answer:1. **Evaluating the sum of the smallest and largest integers**: - The average of the two smallest integers in ( S ) is 5. Thus, [ text{Sum of the two smallest integers} = 2 times 5 = 10 ] - The average of the two largest integers in ( S ) is 22. Thus, [ text{Sum of the two largest integers} = 2 times 22 = 44 ] 2. **Setting up the notation for the set ( S )**: - Let the other five integers in the set ( S ) be ( p, q, r, t, ) and ( u ) where ( p < q < r < t < u ). 3. **Expression for the average of the 9 integers**: - The total sum of all integers in ( S ) can be written as: [ 10 + 44 + p + q + r + t + u ] - The average of these 9 integers is: [ frac{10 + 44 + p + q + r + t + u}{9} = frac{54 + p + q + r + t + u}{9} ] 4. **Maximizing the average**: - To maximize the average, we need to maximize ( p + q + r + t + u ). 5. **Determination of values**: - From the sum of the largest integers being 44, let the two largest integers be ( x ) and ( y ) with ( x < y ). Since ( x ) and ( y ) are integers and ( x + y = 44 ): [ x + y = 44 quad text{and} quad x < y ] - The highest ( x ) can be, given it must be an integer less than ( y ), is 21: [ x = 21, quad y = 23 ] - This makes our two largest integers 21 and 23. 6. **Assigning the other integers**: - To maximize ( p, q, r, t, ) and ( u ), choose the largest possible distinct integers less than 21. - Set ( u = 20 ), ( t = 19 ), ( r = 18 ), ( q = 17 ), and ( p = 16 ). 7. **Summing the chosen integers**: - The sum of these integers is: [ p + q + r + t + u = 16 + 17 + 18 + 19 + 20 = 90 ] 8. **Final calculation of the average**: - Substituting back into the average formula: [ frac{54 + 90}{9} = frac{144}{9} = 16 ] # Conclusion: [ boxed{16} ]

answer:We are looking to count intersections of y = cos x and y = sin nx over [0, 2pi] for each n > 2. 1. **Functions and Common Solutions**: The function sin nx has a period of frac{2pi}{n}, and within each period on [0, 2pi], it will intersect cos x at 2n points since cos x has two cycles within [0, 2pi]. 2. **Number of solutions for each n**: Each intersection count of cos x = sin nx over [0, 2pi] for n > 2 is 2n. 3. **Summing up**: [ sum_{n=3}^{100} G(n) = sum_{n=3}^{100} 2n = 2left(sum_{n=3}^{100} nright) = 2left(frac{100(101)}{2} - frac{3(2)}{2}right) = 2left(5050 - 3right) = 10094. ] Therefore, boxed{10094}.

answer:1. **Base Case:** We start with the base case ( n = 2 ). We need to show that the last digit of ( 2^{2^2} + 1 ) is 7. [ 2^{2^2} = 2^4 = 16 ] The last digit of ( 16 + 1 ) is 7, so the base case holds. 2. **Inductive Step:** Assume that for some ( k geq 2 ), the last digit of ( 2^{2^k} + 1 ) is 7. This means: [ 2^{2^k} equiv 6 pmod{10} ] We need to show that the last digit of ( 2^{2^{k+1}} + 1 ) is also 7. Note that: [ 2^{2^{k+1}} = 2^{2 cdot 2^k} = (2^{2^k})^2 ] By the induction hypothesis: [ 2^{2^k} equiv 6 pmod{10} ] Therefore: [ (2^{2^k})^2 equiv 6^2 pmod{10} ] Simplifying the right-hand side: [ 6^2 = 36 implies 36 equiv 6 pmod{10} ] Thus: [ 2^{2^{k+1}} equiv 6 pmod{10} ] Adding 1 to both sides: [ 2^{2^{k+1}} + 1 equiv 6 + 1 equiv 7 pmod{10} ] This completes the inductive step. By mathematical induction, we have shown that for all positive integers ( n geq 2 ), the last digit of ( 2^{2^n} + 1 ) is 7. (blacksquare)

answer:Let's denote the cost price of the bicycle for A as CP_A. B sold the bicycle to C at a profit of 25%, and the final selling price (SP_C) was Rs. 225. So, we can write the following equation for B's selling price to C: SP_C = CP_B * (1 + Profit_B%) 225 = CP_B * (1 + 25/100) 225 = CP_B * (1.25) CP_B = 225 / 1.25 CP_B = 180 Now, we know that B bought the bicycle from A at a cost price of Rs. 180 (CP_B), and A made a profit of 60%. So, we can write the following equation for A's selling price to B (which is also B's cost price): CP_B = CP_A * (1 + Profit_A%) 180 = CP_A * (1 + 60/100) 180 = CP_A * (1.6) CP_A = 180 / 1.6 CP_A = 112.5 Therefore, the cost price of the bicycle for A was Rs. boxed{112.5} .