1. **State the problem:** XYZ Company bought a machine for 50000 and profits start at 7876 in March, increasing by 2% each month. We want to find the month when total profits equal or exceed 50000. 2. **Define variables:** Let $P_0 = 7876$ be the profit in March. 3. **Monthly profit formula:** Each month profit increases by 2%, so profit in month $n$ (starting from March as month 0) is: $$P_n = P_0 \times (1.02)^n$$ 4. **Total profit after $n$ months:** Sum of a geometric series: $$S_n = P_0 \times \frac{(1.02)^{n+1} - 1}{1.02 - 1} = 7876 \times \frac{(1.02)^{n+1} - 1}{0.02}$$ 5. **Find $n$ such that $S_n \geq 50000$:** $$7876 \times \frac{(1.02)^{n+1} - 1}{0.02} \geq 50000$$ 6. **Simplify inequality:** $$\frac{(1.02)^{n+1} - 1}{0.02} \geq \frac{50000}{7876} \approx 6.35$$ 7. Multiply both sides by 0.02: $$(1.02)^{n+1} - 1 \geq 0.127$$ 8. Add 1: $$(1.02)^{n+1} \geq 1.127$$ 9. Take natural logarithm: $$(n+1) \ln(1.02) \geq \ln(1.127)$$ 10. Calculate values: $$\ln(1.02) \approx 0.0198, \quad \ln(1.127) \approx 0.1197$$ 11. Solve for $n$: $$n+1 \geq \frac{0.1197}{0.0198} \approx 6.05$$ $$n \geq 5.05$$ 12. Since $n$ counts months after March (March is $n=0$), the machine is paid off after 6 months (rounding up). 13. **Months:** - March: $n=0$ - April: $n=1$ - May: $n=2$ - June: $n=3$ - July: $n=4$ - August: $n=5$ - September: $n=6$ 14. Therefore, the machine is fully paid off in **September**. **Final answer:** September