Difference between revisions of "2012 AMC 10A Problems/Problem 12"
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<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> | <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 == | + | == Solution 1 == |
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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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| + | == Solution ? == | ||
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. | 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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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>. | 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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== 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 19:44, 9 February 2012
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 1
Each year we go back is one day back, because 
. Each leap year we go back is two days back, since 
. A leap year is GENERALLY every four years, so 200 years would have 
= 
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 
= 
days back. Since 
, four days back from Tuesday would be 
Solution ?
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.
The number of days in a regular year (365) is 
and the number of days in a leap year (366) is 
. Every four years, we go back the same number of days of the week, which is 
days. Every thirty-five years, we go back 
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 
Because there are four years to 1812, we go back 5 days of the week from Tuesday, which 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 | ||
