Difference between revisions of "2012 AMC 10A Problems/Problem 12"
m (→Solution) |
Wwilliamtang (talk | contribs) m (→Solution) |
||
Line 9: | Line 9: | ||
== Solution == | == Solution == | ||
− | + | Every year we go back is moving one day back, because <math>365 = 1\ (\text{mod}\ 7)</math>. Every leap year we go back two days, since <math>366 = 2\ (\text{mod}\ 7)</math>. A leap year is usually every four years, so 200 years would have <math>\frac{200}{4}</math> = <math>50</math> leap years, but the problem says that 1900 does not count as a leap year. | |
− | + | Therefore there would be 151 regular years and 49 leap years, so <math>1(151)+2(49)</math> = <math>249</math> days back. Since <math>249 = 4\ (\text{mod}\ 7)</math>, four days back from Tuesday would be <math>\boxed{\textbf{(A)}\ \text{Friday}}</math>. | |
== See Also == | == See Also == |
Revision as of 17:26, 17 July 2017
- The following problem is from both the 2012 AMC 12A #9 and 2012 AMC 10A #12, so both problems redirect to this page.
Problem
A year is a leap year if and only if the year number is divisible by 400 (such as 2000) or is divisible by 4 but not 100 (such as 2012). The 200th anniversary of the birth of novelist Charles Dickens was celebrated on February 7, 2012, a Tuesday. On what day of the week was Dickens born?
Solution
Every year we go back is moving one day back, because . Every leap year we go back two days, since . A leap year is usually every four years, so 200 years would have = leap years, but the problem says that 1900 does not count as a leap year.
Therefore there would be 151 regular years and 49 leap years, so = days back. Since , four days back from Tuesday would be .
See Also
2012 AMC 10A (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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 |
2012 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 8 |
Followed by Problem 10 |
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.