answer:1. **Understanding the structure of voting:** - Total number of voters: 105. - The 105 voters are divided into 5 districts: [ text{Number of voters per district} = frac{105}{5} = 21 ] - Each district is divided into 7 sections: [ text{Number of voters per section} = frac{21}{7} = 3 ] 2. **Determining the requirements for winning:** - To win the final, the giraffe "Vyisokiy" must win in the majority of the districts, i.e., at least 3 out of 5 districts. - To win a district, "Vyisokiy" must win in the majority of the sections in that district, i.e., at least 4 out of 7 sections. 3. **Calculating the number of sections needed:** - To secure a district: [ 4 text{ sections} ] - To secure 3 districts: [ 3 cdot 4 = 12 text{ sections} ] 4. **Number of voters required per section:** - To win in a section, at least 2 out of 3 voters must vote for "Vyisokiy." 5. **Calculating the minimal number of voters needed:** - For 12 sections: [ 12 cdot 2 = 24 text{ voters} ] 6. **Verification by valid example generation:** - The calculation ensures that with 24 voters, strategically placed, "Vyisokiy" can win the required 12 sections across 3 districts to become the final winner. Concluding from the above reasoning, the minimum number of voters needed to ensure "Vyisokiy's" victory is: [ boxed{24} ]

answer:Let's denote the weights of A, B, and C as ( W_A, W_B, ) and ( W_C ) respectively. According to the information given: 1. The average weight of A, B, and C is 84 kg: [ frac{W_A + W_B + W_C}{3} = 84 ] [ W_A + W_B + W_C = 84 times 3 ] [ W_A + W_B + W_C = 252 ] 2. When D joins the group, the average weight becomes 80 kg: [ frac{W_A + W_B + W_C + W_D}{4} = 80 ] [ W_A + W_B + W_C + W_D = 80 times 4 ] [ W_A + W_B + W_C + W_D = 320 ] From the first equation, we know that ( W_A + W_B + W_C = 252 ), so we can find the weight of D by subtracting 252 from 320: [ W_D = 320 - 252 ] [ W_D = 68 ] 3. If E replaces A, the average weight of B, C, D, and E becomes 79 kg: [ frac{W_B + W_C + W_D + W_E}{4} = 79 ] [ W_B + W_C + W_D + W_E = 79 times 4 ] [ W_B + W_C + W_D + W_E = 316 ] We know that ( W_B + W_C = 252 - W_A ), and since the weight of A is 77 kg: [ W_B + W_C = 252 - 77 ] [ W_B + W_C = 175 ] Now we can find the weight of E by substituting ( W_B + W_C ) and ( W_D ) into the equation: [ 175 + 68 + W_E = 316 ] [ 243 + W_E = 316 ] [ W_E = 316 - 243 ] [ W_E = 73 ] Now we can find out how much more E weighs than D: [ W_E - W_D = 73 - 68 ] [ W_E - W_D = 5 ] So, E weighs boxed{5} kg more than D.

answer:Step by step, we handle the function: 1. Since ( 1+i ) is non-real, ( f(1+i) = (1+i)^2 = 1 + 2i - 1 = 2i ). 2. ( 2i ) is non-real, so ( f(2i) = (2i)^2 = -4 ). 3. ( -4 ) is negative real, applying the new function rule, ( f(-4) = (-4)^3 = -64 ). 4. ( -64 ) is negative real, so ( f(-64) = (-64)^3 = -262144 ). In each case, the correct rule was applied based on the new definition of ( f ). Therefore: (boxed{-262144})

answer:1. **Definition and Setup**: Let a triple (a, b, c) be termed **good** if and only if (a, b), (a, c), and (b, c) are all pairs in the set S. Define d_i as the number of pairs in S containing the integer i and let D_i be the set of numbers paired with i in S (so left|D_i right| = d_i). 2. **Counting Triple Intersections**: For any pair (i, j) in S, we want to estimate the number of integers k such that {i, j, k} is good. This is equivalent to finding [ left| D_i cap D_j right| ] Note that i notin D_i and j notin D_j. Consequently, neither i nor j can be in D_i cap D_j, ensuring that any k in D_i cap D_j is different from both i and j. 3. **Bounding the Intersection**: Since D_i cup D_j subseteq {1, 2, ldots, n}, we use this to find: [ left| D_i cap D_j right| = left| D_i right| + left| D_j right| - left| D_i cup D_j right| ] We have: [ left| D_i cup D_j right| leq n ] Thus, [ left| D_i cap D_j right| leq d_i + d_j - n ] 4. **Summing Over All Pairs**: Consider all pairs (i, j) in S. Each good triple (a, b, c) will be counted three times (once for each pair of the three elements in the triple). Let T be the total count of good triples. We sum up over the pairs: [ T geq frac{1}{3} sum_{(i, j) in S} left( d_i + d_j - n right) ] 5. **Using Cauchy-Schwartz Inequality**: Rewriting the sum, we see that each d_i term appears d_i times (since it counts how many pairs include i). We write: [ T geq frac{1}{3} left( sum_{i=1}^{n} d_i^2 - m cdot n right) ] 6. **Applying Cauchy-Schwartz**: Using the Cauchy-Schwarz inequality: [ sum_{i=1}^{n} d_i^2 geq frac{left( sum_{i=1}^{n} d_i right)^2}{n} ] Thus: [ T geq frac{1}{3} left( frac{left( sum_{i=1}^{n} d_i right)^2}{n} - m cdot n right) ] 7. **Sum of d_i**: Observe that: [ sum_{i=1}^{n} d_i = 2m ] Each pair (a, b) contributes to both d_a and d_b, hence the total sum of d_i counts twice the number of pairs. 8. **Final Calculation**: Substitute this back in: [ T geq frac{1}{3} left( frac{(2m)^2}{n} - mn right) = frac{1}{3} left( frac{4m^2}{n} - mn right) ] Simplifying: [ T geq 4m frac{ m - frac{n^2}{4} }{3n} ] 9. **Conclusion**: [ boxed{ T geq 4m frac{ m - frac{n^2}{4} }{3n} } ] This concludes the detailed step-by-step solution, demonstrating that the number of good triples (a, b, c) is at least (boxed{4m frac{ m - frac{n^2}{4} }{3n}}).