Difference between revisions of "2006 AMC 10B Problems/Problem 16"
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== Problem == | == Problem == | ||
| − | Leap Day, February 29, | + | Leap Day, February <math>29</math>, <math>2024</math>, will occur on a Thursday. On what day of the week will Leap Day, February <math>29</math>, <math>2028</math>, occur? |
<math> \textbf{(A) } \textrm{Tuesday} \qquad \textbf{(B) } \textrm{Wednesday} \qquad \textbf{(C) } \textrm{Thursday} \qquad \textbf{(D) } \textrm{Friday} \qquad \textbf{(E) } \textrm{Saturday} </math> | <math> \textbf{(A) } \textrm{Tuesday} \qquad \textbf{(B) } \textrm{Wednesday} \qquad \textbf{(C) } \textrm{Thursday} \qquad \textbf{(D) } \textrm{Friday} \qquad \textbf{(E) } \textrm{Saturday} </math> | ||
Revision as of 01:40, 1 January 2024
Problem
Leap Day, February 
, 
, will occur on a Thursday. On what day of the week will Leap Day, February , 
, occur?
Solution
There are 
days in a year, plus 
extra day if there is a Leap Day, which occurs on years that are multiples of 
(with a few exceptions that don't affect this problem).
Therefore, the number of days between Leap Day 
and Leap Day 
is:
Since the days of the week repeat every 
days and 
, the day of the week Leap Day occurs is the day of the week the day before Leap Day occurs, which is 
.
Solution 2 (Feasible Shortcuts)
Since every non-leap year there is one day extra after the 
weeks, we can deduce that if we were to travel forward 
amount of non-leap years, then the answer would be "Sunday + " days.
However, we also have leap-years in our pool of years traveled forward, and for every leap-year that passes, we have two extra days after the 52 weeks. So we would travel "Sunday + 
" days.
Mixing these two together, we get leap years 
, and 
non-leap years. We thus get 
"extra days" after Sunday. Since seven days are in a week, we can get 
, and six days after Sunday is .
- Note how we did not require any complex or time-consuming computation, and thus it will save crucial time, especially during a test like the AMC 10.
See Also
| 2006 AMC 10B (Problems • Answer Key • Resources) | ||
| Preceded by Problem 15 |
Followed by Problem 17 | |
| 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 10 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
