Difference between revisions of "2016 AMC 12B Problems/Problem 5"
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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]] | 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]] | ||
+ | |||
+ | ==Solution 3 (Doomsday)== | ||
+ | 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
Contents
[hide]Problem
The War of started with a declaration of war on Thursday, June , . The peace treaty to end the war was signed days later, on December , . On what day of the week was the treaty signed?
Solution
By: dragonfly
To find what day of the week it is in days, we have to divide by to see the remainder, and then add the remainder to the current day. We get that has a remainder of 2, so we increase the current day by to get .
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)
is a leap year (divisible by , but not divisible by unless divisible by ), but June is after February so there are no leap days that need to be accounted for.
Since , the same day after year is weekday ahead. Since and , the same day after / days is / weekdays ahead.
- years ( weekdays) passed, reaching June , .
- June - and July - ( days or weekdays) passed, reaching July .
- July - and August - ( days or weekdays) passed, reaching August .
- August-September ( weekdays) passed.
- September-October ( weekdays) passed.
- October-November ( weekdays) passed.
- November-December ( weekdays) passed, reaching December .
- days passed, reaching December .
This gives a total of weekdays passing, which gets from Thursday to . ~emerald_block
Solution 3 (Doomsday)
Since doomsday of is (Tuesday) and is a multiple of , counting down by -year intervals gets that doomsday of is . Counting up by -year intervals, doomsday of is . Doomsday of is then , so December 12 is weekday . December 24 is then weekday . ~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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.