answer:1. **First Step:** We know each counterfeiter can exchange any amount from 1 ruble to 25 rubles with another counterfeiter. Let's start by analyzing the scenario where the first counterfeiter upholds exactly 25 rubles to the second counterfeiter. 2. **Detailed Breakdown:** If the first counterfeiter gives ( x ) rubles to the second counterfeiter and receives ( y ) rubles as change, we have: [ x - y = 25 ] Since ( x ) and ( y ) can each be at most 100 rubles (assuming the total any counterfeiter can possess is capped at 100 rubles due to the nature of producing sums from 1 to 25 rubles), let’s consider various feasible values for ( x ) and ( y ): - If ( x geq 75 ), then ( y ) must be at least 25 because ( x - y = 25 ). - Similarly, ( x leq 100 ) implies ( y geq 0 ) but ensuring non-negative change and keeping within allowable rubles. Therefore, one of the values, either ( x ) or ( y ) (say ( x )), will fall within the range of 25 to 75 rubles. 3. **Combining Values:** Now considering ( x leq 75 ), ( y ) maintains the upper limit ( 100 geq y geq 0 ), but since it’s impossible ( y ) to be negative and as we structured ( x + y leq 100 ). 4. **Expand Scenario:** If ( x > 50 ), then consider those exact rubles leftover: - If ( x geq 50 ) remains valid, collect summary amounts from the first counterfeiter to just still remain within limits. 5. **Formulating Solution for Combined Sum:** Let's identify the true needful sum, ( S ). Assume the first counterfeiter holds the sum ( S ) exactly. The second counterfeiter can provide a set value within the span of 25 rubles directly. 6. **Final Arrangements:** If the second counterfeiter manages any span value 0 to 25 rubles, the total combination for first and second counterfeiter span covers 75 to 125 rubles. 7. **Using Combined Capabilities:** By pooling these and the first's balance providing coverage ( S ) covers immediately all needed sum from 100 to 150 rubles. 8. **Final Finalization:** All set combinations valid through combinations below 75 and immediate vacancies summed. [ S + (0 to 25 for second) rightarrow all sum validation akin spans 100 to 150 text{ rubles} ] 9. **Conclusion:** Subtracting any amalgamation should valid scenarios compose the rubles span directly completing direct need summation 100 to 200 rubles precisely. Hence verified: [ 100 -200 ] So, combining values cover total interval covers valid expected occupancy. (blacksquare)

answer:Let's denote the time it takes for the second tap to empty the cistern as ( T ) hours. The filling rate of the first tap is ( frac{1}{4} ) of the cistern per hour, since it fills the cistern in 4 hours. The emptying rate of the second tap is ( frac{1}{T} ) of the cistern per hour. When both taps are opened simultaneously, the net rate at which the cistern gets filled is the filling rate of the first tap minus the emptying rate of the second tap. This net rate fills the cistern in ( frac{20}{3} ) hours (which is 6.666666666666667 hours in decimal form). So, we have the equation: [ frac{1}{4} - frac{1}{T} = frac{1}{frac{20}{3}} ] Solving for ( T ): [ frac{1}{4} - frac{1}{T} = frac{3}{20} ] [ frac{1}{T} = frac{1}{4} - frac{3}{20} ] To subtract these fractions, we need a common denominator, which is 20: [ frac{1}{T} = frac{5}{20} - frac{3}{20} ] [ frac{1}{T} = frac{2}{20} ] [ frac{1}{T} = frac{1}{10} ] So, ( T = 10 ) hours. It takes boxed{10} hours for the second tap to empty the cistern.

answer:To find Marie's speed in miles per hour, we need to divide the total distance she bikes by the total time it takes her to bike that distance. Speed = Distance / Time Given that Marie bikes 372 miles in 31 hours, her speed would be: Speed = 372 miles / 31 hours = 12 miles per hour So, Marie's speed is boxed{12} miles per hour.

answer:Solution: (1) Since S_n= frac {5}{6}n+ frac {n(n-1)}{2} times (- frac {1}{6}) =-5, we have n^2-11n-60=0. Solving this, we get n=15 or n=-4 (discard the latter), thus a_n=a_{15}= frac {5}{6} -14 times frac {1}{6} = - frac {3}{2}; (2) Since a_n=a_1+14 times 2=-10, we have a_1=-38, S_n=15 times (-38) + frac {15 times 14}{2} times 2 =-360. Therefore, the answers are n=15, a_n= - frac {3}{2} for the first part, and a_1=-38, S_n=-360 for the second part. Thus, the final answers are boxed{n=15, a_n= - frac {3}{2}} for the first question, and boxed{a_1=-38, S_n=-360} for the second question.