Difference between revisions of "2023 AMC 12A Problems/Problem 7"

Revision as of 09:56, 10 November 2023

The following problem is from both the 2022 AMC 10A #9 and 2022 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

Do careful casework by each month. In the month and the date, we need a $0$ , a $3$ , and two digits repeated (which have to be $1$ and $2$ after consideration). After the case work, we get $9$ , meaning the answer $\boxed{\textbf{(E)}~9}$ . For those who are wondering, the numbers 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

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.

@@ Line 1: / Line 1: @@
+{{duplicate|[[2022 AMC 10A Problems/Problem 9|2022 AMC 10A #9]] and [[2022 AMC 12A Problems/Problem 7|2022 AMC 12A #7]]}}
 ==Problem==
 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?

Page

Toolbox

Search