answer:1. Let's analyze the case when ( q < n ): - Given ( q leq n ), we have ( q mid (n-1)! ) because any integer ( q leq n ) will have its prime factors as a subset of the prime factors of ((n-1)!). - Since ( q ) is a factor of ((n-1)!), it follows that ( q(q-1) mid (n-1)! ). Therefore, [ q - 1 mid frac{(n-1)!}{q} ] By definition, we have [ q - 1 mid leftlfloor frac{(n-1)!}{q} rightrfloor ] because ( leftlfloor frac{(n-1)!}{q} rightrfloor ) is the greatest integer less than or equal to ( frac{(n-1)!}{q} ). 2. Now, consider the case when ( q = n ): - We need to discuss two scenarios: when ( n ) (and thus ( q )) is a prime number and when it is composite. 3. When ( n ) is a prime number: - According to Wilson's Theorem, for a prime ( n ): [ (n-1)! equiv -1 pmod{n} ] This implies that [ (n-1)! = kn + (n-1) text{ for some integer } k ] Thus: [ frac{(n-1)!}{n} = k + frac{n-1}{n} ] And thus: [ leftlfloor frac{(n-1)!}{n} rightrfloor = k ] Notice that since ( k ) is an integer, it implies ( k ) is divisible by ( q - 1 = n - 1 ). 4. When ( n ) is composite: - Let ( n = pm ), where ( p ) is the largest prime factor of ( n ) such that ( 1 < m < n ). Hence, ( p ) and ( m ) are both less than ( n ). - Since ( n-1 ) and ( n ) are coprime, ( 1 < m leq n-2 ) and so ( p leq n-2 ) by similar logic. - If ( p ) and ( m ) are distinct, we will have: [ p mid (n-2)! quad text{ and } quad m mid (n-2)! ] Hence, ( n = pm mid (n-2)! ), implying [ n(n-1) mid (n-1)! ] Therefore, [ q-1 = n-1 mid leftlfloor frac{(n-1)!}{n} rightrfloor ] 5. If ( p = m ), then ( n = p^2 ), and we know ( n > 4 ) implying ( p > 2 ): - Since ( (2p, n) = p ) and thus, [ 2p neq n-1 Rightarrow 2p leq n-2 ] Since both ( p ) and ( 2p ) must divide ( (n-2)! ), it follows that: [ 2p^2 mid (n-2)! ] Thus, [ n(n-1) mid (n-1)! ] This completes the proof. Hence [ q-1 mid leftlfloor frac{(n-1)!}{q} rightrfloor ] holds true in all cases. (blacksquare)

answer:Solution: Using 1, 2, 3, 4 to represent rain, the probability of rain on any given day is 0.4; According to the problem, to simulate the result of having exactly two days of rain in three days, the following 20 groups of random numbers were generated through random simulation. Among the 20 groups of random numbers, those representing exactly two days of rain in three days are: 191, 271, 932, 812, 393, totaling 5 groups of random numbers. The probability sought is frac{5}{20} = 0.25, Therefore, the answers are: 0.4, 0.25. According to the problem, using 1, 2, 3, 4 to represent rain, the probability of rain on any given day is 0.4; to simulate the result of having exactly two days of rain in three days, the following 20 groups of random numbers were generated through random simulation. Among these 20 groups of random numbers, those representing exactly two days of rain in three days can be determined by listing, totaling 5 groups of random numbers. According to the probability formula, the result is obtained. This problem examines the estimation of probability using simulation methods. The solution mainly relies on the probability of equally likely events, noting the application of the listing method in this problem. Thus, the probability of rain on any given day is boxed{0.4}, and the probability of having exactly two days of rain in three days is boxed{0.25}.

answer:1. Given that the polynomial ( P ) has integer coefficients and ( P(x) = 5 ) for five different integers ( x ), let's denote these integers as ( x_1, x_2, x_3, x_4, ) and ( x_5 ). 2. Based on this information, we know that: [ P(x_k) = 5 quad text{for} quad k = 1, 2, 3, 4, 5 ] 3. We can express the polynomial ( P(x) ) in the form: [ P(x) - 5 = (x - x_1)(x - x_2)(x - x_3)(x - x_4)(x - x_5)Q(x) ] where ( Q(x) ) is a polynomial with integer coefficients. 4. Now, let's assume that there exists an integer ( x_0 ) such that ( P(x_0) = n ). From this, we have: [ P(x_0) - 5 = n - 5 ] Substituting this into our earlier expression for ( P(x) - 5 ), we get: [ n - 5 = (x_0 - x_1)(x_0 - x_2)(x_0 - x_3)(x_0 - x_4)(x_0 - x_5)Q(x_0) ] This means ( n - 5 ) must be a product of six non-zero integers, five of which are distinct (because the ( x_k ) are distinct). 5. We need to check the condition that ( P(x) ) should not fall within the intervals ([-6, 4]) or ([6, 16]). We express this mathematically considering ( P(x_0) = n ): [-6 leq n leq 4 quad text{or} quad 6 leq n leq 16] 6. Next, we analyze the possible products. ( |n - 5| leq 11 ), since within the specified ranges for ( n ): [-6 leq n leq 4 implies -6 - 5 leq n - 5 leq 4 - 5 implies -11 leq n - 5 leq -1] [6 leq n leq 16 implies 6 - 5 leq n - 5 leq 16 - 5 implies 1 leq n - 5 leq 11] 7. The absolute value of ( n - 5 ) (denoted ( |n - 5| )) therefore lies between 1 and 11: [ |n - 5| leq 11 quad Rightarrow -11 leq n - 5 leq 11 ] 8. The value ( n - 5 ) must be the product of six integers, five of which are distinct (i.e., ( (x_0 - x_1), (x_0 - x_2), (x_0 - x_3), (x_0 - x_4), (x_0 - x_5) )) and non-zero. 9. Considering the absolute value of this product: - The smallest absolute value for a product of five distinct non-zero integers is ( 1 times 2 times 3 times 4 times 5 = 120 ). - This means the smallest product in absolute value for six non-zero integers would be ( 120 ) multiplied by the smallest integer greater than ( 1 ), for example ( 1 ): ( 1 times 120 geq 120). 10. Given that ( 120 ) is much larger than ( 11 ), it is impossible for ( |n - 5| ) to be as low as ( 11 ), making it impossible for ( n - 5 ) to satisfy the required conditions. Hence, ( P(x) ) cannot fall within the intervals ([-6, 4]) or ([6, 16]). # Conclusion: [ boxed{} ]

answer:Given the new operation defined by a @ b = frac{5a - 2b}{3}, we substitute a = 7 and b = 8: 1. Calculate the numerator: 5 times 7 - 2 times 8 = 35 - 16 = 19. 2. Divide by the denominator: frac{19}{3}. Thus, the value of 7 @ 8 is boxed{frac{19}{3}}.