💡First, I need to determine how much work A can complete in one day. Since A can finish the work in 15 days, A's work rate is ( frac{1}{15} ) of the work per day.Next, I know that A and B together can complete the work in 10 days, which means their combined work rate is ( frac{1}{10} ) of the work per day.To find out how much work B can do alone, I'll subtract A's work rate from the combined work rate:[frac{1}{10} - frac{1}{15} = frac{3}{30} - frac{2}{30} = frac{1}{30}]This means B's work rate is ( frac{1}{30} ) of the work per day.Finally, to find out how many days B alone will take to complete the work, I take the reciprocal of B's work rate:[x = 30]

💡Alright, so I have this probability problem here about a team answering questions in a contest. Let me try to figure it out step by step.First, the problem says that a unit selects three people, A, B, and C, to form a team. When they answer a question together, we know a few probabilities:1. The probability that A answers correctly is 3/4.2. The probability that both A and C answer incorrectly is 1/12.3. The probability that both B and C answer correctly is 1/4.And the team is considered to have answered correctly if at least one person answers correctly. So, part (I) is asking for the probability that the team answers correctly.Okay, let's break this down. I think I need to find the probability that at least one of A, B, or C answers correctly. To find that, it might be easier to first find the probability that none of them answer correctly and then subtract that from 1.So, let me denote the probabilities of A, B, and C answering correctly as P(A), P(B), and P(C) respectively. We already know P(A) = 3/4.Next, the problem says that the probability of both A and C answering incorrectly is 1/12. So, that's P(not A and not C) = 1/12. Since A and C might not be independent, but I think in this case, we can assume independence unless stated otherwise. Wait, actually, the problem doesn't specify if their answers are independent. Hmm, that might complicate things.But let's see. If we assume that A and C's answers are independent, then P(not A and not C) = P(not A) * P(not C). Since P(A) = 3/4, P(not A) = 1 - 3/4 = 1/4. So, P(not A) * P(not C) = 1/4 * P(not C) = 1/12. Therefore, P(not C) = (1/12) / (1/4) = 1/3. So, P(C) = 1 - P(not C) = 1 - 1/3 = 2/3.Okay, so P(C) is 2/3.Now, the problem also says that the probability of both B and C answering correctly is 1/4. So, P(B and C) = 1/4. If we assume that B and C are independent, then P(B and C) = P(B) * P(C). We already know P(C) = 2/3, so P(B) = (1/4) / (2/3) = (1/4) * (3/2) = 3/8.So, P(B) is 3/8.Now, we have all three probabilities:- P(A) = 3/4- P(B) = 3/8- P(C) = 2/3Assuming that A, B, and C answer independently, the probability that all three answer incorrectly is P(not A) * P(not B) * P(not C). Let's compute that.P(not A) = 1 - 3/4 = 1/4P(not B) = 1 - 3/8 = 5/8P(not C) = 1 - 2/3 = 1/3So, P(all incorrect) = (1/4) * (5/8) * (1/3) = (1 * 5 * 1) / (4 * 8 * 3) = 5 / 96Therefore, the probability that at least one answers correctly is 1 - P(all incorrect) = 1 - 5/96 = 91/96.So, the answer to part (I) is 91/96.Now, moving on to part (II). The competition has 10 compulsory questions. For each correct answer, the team gets 20 points, and for each incorrect answer, they lose 10 points. There's also a mention of "not scoring for that question," but I think that might just mean that if they don't answer, they get zero, but in this case, since they have to answer, it's either correct or incorrect.We need to find the expected score of the team for these 10 questions. The probability of answering each question correctly is the same as we found in part (I), which is 91/96. Also, the correctness of each question is independent of the others.So, let's model this. Let’s denote ξ as the number of questions answered correctly. Since each question is independent and has the same probability, ξ follows a binomial distribution with parameters n = 10 and p = 91/96.The score η is calculated as 20 points for each correct answer and -10 points for each incorrect answer. Since there are 10 questions, the total score can be expressed as:η = 20ξ - 10(10 - ξ) = 20ξ - 100 + 10ξ = 30ξ - 100So, η = 30ξ - 100.To find the expected score E[η], we can use the linearity of expectation:E[η] = E[30ξ - 100] = 30E[ξ] - 100We know that for a binomial distribution, E[ξ] = n * p = 10 * (91/96).So, E[ξ] = 10 * (91/96) = 910/96.Then, E[η] = 30 * (910/96) - 100.Let me compute that:First, compute 30 * (910/96):30 * 910 = 27,30027,300 / 96 = Let's divide 27,300 by 96.96 * 284 = 27,264 (since 96*280=26,880; 96*4=384; 26,880+384=27,264)So, 27,300 - 27,264 = 36So, 27,300 / 96 = 284 + 36/96 = 284 + 3/8 = 284.375Then, subtract 100:284.375 - 100 = 184.375But let me express this as a fraction. 36/96 simplifies to 3/8, so 284 3/8 - 100 = 184 3/8.So, 184 3/8 is equal to 1475/8 because 184 * 8 = 1472, plus 3 is 1475.Therefore, E[η] = 1475/8.So, the expected score is 1475/8.Let me just double-check my calculations to make sure I didn't make a mistake.First, P(not A) = 1/4, P(not C) = 1/3, so P(not A and not C) = 1/4 * 1/3 = 1/12, which matches the given information. Then, P(B and C) = 1/4, and since P(C) = 2/3, P(B) = (1/4)/(2/3) = 3/8, which seems correct.Then, P(all incorrect) = 1/4 * 5/8 * 1/3 = 5/96, so 1 - 5/96 = 91/96, which is correct.For the expected score, η = 30ξ - 100. E[ξ] = 10 * 91/96 = 910/96. Then, 30 * 910/96 = 27,300/96 = 284.375. Subtract 100 gives 184.375, which is 1475/8. That seems correct.So, I think my answers are correct.**Final Answer**(I) The probability of the team answering correctly is boxed{dfrac{91}{96}}.(II) The expected score of the team is boxed{dfrac{1475}{8}}.

💡Okay, so I have this polynomial P(x) = x⁴ + x³ + x² + x + 1, and I need to prove that there exist polynomials Q(y) and R(y) with integer coefficients and positive degree such that Q(y) · R(y) = P(5y²). Hmm, that sounds a bit complicated, but let me try to break it down step by step.First, I should understand what P(5y²) actually is. Since P(x) is a quartic polynomial, substituting 5y² for x will give me another polynomial in terms of y. Let me compute that:P(5y²) = (5y²)⁴ + (5y²)³ + (5y²)² + 5y² + 1.Calculating each term:- (5y²)⁴ = 5⁴ y⁸ = 625y⁸- (5y²)³ = 5³ y⁶ = 125y⁶- (5y²)² = 5² y⁴ = 25y⁴- 5y² = 5y²- 1 = 1So, putting it all together:P(5y²) = 625y⁸ + 125y⁶ + 25y⁴ + 5y² + 1.Alright, so now I have this polynomial in y. The task is to factor this into two polynomials Q(y) and R(y) with integer coefficients. Both Q and R should have positive degrees, which means they can't just be constants; they have to be actual polynomials.Looking at the structure of P(5y²), it seems like a palindromic polynomial because the coefficients read the same forwards and backwards: 625, 125, 25, 5, 1. Wait, actually, it's not exactly palindromic because the coefficients aren't symmetric in the usual sense, but they do follow a pattern of decreasing powers of 5. Maybe that can help in factoring.I remember that sometimes polynomials can be factored into products of lower-degree polynomials, especially if they have some symmetry or follow a particular pattern. Since P(x) is a cyclotomic polynomial, specifically the 5th cyclotomic polynomial, which factors as (x⁵ - 1)/(x - 1). But I'm not sure if that helps directly here.Alternatively, maybe I can factor P(5y²) by grouping terms or looking for patterns. Let me see:625y⁸ + 125y⁶ + 25y⁴ + 5y² + 1.I notice that each term is a multiple of 5 raised to a power, multiplied by y raised to an even power. Maybe I can factor this as a product of two quartic polynomials (degree 4). Let me assume that:P(5y²) = (a y⁴ + b y³ + c y² + d y + e)(f y⁴ + g y³ + h y² + k y + m).Since the leading term is 625y⁸, the product of the leading coefficients a and f must be 625. Similarly, the constant term is 1, so e and m must multiply to 1. Since we're dealing with integer coefficients, e and m are both 1 or both -1. But since all coefficients in P(5y²) are positive, I can assume e = m = 1.So, now I have:( a y⁴ + b y³ + c y² + d y + 1 )( f y⁴ + g y³ + h y² + k y + 1 ) = 625y⁸ + 125y⁶ + 25y⁴ + 5y² + 1.Multiplying these two quartic polynomials will give me an octic polynomial. Let me think about the coefficients:- The leading term is a*f y⁸ = 625 y⁸, so a*f = 625.- The next term is (a*g + b*f) y⁷. But in P(5y²), the coefficient of y⁷ is 0, so a*g + b*f = 0.- The coefficient of y⁶ is (a*h + b*g + c*f) = 125.- The coefficient of y⁵ is (a*k + b*h + c*g + d*f) = 0 (since there's no y⁵ term in P(5y²)).- The coefficient of y⁴ is (a*1 + b*k + c*h + d*g + e*f) = 25.- The coefficient of y³ is (b*1 + c*k + d*h + e*g) = 0.- The coefficient of y² is (c*1 + d*k + e*h) = 5.- The coefficient of y is (d*1 + e*k) = 0.- The constant term is e*m = 1, which we already have.This gives me a system of equations:1. a*f = 6252. a*g + b*f = 03. a*h + b*g + c*f = 1254. a*k + b*h + c*g + d*f = 05. a + b*k + c*h + d*g + f = 256. b + c*k + d*h + e*g = 07. c + d*k + e*h = 58. d + e*k = 09. e = 1, m = 1Since e = 1, equation 8 becomes d + k = 0, so k = -d.From equation 2: a*g + b*f = 0. Since a and f are positive integers (because all coefficients in P(5y²) are positive), and a*f = 625, which is 5⁴, possible pairs for (a, f) are (1,625), (5,125), (25,25), (125,5), (625,1). Let's try a = 25 and f = 25 because it's symmetric and might simplify things.So, a = 25, f = 25.Then equation 2: 25*g + b*25 = 0 => 25(g + b) = 0 => g + b = 0 => g = -b.Equation 3: 25*h + b*g + c*25 = 125. Substitute g = -b:25*h + b*(-b) + 25*c = 125 => 25h - b² + 25c = 125.Equation 4: 25*k + b*h + c*g + d*25 = 0. Substitute g = -b and k = -d:25*(-d) + b*h + c*(-b) + 25*d = 0 => -25d + b*h - b*c + 25d = 0 => b*h - b*c = 0 => b(h - c) = 0.So either b = 0 or h = c.If b = 0, then from equation 2, g = 0. Let's see if that works.If b = 0, then equation 3 becomes 25h + 25c = 125 => h + c = 5.Equation 5: 25 + 0 + c*h + d*g + 25 = 25 => 50 + c*h + d*g = 25. But d*g = d*(-b) = d*0 = 0, so 50 + c*h = 25 => c*h = -25. But c and h are integers, and since h + c = 5, we have c*(5 - c) = -25 => 5c - c² = -25 => c² - 5c -25 = 0. The discriminant is 25 + 100 = 125, which is not a perfect square, so c is not integer. Thus, b cannot be 0.Therefore, h = c.So, h = c. Then equation 3: 25h - b² + 25c = 125. Since h = c, this becomes 25c - b² + 25c = 125 => 50c - b² = 125 => 50c = b² + 125.Equation 5: 25 + b*k + c*h + d*g + 25 = 25. Substitute h = c, g = -b, k = -d:25 + b*(-d) + c*c + d*(-b) + 25 = 25 => 50 - b*d + c² - b*d = 25 => 50 - 2b*d + c² = 25 => -2b*d + c² = -25.Equation 7: c + d*k + e*h = 5. Substitute k = -d, e =1, h = c:c + d*(-d) + c = 5 => 2c - d² = 5.Equation 6: b + c*k + d*h + e*g = 0. Substitute k = -d, h = c, e =1, g = -b:b + c*(-d) + d*c + (-b) = 0 => b - c*d + c*d - b = 0 => 0 = 0. So equation 6 is automatically satisfied.So, summarizing the equations we have:From equation 3: 50c = b² + 125 => b² = 50c - 125.From equation 5: -2b*d + c² = -25.From equation 7: 2c - d² = 5.We need to find integers b, c, d such that these equations are satisfied.Let me try to express d from equation 7:2c - d² = 5 => d² = 2c -5.Since d² must be non-negative, 2c -5 ≥ 0 => c ≥ 3.Also, from equation 3: b² = 50c - 125. Since b² must be non-negative, 50c -125 ≥ 0 => c ≥ 2.5. Since c is integer, c ≥ 3.So c starts from 3.Let me try c = 3:d² = 2*3 -5 = 6 -5 =1 => d = ±1.From equation 3: b² = 50*3 -125 =150 -125=25 => b=±5.From equation 5: -2b*d + c² = -25 => -2b*d +9 = -25 => -2b*d = -34 => b*d =17.But b=±5, d=±1. So b*d=±5. But 17 is prime, so no solution here.Next, c=4:d²=2*4 -5=8-5=3. Not a square. Discard.c=5:d²=2*5 -5=10-5=5. Not a square. Discard.c=6:d²=12-5=7. Not a square.c=7:d²=14-5=9 => d=±3.From equation 3: b²=50*7 -125=350-125=225 => b=±15.From equation 5: -2b*d + c² = -25 => -2b*d +49 = -25 => -2b*d = -74 => b*d=37.But b=±15, d=±3. So b*d=±45 or ±45. 37 is prime, no solution.c=8:d²=16-5=11. Not a square.c=9:d²=18-5=13. Not a square.c=10:d²=20-5=15. Not a square.c=11:d²=22-5=17. Not a square.c=12:d²=24-5=19. Not a square.c=13:d²=26-5=21. Not a square.c=14:d²=28-5=23. Not a square.c=15:d²=30-5=25 => d=±5.From equation 3: b²=50*15 -125=750-125=625 => b=±25.From equation 5: -2b*d + c² = -25 => -2b*d +225 = -25 => -2b*d = -250 => b*d=125.With b=±25, d=±5. So b*d=±125. So possible if b=25, d=5 or b=-25, d=-5.Let's check b=25, d=5:From equation 7: 2c -d²=5 => 2*15 -25=30-25=5. Correct.From equation 5: -2*25*5 +15²= -250 +225= -25. Correct.So this works.Thus, c=15, d=5, b=25.So, now we have:a=25, f=25b=25, g=-25c=15, h=15d=5, k=-5e=1, m=1So, the polynomials are:Q(y) = a y⁴ + b y³ + c y² + d y + e =25 y⁴ +25 y³ +15 y² +5 y +1R(y) = f y⁴ +g y³ +h y² +k y +m=25 y⁴ -25 y³ +15 y² -5 y +1Let me check if their product is indeed P(5y²):Multiply Q(y) and R(y):(25y⁴ +25y³ +15y² +5y +1)(25y⁴ -25y³ +15y² -5y +1)Let me compute term by term:First, 25y⁴ * 25y⁴ =625y⁸25y⁴*(-25y³)= -625y⁷25y⁴*15y²=375y⁶25y⁴*(-5y)= -125y⁵25y⁴*1=25y⁴25y³*25y⁴=625y⁷25y³*(-25y³)= -625y⁶25y³*15y²=375y⁵25y³*(-5y)= -125y⁴25y³*1=25y³15y²*25y⁴=375y⁶15y²*(-25y³)= -375y⁵15y²*15y²=225y⁴15y²*(-5y)= -75y³15y²*1=15y²5y*25y⁴=125y⁵5y*(-25y³)= -125y⁴5y*15y²=75y³5y*(-5y)= -25y²5y*1=5y1*25y⁴=25y⁴1*(-25y³)= -25y³1*15y²=15y²1*(-5y)= -5y1*1=1Now, let's collect like terms:y⁸: 625y⁸y⁷: -625y⁷ +625y⁷ =0y⁶:375y⁶ -625y⁶ +375y⁶ = (375 -625 +375)y⁶=125y⁶y⁵:-125y⁵ +375y⁵ -375y⁵ +125y⁵= (-125 +375 -375 +125)y⁵=0y⁴:25y⁴ -125y⁴ +225y⁴ -125y⁴ +25y⁴= (25 -125 +225 -125 +25)y⁴=25y⁴y³:25y³ -75y³ -25y³= (25 -75 -25)y³= -75y³y²:15y² -25y² +15y²= (15 -25 +15)y²=5y²y:5y -5y=0Constants:1Wait, but in P(5y²), we have 625y⁸ +125y⁶ +25y⁴ +5y² +1. However, in the product, I have:625y⁸ +125y⁶ +25y⁴ +5y² +1 -75y³.But P(5y²) doesn't have a y³ term. That means I must have made a mistake in my calculation.Wait, let me check the coefficients again.Looking back at the multiplication:When I collected y³ terms:25y³ -75y³ -25y³= (25 -75 -25)= -75y³. But in P(5y²), there is no y³ term, so this suggests that my factorization is incorrect.Hmm, that's a problem. Maybe I made a mistake in the setup.Wait, let me double-check the equations. I assumed that both Q and R are quartic polynomials with leading coefficient 25. Maybe that's not the only possibility. Let me try a different pair for a and f.Earlier, I tried a=25, f=25. Maybe I should try a=5, f=125 or a=125, f=5.Let me try a=5, f=125.Then, equation 2: a*g + b*f =0 =>5g +125b=0 =>5g = -125b =>g= -25b.Equation 3: a*h + b*g +c*f=125 =>5h +b*(-25b) +c*125=125 =>5h -25b² +125c=125.Equation 4: a*k + b*h +c*g +d*f=0 =>5k +b*h +c*(-25b) +d*125=0 =>5k +b*h -25b² +125d=0.Equation 5: a +b*k +c*h +d*g +f=25 =>5 +b*k +c*h +d*(-25b) +125=25 =>5 +b*k +c*h -25b*d +125=25 =>b*k +c*h -25b*d= -105.Equation 6: b +c*k +d*h +e*g=0 =>b +c*k +d*h +1*(-25b)=0 =>b +c*k +d*h -25b=0 =>c*k +d*h -24b=0.Equation 7: c +d*k +e*h=5 =>c +d*k +1*h=5.Equation 8: d +e*k=0 =>d +1*k=0 =>k= -d.So, k= -d.From equation 6: c*(-d) +d*h -24b=0 => -c*d +d*h -24b=0 =>d*(h -c) -24b=0.From equation 7: c +d*(-d) +h=5 =>c -d² +h=5.From equation 3:5h -25b² +125c=125 =>5h +125c=25b² +125 =>Divide both sides by 5: h +25c=5b² +25 =>h=5b² +25 -25c.From equation 7: c -d² +h=5. Substitute h:c -d² +5b² +25 -25c=5 => -24c -d² +5b² +25=5 => -24c -d² +5b²= -20.From equation 6: d*(h -c) -24b=0. Substitute h=5b² +25 -25c:d*(5b² +25 -25c -c) -24b=0 =>d*(5b² +25 -26c) -24b=0.This is getting complicated. Maybe I can make some assumptions.Let me assume that b is small, say b=1.Then, from equation 3: h=5(1)² +25 -25c=5 +25 -25c=30 -25c.From equation 7: c -d² +h=5 =>c -d² +30 -25c=5 =>-24c -d² +30=5 =>-24c -d²= -25 =>24c +d²=25.Since c and d are integers, 24c +d²=25.Possible c:c=0: d²=25 =>d=±5c=1:24 +d²=25 =>d²=1 =>d=±1c=2:48 +d²=25 =>d² negative, invalid.So c=0 or c=1.If c=0:From equation 3: h=30 -0=30.From equation 7:0 -d² +30=5 =>-d²= -25 =>d²=25 =>d=±5.From equation 6: d*(h -c) -24b= d*(30 -0) -24*1=30d -24=0 =>30d=24 =>d=24/30=4/5. Not integer. Discard.If c=1:From equation 3: h=30 -25=5.From equation 7:1 -d² +5=5 =>1 -d²=0 =>d²=1 =>d=±1.From equation 6: d*(h -c) -24b= d*(5 -1) -24=4d -24=0 =>4d=24 =>d=6. But d²=36≠1. Contradiction. So b=1 doesn't work.Try b=2:From equation 3: h=5*(4) +25 -25c=20 +25 -25c=45 -25c.From equation 7: c -d² +45 -25c=5 =>-24c -d² +45=5 =>-24c -d²= -40 =>24c +d²=40.Possible c:c=0: d²=40. Not square.c=1:24 +d²=40 =>d²=16 =>d=±4.c=2:48 +d²=40 =>d² negative. Invalid.So c=1, d=±4.From equation 6: d*(h -c) -24b= d*(45 -25*1 -1)=d*(45 -25 -1)=d*19 -24*2=19d -48=0 =>19d=48 =>d=48/19. Not integer. Discard.Next, b=3:From equation 3: h=5*9 +25 -25c=45 +25 -25c=70 -25c.From equation 7: c -d² +70 -25c=5 =>-24c -d² +70=5 =>-24c -d²= -65 =>24c +d²=65.Possible c:c=0: d²=65. Not square.c=1:24 +d²=65 =>d²=41. Not square.c=2:48 +d²=65 =>d²=17. Not square.c=3:72 +d²=65 =>d² negative. Invalid.No solution.b=4:From equation 3: h=5*16 +25 -25c=80 +25 -25c=105 -25c.From equation 7: c -d² +105 -25c=5 =>-24c -d² +105=5 =>-24c -d²= -100 =>24c +d²=100.Possible c:c=0: d²=100 =>d=±10.c=1:24 +d²=100 =>d²=76. Not square.c=2:48 +d²=100 =>d²=52. Not square.c=3:72 +d²=100 =>d²=28. Not square.c=4:96 +d²=100 =>d²=4 =>d=±2.So c=0 or c=4.If c=0:From equation 3: h=105 -0=105.From equation 7:0 -d² +105=5 =>-d²= -100 =>d²=100 =>d=±10.From equation 6: d*(h -c) -24b= d*(105 -0) -24*4=105d -96=0 =>105d=96 =>d=96/105=32/35. Not integer.If c=4:From equation 3: h=105 -25*4=105 -100=5.From equation 7:4 -d² +5=5 =>4 -d²=0 =>d²=4 =>d=±2.From equation 6: d*(h -c) -24b= d*(5 -4) -24*4= d*1 -96=0 =>d=96. But d²=9216≠4. Contradiction.No solution for b=4.This is getting too time-consuming. Maybe my initial assumption of a=25, f=25 was correct, but I must have made a mistake in the multiplication.Wait, let me re-examine the multiplication. I think I might have missed some terms or miscalculated.Let me try multiplying Q(y) and R(y) again:Q(y) =25y⁴ +25y³ +15y² +5y +1R(y)=25y⁴ -25y³ +15y² -5y +1Multiply term by term:First, 25y⁴ *25y⁴=625y⁸25y⁴*(-25y³)= -625y⁷25y⁴*15y²=375y⁶25y⁴*(-5y)= -125y⁵25y⁴*1=25y⁴25y³*25y⁴=625y⁷25y³*(-25y³)= -625y⁶25y³*15y²=375y⁵25y³*(-5y)= -125y⁴25y³*1=25y³15y²*25y⁴=375y⁶15y²*(-25y³)= -375y⁵15y²*15y²=225y⁴15y²*(-5y)= -75y³15y²*1=15y²5y*25y⁴=125y⁵5y*(-25y³)= -125y⁴5y*15y²=75y³5y*(-5y)= -25y²5y*1=5y1*25y⁴=25y⁴1*(-25y³)= -25y³1*15y²=15y²1*(-5y)= -5y1*1=1Now, collect like terms:y⁸:625y⁸y⁷:-625y⁷ +625y⁷=0y⁶:375y⁶ -625y⁶ +375y⁶= (375 -625 +375)y⁶=125y⁶y⁵:-125y⁵ +375y⁵ -375y⁵ +125y⁵= (-125 +375 -375 +125)y⁵=0y⁴:25y⁴ -125y⁴ +225y⁴ -125y⁴ +25y⁴= (25 -125 +225 -125 +25)y⁴=25y⁴y³:25y³ -75y³ -25y³= (25 -75 -25)y³= -75y³y²:15y² -25y² +15y²= (15 -25 +15)y²=5y²y:5y -5y=0Constants:1Wait, so the product is 625y⁸ +125y⁶ +25y⁴ -75y³ +5y² +1. But P(5y²)=625y⁸ +125y⁶ +25y⁴ +5y² +1. So there's an extra -75y³ term. That means my factorization is incorrect.Hmm, so my initial assumption that a=25, f=25 leads to an incorrect product. Maybe I need to adjust the coefficients.Alternatively, perhaps I should consider that P(x) is related to the 5th roots of unity. Since P(x) = (x⁵ -1)/(x -1), which factors as a product of cyclotomic polynomials. But I'm not sure how that helps with factoring P(5y²).Wait, another approach: Maybe P(5y²) can be written as a product of two quadratics in y², but that might not lead to integer coefficients.Alternatively, perhaps I can factor P(5y²) as a product of two quartic polynomials with integer coefficients by using the fact that P(x) is symmetric.Wait, P(x) is symmetric, meaning P(x) = x⁴ +x³ +x² +x +1 = (x⁴ +x³ +x² +x +1). So P(5y²) = (5y²)^4 + (5y²)^3 + (5y²)^2 +5y² +1.Since P(x) is symmetric, maybe P(5y²) can be factored into two polynomials that are symmetric in some way.Alternatively, perhaps I can use the identity that x⁴ +x³ +x² +x +1 = (x² + (1 + sqrt(5))/2 x +1)(x² + (1 - sqrt(5))/2 x +1). But these factors have irrational coefficients, so that won't help with integer coefficients.Wait, but maybe over integers, P(x) is irreducible, but when composed with 5y², it becomes reducible. So perhaps P(5y²) factors into two quartic polynomials with integer coefficients.Given that, maybe I can use the fact that P(x) is the 5th cyclotomic polynomial, and when composed with 5y², it might factor into products of polynomials corresponding to certain roots.Alternatively, perhaps I can use the fact that P(x) divides x⁵ -1, so P(5y²) divides (5y²)^5 -1=5⁵ y¹⁰ -1=3125y¹⁰ -1.But I'm not sure if that helps directly.Wait, another idea: Maybe P(5y²) can be written as (25y⁴ + ay³ + by² + cy +1)(25y⁴ -ay³ + by² -cy +1). Let me try this form, assuming symmetry.Let me denote Q(y) =25y⁴ + ay³ + by² + cy +1R(y)=25y⁴ -ay³ + by² -cy +1Then, Q(y)*R(y)= (25y⁴ +1)^2 + (by²)^2 - (ay³)^2 - (cy)^2 + cross terms.Wait, actually, let me compute the product:(25y⁴ + ay³ + by² + cy +1)(25y⁴ -ay³ + by² -cy +1)Multiply term by term:25y⁴*25y⁴=625y⁸25y⁴*(-ay³)= -25a y⁷25y⁴*by²=25b y⁶25y⁴*(-cy)= -25c y⁵25y⁴*1=25y⁴ay³*25y⁴=25a y⁷ay³*(-ay³)= -a² y⁶ay³*by²=ab y⁵ay³*(-cy)= -ac y⁴ay³*1= a y³by²*25y⁴=25b y⁶by²*(-ay³)= -ab y⁵by²*by²= b² y⁴by²*(-cy)= -bc y³by²*1= b y²cy*25y⁴=25c y⁵cy*(-ay³)= -ac y⁴cy*by²= bc y³cy*(-cy)= -c² y²cy*1= c y1*25y⁴=25y⁴1*(-ay³)= -a y³1*by²= b y²1*(-cy)= -c y1*1=1Now, collect like terms:y⁸:625y⁸y⁷:-25a y⁷ +25a y⁷=0y⁶:25b y⁶ -a² y⁶ +25b y⁶= (50b -a²)y⁶y⁵:-25c y⁵ +ab y⁵ -ab y⁵ +25c y⁵=0y⁴:25y⁴ -ac y⁴ +b² y⁴ -ac y⁴ +25y⁴= (50 -2ac +b²)y⁴y³:a y³ -bc y³ +bc y³ -a y³=0y²:b y² -c² y² +b y²= (2b -c²)y²y:c y -c y=0Constants:1So, the product is:625y⁸ + (50b -a²)y⁶ + (50 -2ac +b²)y⁴ + (2b -c²)y² +1.We want this to equal P(5y²)=625y⁸ +125y⁶ +25y⁴ +5y² +1.So, set up the equations:1. 50b -a² =1252. 50 -2ac +b²=253. 2b -c²=5We have three equations with three variables a, b, c.Let me solve them step by step.From equation 1:50b -a²=125 =>a²=50b -125.Since a² must be non-negative, 50b -125 ≥0 =>b≥2.5. So b≥3.From equation 3:2b -c²=5 =>c²=2b -5. Since c² must be non-negative, 2b -5 ≥0 =>b≥2.5. So b≥3.Let me express c²=2b -5.From equation 2:50 -2ac +b²=25 =>b² -2ac= -25.From equation 1:a²=50b -125.Let me express a in terms of b: a=√(50b -125). Since a must be integer, 50b -125 must be a perfect square.Let me denote k²=50b -125 =>50b =k² +125 =>b=(k² +125)/50.Since b must be integer, k² +125 must be divisible by 50. Let me find k such that k² ≡ -125 mod 50.But -125 mod 50 is 25, so k² ≡25 mod50.Looking for integers k where k² ≡25 mod50.Possible k: numbers ending with 5, because 5²=25, 15²=225≡25 mod50, 25²=625≡25 mod50, etc.So k=5m +5, where m is integer.Let me try k=5:k=5: k²=25. Then b=(25 +125)/50=150/50=3.So b=3.Then, from equation 3:c²=2*3 -5=6-5=1 =>c=±1.From equation 1:a²=50*3 -125=150-125=25 =>a=±5.From equation 2: b² -2ac= -25 =>9 -2a c= -25 =>-2ac= -34 =>ac=17.But a=±5, c=±1. So a*c=±5. 17 is prime, so no solution.Next, k=15:k=15: k²=225. Then b=(225 +125)/50=350/50=7.So b=7.From equation 3:c²=2*7 -5=14-5=9 =>c=±3.From equation 1:a²=50*7 -125=350-125=225 =>a=±15.From equation 2: b² -2ac= -25 =>49 -2a c= -25 =>-2ac= -74 =>ac=37.But a=±15, c=±3. So a*c=±45 or ±45. 37 is prime, no solution.Next, k=25:k=25: k²=625. Then b=(625 +125)/50=750/50=15.So b=15.From equation 3:c²=2*15 -5=30-5=25 =>c=±5.From equation 1:a²=50*15 -125=750-125=625 =>a=±25.From equation 2: b² -2ac= -25 =>225 -2a c= -25 =>-2ac= -250 =>ac=125.With a=±25, c=±5. So a*c=±125. So possible if a=25, c=5 or a=-25, c=-5.Let's check a=25, c=5:From equation 2:225 -2*25*5=225 -250= -25. Correct.So this works.Thus, a=25, b=15, c=5.Therefore, the polynomials are:Q(y)=25y⁴ +25y³ +15y² +5y +1R(y)=25y⁴ -25y³ +15y² -5y +1Wait, but earlier when I multiplied these, I got an extra -75y³ term. Did I make a mistake?Wait, no, in this case, the product should be:625y⁸ + (50b -a²)y⁶ + (50 -2ac +b²)y⁴ + (2b -c²)y² +1.Plugging in a=25, b=15, c=5:50b -a²=50*15 -625=750 -625=12550 -2ac +b²=50 -2*25*5 +225=50 -250 +225=252b -c²=30 -25=5So the product is indeed 625y⁸ +125y⁶ +25y⁴ +5y² +1, which matches P(5y²).Wait, but earlier when I multiplied Q(y) and R(y), I got an extra -75y³ term. That must mean I made a mistake in the multiplication.Let me try multiplying again carefully:Q(y)=25y⁴ +25y³ +15y² +5y +1R(y)=25y⁴ -25y³ +15y² -5y +1Multiply term by term:25y⁴*25y⁴=625y⁸25y⁴*(-25y³)= -625y⁷25y⁴*15y²=375y⁶25y⁴*(-5y)= -125y⁵25y⁴*1=25y⁴25y³*25y⁴=625y⁷25y³*(-25y³)= -625y⁶25y³*15y²=375y⁵25y³*(-5y)= -125y⁴25y³*1=25y³15y²*25y⁴=375y⁶15y²*(-25y³)= -375y⁵15y²*15y²=225y⁴15y²*(-5y)= -75y³15y²*1=15y²5y*25y⁴=125y⁵5y*(-25y³)= -125y⁴5y*15y²=75y³5y*(-5y)= -25y²5y*1=5y1*25y⁴=25y⁴1*(-25y³)= -25y³1*15y²=15y²1*(-5y)= -5y1*1=1Now, collect like terms:y⁸:625y⁸y⁷:-625y⁷ +625y⁷=0y⁶:375y⁶ -625y⁶ +375y⁶=125y⁶y⁵:-125y⁵ +375y⁵ -375y⁵ +125y⁵=0y⁴:25y⁴ -125y⁴ +225y⁴ -125y⁴ +25y⁴=25y⁴y³:25y³ -75y³ -25y³= -75y³y²:15y² -25y² +15y²=5y²y:5y -5y=0Constants:1Wait, again, I get a -75y³ term. But according to the earlier calculation, it should not be there. This suggests that either my factorization is wrong or I'm making a consistent mistake in multiplication.But according to the equations, the product should be P(5y²). So perhaps I'm missing something in the multiplication.Wait, no, the equations only account for the coefficients up to y⁶, y⁴, y², and constants. The y³ term is not accounted for, which suggests that my assumption of the form Q(y) and R(y) might be missing something.Wait, no, in the earlier setup, when I assumed Q(y) and R(y) to be of the form (25y⁴ + ay³ + by² + cy +1)(25y⁴ -ay³ + by² -cy +1), the product indeed doesn't have a y³ term because the y³ terms cancel out. But in my manual multiplication, I'm getting a -75y³ term. That must mean I made a mistake in the multiplication.Wait, let me check the y³ terms again:From Q(y)*R(y):25y³*1=25y³15y²*(-5y)= -75y³5y*15y²=75y³1*(-25y³)= -25y³So total y³ terms:25y³ -75y³ +75y³ -25y³=0.Wait, that's different from what I thought earlier. So actually, the y³ terms cancel out.Similarly, y terms:5y*1=5y1*(-5y)= -5yTotal y terms:5y -5y=0.So, in reality, the product is:625y⁸ +125y⁶ +25y⁴ +5y² +1.Which matches P(5y²). So my earlier mistake was in the manual multiplication where I incorrectly added the y³ terms. The correct product indeed has no y³ term.Therefore, the factorization is correct.So, the polynomials Q(y) and R(y) are:Q(y)=25y⁴ +25y³ +15y² +5y +1R(y)=25y⁴ -25y³ +15y² -5y +1Both have integer coefficients and positive degree (degree 4), and their product is P(5y²).Thus, we have found such polynomials Q(y) and R(y).

💡Okay, so I have this problem where I'm given that sin x equals 1, and sin(x + z) equals -1/2. I need to find the smallest positive value of z in radians. Hmm, let me think about how to approach this step by step.First, I know that sin x = 1. When does sine equal 1? Well, sine reaches its maximum value of 1 at π/2 radians. So, x must be π/2 plus any multiple of 2π, right? Because sine has a period of 2π, so adding 2π to x won't change the value of sine. So, x = π/2 + 2πn, where n is any integer. But since we're looking for the smallest positive z, I think I can just take x = π/2 as the principal value.Now, the second part is sin(x + z) = -1/2. I need to find z such that when I add it to x, the sine of that angle is -1/2. Let me write that down: sin(π/2 + z) = -1/2.Hmm, how do I solve for z here? Maybe I can use the sine addition formula. The sine of (a + b) is sin a cos b + cos a sin b. So, applying that here:sin(π/2 + z) = sin(π/2)cos(z) + cos(π/2)sin(z).I know that sin(π/2) is 1 and cos(π/2) is 0. So, substituting those values in:1 * cos(z) + 0 * sin(z) = cos(z).So, sin(π/2 + z) simplifies to cos(z). Therefore, the equation becomes cos(z) = -1/2.Alright, so now I need to find z such that cos(z) = -1/2. I remember that cosine equals -1/2 at specific angles in the unit circle. Let me recall where those are.Cosine is negative in the second and third quadrants. The reference angle for which cosine is 1/2 is π/3, because cos(π/3) = 1/2. So, in the second quadrant, the angle would be π - π/3 = 2π/3. In the third quadrant, it would be π + π/3 = 4π/3. So, the general solutions for z would be z = 2π/3 + 2πn or z = 4π/3 + 2πn, where n is any integer.But I need the smallest positive value of z. Let's list out the possible solutions:For n = 0: z = 2π/3 and z = 4π/3.For n = 1: z = 2π/3 + 2π = 8π/3 and z = 4π/3 + 2π = 10π/3.Clearly, 2π/3 is smaller than 4π/3, which is smaller than 8π/3, and so on. So, the smallest positive z is 2π/3.Wait, let me double-check that. If I plug z = 2π/3 back into the original equation, does it satisfy sin(x + z) = -1/2?Given x = π/2, then x + z = π/2 + 2π/3. Let me compute that:π/2 is 3π/6, and 2π/3 is 4π/6. So, adding them together: 3π/6 + 4π/6 = 7π/6.What's sin(7π/6)? That's in the third quadrant, and sine is negative there. The reference angle is π/6, so sin(7π/6) = -1/2. Perfect, that matches the given equation.So, z = 2π/3 is indeed a solution. Is there a smaller positive z? Let me think. The next smallest solution would be z = 4π/3, but that's larger than 2π/3. If I go to negative n, like n = -1, then z would be negative, which isn't positive. So, 2π/3 is the smallest positive solution.Therefore, the smallest possible positive value of z is 2π/3 radians.

💡Alright, so I have this problem where I need to find the value of ( a + b ) for the line ( y = ax + b ) that is a common tangent to the curves ( f(x) = e^x ) and ( g(x) = ln x + 2 ). Hmm, okay, let's break this down step by step.First, I remember that a tangent line to a curve at a particular point has the same slope as the derivative of the function at that point. So, for both curves, I need to find points where the tangent lines have the same slope and the same equation.Let's start with the curve ( f(x) = e^x ). The derivative of this function is ( f'(x) = e^x ). So, at any point ( x = t ), the slope of the tangent line is ( e^t ). The equation of the tangent line at this point can be written using the point-slope form:[y - e^t = e^t (x - t)]Simplifying this, we get:[y = e^t x + (1 - t)e^t]Okay, so that's the equation of the tangent line for ( f(x) ) at ( x = t ).Now, moving on to the second curve ( g(x) = ln x + 2 ). The derivative of this function is ( g'(x) = frac{1}{x} ). For the line to be a common tangent, the slope of the tangent line must be the same for both curves. So, the slope ( a ) must satisfy:[a = e^t = frac{1}{x}]Wait, so ( e^t = frac{1}{x} ). That means ( x = e^{-t} ). Okay, so the point of tangency on ( g(x) ) is at ( x = e^{-t} ).Let's find the corresponding ( y )-coordinate on ( g(x) ):[g(e^{-t}) = ln(e^{-t}) + 2 = -t + 2]So, the point of tangency on ( g(x) ) is ( (e^{-t}, 2 - t) ).Now, the tangent line to ( g(x) ) at this point should have the same equation as the tangent line to ( f(x) ) at ( x = t ). So, let's write the equation of the tangent line for ( g(x) ) using the point-slope form:[y - (2 - t) = e^t (x - e^{-t})]Simplifying this, we get:[y = e^t x + (2 - t) - e^t cdot e^{-t}]Since ( e^t cdot e^{-t} = 1 ), this simplifies to:[y = e^t x + (2 - t - 1) = e^t x + (1 - t)]Wait a minute, so the equation of the tangent line for ( g(x) ) is ( y = e^t x + (1 - t) ). Comparing this with the equation of the tangent line for ( f(x) ), which was ( y = e^t x + (1 - t)e^t ), we can set them equal to each other:[e^t x + (1 - t) = e^t x + (1 - t)e^t]Subtracting ( e^t x ) from both sides, we get:[1 - t = (1 - t)e^t]Hmm, so ( 1 - t = (1 - t)e^t ). Let's solve for ( t ).First, if ( 1 - t neq 0 ), we can divide both sides by ( 1 - t ):[1 = e^t]Taking the natural logarithm of both sides:[ln 1 = t implies t = 0]Alternatively, if ( 1 - t = 0 ), then ( t = 1 ). But let's check what happens when ( t = 1 ).If ( t = 1 ), then the equation of the tangent line for ( f(x) ) is:[y = e^1 x + (1 - 1)e^1 = e x + 0 = e x]And for ( g(x) ), the tangent line would be:[y = e^1 x + (1 - 1) = e x + 0 = e x]So, both tangent lines are ( y = e x ). But wait, the problem states that ( b > 0 ). In this case, ( b = 0 ), which doesn't satisfy the condition. Therefore, ( t = 1 ) is not a valid solution.So, the only valid solution is ( t = 0 ).Let's substitute ( t = 0 ) back into the equation of the tangent line for ( f(x) ):[y = e^0 x + (1 - 0)e^0 = 1 cdot x + 1 cdot 1 = x + 1]So, the tangent line is ( y = x + 1 ). Therefore, ( a = 1 ) and ( b = 1 ).Adding them together, ( a + b = 1 + 1 = 2 ).Wait, let me double-check to make sure I didn't make any mistakes. For ( t = 0 ), the point on ( f(x) ) is ( (0, e^0) = (0, 1) ), and the slope is ( e^0 = 1 ). So, the tangent line is indeed ( y = x + 1 ).For ( g(x) ), when ( t = 0 ), ( x = e^{-0} = 1 ), and ( g(1) = ln 1 + 2 = 0 + 2 = 2 ). The slope is ( e^0 = 1 ), so the tangent line is ( y - 2 = 1(x - 1) ), which simplifies to ( y = x + 1 ). That matches.And since ( b = 1 ), which is greater than 0, it satisfies the condition.So, I think I did everything correctly. The value of ( a + b ) is 2.**Final Answer**The value of ( a + b ) is boxed{2}.

💡First, determine the time it took Matt to reach 220 steps by dividing the total steps by his rate: 220 ÷ 20 = 11 minutes.Next, calculate Tom's rate by adding 5 steps per minute to Matt's rate: 20 + 5 = 25 steps per minute.Finally, find out how many steps Tom climbed by multiplying his rate by the time: 25 × 11 = 275 steps.