2000 AMC 10 Problem 25

From day $300$ of year $N$ to day $200$ of year $N+1$ is $(L_N - 300) + 200 = L_N - 100$ days, where $L_N$ is the length of year $N$: $265$ days if $L_N = 365$, or $266$ if year $N$ is a leap year. Both dates are Tuesdays, so the gap must be a multiple of $7$. Since $266 = 7 \cdot 38$ and $265$ is not divisible by $7$, year $N$ is a leap year.

Consecutive years are never both leap years, so year $N-1$ has $365$ days. From day $100$ of year $N-1$ to day $300$ of year $N$ is $(365 - 100) + 300 = 565$ days, and $565 = 7 \cdot 80 + 5$. The earlier date is therefore $5$ weekdays before Tuesday: Monday, Sunday, Saturday, Friday, Thursday.

The answer is $\boxed{\textbf{(A)}\ \text{Thursday}}$.

Weekday problems are arithmetic mod $7$ on day counts. The only subtlety is the length of each year, and the problem gives you exactly enough information to pin that down: two dates with the same weekday must be separated by a multiple of $7$. Always decide leap-year status first, then count.

• Counting 5 days forward from Tuesday (getting Sunday) when the target date is earlier, or ignoring leap years entirely.
• Assuming year $N+1$ (rather than $N$) is the leap year; the leap day must fall between the two given dates, and Feb 29 of year $N+1$ does.