2012 AMC 10A Problems/Problem 12: Difference between revisions
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Each year we go back is one day back, because <math>365 = 1\ (\text{mod}\ 7)</math>. Each leap year we go back is two days back, since <math>366 = 2\ (\text{mod}\ 7)</math>. A leap year is GENERALLY every four years, so 200 years would have <math>\frac{200}{4}</math> = <math>50</math> leap years, but the problem points out that 1900 does not count as a leap year. | Each year we go back is one day back, because <math>365 = 1\ (\text{mod}\ 7)</math>. Each leap year we go back is two days back, since <math>366 = 2\ (\text{mod}\ 7)</math>. A leap year is GENERALLY every four years, so 200 years would have <math>\frac{200}{4}</math> = <math>50</math> leap years, but the problem points out that 1900 does not count as a leap year. | ||
This would mean a total of | This would mean a total of 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 23:24, 13 November 2012
- 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
Each year we go back is one day back, because 
. Each leap year we go back is two days back, since 
. A leap year is GENERALLY every four years, so 200 years would have 
= 
leap years, but the problem points out that 1900 does not count as a leap year.
This would mean a total of 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 | |
