2010 AMC 10A Problem 10

First figure out how the weekday moves from one birthday to the next.

1. A normal year has $365$ days. That is $52$ full weeks ($364$ days) plus $1$ extra day. So next year's birthday falls $1$ weekday later.
2. If a February 29 comes between two birthdays, there are $366$ days, which is $52$ weeks plus $2$ days. Then the birthday moves $2$ weekdays later. This happens when the later year is a leap year: 2012 and 2016. (February 29, 2008 is before May 27, 2008, so it does not count.)
3. Now step year by year from Tuesday in 2008:

1. Notice that in 2012 the birthday jumps from Friday straight to Sunday and skips Saturday. The first Saturday is in 2017.

The answer is $\boxed{\textbf{(E)}\ 2017}$.

Weekdays repeat every $7$ days, so only the leftover days after full weeks matter: $1$ for a normal year, $2$ for a leap year. The danger is the leap-year jump, which can skip right over the day you are looking for. Writing a small table keeps track of it.

From one May 27 to the next is $365$ days, which is $52$ weeks plus $1$ day, so the weekday advances by $1$. If a February 29 lies in between (that is, the later year is a leap year), the gap is $366$ days and the weekday advances by $2$.

Saturday is $4$ days after Tuesday, so we need the total shift to be $4$, or $11$, or $18$, and so on. Tally year by year, starting from Tuesday in $2008$:

• $2009$: $+1$, Wednesday
• $2010$: $+1$, Thursday
• $2011$: $+1$, Friday
• $2012$ (leap): $+2$, Sunday
• $2013$: $+1$, Monday
• $2014$: $+1$, Tuesday
• $2015$: $+1$, Wednesday
• $2016$ (leap): $+2$, Friday
• $2017$: $+1$, Saturday

The birthday first falls on a Saturday in $2017$.

The answer is $\boxed{\textbf{(E)}\ 2017}$.