answer:1. **Factorization and GCD Analysis:** We start by factoring the expression ( n^5 + n^4 + 1 ): [ n^5 + n^4 + 1 = (1 + n + n^2)(1 - n + n^3). ] Let ( d ) be the greatest common divisor (g.c.d.) of the factors ( 1 + n + n^2 ) and ( 1 - n + n^3 ). We need to determine the possible values of ( d ). 2. **Finding the GCD:** To find ( d ), we use the Euclidean algorithm: [ gcd(1 + n + n^2, 1 - n + n^3). ] Notice that: [ 1 - n + n^3 = n^3 - n + 1. ] We need to find the common divisors of ( 1 + n + n^2 ) and ( n^3 - n + 1 ). By substituting small values of ( n ), we can observe that ( d ) can be either ( 1 ) or ( 7 ). 3. **Divisor Count Analysis:** Since ( n^5 + n^4 + 1 ) has exactly 6 divisors, it must be of the form ( p^5 ) or ( pq^2 ), where ( p ) and ( q ) are primes. The form ( p^5 ) is impossible because ( d in {1, 7} ) and ( n > 1 ). Therefore, ( n^5 + n^4 + 1 ) must be of the form ( pq^2 ). 4. **Case ( d = 1 ):** If ( d = 1 ), then ( 1 + n + n^2 ) and ( n^3 - n + 1 ) are coprime. Since ( n^5 + n^4 + 1 ) has 6 divisors, it must be of the form ( pq^2 ). This implies: [ n^3 - n + 1 = q^2, ] where ( q ) is a prime number. 5. **Case ( d = 7 ):** If ( d = 7 ), then ( 1 + n + n^2 = 7 ). Solving for ( n ): [ n^2 + n + 1 = 7 implies n^2 + n - 6 = 0. ] Solving this quadratic equation: [ n = frac{-1 pm sqrt{1 + 24}}{2} = frac{-1 pm 5}{2}. ] This gives: [ n = 2 quad text{or} quad n = -3. ] Since ( n ) is a natural number, ( n = 2 ). Substituting ( n = 2 ) into ( n^3 - n + 1 ): [ 2^3 - 2 + 1 = 8 - 2 + 1 = 7, ] which is not a perfect square. Therefore, ( d = 7 ) does not yield a valid solution. 6. **Conclusion:** Since ( d = 7 ) does not work, we conclude that ( n^3 - n + 1 = q^2 ) for some prime ( q ). (blacksquare)

answer:1. Let (n) be the number of teams participating in the tournament. 2. In this tournament, each team plays one match against every other team. Therefore, the total number of matches played in the tournament is given by the combination formula for choosing 2 teams out of (n): [ text{Total number of matches} = binom{n}{2} = frac{n(n-1)}{2}. ] Since there is one point awarded for each match (to the winning team), the total number of points distributed is also (frac{n(n-1)}{2}). 3. Given that at one point in the tournament, all teams had a different number of points, it means that in the end, the points each team accumulated must be from the set ({0, 1, 2, ldots, n-1}). 4. If each team finishes with a different number of points, we must assume that one team finishes with 0 points, the next team with 1 point, the next with 2 points, and so on up to (n-1) points for the team that accumulated the most points. 5. We are tasked with finding the number of points the second-last team accumulated. Given the arrangement ({0, 1, 2, ldots, n-1}), the second-last team has (n-2) points. 6. This also implies that the team that won the tournament (with (n-1) points) has won all its matches except the one against the second-last team, who must hence have fetched one victory point, making its total points exactly 1. 7. Conclusion: [ boxed{1}. ]

answer:Let ( z = -x - frac{pi}{4} ). Then ( frac{pi}{4} le z le frac{7 pi}{12} ), and ( frac{pi}{2} le 2z le frac{7 pi}{6} ). Rewriting the trigonometric terms with ( z ): [ tan left( x + frac{3 pi}{4} right) = tan left( frac{pi}{2} - z right) = cot z ] [ tan left( x + frac{pi}{3} right) = tan left( -frac{pi}{12} - z right) = -tan left( z + frac{pi}{12} right) ] [ cos left( x + frac{pi}{3} right) = cos left( -frac{pi}{12} - z right) = cos left( z + frac{pi}{12} right) ] [ sin left( x + frac{pi}{4} right) = sin left( -z right) = -sin z ] Combining these: [ y = cot z - tan left( z + frac{pi}{12} right) + cos left( z + frac{pi}{12} right) - sin z ] Employing the identity (cot z = frac{1}{tan z}) and noting that ( cos left( z + frac{pi}{12} right) ) and ( sin z ) decrease in the given range, while ( tan left( z + frac{pi}{12} right) ) increases, ( y ) is likely to decrease in this interval. The maximum value occurs at the left endpoint ( z = frac{pi}{4} ): [ y = cot frac{pi}{4} - tan left( frac{pi}{4} + frac{pi}{12} right) + cos left( frac{pi}{4} + frac{pi}{12} right) - sin frac{pi}{4} ] [ y = 1 - tan frac{pi}{3} + cos frac{pi}{3} - frac{sqrt{2}}{2} ] [ y = 1 - sqrt{3} + frac{1}{2} - frac{sqrt{2}}{2} = frac{3}{2} - sqrt{3} - frac{sqrt{2}}{2} ] [ boxed{frac{3}{2} - sqrt{3} - frac{sqrt{2}}{2}} ]

answer:# Problem: text{Given a function } f: mathbf{N}^{*} rightarrow mathbf{N}^{*} text{ that satisfies the equation:} fleft(t^{2} f(s)right)=s(f(t))^{2}, forall s,t in mathbf{N}^{*}. Determine the minimum possible value of ( f(1998) ). 1. **Rewrite the Given Equation:** The functional equation given is: fleft(t^{2} f(s)right) = s(f(t))^{2}. Let's explore this equation for special values of ( s ) and ( t ). 2. **Set ( t = 1 ):** Substitute ( t = 1 ) into the original equation: fleft(1^{2} f(s)right) = s (f(1))^{2}. Simplifies to: f(f(s)) = s (f(1))^{2}. eqno{(1)} 3. **Set ( s = 1 ):** Substitute ( s = 1 ) into the original equation: fleft(t^{2} f(1)right) = 1 (f(t))^{2}. Simplifies to: f(t^2 f(1)) = (f(t))^2. eqno{(2)} 4. **Consider ( f(s t) ):** Let's find ( f(s t) ) by combining the results of equations (1) and (2): f(s t) = frac{f(s) f(t)}{f(1)}. eqno{(3)} 5. **Deriving ( f(t^2) ):** Use equation (2) for ( t = t ): f(t^2 f(1)) = (f(t))^2. Rearrange: f(t^2) = frac{(f(t))^2}{f(1)}. eqno{(4)} 6. **Generalize ( f(t^n) ):** By induction, from ( f(t^2) ): f(t^3) = f(t cdot t^2) = frac{f(t) f(t^2)}{f(1)} = frac{f(t) cdot frac{(f(t))^2}{f(1)}}{f(1)} = frac{(f(t))^3}{(f(1))^2}, and in general: f(t^n) = left(frac{f(t)}{f(1)}right)^{n-1} cdot f(t). eqno{(5)} 7. **Check ( f(t) div f(1) ):** For ( f(t) ) to be an integer, ( f(1) ) must divide ( f(t) ): f(1) mid f(t). 8. **Find the Minimum f(n):** We need ( f(1998) ). Factorize 1998: 1998 = 2 times 3^3 times 37. Using ( f ): f(1998) = fleft(2 times 3^3 times 37right). 9. **Calculate using (3) and (5):** Expand using factors: f(1998) = frac{f(2) (f(3^3)) f(37)}{(f(1))^4}. Using ( f(3^3) = frac{f(3)^3}{f(1)^2} ): f(1998) = frac{f(2) left(frac{f(3)^3}{f(1)^2}right) f(37)}{(f(1))^4} = frac{f(2) f(3)^3 f(37)}{(f(1))^4}. 10. **Define ( f ) for specific primes:** We must ensure distinct values for each prime: f(1) = 1, f(2) = 3, f(3) = 2, f(37) = 5. So, f(1998) = frac{3 cdot 2^3 cdot 5}{1^4} = 3 cdot 8 cdot 5 = 120. # Conclusion: The minimum value of ( f(1998) ) is boxed{120}.