# 2023 AMC 12A Problems/Problem 7

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 (Casework)

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)

## Solution 3

We start with $2023----$ we need an extra $0$ and an extra $3$. So we have at least one of those extras in the days, except we can have the month $03$. We now have $6$ possible months $01,02,03,10,11,12$. For month $1$ we have two cases, we now have to add in another 1, and the possible days are $13,31$. For month $2$ we need an extra $2$ so we can have the day $23$ note that we can't use $32$ because it is to large. Now for month $3$ we can have any number and multiply it by $11$ so we have the solution $11,22$. For October we need a $1$ and a $3$ so we have $13,31$ as our choices. For November we have two choices which are $03,30$.Now for December we have $0$ options. Summing $2+1+2+2+2$ we get $\boxed{\textbf{(E)}~9}$ solutions.

~kyogrexu

## Video Solution

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)