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Difference between revisions of "2000 AMC 10 Problems/Problem 25"
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Therefore year <math>N</math> must be a leap year. (Then <math>B</math> is <math>266</math> days after <math>A</math>.)
 
Therefore year <math>N</math> must be a leap year. (Then <math>B</math> is <math>266</math> days after <math>A</math>.)
  
−
As there can not be two leap years after each other, <math>N-1</math> is not a leap year. Therefore day <math>A</math> is <math>265 + 300 = 565</math> days after <math>C</math>. We have <math>565\bmod 7 = 5</math>. Therefore <math>C</math> is <math>5</math> weekdays before <math>A</math>, i.e., <math>C</math> is a <math>\boxed{\text{Thursday}}</math>.
+
As there can not be two leap years after each other, <math>N-1</math> is not a leap year. Therefore day <math>A</math> is <math>265 + 300 = 565</math> days after <math>C</math>. We have <math>565\bmod 7 = 5</math>. Therefore <math>C</math> is <math>5</math> weekdays before <math>A</math>, i.e., <math>C</math> is a <math>\boxed{\text{Saturday}}</math>.
  
 
(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for <math>N=2004</math> we have <math>A</math>=Tuesday, October 26th 2004, <math>B</math>=Tuesday, July 19th, 2005 and <math>C</math>=Thursday, April 10th 2003.)
 
(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for <math>N=2004</math> we have <math>A</math>=Tuesday, October 26th 2004, <math>B</math>=Tuesday, July 19th, 2005 and <math>C</math>=Thursday, April 10th 2003.)
 +
 
==See Also==
 
==See Also==
  
 
{{AMC10 box|year=2000|num-b=24|after=Last Question}}
 
{{AMC10 box|year=2000|num-b=24|after=Last Question}}

Revision as of 22:13, 22 November 2011

Problem

In year $N$, the $300^\text{th}$ day of the year is a Tuesday. In year $N+1$, the $200^\text{th}$ day is also a Tuesday. On what day of the week did the $100^\text{th}$ day of year $N-1$ occur?

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

Solution

Clearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.

Let $A$ be the $300^\text{th}$ day of year $N$, $B$ the $200^\text{th}$ day of year $N+1$ and $C$ the $100^\text{th}$ day of year $N-1$.

If year $N$ is not a leap year, the day $B$ will be $(365-300) + 200 = 265$ days after $A$. As $265 \bmod 7 = 6$, that would be a Monday.

Therefore year $N$ must be a leap year. (Then $B$ is $266$ days after $A$.)

As there can not be two leap years after each other, $N-1$ is not a leap year. Therefore day $A$ is $265 + 300 = 565$ days after $C$. We have $565\bmod 7 = 5$. Therefore $C$ is $5$ weekdays before $A$, i.e., $C$ is a $\boxed{\text{Saturday}}$.

(Note that the situation described by the problem statement indeed occurs in our calendar. For example, for $N=2004$ we have $A$=Tuesday, October 26th 2004, $B$=Tuesday, July 19th, 2005 and $C$=Thursday, April 10th 2003.)

See Also

2000 AMC 10 (Problems • Answer Key • Resources)
Preceded by
Problem 24 		Followed by
Last Question
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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