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

Latest revision as of 06:26, 5 November 2024

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

1 Problem
2 Solution 1 (Casework)
3 Solution 2
4 Solution 3
5 Video Solution by Little Fermat
6 Video Solution by Math-X (First understand the problem!!!)
7 Video Solution (easy to digest) by Power Solve
8 Video Solution 1 by OmegaLearn
9 Video Solution by CosineMethod [🔥Fast and Easy🔥]
10 Video Solution
11 See Also

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:

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 by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=RjE2si8uG_p5xO_2&t=1670 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=iDDNwV-Ut5UZSoQu&t=2010

~Math-X

Video Solution (easy to digest) by Power Solve

https://www.youtube.com/watch?v=4TPsTOHKQTw

Video Solution 1 by OmegaLearn

https://youtu.be/xguAy0PV7EA

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=a5w_1lN3H4s

Video Solution

https://youtu.be/ShFMyFBxMcY

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

@@ Line 1: / Line 1: @@
+{{duplicate|[[2023 AMC 10A Problems/Problem 9|2023 AMC 10A #9]] and [[2023 AMC 12A Problems/Problem 7|2023 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?
+<math>\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9</math>
+==Solution 1 (Casework)==
+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>.
+For curious readers, the numbers (in chronological order) are:
+ 20230113
+ 20230131
+ 20230223
+ 20230311
+ 20230322
+ 20231013
+ 20231031
+ 20231103
+ 20231130
+==Solution 2==
+There is one <math>3</math>, so we need one more (three more means that either the month or units digit of the day is <math>3</math>). For the same reason, we need one more <math>0</math>.
+If <math>3</math> is the units digit of the month, then the <math>0</math> 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 (<math>11, 22</math>). For the second (tens digit of the day), we must have the other two be <math>1</math>, as a month can't start with <math>2</math> or <math>0</math>. There are <math>3</math> successes this way.
+If <math>3</math> is the tens digit of the day, then <math>0</math> can be either the tens digit of the month or the units digit of the day. For the first case, <math>1</math> must go in the other slots. For the second, the other two slots must be <math>1</math> as well. There are <math>2</math> successes here.
+If <math>3</math> is the units digit of the day, then <math>0</math> could go in any of the <math>3</math> remaining slots again. If it's the tens digit of the day, then the other digits must be <math>1</math>. If <math>0</math> is the units digit of the day, then the other two slots must both be <math>1</math>. If <math>0</math> is the tens digit of the month, then the other two slots can be either both <math>1</math> or both <math>2</math>. In total, there are <math>4</math> successes here.
+Summing through all cases, there are <math>3 + 2 + 4 = \boxed{\textbf{(E)}~9}</math> dates.
+-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.
+~kyogrexu
-Janet rolls a standard <math>6</math>-sided die <math>4</math> times and keeps a running total of the numbers
+==Video Solution by Little Fermat==
-she rolls. What is the probability that at some point, her running total will equal <math>3</math>?
+https://youtu.be/h2Pf2hvF1wE?si=RjE2si8uG_p5xO_2&t=1670
+~little-fermat
+==Video Solution by Math-X (First understand the problem!!!)==
+https://youtu.be/GP-DYudh5qU?si=iDDNwV-Ut5UZSoQu&t=2010
+ ~Math-X
-==Solution 1==
+==Video Solution (easy to digest) by Power Solve==
+https://www.youtube.com/watch?v=4TPsTOHKQTw
-There are <math>4</math> cases where her running total can equal <math>3</math>:
+== Video Solution 1 by OmegaLearn ==
+https://youtu.be/xguAy0PV7EA
-. She rolled <math>1</math> for three times consecutively from the beginning. Probability: <math>\frac{1}{6^3} = \frac{1}{216}</math>
-.  She rolled a <math>1</math>, then <math>2</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
+== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
-.  She rolled a <math>2</math>, then <math>1</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
+https://www.youtube.com/watch?v=a5w_1lN3H4s
-.  She rolled a <math>3</math> at the beginning. Probability: <math>\frac{1}{6}</math>
+==Video Solution==
-Add them together to get <math>\boxed{\textbf{(B)} \frac{49}{216}}.</math>
+https://youtu.be/ShFMyFBxMcY
-~d_code
+~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 ==See Also==
-{{AMC12 box|year=2023|ab=A|num-b=6|num-a=4}}
+{{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}}
-{{AMC10 box|year=2023|ab=A|num-b=8|num-a=4}}
+{{AMC12 box|year=2023|ab=A|num-b=6|num-a=8}}
 {{MAA Notice}}

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

Page

Toolbox

Search