2010 AMC 10A Problem 10

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 $\boxed{\textbf{(E)}\ 2017}$.

Weekday drift is arithmetic mod $7$: each ordinary year contributes $1$, each leap year crossed contributes $2$. The leap day of $2008$ itself is before May 27, so it never matters; the leap days that count are Feb 29 of $2012$ and $2016$. The skipped Saturday in $2011$-$2012$ (Friday jumps to Sunday) is exactly why the answer is not $2012$.

• Counting 2008 itself as adding a leap-day shift, or forgetting the extra shift in 2012 and 2016 entirely.
• Assuming the weekday advances by exactly one every year and answering $2012$.