answer:1. Given the problem of finding two positive irreducible fractions with denominators not exceeding 100, whose sum is (frac{86}{111}), we start by analyzing the denominator 111. 2. Since (111 = 37 times 3), we consider one fraction to have the denominator 37 and the other to have the denominator 3. This is because we want the sum of these fractions to have a common denominator equal to the product of these numbers. 3. Let the two fractions be (frac{a}{37}) and (frac{b}{3}). We are given that: [ frac{a}{37} + frac{b}{3} = frac{86}{111} ] 4. To combine these fractions, convert them to a common denominator, which is (37 times 3 = 111). [ frac{a cdot 3}{37 cdot 3} + frac{b cdot 37}{3 cdot 37} = frac{86}{111} ] Hence, [ frac{3a}{111} + frac{37b}{111} = frac{86}{111} ] 5. Since the denominators are now the same, we combine the numerators: [ 3a + 37b = 86 ] 6. We need integer solutions for (a) and (b) that make both fractions irreducible. By inspection, or testing small values for (b): - For (b = 2): [ 3a + 37 cdot 2 = 86 Rightarrow 3a + 74 = 86 Rightarrow 3a = 12 Rightarrow a = 4 ] Thus, the fractions are: [ frac{a}{37} = frac{4}{37} quad text{and} quad frac{b}{3} = frac{2}{3} ] 7. Verify their sum: [ frac{4}{37} + frac{2}{3} = frac{4 cdot 3}{37 cdot 3} + frac{2 cdot 37}{3 cdot 37} = frac{12}{111} + frac{74}{111} = frac{12 + 74}{111} = frac{86}{111} ] 8. Since the conditions are satisfied, the required fractions are: [ boxed{frac{2}{3} + frac{4}{37}} ]

answer:z=icdot(1+i)=-1+i, Therefore, the point corresponding to the complex number z is (-1, 1), which is in the second quadrant of the complex plane, Thus, the correct choice is boxed{text{B}}.

answer:1. **Calculate the cost price per orange including the free benefit:** The farmer buys 8 oranges, receives 1 free, effectively getting 9 oranges for 15 + 15 = 30 cents (since 4 oranges cost 15 cents, 8 cost 30 and 9th is free). [ text{Cost price per orange} = frac{30 text{ cents}}{9} approx 3.33 text{ cents/orange} ] 2. **Calculate the selling price per orange:** He sells 7 oranges for 35 cents. [ text{Selling price per orange} = frac{35 text{ cents}}{7} = 5 text{ cents/orange} ] 3. **Calculate the profit per orange:** The profit per orange is now: [ text{Profit per orange} = 5 text{ cents/orange} - 3.33 text{ cents/orange} approx 1.67 text{ cents/orange} ] 4. **Determine the number of oranges needed to make 200 cents of profit:** To achieve 200 cents of profit, we divide the total desired profit by the profit per orange: [ text{Number of oranges} = frac{200 text{ cents}}{1.67 text{ cents/orange}} approx 119.76 ] Rounding up to the nearest whole number, he needs about 120 oranges. 5. **Conclusion:** Therefore, the farmer needs to sell 120 oranges to make a profit of 200 cents. The final answer is boxed{C) 120 oranges}

answer:**Analysis with Modular Arithmetic:** 1. **Observing Powers Modulo 5:** Start by calculating individual powers of 7 modulo 5: - 7^0 equiv 1 pmod{5} - 7^1 equiv 2 pmod{5} - 7^2 equiv 4 pmod{5} - 7^3 equiv 3 pmod{5} - 7^4 equiv 1 pmod{5} (this shows that 7^4 equiv 7^0 pmod{5}, starting a repeating cycle every 4 terms) 2. **Using the cyclic property:** Since the cycle repeats every 4 terms: - 7^5 equiv 2 pmod{5}, 7^6 equiv 4 pmod{5}, 7^7 equiv 3 pmod{5}, 7^8 equiv 1 pmod{5} - 7^9 equiv 2 pmod{5}, 7^{10} equiv 4 pmod{5} 3. **Summing the series:** Add up the results of these powers modulo 5: - (1 + 2 + 4 + 3) + (1 + 2 + 4 + 3) + (1 + 2 + 4)} equiv 10 + 10 + 7 equiv 27 pmod{5} 4. **Final calculation of remainder:** - 27 equiv 2 pmod{5} **Conclusion:** The remainder when 7^0 + 7^1 + 7^2 + cdots + 7^{10} is divided by 5 is 2. The final answer is The final answer, given the choices, is boxed{mathrm{(C)} 2}.