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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 18]] |
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| − | In year <math>N</math>, the <math>300^\text{th}</math> day of the year is a Tuesday. In year <math>N+1</math>, the <math>200^\text{th}</math> day is also a Tuesday. On what day of the week did the <math>100^\text{th}</math> day of year <math>N-1</math> occur?
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| − | <math>\mathrm{(A)}\ \text{Thursday} \qquad\mathrm{(B)}\ \text{Friday} \qquad\mathrm{(C)}\ \text{Saturday} \qquad\mathrm{(D)}\ \text{Sunday} \qquad\mathrm{(E)}\ \text{Monday}</math>
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| − | ==Solution==
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| − | Clearly, identifying what of these years may/must/may not be a leap year will be key in solving the problem.
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| − | Let <math>A</math> be the <math>300^\text{th}</math> day of year <math>N</math>, <math>B</math> the <math>200^\text{th}</math> day of year <math>N+1</math> and <math>C</math> the <math>100^\text{th}</math> day of year <math>N-1</math>.
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| − | If year <math>N</math> is not a leap year, the day <math>B</math> will be <math>(365-300) + 200 = 265</math> days after <math>A</math>. As <math>265 \bmod 7 = 6</math>, that would be a Monday.
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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>.)
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| − | 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>.
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| − | (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.)
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=24|after=Last Question}}
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