2000 AMC 10 Problem 25

1. Key fact. Two dates on the same day of the week are a whole number of weeks apart, so the number of days between them divides evenly by $7$.
2. Is year $N$ a leap year? Count from day $300$ of year $N$ to day $200$ of year $N+1$.
- If year $N$ has $365$ days: $65$ days are left in year $N$, then $200$ more, for $265$ days.
- If year $N$ has $366$ days: $66 + 200 = 266$ days.

Both dates are Tuesdays, so the gap must divide by $7$. Now $266 = 7 \cdot 38$, but $265$ leaves remainder $6$. So year $N$ is a leap year.
3. Year $N-1$. Leap years are $4$ years apart, so the year right before a leap year is not one. Year $N-1$ has $365$ days.
4. Count from day $100$ of year $N-1$ to day $300$ of year $N$. There are $365 - 100 = 265$ days to the end of year $N-1$, then $300$ more:
$$265 + 300 = 565 \text{ days}.$$
5. Divide by $7$. $565 = 7 \cdot 80 + 5$. So after $80$ full weeks there are $5$ extra days.
6. Go backward. Day $100$ of year $N-1$ comes before the Tuesday, so count $5$ days back from Tuesday: Monday, Sunday, Saturday, Friday, Thursday.

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

Day-of-the-week problems are about remainders after dividing by $7$. The only tricky part is how long each year is, and the problem gives just enough information to decide that: two Tuesdays must be a whole number of weeks apart. Decide which years are leap years first, then count, and be careful to count backward when the date you want comes earlier.

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