Difference between revisions of "2006 AMC 10B Problems/Problem 16"
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<math> 16 \cdot 365 + 4 \cdot 1 = 5844 </math> | <math> 16 \cdot 365 + 4 \cdot 1 = 5844 </math> | ||
| − | Since the days of the week repeat every <math>7</math> days and <math> 5844 \equiv -1 \bmod{7}</math>, the day of the week Leap Day 2020 occurs is the day of the week the day before Leap Day 2004 occurs, which is <math>\textrm{Saturday} | + | Since the days of the week repeat every <math>7</math> days and <math> 5844 \equiv -1 \bmod{7}</math>, the day of the week Leap Day 2020 occurs is the day of the week the day before Leap Day 2004 occurs, which is <math>\boxed{\textbf{(E) }\textrm{Saturday}}</math>. |
== Solution 2 (Feasible Shortcuts) == | == Solution 2 (Feasible Shortcuts) == | ||
Revision as of 10:47, 7 March 2022
Problem
Leap Day, February 29, 2004, occurred on a Sunday. On what day of the week will Leap Day, February 29, 2020, 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 4 (with a few exceptions that don't affect this problem).
Therefore, the number of days between Leap Day 2004 and Leap Day 2020 is:
Since the days of the week repeat every 
days and 
, the day of the week Leap Day 2020 occurs is the day of the week the day before Leap Day 2004 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. 
