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| − | == Problem==
| + | #redirect [[2010 AMC 12B Problems/Problem 4]] |
| − | A month with <math>31</math> days has the same number of Mondays and Wednesdays. How many of the seven days of the week could be the first day of this month?
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| − | <math>\textbf{(A)}\ 2 \qquad \textbf{(B)}\ 3 \qquad \textbf{(C)}\ 4 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ 6</math>
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| − | ==Solution==
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| − | In this month there are four weeks and three remaining days. Any 7 days must have exactly one Monday and one Wednesday, so it works if the last <math>31 - 4\cdot 7 = 3</math> days have the same number of Mondays and Wednesdays. We have three choices: Monday, Tuesday, Wednesday; Thursday, Friday, Saturday; Friday, Saturday, Sunday. The number of days the month can start on are Monday, Thursday, and Friday, for a final answer of <math>\boxed{\textbf{(B)}\ 3}.</math>
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| − | ==Solution 2==
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| − | Let's make a calendar to visualize the situation better.
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| − | \begin{table}[]
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| − | \begin{tabular}{lllllll}
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| − | \hline
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| − | \multicolumn{1}{|l|}{1} & \multicolumn{1}{l|}{2} & \multicolumn{1}{l|}{3} & \multicolumn{1}{l|}{4} & \multicolumn{1}{l|}{5} & \multicolumn{1}{l|}{6} & \multicolumn{1}{l|}{7} \\ \hline
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| − | \multicolumn{1}{|l|}{8} & \multicolumn{1}{l|}{9} & \multicolumn{1}{l|}{10} & \multicolumn{1}{l|}{11} & \multicolumn{1}{l|}{12} & \multicolumn{1}{l|}{13} & \multicolumn{1}{l|}{14} \\ \hline
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| − | \multicolumn{1}{|l|}{15} & \multicolumn{1}{l|}{16} & \multicolumn{1}{l|}{17} & \multicolumn{1}{l|}{18} & \multicolumn{1}{l|}{19} & \multicolumn{1}{l|}{20} & \multicolumn{1}{l|}{21} \\ \hline
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| − | \multicolumn{1}{|l|}{22} & \multicolumn{1}{l|}{23} & \multicolumn{1}{l|}{24} & \multicolumn{1}{l|}{25} & \multicolumn{1}{l|}{26} & \multicolumn{1}{l|}{27} & \multicolumn{1}{l|}{28} \\ \hline
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| − | 29 & 30 & 31 & & & &
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| − | \end{tabular}
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| − | \end{table}
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| − | ==See Also==
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| − | {{AMC10 box|year=2010|ab=B|num-b=4|num-a=6}}
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| − | {{MAA Notice}}
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