answer:To find when all four coaches will be together next, we need to compute the least common multiple (LCM) of the days they work, which are 5, 9, 8, and 11. These numbers are 5, 9 = 3^2, 8 = 2^3, and 11: - The LCM of these numbers is obtained by taking the highest powers of all prime factors involved. - 5 is prime, 3^2 from 9 is the highest power of 3, 2^3 from 8 is the highest power of 2, and 11 is prime. - Thus, LCM(5, 9, 8, 11) = 5 times 3^2 times 2^3 times 11 = 5 times 9 times 8 times 11 = 3960. Therefore, Ella, Felix, Greta, and Harry will all be together again coaching at the sports club in boxed{3960} days.

answer:We are given the equation ( p^x = y^3 + 1 ) to be solved for all prime numbers ( p ) where ( x ) and ( y ) are positive integers. We can rewrite this equation in terms of its factors. 1. **Rewriting the Equation:** [ p^x = y^3 + 1 = (y + 1)(y^2 - y + 1) ] Since ( y ) is a positive integer, we have ( y + 1 geq 2 ). 2. **Factoring and Considering Values:** We can express ( y + 1 ) as a power of ( p ): [ y + 1 = p^t quad text{for some integer } t text{ where } 1 leq t leq x ] Isolating ( y ), we get: [ y = p^t - 1 ] 3. **Substitution:** Substituting ( y = p^t - 1 ) back into the equation, we obtain: [ y^2 - y + 1 = (p^t - 1)^2 - (p^t - 1) + 1 ] Simplifying, we have: [ (p^t - 1)^2 - (p^t - 1) + 1 = p^{2t} - 3p^t + 3 ] Thus, we need: [ p^{2t} - 3p^t + 3 = p^{x-t} ] 4. **Rearranging the Equation:** Equating powers of ( p ), we get: [ p^{x-1}(p^{3t-x} - 1) = 3(p^t - 1) ] Now consider the cases for ( p ): 5. **Case ( p = 2 ):** Then for the equation ( p^{3t-x} - 1 ) and ( p^t - 1 ), both are odd numbers. This implies ( p^{x-t} ) must also be odd, therefore: [ x = t quad text{and} quad y^2 - y + 1 = 1 ] This gives: [ y = 1, quad p = 2, quad x = 1 ] So, one solution is: [ (p, x, y) = (2, 1, 1) ] 6. **Case ( p neq 2 ):** Since ( p ) is odd, both ( p^{3t-x} - 1 ) and ( p^t - 1 ) are even, implying ( p^{x-t} ) is odd. Thus: [ 3 mid p^{x-t} quad text{or} quad 3 mid (p^{3t-x} - 1) ] - When ( 3 mid p^{x-t} ), then ( p = 3 ) and ( x = t+1 ): [ y^2 - y + 1 = 3 ] Solving this: [ y = 2, quad x = 2 ] Therefore, another solution is: [ (p, x, y) = (3, 2, 2) ] - When ( 3 mid (p^{3t-x} - 1) ): [ p^{x-t} mid (p^t - 1) quad Rightarrow quad x = t ] From our previous analysis, this is effectively case (1) where ( y = 1, p = 2 ), which we have already solved. 7. **Conclusion:** Combining our solutions, we have two sets of solutions: [ (p, x, y) = (2, 1, 1) quad text{and} quad (p, x, y) = (3, 2, 2) ] Thus the solutions are: [ boxed{(p, x, y) = (2, 1, 1) text{ and } (3, 2, 2)} ]

answer:1. **Consider a regular n-gon inscribed in a circle:** - By definition, a regular n-gon has n equal sides and all its interior angles are equal. - Let's label the vertices of the n-gon as (A_1, A_2, ldots, A_n). 2. **Divide the circle into n equal arcs:** - Each side of the n-gon subtends an angle at the center of the circle. - Since the n-gon is regular, these angles are equal. - The total angle at the center of the circle is (360^circ). 3. **Calculate the angle subtended by each side at the center:** - The angle subtended by each side at the center is given by: [ frac{360^circ}{n} ] 4. **Analyze the angles formed by sides and diagonals:** - Consider any vertex (A_i) of the n-gon. A diagonal will connect this vertex to another vertex (A_j). - The angle at (A_i) will be formed by the two sides of the n-gon that meet at (A_i) and the diagonal (A_iA_j). 5. **Determine the angle between the sides and diagonals:** - Let the angle subtended at the center by the arc (A_iA_j) be (theta = k cdot frac{360^circ}{n}), where (k) is an integer. - When (A_i) and (A_j) are not adjacent, the line segments forming the angle include non-adjacent arcs. 6. **Use the property of regular polygons and circle geometry:** - The interior angle (alpha) at any vertex of the n-gon can be determined by: [ alpha = 180^circ - frac{360^circ}{n} ] - For the specific angles involving diagonals, we observe that moving clockwise from (A_i) to (A_j) via intermediate vertices introduces additional multiples of (frac{360^circ}{n}) at the center. 7. **Confirm the angles formed:** - The resulting angle formed between the sides and the diagonals when projected back to any vertex will thus follow multiples of: [ 180^circ cdot frac{k}{n} ] - Since (frac{k}{n}) is a ratio representing the fraction of the entire central angle (360^circ) divided by (n). # Conclusion: The detailed proof shows that all angles formed by the sides and diagonals of a regular n-gon are indeed multiples of frac{180^circ}{n}. Therefore, we conclude: [ boxed{theta = frac{180^circ k}{n}} ] where (k) is an integer determined by the specific positions of the sides and diagonals in question.

answer:Let c_1, c_2, and c_3 be the number of cashews initially taken by the first, second, and third monkeys, respectively. The total is c_1 + c_2 + c_3. - The first monkey ends with frac{2}{3}c_1 + frac{1}{6}c_2 + frac{4}{18}c_3, - The second monkey ends with frac{1}{6}c_1 + frac{1}{3}c_2 + frac{4}{18}c_3, - The third monkey ends with frac{1}{6}c_1 + frac{1}{3}c_2 + frac{1}{9}c_3. From the ratio 3:2:1, we have: - 3(y) = frac{2}{3}c_1 + frac{1}{6}c_2 + frac{4}{18}c_3 - 2(y) = frac{1}{6}c_1 + frac{1}{3}c_2 + frac{4}{18}c_3 - 1(y) = frac{1}{6}c_1 + frac{1}{3}c_2 + frac{1}{9}c_3 Solving these equations, we find: - frac{c_1}{7} = frac{c_2}{10} = frac{c_3}{18}, with each fraction needing to be whole numbers. - The minimal value occurs when each fraction equals 18 (the least common multiple of the denominators and ensuring divisibility by 6 for whole number distributions). Thus, the smallest total number of cashews is 18(7 + 10 + 18) = boxed{630}.