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

Revision as of 14:54, 11 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 (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.

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

Video Solution

https://youtu.be/ShFMyFBxMcY

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

@@ Line 31: / Line 31: @@
 -Benedict T (countmath1)
+==Solution 3==
+We start with <math>2023----</math> we need an extra <math>0</math> and an extra <math>3</math>. So we have at least one of those extras in the days, except we can have the month <math>03</math>. We now have <math>6</math> possible months <math>01,02,03,10,11,12</math>. For month <math>1</math> we have two cases, we now have to add in another 1, and the possible days are <math>13,31</math>. For month <math>2</math> we need an extra <math>2</math> so we can have the day <math>23</math> note that we can't use <math>32</math> because it is to large. Now for month <math>3</math> we can have any number and multiply it by <math>11</math> so we have the solution <math>11,22</math>. For October we need a <math>1</math> and a <math>3</math> so we have <math>13,31</math> as our choices. For November we have two choices which are <math>03,30</math>.Now for December we have <math>0</math> options. Summing <math>2+1+2+2+2</math> we get <math>\boxed{\textbf{(E)}~9}</math> solutions.
 == Video Solution 1 by OmegaLearn ==

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

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