Difference between revisions of "2012 AMC 10A Problems/Problem 12"

Line 1: Line 1:
 +
{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #9]] and [[2012 AMC 10A Problems|2012 AMC 10A #12]]}}
 +
 
== Problem ==
 
== Problem ==
  
Line 14: Line 16:
  
 
{{AMC10 box|year=2012|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2012|ab=A|num-b=11|num-a=13}}
 +
{{AMC12 box|year=2012|ab=A|num-b=8|num-a=10}}

Revision as of 14:16, 12 February 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?

$\textbf{(A)}\ \text{Friday}\qquad\textbf{(B)}\ \text{Saturday}\qquad\textbf{(C)}\ \text{Sunday}\qquad\textbf{(D)}\ \text{Monday}\qquad\textbf{(E)}\ \text{Tuesday}$

Solution 1

Each year we go back is one day back, because $365 = 1\ (\text{mod}\ 7)$. Each leap year we go back is two days back, since $366 = 2\ (\text{mod}\ 7)$. A leap year is GENERALLY every four years, so 200 years would have $\frac{200}{4}$ = $50$ leap years, but the problem points out that 1900 does not count as a leap year.

This would mean a total of 150 regular years and 49 leap years, so $1(151)+2(49)$ = $249$ days back. Since $249 = 4\ (\text{mod}\ 7)$, four days back from Tuesday would be $\boxed{\textbf{(A)}\ \text{Friday}}$

See Also

2012 AMC 10A (ProblemsAnswer KeyResources)
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 (ProblemsAnswer KeyResources)
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