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Difference between revisions of "2016 AMC 12B Problems/Problem 5"

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By: dragonfly
 
By: dragonfly
  
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To find what day of the week it is in <math>919</math> days, we have to divide <math>919</math> by <math>7</math> to see the remainder, and then add the remainder to the current day. We get that <math>\frac{919}{7}</math> has a remainder of 2, so we increase the current day by <math>2</math> to get <math>\boxed{\textbf{(B)} \text{Saturday}}</math>
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To find what day of the week it is in <math>919</math> days, we have to divide <math>919</math> by <math>7</math> to see the remainder, and then add the remainder to the current day. We get that <math>\frac{919}{7}</math> has a remainder of 2, so we increase the current day by <math>2</math> to get <math>\boxed{\textbf{(B)}\ \text{Saturday}}</math>.
  
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Question: Why don't we have to account for leap years?
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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.
  
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Answer: Although 1812 is a leap year, June 18 is after February 29. Plus, it gives you the number of days...so...
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==Solution 2 (Highly not recommended)==
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<math>1812</math> is a leap year (divisible by <math>4</math>, but not divisible by <math>100</math> unless divisible by <math>400</math>), but June is after February so there are no leap days that need to be accounted for.
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Since <math>365 \equiv 1 \pmod{7}</math>, the same day after <math>1</math> year is <math>1</math> weekday ahead. Since <math>30 \equiv 2 \pmod{7}</math> and <math>31 \equiv 3 \pmod{7}</math>, the same day after <math>30</math>/<math>31</math> days is <math>2</math>/<math>3</math> weekdays ahead.
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* <math>2</math> years (<math>2</math> weekdays) passed, reaching June <math>18</math>, <math>1812</math>.
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* June <math>18</math>-<math>30</math> and July <math>1</math>-<math>17</math> (<math>30</math> days or <math>2</math> weekdays) passed, reaching July <math>18</math>.
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* July <math>18</math>-<math>31</math> and August <math>1</math>-<math>17</math> (<math>31</math> days or <math>3</math> weekdays) passed, reaching August <math>18</math>.
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* August-September (<math>3</math> weekdays) passed.
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* September-October (<math>2</math> weekdays) passed.
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* October-November (<math>3</math> weekdays) passed.
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* November-December (<math>2</math> weekdays) passed, reaching December <math>18</math>.
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* <math>6</math> days passed, reaching December <math>24</math>.
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This gives a total of <math>2+2+3+3+2+3+2+6 \equiv 2 \pmod{7}</math> weekdays passing, which gets from Thursday to <math>\boxed{\textbf{(B)}\ \text{Saturday}}</math>. ~[[User:emerald_block|emerald_block]]
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==Solution 3 (Doomsday)==
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Since [https://en.wikipedia.org/wiki/Doomsday_rule doomsday] of <math>2000</math> is <math>2</math> (Tuesday) and <math>2000</math> is a multiple of <math>400</math>, counting down by <math>100</math>-year intervals gets that doomsday of <math>1800</math> is <math>2+1+2 \equiv 5</math>. Counting up by <math>4</math>-year intervals, doomsday of <math>1812</math> is <math>5-3\cdot2 \equiv 6</math>. Doomsday of <math>1814</math> is then <math>6+2\cdot1 \equiv 1</math>, so December 12 is weekday <math>1</math>. December 24 is then weekday <math>1+12 \equiv 6 = \boxed{\textbf{(B)}\ \text{Saturday}}</math>. ~[[User:emerald_block|emerald_block]]
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=B|num-b=4|num-a=6}}
 
{{AMC12 box|year=2016|ab=B|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 22:45, 6 November 2021

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

See Also

2016 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

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