). Since this must hold for all nonzero real ( x ), the equation must hold as an identity for polynomials. That suggests that the polynomials on both sides of the equation must be equal for all ( x neq 0 ), and hence equal as polynomials (since polynomials that agree infinitely many points must be identical).Given that ( P(x) ) is a polynomial, I might consider the structure of such polynomials. Let me note that if ( P(x) ) is a solution, then so might be polynomials of certain forms, maybe symmetric or antisymmetric in some way. Let me test some simple cases first.Case 1: Constant PolynomialSuppose ( P(x) = c ), where ( c ) is a constant. Then the left-hand side (LHS) becomes ( c + c = 2c ). The right-hand side (RHS) becomes ( frac{c + c}{2} = frac{2c}{2} = c ). So ( 2c = c ) implies ( c = 0 ). So the zero polynomial is a solution. But are there other solutions?Case 2: Linear PolynomialLet me try ( P(x) = ax + b ). Then:LHS: ( ax + b + aleft(frac{1}{x}right) + b = aleft(x + frac{1}{x}right) + 2b ).RHS: ( frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2} ).Calculating each term:( Pleft(x + frac{1}{x}right) = aleft(x + frac{1}{x}right) + b )( Pleft(x - frac{1}{x}right) = aleft(x - frac{1}{x}right) + b )Average: ( frac{aleft(x + frac{1}{x}right) + b + aleft(x - frac{1}{x}right) + b}{2} = frac{2ax + 2b}{2} = ax + b ).So RHS is ( ax + b ). Comparing LHS and RHS:( aleft(x + frac{1}{x}right) + 2b = ax + b ).Simplify: ( a x + frac{a}{x} + 2b = a x + b ).Subtract ( ax + b ) from both sides: ( frac{a}{x} + b = 0 ).This must hold for all ( x neq 0 ). However, ( frac{a}{x} + b ) is a function that is only identically zero if both coefficients ( a ) and ( b ) are zero. Thus, ( a = 0 ) and ( b = 0 ). Hence, only the zero polynomial works here as well. So linear polynomials don't contribute new solutions.Case 3: Quadratic PolynomialLet me try ( P(x) = ax^2 + bx + c ). Let's compute both sides.LHS: ( ax^2 + bx + c + aleft(frac{1}{x^2}right) + bleft(frac{1}{x}right) + c ).Simplify: ( aleft(x^2 + frac{1}{x^2}right) + bleft(x + frac{1}{x}right) + 2c ).RHS: ( frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2} ).Compute ( Pleft(x + frac{1}{x}right) ):First, ( left(x + frac{1}{x}right)^2 = x^2 + 2 + frac{1}{x^2} ).Thus,( Pleft(x + frac{1}{x}right) = aleft(x^2 + 2 + frac{1}{x^2}right) + bleft(x + frac{1}{x}right) + c ).Similarly, ( Pleft(x - frac{1}{x}right) ):( left(x - frac{1}{x}right)^2 = x^2 - 2 + frac{1}{x^2} ).Thus,( Pleft(x - frac{1}{x}right) = aleft(x^2 - 2 + frac{1}{x^2}right) + bleft(x - frac{1}{x}right) + c ).Average these two:( frac{1}{2} left[ aleft(x^2 + 2 + frac{1}{x^2}right) + bleft(x + frac{1}{x}right) + c + aleft(x^2 - 2 + frac{1}{x^2}right) + bleft(x - frac{1}{x}right) + c right] ).Simplify term by term:For the ( a ) terms:( frac{1}{2} [ a(x^2 + 2 + 1/x^2 + x^2 - 2 + 1/x^2) ] = frac{1}{2} [ a(2x^2 + 2/x^2) ] = a(x^2 + 1/x^2) ).For the ( b ) terms:( frac{1}{2} [ b(x + 1/x + x - 1/x) ] = frac{1}{2} [ b(2x) ] = bx ).For the constant terms:( frac{1}{2} [ c + c ] = c ).Thus, RHS simplifies to:( a(x^2 + 1/x^2) + bx + c ).Compare to LHS:LHS was ( a(x^2 + 1/x^2) + b(x + 1/x) + 2c ).So, setting LHS = RHS:( a(x^2 + 1/x^2) + b(x + 1/x) + 2c = a(x^2 + 1/x^2) + bx + c ).Subtract RHS from LHS:( b(x + 1/x) + 2c - bx - c = b/x + c = 0 ).This must hold for all ( x neq 0 ), so the coefficients of ( 1/x ) and the constant term must be zero:Thus, ( b = 0 ) and ( c = 0 ). Therefore, the quadratic polynomial reduces to ( P(x) = ax^2 ).But let's check if this works. If ( P(x) = ax^2 ), then LHS is ( ax^2 + a(1/x^2) ), and RHS is ( a(x^2 + 1/x^2) ). So indeed, LHS = RHS. Therefore, any quadratic polynomial of the form ( ax^2 ) satisfies the equation. So quadratic polynomials provide non-trivial solutions.Case 4: Cubic PolynomialLet me try a cubic polynomial ( P(x) = ax^3 + bx^2 + cx + d ). This might get more complicated, but let's see.First, compute LHS:( P(x) + P(1/x) = ax^3 + bx^2 + cx + d + a(1/x^3) + b(1/x^2) + c(1/x) + d ).Simplify: ( a(x^3 + 1/x^3) + b(x^2 + 1/x^2) + c(x + 1/x) + 2d ).RHS: Average of ( P(x + 1/x) ) and ( P(x - 1/x) ).Compute ( P(x + 1/x) ):First, ( (x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3 ).So,( P(x + 1/x) = a(x + 1/x)^3 + b(x + 1/x)^2 + c(x + 1/x) + d ).Expanding term by term:- Cubic term: ( a(x^3 + 3x + 3/x + 1/x^3) )- Quadratic term: ( b(x^2 + 2 + 1/x^2) )- Linear term: ( c(x + 1/x) )- Constant term: ( d )Similarly, ( P(x - 1/x) ):First, ( (x - 1/x)^3 = x^3 - 3x + 3/x - 1/x^3 ).So,( P(x - 1/x) = a(x - 1/x)^3 + b(x - 1/x)^2 + c(x - 1/x) + d ).Expanding term by term:- Cubic term: ( a(x^3 - 3x + 3/x - 1/x^3) )- Quadratic term: ( b(x^2 - 2 + 1/x^2) )- Linear term: ( c(x - 1/x) )- Constant term: ( d )Now, average these two expressions:First, let's compute the average for each term:Cubic terms:Average of ( a(x^3 + 3x + 3/x + 1/x^3) ) and ( a(x^3 - 3x + 3/x - 1/x^3) ):( frac{a}{2} [ (x^3 + 3x + 3/x + 1/x^3) + (x^3 - 3x + 3/x - 1/x^3) ] = frac{a}{2} [2x^3 + 6/x] = a(x^3 + 3/x) ).Quadratic terms:Average of ( b(x^2 + 2 + 1/x^2) ) and ( b(x^2 - 2 + 1/x^2) ):( frac{b}{2} [ (x^2 + 2 + 1/x^2) + (x^2 - 2 + 1/x^2) ] = frac{b}{2} [2x^2 + 2/x^2] = b(x^2 + 1/x^2) ).Linear terms:Average of ( c(x + 1/x) ) and ( c(x - 1/x) ):( frac{c}{2} [ (x + 1/x) + (x - 1/x) ] = frac{c}{2} (2x) = c x ).Constant terms:Average of ( d ) and ( d ): ( d ).Putting it all together, the RHS is:( a(x^3 + 3/x) + b(x^2 + 1/x^2) + c x + d ).Compare this to the LHS, which was:( a(x^3 + 1/x^3) + b(x^2 + 1/x^2) + c(x + 1/x) + 2d ).Set LHS equal to RHS:( a(x^3 + 1/x^3) + b(x^2 + 1/x^2) + c(x + 1/x) + 2d = a(x^3 + 3/x) + b(x^2 + 1/x^2) + c x + d ).Subtract RHS from both sides:Left side minus right side:( a(x^3 + 1/x^3 - x^3 - 3/x) + b(x^2 + 1/x^2 - x^2 -1/x^2) + c(x + 1/x - x) + (2d - d) = 0 ).Simplify term by term:- For ( a ): ( a(1/x^3 - 3/x) )- For ( b ): ( 0 )- For ( c ): ( c(1/x) )- For constants: ( d )Thus, overall:( a left( frac{1}{x^3} - frac{3}{x} right) + frac{c}{x} + d = 0 ).Multiply through by ( x^3 ) to eliminate denominators:( a(1 - 3x^2) + c x^2 + d x^3 = 0 ).But this must hold for all ( x neq 0 ), which implies that the polynomial ( d x^3 + (c - 3a) x^2 + a = 0 ) must be identically zero. Therefore, each coefficient must be zero:- Coefficient of ( x^3 ): ( d = 0 )- Coefficient of ( x^2 ): ( c - 3a = 0 Rightarrow c = 3a )- Constant term: ( a = 0 )If ( a = 0 ), then ( c = 0 ), and ( d = 0 ). So the cubic polynomial reduces to ( P(x) = bx^2 ). Wait, but that's a quadratic polynomial. So essentially, the only cubic solutions are actually quadratic? That seems odd. Wait, maybe I made a miscalculation.Wait, when we set ( a = 0 ), then from ( c = 3a ), ( c = 0 ), and ( d = 0 ). So the polynomial would be ( P(x) = bx^2 ). But that's a quadratic polynomial, not cubic. Therefore, the conclusion is that there are no cubic solutions beyond the quadratic ones. So even if we started with a cubic polynomial, the only solution reduces to a quadratic. Therefore, cubic polynomials do not contribute new solutions.Case 5: Higher Degree PolynomialsPerhaps this pattern continues. Let's test quartic (degree 4) polynomial.Let ( P(x) = ax^4 + bx^3 + cx^2 + dx + e ). Computing LHS and RHS would be even more tedious, but maybe there's a pattern. However, given that quadratic works and cubic doesn't give anything new, maybe only even degree polynomials work?Alternatively, maybe all polynomials of the form ( P(x) = kx^{2n} ), where ( n ) is a non-negative integer. Let's test a quartic monomial ( P(x) = ax^4 ).Compute LHS: ( ax^4 + a(1/x^4) ).Compute RHS: Average of ( P(x + 1/x) ) and ( P(x - 1/x) ).First, ( (x + 1/x)^4 = x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4 ).So ( P(x + 1/x) = a(x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4) ).Similarly, ( (x - 1/x)^4 = x^4 - 4x^2 + 6 - 4/x^2 + 1/x^4 ).So ( P(x - 1/x) = a(x^4 - 4x^2 + 6 - 4/x^2 + 1/x^4) ).Average of these two:( frac{a}{2} [ (x^4 + 4x^2 + 6 + 4/x^2 + 1/x^4) + (x^4 - 4x^2 + 6 - 4/x^2 + 1/x^4) ] ).Simplify:( frac{a}{2} [2x^4 + 12 + 2/x^4] = a(x^4 + 6 + 1/x^4) ).Compare to LHS, which is ( a(x^4 + 1/x^4) ).Setting LHS = RHS:( a(x^4 + 1/x^4) = a(x^4 + 6 + 1/x^4) ).Subtract LHS from RHS:( 6a = 0 Rightarrow a = 0 ).So the quartic monomial doesn't work unless ( a = 0 ). Therefore, quartic polynomials of the form ( ax^4 ) do not satisfy the equation unless they are zero. Hmm, that's different from the quadratic case. So maybe higher even degrees don't work? Wait, but quadratic works. Let me check degree 6?Wait, perhaps there's a different structure here. Let me think.Alternatively, maybe only quadratic polynomials of the form ( ax^2 ) work. But why?Wait, let's consider a general even-degree monomial ( P(x) = ax^{2n} ). Let's compute LHS and RHS.LHS: ( ax^{2n} + a/x^{2n} ).RHS: Average of ( P(x + 1/x) ) and ( P(x - 1/x) ).Compute ( P(x + 1/x) = a(x + 1/x)^{2n} ).Similarly, ( P(x - 1/x) = a(x - 1/x)^{2n} ).Average: ( frac{a}{2} [ (x + 1/x)^{2n} + (x - 1/x)^{2n} ] ).Note that when we expand ( (x + 1/x)^{2n} + (x - 1/x)^{2n} ), all the odd-powered terms cancel out, leaving only even-powered terms. Specifically, the expansion will have terms like ( binom{2n}{k} x^{2n - 2k} ), where ( k ) is from 0 to ( n ).However, for the average to equal ( ax^{2n} + a/x^{2n} ), we need all the other terms in the expansion to cancel out. But this is only possible if ( n = 1 ), as we saw in the quadratic case.For example, with ( n = 2 ) (quartic), the expansion of ( (x + 1/x)^4 + (x - 1/x)^4 ) gives ( 2x^4 + 12x^0 + 2x^{-4} ), so the average is ( x^4 + 6 + x^{-4} ), which includes a constant term. Therefore, unless ( a = 0 ), the constant term remains, which isn't present in the LHS. Hence, higher even-degree monomials don't work.So perhaps only quadratic monomials work? Let's verify with ( P(x) = ax^2 ). Then as we saw earlier, LHS is ( a(x^2 + 1/x^2) ), and RHS is the average of ( a(x + 1/x)^2 ) and ( a(x - 1/x)^2 ), which is ( a(x^2 + 1/x^2) ). Thus, LHS equals RHS. So quadratic monomials work.Similarly, if we take a linear combination of such quadratics, but wait, if ( P(x) = ax^2 + bx^2 = (a + b)x^2 ), which is still a single quadratic term. Hence, the only monomial solutions are quadratic. Therefore, the general solution could be any quadratic polynomial of the form ( ax^2 ), but when we tried the general quadratic (with linear and constant terms), we found that only the quadratic term survives if ( b = 0 ) and ( c = 0 ). Wait, in the quadratic case earlier, when we considered ( P(x) = ax^2 + bx + c ), we ended up with the requirement that ( b = 0 ) and ( c = 0 ), leaving only ( ax^2 ). So the only quadratic solutions are ( ax^2 ).Wait, but maybe if we have higher-degree polynomials that can be expressed in terms of ( x^2 )? For example, maybe polynomials in ( x + 1/x ), or ( x - 1/x ), but since we need the polynomial to be in ( x ), perhaps only even-powered terms. Wait, but if we have a polynomial in ( x^2 ), like ( P(x) = a(x^2)^n ), then maybe such polynomials could work?Wait, let me check ( P(x) = a(x^2)^n ). Let's compute LHS and RHS.LHS: ( a x^{2n} + a/x^{2n} ).RHS: Average of ( a(x + 1/x)^{2n} ) and ( a(x - 1/x)^{2n} ).But as before, the average of these two expansions will contain terms other than ( x^{2n} ) and ( x^{-2n} ). For example, for ( n = 2 ), we get the average as ( a(x^4 + 6x^0 + x^{-4}) ), which has a constant term. Therefore, unless ( a = 0 ), there's an extra term. Hence, only ( n = 1 ) works, giving ( ax^2 ).Therefore, the only monomial solutions are quadratic, and linear combinations might not work unless they are scalar multiples of ( x^2 ). But since in the quadratic case, when we considered a general quadratic, only ( ax^2 ) survived. So linear combinations of such terms (i.e., scalar multiples) are allowed, but higher-degree terms do not.Case 6: Combining SolutionsSince the problem is linear in ( P ), if ( P_1 ) and ( P_2 ) are solutions, then so is ( P_1 + P_2 ). We already found that the zero polynomial is a solution (trivially), and quadratic monomials ( ax^2 ) are solutions. Are there any other solutions?Suppose ( P(x) = ax^2 + Q(x) ), where ( Q(x) ) is another polynomial. Then the equation becomes:( ax^2 + Q(x) + a/x^2 + Q(1/x) = frac{ a(x + 1/x)^2 + Q(x + 1/x) + a(x - 1/x)^2 + Q(x - 1/x) }{2} ).We already know that the terms involving ( ax^2 ) cancel out as before, leaving:( Q(x) + Q(1/x) = frac{ Q(x + 1/x) + Q(x - 1/x) }{2} ).Therefore, if ( Q(x) ) is another solution, then the entire equation reduces to the same functional equation for ( Q(x) ). Therefore, the general solution is a linear combination of the quadratic monomial solutions and solutions to the homogeneous equation ( Q(x) + Q(1/x) = frac{ Q(x + 1/x) + Q(x - 1/x) }{2} ). However, from previous cases, we saw that constant polynomials only work if zero, linear only zero, cubic only zero, etc. So perhaps the only solutions are quadratic polynomials ( ax^2 ).Wait, but maybe there are other polynomials ( Q(x) ) that satisfy the homogeneous equation. Let's suppose ( Q(x) ) is such a polynomial. Let's analyze its degree.Suppose ( Q(x) ) is a polynomial of degree ( n ). Then ( Q(1/x) ) is a polynomial of degree ( n ) in ( 1/x ), so when combined with ( Q(x) ), the left-hand side ( Q(x) + Q(1/x) ) has terms of degree from ( -n ) to ( n ). On the right-hand side, ( Q(x + 1/x) ) and ( Q(x - 1/x) ) are polynomials in ( x + 1/x ) and ( x - 1/x ), which when averaged and expanded, will have terms involving various powers of ( x ) and ( 1/x ).For the equation to hold, the coefficients of corresponding powers of ( x ) must match on both sides. Let's analyze the degrees.Suppose ( Q(x) ) is a monomial ( Q(x) = cx^k ). Then:LHS: ( cx^k + c x^{-k} ).RHS: Average of ( c(x + 1/x)^k ) and ( c(x - 1/x)^k ).Let’s compute this for different ( k ).If ( k ) is even:Let ( k = 2m ). Then:( (x + 1/x)^{2m} + (x - 1/x)^{2m} ) will expand to a sum of even-powered terms. For example, ( (x + 1/x)^2 = x^2 + 2 + 1/x^2 ), so when averaged, gives ( x^2 + 1/x^2 + 2 ). Wait, but in that case, the average would be:( frac{(x + 1/x)^{2m} + (x - 1/x)^{2m}}{2} ).If ( k = 2m ), then this average will contain terms like ( x^{2m} + x^{-2m} ) plus other even-powered terms. For example, when ( m = 1 ), as we saw, the average is ( x^2 + 1/x^2 + 2 ), which introduces a constant term. Therefore, unless ( c = 0 ), this introduces extra terms not present on the LHS, which only has ( x^{2m} + x^{-2m} ). Therefore, for ( k = 2m ), unless ( m = 1 ) (quadratic), but even then, the quadratic case had ( P(x) = ax^2 ), which when averaged with the quadratic term gave the correct terms. Wait, but when we considered ( k = 2 ), the average introduced a constant term. Wait, no:Wait, in the quadratic case, when we had ( Q(x) = ax^2 ), then the equation was:LHS: ( ax^2 + a/x^2 )RHS: average of ( a(x + 1/x)^2 ) and ( a(x - 1/x)^2 ), which is ( a(x^2 + 1/x^2) ). Wait, but ( (x + 1/x)^2 = x^2 + 2 + 1/x^2 ), and ( (x - 1/x)^2 = x^2 - 2 + 1/x^2 ). The average is indeed ( x^2 + 1/x^2 ). Therefore, in that case, the constants canceled out. Therefore, for ( k = 2 ), the average cancels the middle terms. But for ( k = 4 ), when we expand ( (x + 1/x)^4 + (x - 1/x)^4 ), we get ( 2x^4 + 12 + 2/x^4 ), so the average is ( x^4 + 6 + 1/x^4 ), which introduces a constant term. Therefore, unless ( c = 0 ), the equation isn't satisfied. Therefore, only ( k = 2 ) works if we set the coefficients appropriately. Wait, but in the quadratic case, even with ( k = 2 ), the constants canceled out. So maybe only ( k = 2 ) works?Wait, let's check ( k = 0 ), which is a constant. Then LHS is ( c + c = 2c ), and RHS is ( frac{c + c}{2} = c ). Therefore, 2c = c ⇒ c = 0. So constants only work if zero.For ( k = 1 ), linear, as before, we get that only the zero polynomial works.For ( k = 3 ), cubic:LHS: ( cx^3 + c/x^3 ).RHS: Average of ( c(x + 1/x)^3 ) and ( c(x - 1/x)^3 ).Expanding:( (x + 1/x)^3 = x^3 + 3x + 3/x + 1/x^3 )( (x - 1/x)^3 = x^3 - 3x + 3/x - 1/x^3 )Average: ( frac{1}{2}(2x^3 + 6/x) = x^3 + 3/x ).Thus, RHS is ( c(x^3 + 3/x) ). Comparing to LHS ( c(x^3 + 1/x^3) ), we get:( c(x^3 + 1/x^3) = c(x^3 + 3/x) ).Which implies:( c(1/x^3 - 3/x) = 0 ).Which for non-zero ( c ) requires ( 1/x^3 - 3/x = 0 ), which isn't true for all ( x ). Hence, ( c = 0 ).Similarly, for ( k = 4 ):LHS: ( cx^4 + c/x^4 ).RHS: Average of ( c(x + 1/x)^4 ) and ( c(x - 1/x)^4 ).As calculated before, the average is ( c(x^4 + 6 + 1/x^4) ).Setting equal to LHS:( c(x^4 + 1/x^4) = c(x^4 + 6 + 1/x^4) Rightarrow 6c = 0 Rightarrow c = 0 ).Therefore, only ( k = 2 ) works for monomials. Thus, the only monomial solutions are quadratic.Therefore, if we suppose that any solution must be a linear combination of monomial solutions, but since higher-degree monomials do not work, the only solutions are scalar multiples of ( x^2 ).But wait, the equation is linear, so if there are multiple solutions, their linear combinations are also solutions. But in our previous analysis, even when considering a general quadratic polynomial ( ax^2 + bx + c ), only ( ax^2 ) survived. So even if we combine ( x^2 ) terms with other terms, the other terms must be zero.Hence, the only non-trivial solutions are quadratic polynomials of the form ( P(x) = ax^2 ).But let's check another quadratic polynomial. Suppose ( P(x) = ax^2 + bx^2 = (a + b)x^2 ), which is still a quadratic term, so the space of solutions is one-dimensional, spanned by ( x^2 ). So the general solution is ( P(x) = kx^2 ), where ( k ) is a real constant.But wait, in the quadratic case, when we had ( P(x) = ax^2 + bx + c ), the linear and constant terms forced ( b = 0 ) and ( c = 0 ). Therefore, even if we tried adding higher-degree terms, they would have to vanish. Therefore, the only solutions are scalar multiples of ( x^2 ).Verification:Let me verify once more that ( P(x) = kx^2 ) satisfies the equation.LHS: ( kx^2 + k(1/x^2) ).RHS: Average of ( k(x + 1/x)^2 ) and ( k(x - 1/x)^2 ).Compute each:( (x + 1/x)^2 = x^2 + 2 + 1/x^2 ).( (x - 1/x)^2 = x^2 - 2 + 1/x^2 ).Average: ( frac{(x^2 + 2 + 1/x^2) + (x^2 - 2 + 1/x^2)}{2} = frac{2x^2 + 2/x^2}{2} = x^2 + 1/x^2 ).Multiply by ( k ): ( kx^2 + k/x^2 ).Which matches LHS. Therefore, verified.Conclusion:After testing constant, linear, quadratic, cubic, and quartic polynomials, and considering the structure of the equation, the only polynomials that satisfy the given condition are scalar multiples of ( x^2 ). Therefore, all solutions are of the form ( P(x) = kx^2 ) where ( k ) is a real constant.Final Answerboxed{P(x) = ax^2} where (a) is a real constant.</think>To find all polynomials (P(x)) with real coefficients such that for all nonzero real numbers (x),[P(x) + Pleft(frac{1}{x}right) = frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2},]we analyze the equation step by step:1. Constant Polynomials: - If (P(x) = c), then the left-hand side (LHS) is (2c) and the right-hand side (RHS) is (c). This implies (c = 0). Thus, the zero polynomial is a solution.2. Linear Polynomials: - If (P(x) = ax + b), the LHS becomes (aleft(x + frac{1}{x}right) + 2b) and the RHS simplifies to (ax + b). This leads to (a = 0) and (b = 0), so only the zero polynomial works.3. Quadratic Polynomials: - For (P(x) = ax^2 + bx + c), the LHS is (aleft(x^2 + frac{1}{x^2}right) + bleft(x + frac{1}{x}right) + 2c). The RHS simplifies to (aleft(x^2 + frac{1}{x^2}right) + bx + c). Equating both sides, we find (b = 0) and (c = 0), leaving (P(x) = ax^2).4. Cubic Polynomials: - Testing (P(x) = ax^3 + bx^2 + cx + d) leads to terms that cannot be balanced unless (a = 0), (b = 0), (c = 0), and (d = 0), reducing to the zero polynomial.5. Higher-Degree Polynomials: - For higher even degrees like quartic ((ax^4)), the expansion introduces additional terms (e.g., constants) that cannot be canceled unless (a = 0). Similar issues arise for higher degrees, confirming only quadratic terms work.6. General Case: - Considering polynomials in (x + frac{1}{x}) and (x - frac{1}{x}) shows that only quadratic polynomials of the form (ax^2) satisfy the equation without introducing extraneous terms.Verification confirms that (P(x) = ax^2) satisfies the given functional equation. Thus, the only solutions are quadratic polynomials of the form (ax^2).[boxed{P(x) = ax^2}]where (a) is a real constant.]

🔑:1. Given the functional equation for the polynomial (P(x)): [ P(x) + Pleft(frac{1}{x}right) = frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2} ] we start by replacing (x) with (frac{1}{x}): [ Pleft(frac{1}{x}right) + P(x) = frac{Pleft(frac{1}{x} + xright) + Pleft(frac{1}{x} - xright)}{2} ] This is the same as the original equation, indicating symmetry.2. Next, we observe that: [ Pleft(x - frac{1}{x}right) = Pleft(frac{1}{x} - xright) ] Since (z = x - frac{1}{x}) can take all real values for (x in mathbb{R}), we conclude: [ P(z) = P(-z) quad forall z in mathbb{R} ] This implies that (P(x)) is an even function, and thus can be written as: [ P(x) = Q(x^2) ] for some polynomial (Q(x)) with real coefficients.3. Let us assume (deg(P(x)) = 2n), where (n in mathbb{Z}^{+}). We claim that (n leq 2).4. Assume for contradiction that (n > 2). Let: [ P(x) = a_{2n}x^{2n} + a_{2n-2}x^{2n-2} + a_{2n-4}x^{2n-4} + cdots + a_0 ] where (a_{2i} in mathbb{R}) and (a_{2n} neq 0).5. Compare the degrees of (x^{2n-4}) in the original equation: [ 2a_{2n-4} = 2 left( binom{2n-2}{2n} a_{2n} + binom{2n-3}{2n-2} a_{2n-2} + binom{2n-4}{2n-4} a_{2n-4} right) ] Simplifying, we get: [ 2a_{2n-4} = 2 left( 0 + 0 + a_{2n-4} right) implies a_{2n-4} = 0 ]6. Compare the degrees of (x^{2n-2}) in the given relation: [ 2n(2n-1)a_{2n} + (2n-2)a_{2n-2} = 0 ] Since (a_{2n-2} = 0), we get: [ 2n(2n-1)a_{2n} = 0 implies a_{2n} = 0 ] This is a contradiction since (a_{2n} neq 0). Hence, (n leq 2).7. Therefore, the degree of (P(x)) is at most 4. Let: [ P(x) = ax^4 + bx^2 + c ] Plugging this into the original equation, we get: [ ax^4 + bx^2 + c + aleft(frac{1}{x}right)^4 + bleft(frac{1}{x}right)^2 + c = frac{Pleft(x + frac{1}{x}right) + Pleft(x - frac{1}{x}right)}{2} ] Simplifying, we find: [ c = 6a ] Thus, the polynomial (P(x)) that satisfies the given relation is: [ P(x) = ax^4 + bx^2 + 6a ]The final answer is (boxed{ax^4 + bx^2 + 6a}).