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Difference between revisions of "2012 AMC 10A Problems/Problem 12"
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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>
 
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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== Solution ? ==
 
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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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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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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>.
 
  
 
== 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}}

Revision as of 03:42, 10 February 2012

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 (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
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