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Difference between revisions of "2012 AMC 10A Problems/Problem 12"
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{{duplicate|[[2012 AMC 12A Problems|2012 AMC 12A #9]] and [[2012 AMC 10A Problems|2012 AMC 10A #12]]}}
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== Problem ==
 
== Problem ==
  
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== Solution ==
 
== Solution ==
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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.
  
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Ignore their over-complicated definition of a leap year because it is the same as we know it; every year that is a multiple of 4.
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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>.
  
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The number of days in a regular year (365) is <math>1\ (\text{mod}\ 7)</math> and the number of days in a leap year (366) is <math>2\ (\text{mod}\ 7)</math>. Every four years, we go back the same number of days of the week, which is <math>1+1+1+2=5</math> days. Every thirty-five years, we go back <math>5 \cdot 7=35</math> days of the week, or no days of the week at all. Therefore, no matter how many times we subtract 28 years from February 7, 2012, it will always be a Tuesday. The number closest to 1812 (200 years back) that follows that is <math>2012-28\cdot7=1816.</math>
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== Solution 2==  
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Because there are four years to 1812, we go back 5 days of the week from Tuesday, which is <math>\boxed{\textbf{(A)}\ \text{Friday}}</math>.
 
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== Solution 2 ==
 
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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.
 
  
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This would mean a total of 150 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>
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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 exclude <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 ==
 
== See Also ==
  
 
{{AMC10 box|year=2012|ab=A|num-b=11|num-a=13}}
 
{{AMC10 box|year=2012|ab=A|num-b=11|num-a=13}}
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{{AMC12 box|year=2012|ab=A|num-b=8|num-a=10}}
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{{MAA Notice}}

Latest revision as of 18:44, 22 September 2024

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

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 $365\equiv 1\pmod 7$. Every leap year we move to the right results in moving $2$ days to the right since $366\equiv 2\pmod 7$. A leap year is usually every four years, so 200 years would have $\frac{200}{4}$ = $50$ 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 $1(151)+2(49)$ = $249$ days back. Since $249 \equiv 4\ (\text{mod}\ 7)$, four days back from Tuesday would be $\boxed{\textbf{(A)}\ \text{Friday}}$.

Solution 2

Since they say that February $7$th, $2012$ is the $200$th anniversary of Charles dickens birthday, that means that the birth of Charles Dickens is on February $7$th, $1812$. We then see that there is a leap year on $1812, 1816, ...., 2008$ but we must exclude $1900$ which equates to $49$ leap years. So, the amount of days we have to go back is $200(365) + 49$ days which in $\text{mod }7$ gives us 4. Thus, $4$ days back from Tuesday is $\boxed{\textbf{(A)}\ \text{Friday}}$.

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

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