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

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== Solution ==
 
== Solution ==
 
   
 
   
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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 usually 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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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.
  
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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>.
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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 = 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 16:26, 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?

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

Solution

Every year we go back is moving one day back, because $365 = 1\ (\text{mod}\ 7)$. Every leap year we go back two days, since $366 = 2\ (\text{mod}\ 7)$. 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 $249 = 4\ (\text{mod}\ 7)$, four days back from Tuesday would be $\boxed{\textbf{(A)}\ \text{Friday}}$.

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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