1. **Problem statement:** A student bought five online courses with durations 131, 46, 31, 22, and 15 hours, totaling 245 hours. He watches exactly 3 or 4 hours per day of some course, always at the same time and continuously, finishing all courses in 74 days. The goal is to find the number of ways he could have completed the courses, minimizing the days spent on the 131-hour course. 2. **Understanding the problem:** - Total hours: $131 + 46 + 31 + 22 + 15 = 245$ hours. - Total days: 74. - Each day, he watches either 3 or 4 hours of exactly one course. - The 131-hour course should have the minimum possible days. 3. **Key formulas and rules:** - Number of days spent on each course $d_i = \frac{\text{hours of course } i}{h_i}$ where $h_i$ is either 3 or 4 hours per day. - Total days sum: $\sum d_i = 74$. - Minimize $d_1$ for the 131-hour course. 4. **Finding days for the 131-hour course:** - If $h_1=4$, days for course 1: $\frac{131}{4} = 32.75$ (not integer). - If $h_1=3$, days for course 1: $\frac{131}{3} \approx 43.67$ (not integer). Since days must be integer, the student must mix 3 and 4 hour days for the 131-hour course. 5. **Let $x$ be the number of 4-hour days and $y$ the number of 3-hour days for the 131-hour course:** $$4x + 3y = 131$$ $$x + y = d_1$$ We want to minimize $d_1 = x + y$. 6. **Solve for integer solutions:** From $4x + 3y = 131$, express $y$: $$y = \frac{131 - 4x}{3}$$ For $y$ to be integer, $131 - 4x$ must be divisible by 3. Check values of $x$ to find minimal $d_1 = x + y$: - Try $x=20$: $131 - 80 = 51$, $51/3=17$, $d_1=20+17=37$ - Try $x=21$: $131 - 84=47$, $47/3$ not integer - Try $x=19$: $131 - 76=55$, $55/3$ not integer Minimal $d_1=37$ days. 7. **Days for other courses:** Since total days = 74, $$d_2 + d_3 + d_4 + d_5 = 74 - 37 = 37$$ Each other course hours divided by 3 or 4 must be integer days. 8. **Check other courses with 3 or 4 hours per day:** - 46 hours: $46/3 \approx 15.33$, $46/4=11.5$ no integer - 31 hours: $31/3 \approx 10.33$, $31/4=7.75$ no integer - 22 hours: $22/3 \approx 7.33$, $22/4=5.5$ no integer - 15 hours: $15/3=5$, $15/4=3.75$ So for these courses, days must be split between 3 and 4 hours as well. 9. **From the options, only option B matches the days distribution:** $$\frac{74!}{37! \cdot 15! \cdot 10! \cdot 7! \cdot 5!}$$ Where days are 37, 15, 10, 7, and 5 respectively. 10. **Final answer:** Option B.