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Difference between revisions of "2023 AMC 12A Problems/Problem 7"
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==Solution==
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==Solution 1==
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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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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 has to be <math>1</math> and <math>2</math> after consideration). After the casework, we get <math>\boxed{\textbf{(E)}~9}</math>.
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For those who are wondering, the numbers are:
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For curious readers, the numbers (in chronological order) are:
 
<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>.
 
<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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==Solution 2==
 
==Solution 2==

Revision as of 14:43, 10 November 2023

The following problem is from both the 2023 AMC 10A #9 and 2023 AMC 12A #7, so both problems redirect to this page.

Problem

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 1

Do careful casework by each month. In the month and the date, we need a $0$, a $3$, and two digits repeated (which has to be $1$ and $2$ after consideration). After the casework, we get $\boxed{\textbf{(E)}~9}$. For curious readers, the numbers (in chronological order) are: $20230113$, $20230131$, $20230223$, $20230311$, $20230322$, $20231013$, $20231031$, $20231103$, $20231130$.

Solution 2

There is one $3$, so we need one more (three more means that either the month or units digit of the day is $3$). For the same reason, we need one more $0$.


If $3$ is the units digit of the month, then the $0$ can be in either of the three remaining slots. For the first case (tens digit of the month), then the last two digits must match ($11, 22$). For the second (tens digit of the day), we must have the other two be $1$, as a month can't start with $2$ or $0$. There are $3$ successes this way.


If $3$ is the tens digit of the day, then $0$ can be either the tens digit of the month or the units digit of the day. For the first case, $1$ must go in the other slots. For the second, the other two slots must be $1$ as well. There are $2$ successes here.


If $3$ is the units digit of the day, then $0$ could go in any of the $3$ remaining slots again. If it's the tens digit of the day, then the other digits must be $1$. If $0$ is the units digit of the day, then the other two slots must both be $1$. If $0$ is the tens digit of the month, then the other two slots can be either both $1$ or both $2$. In total, there are $4$ successes here.

Summing through all cases, there are $3 + 2 + 4 = \boxed{\textbf{(E)}~9}$ dates.

-Benedict T (countmath1)


Video Solution 1 by OmegaLearn

https://youtu.be/xguAy0PV7EA

Video Solution by Math-X (First understand the problem!!!)

https://youtu.be/cMgngeSmFCY?si=2iAoiLoyeAVrVoDf&t=1930 ~Math-X


See Also

2023 AMC 10A (Problems • Answer Key • Resources)
Preceded by
Problem 8 		Followed by
Problem 10
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
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 6 		Followed by
Problem 8
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 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

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