2000 AMC 12 Problems/Problem 18
- The following problem is from both the 2000 AMC 12 #18 and 2000 AMC 10 #25, so both problems redirect to this page.
Problem
In year , the day of the year is a Tuesday. In year , the day is also a Tuesday. On what day of the week did the ^{th} day of year occur?
Solution 1
There are either or days between the first two dates depending upon whether or not year is a leap year. Since divides into but not , for both days to be a Tuesday, year must be a leap year.
Hence, year is not a leap year, and so since there are days between the date in years , this leaves a remainder of upon division by . Since we are subtracting days, we count 5 days before Tuesday, which gives us
Solution 2
The day of year and the day of year are Tuesdays. If there were the same number of days in year and year the day of year will be a Tuesday. But year is a leap year because so the day of year is a Tuesday. It follows that the day of year is
~dolphin7
Video Solution
https://www.youtube.com/watch?v=YfiUn8hLlpA ~David
See also
2000 AMC 12 (Problems • Answer Key • Resources) | |
Preceded by Problem 17 |
Followed by Problem 19 |
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 |
2000 AMC 10 (Problems • Answer Key • Resources) | ||
Preceded by Problem 24 |
Followed by Last Problem | |
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.