Difference between revisions of "2023 AMC 10A Problems/Problem 9"

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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>
 
<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. After the case work, we get <math>9</math>, meaning the answer <math>\boxed{E}</math>
  
 
== Video Solution 1 by OmegaLearn ==
 
== Video Solution 1 by OmegaLearn ==
 
https://youtu.be/xguAy0PV7EA
 
https://youtu.be/xguAy0PV7EA

Revision as of 17:46, 9 November 2023

A digital display shows the current date as an $8$-digit integer consisting of a $4$-digit year, followed by a $2$-digit month, followed by a $2$-digit date within the month. For example, Arbor Day this year is displayed as 20230428. For how many dates in $2023$ will each digit appear an even number of times in the 8-digital display for that date?

$\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9$

Solution

Do careful casework by each month. In the month and the date, we need a $0$, a $3$, and two digits repeated. After the case work, we get $9$, meaning the answer $\boxed{E}$

Video Solution 1 by OmegaLearn

https://youtu.be/xguAy0PV7EA