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| − | + | {{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #9]] and [[2012 AMC 10A Problems|2012 AMC 10A #12]]}} | |
| − | + | == Problem == | |
| − | <math> \textbf{(A)}\ | + | 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? |
| + | |||
| + | <math> \textbf{(A)}\ \text{Friday}\qquad\textbf{(B)}\ \text{Saturday}\qquad\textbf{(C)}\ \text{Sunday}\qquad\textbf{(D)}\ \text{Monday}\qquad\textbf{(E)}\ \text{Tuesday} </math> | ||
| + | |||
| + | == Solution == | ||
| + | |||
| + | In this solution we refer to moving to the left as decreasing the year or date number and moving to the right as increasing the year or date number. Every non-leap year we move to the right results in moving one day to the right because <math>365\equiv 1\pmod 7</math>. Every leap year we move to the right results in moving <math>2</math> days to the right since <math>366\equiv 2\pmod 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 \equiv 4\ (\text{mod}\ 7)</math>, four days back from Tuesday would be <math>\boxed{\textbf{(A)}\ \text{Friday}}</math>. | ||
| + | |||
| + | == Solution 2== | ||
| + | |||
| + | Since they say that February <math>7</math>th, <math>2012</math> is the <math>200</math>th anniversary of Charles dickens birthday, that means that the birth of Charles Dickens is on February <math>7</math>th, <math>1812</math>. We then see that there is a leap year on <math>1812, 1816, ...., 2008</math> but we must excluse <math>1900</math> which equates to <math>49</math> leap years. So, the amount of days we have to go back is <math>200(365) + 49</math> days which in <math>\text{mod }7</math> gives us 4. Thus, <math>4</math> days back from Tuesday is <math>\boxed{\textbf{(A)}\ \text{Friday}}</math>. | ||
| + | |||
| + | == See Also == | ||
| + | |||
| + | {{AMC10 box|year=2012|ab=A|num-b=11|num-a=13}} | ||
| + | {{AMC12 box|year=2012|ab=A|num-b=8|num-a=10}} | ||
| + | {{MAA Notice}} | ||
Revision as of 16:31, 27 March 2024
- The following problem is from both the 2012 AMC 12A #9 and 2012 AMC 10A #12, so both problems redirect to this page.
Contents
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
In this solution we refer to moving to the left as decreasing the year or date number and moving to the right as increasing the year or date number. Every non-leap year we move to the right results in moving one day to the right because 
. Every leap year we move to the right results in moving 
days to the right 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 
.
Solution 2
Since they say that February 
th, 
is the 
th anniversary of Charles dickens birthday, that means that the birth of Charles Dickens is on February th, 
. We then see that there is a leap year on 
but we must excluse 
which equates to 
leap years. So, the amount of days we have to go back is 
days which in 
gives us 4. Thus, 
days back from Tuesday is .
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. 
