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| − | ==Problem==
| + | #redirect[[2023 AMC 12A Problems/Problem 7]] |
| − | A digital display shows the current date as an <math>8</math>-digit integer consisting of a <math>4</math>-digit year, followed by a <math>2</math>-digit month, followed by a <math>2</math>-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in <math>2023</math> will each digit appear an even number of times in the 8-digital display for that date?
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| − | <math>\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9</math>
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| − | ==Solution==
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| − | Do careful casework by each month. In the month and the date, we need a <math>0</math>, a <math>3</math>, and two digits repeated (which have to be <math>1</math> and <math>2</math> after consideration). After the case work, we get <math>9</math>, meaning the answer <math>\boxed{\textbf{(E)}~9}</math>.
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| − | For those who are wondering, the numbers are:
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| − | <math>20230113</math>, <math>20230131</math>, <math>20230223</math>, <math>20230311</math>, <math>20230322</math>, <math>20231013</math>, <math>20231031</math>, <math>20231103</math>, <math>20231130</math>.
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| − | == Video Solution 1 by OmegaLearn ==
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| − | https://youtu.be/xguAy0PV7EA
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| − | ==See Also==
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| − | {{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}}
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| − | {{MAA Notice}}
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