2012 AMC 10A Problem 12

Dickens was born on February $7$, $1812$. We know February $7$, $2012$ was a Tuesday, so find how far the weekday moved in those $200$ years.

1. An ordinary year has $365 = 52 \times 7 + 1$ days: $52$ full weeks plus $1$ extra day. So one year later, the same date falls $1$ weekday later. For example, if a date is a Monday, a year later it is a Tuesday.
2. A February $29$ in between adds one more day.
3. From February $7$, $1812$ to February $7$, $2012$ there are $200$ years, so that is $200$ extra days, plus one for each February $29$ in between.
4. Which February $29$'s are in between? February $29$, $1812$ comes after February $7$, $1812$, so it counts. February $29$, $2012$ comes after February $7$, $2012$, so it does not. We need the leap years from $1812$ to $2008$.
5. The multiples of $4$ from $1812$ to $2008$: $(2008 - 1812) \div 4 + 1 = 49 + 1 = 50$ of them.
6. But $1900$ is divisible by $100$ and not by $400$, so it is not a leap year. That leaves $49$ leap days. ($2000$ is divisible by $400$, so it stays.)
7. Total shift: $200 + 49 = 249$ days. Since $249 = 35 \times 7 + 4$, the $35$ full weeks do not change the weekday, and $4$ days are left over.
8. So 2012's date is $4$ weekdays after 1812's date. Count back $4$ from Tuesday: Monday, Sunday, Saturday, Friday.

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

Full weeks never change the day of the week, so only the leftover days matter. Since $365$ is one more than a multiple of $7$, you only need to count years and leap days. The problem spells out the leap-year rule so that you notice the $1900$ exception. Be careful about which February $29$'s fall between the two dates, and about moving backward in time.