# 2016 AMC 12B Problems/Problem 5

## Problem

The War of $1812$ started with a declaration of war on Thursday, June $18$, $1812$. The peace treaty to end the war was signed $919$ days later, on December $24$, $1814$. On what day of the week was the treaty signed? $\textbf{(A)}\ \text{Friday} \qquad \textbf{(B)}\ \text{Saturday} \qquad \textbf{(C)}\ \text{Sunday} \qquad \textbf{(D)}\ \text{Monday} \qquad \textbf{(E)}\ \text{Tuesday}$

## Solution

By: dragonfly

To find what day of the week it is in $919$ days, we have to divide $919$ by $7$ to see the remainder, and then add the remainder to the current day. We get that $\frac{919}{7}$ has a remainder of 2, so we increase the current day by $2$ to get $\boxed{\textbf{(B)}\ \text{Saturday}}$.

Note that the dates themselves (and thus leap years) can be ignored, as we only need the number of days that passed to figure out the day of the week.

## Solution 2 (Highly not recommended) $1812$ is a leap year (divisible by $4$, but not divisible by $100$ unless divisible by $400$), but June is after February so there are no leap days that need to be accounted for.

Since $365 \equiv 1 \pmod{7}$, the same day after $1$ year is $1$ weekday ahead. Since $30 \equiv 2 \pmod{7}$ and $31 \equiv 3 \pmod{7}$, the same day after $30$/ $31$ days is $2$/ $3$ weekdays ahead.

• $2$ years ( $2$ weekdays) passed, reaching June $18$, $1812$.
• June $18$- $30$ and July $1$- $17$ ( $30$ days or $2$ weekdays) passed, reaching July $18$.
• July $18$- $31$ and August $1$- $17$ ( $31$ days or $3$ weekdays) passed, reaching August $18$.
• August-September ( $3$ weekdays) passed.
• September-October ( $2$ weekdays) passed.
• October-November ( $3$ weekdays) passed.
• November-December ( $2$ weekdays) passed, reaching December $18$.
• $6$ days passed, reaching December $24$.

This gives a total of $2+2+3+3+2+3+2+6 \equiv 2 \pmod{7}$ weekdays passing, which gets from Thursday to $\boxed{\textbf{(B)}\ \text{Saturday}}$. ~emerald_block

## Solution 3 (Doomsday)

Since doomsday of $2000$ is $2$ (Tuesday) and $2000$ is a multiple of $400$, counting down by $100$-year intervals gets that doomsday of $1800$ is $2+1+2 \equiv 5$. Counting up by $4$-year intervals, doomsday of $1812$ is $5-3\cdot2 \equiv 6$. Doomsday of $1814$ is then $6+2\cdot1 \equiv 1$, so December 12 is weekday $1$. December 24 is then weekday $1+12 \equiv 6 = \boxed{\textbf{(B)}\ \text{Saturday}}$. ~emerald_block

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 