answer:First, check if 48, 55, and 73 form a right triangle using the Pythagorean Theorem: [ 48^2 + 55^2 = 2304 + 3025 = 5329.] [ 73^2 = 5329.] Therefore, 48, 55, and 73 do form a right triangle, as 73^2 = 48^2 + 55^2. Next, we need to find the multiplicative inverse of 480 modulo 4273. Using the relationship from the triangle: [ 55 + 73 = 128. ] [ 480 = 128 times 3.75. ] Thus, compute inverse of 128 modulo 4273. Using the Extended Euclidean Algorithm: 1. Compute gcd of 128 and 4273, and express as a linear combination: [ 4273 = 128 times 33 + 49 ] [ 128 = 49 times 2 + 30 ] [ 49 = 30 times 1 + 19 ] [ 30 = 19 times 1 + 11 ] [ 19 = 11 times 1 + 8 ] [ 11 = 8 times 1 + 3 ] [ 8 = 3 times 2 + 2 ] [ 3 = 2 times 1 + 1 ] [ 2 = 1 times 2 + 0 ] Thus, gcd is 1, which confirms 128 and 4273 are coprime. 2. Rewriting backwards: [ 1 = 3 - 2 times 1 ] [ 2 = 8 - 3 times 2 ] [ vdots ] [ 1 = 4273 - 128 times 33 + 49 times 67 ] [ 1 = 4273 times 1 + 128 times (-33) ] (We only use the coefficients of 4283 and 128) The inverse of 128 modulo 4273 is -33 equiv 4240 pmod{4273}. Thus, the inverse of 480 modulo 4273 is: [ 480^{-1} equiv (128 times 3.75)^{-1} equiv 4240 times 3.75 pmod{4273} equiv 15900 pmod{4273} equiv 1643 pmod{4273}. ] Therefore, the inverse of 480 modulo 4273 is boxed{1643}.

answer:1. **Compute the total number of ways to choose a group of three people out of ten**: [ binom{10}{3} = frac{10!}{3!(10-3)!} = frac{10!}{3! cdot 7!} ] First, simplify the factorials: [ 10! = 10 times 9 times 8 times 7! ] Thus: [ frac{10 times 9 times 8 times 7!}{3! times 7!} = frac{10 times 9 times 8}{3!} = frac{10 times 9 times 8}{6} = frac{720}{6} = 120 ] So, there are 120 different ways to form a group of three people from the group of ten. 2. **Identify the impact of each conflict on the groups of three**: Each conflict between a pair of people can destroy several groups of three. Specifically, if two people in the pair are in a group of three, that group is considered 'destroyed'. For each specific pairwise conflict, the number of groups of three potentially destroyed is calculated by fixing the pair and then choosing the third person from the remaining eight people: [ binom{8}{1} = 8 ] Hence, each conflict dissolves exactly 8 groups of three. 3. **Calculate the total number of groups dissolved by all conflicts**: Given there are 14 conflicts and each conflict dissolves 8 groups: [ 14 times 8 = 112 ] Thus, the total number of dissolved groups of three is 112. 4. **Determine the remaining number of groups of three that are unaffected**: Start with the total number of groups and subtract the number of dissolved groups: [ 120 - 112 = 8 ] Therefore, there are 8 groups of three that remain unaffected and are still friends. 5. **Conclusion**: There are at least 8 groups of three friends who have not been affected by any of the 14 conflicts. Therefore, it is indeed possible to form a company of three friends. Hence, we have [ boxed{8} ] Thus, it is guaranteed that even with 14 conflicts among 10 people, there are still groups of three friends that can be formed.

answer:Let's denote the total number of trees as T, the number of pure Fuji trees as F, and the number of pure Gala trees as G. According to the problem, 10% of the trees cross-pollinated, so the number of cross-pollinated trees is 0.10T. We are given two pieces of information: 1. The number of pure Fuji trees plus the cross-pollinated ones totals 170: F + 0.10T = 170 2. Three-quarters of all his trees are pure Fuji: F = (3/4)T We can use the second equation to express F in terms of T and substitute it into the first equation: (3/4)T + 0.10T = 170 To solve for T, we combine the terms on the left side: 0.75T + 0.10T = 170 0.85T = 170 Now, we divide both sides by 0.85 to find T: T = 170 / 0.85 T = 200 So, the total number of trees is 200. Now, we can find the number of pure Fuji trees using the second piece of information: F = (3/4)T F = (3/4) * 200 F = 150 Since the total number of trees is 200 and there are 150 pure Fuji trees, the number of pure Gala trees is the difference between the total number of trees and the number of pure Fuji trees: G = T - F G = 200 - 150 G = 50 Therefore, there are boxed{50} pure Gala trees in the orchard.

answer:- For all k where 1 < k leq n+1, k divides (n+1)!. - Hence, (n+1)! + k is divisible by k for 1 < k < n+1, making (n+1)! + k composite for these values of k. - We need to check the primality of (n+1)! + 1 and (n+1)! + (n+1): - (n+1)! + 1 could be prime. It needs checking for specific values of n. - (n+1)! + (n+1) = (n+1)((n+1)! + 1) is clearly composite as it is divisible by (n+1). Thus, all numbers in the range (n+1)! + 1 < a < (n+1)! + (n+1) are composite. The boundaries need individual checking, but typically (n+1)! + 1 might be prime, and (n+1)! + (n+1) is not. Therefore, the number of primes in that range is boxed{0} or boxed{1}.