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

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{{duplicate|[[2023 AMC 10A Problems/Problem 9|2023 AMC 10A #9]] and [[2023 AMC 12A Problems/Problem 7|2023 AMC 12A #7]]}}
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==Problem==
 
==Problem==
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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?
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<math>\textbf{(A)}~5\qquad\textbf{(B)}~6\qquad\textbf{(C)}~7\qquad\textbf{(D)}~8\qquad\textbf{(E)}~9</math>
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 +
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==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>.
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For curious readers, the numbers (in chronological order) are:
 +
 +
20230113
 +
20230131
 +
20230223
 +
20230311
 +
20230322
 +
20231013
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20231031
 +
20231103
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20231130
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==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>.
 +
 +
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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.
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 +
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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.
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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.
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Summing through all cases, there are <math>3 + 2 + 4 = \boxed{\textbf{(E)}~9}</math> dates.
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-Benedict T (countmath1)
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==Solution 3==
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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.
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~kyogrexu
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 +
==Video Solution by Little Fermat==
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https://youtu.be/h2Pf2hvF1wE?si=RjE2si8uG_p5xO_2&t=1670
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~little-fermat
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==Video Solution by Math-X (First understand the problem!!!)==
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https://youtu.be/GP-DYudh5qU?si=iDDNwV-Ut5UZSoQu&t=2010
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~Math-X
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 +
==Video Solution (easy to digest) by Power Solve==
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https://www.youtube.com/watch?v=4TPsTOHKQTw
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 +
== Video Solution 1 by OmegaLearn ==
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https://youtu.be/xguAy0PV7EA
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 +
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== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
  
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Janet rolls a standard <math>6</math>-sided die <math>4</math> times and keeps a running total of the numbers
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https://www.youtube.com/watch?v=a5w_1lN3H4s
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she rolls. What is the probability that at some point, her running total will equal <math>3</math>?
 
  
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==Solution 1==
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==Video Solution==
  
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There are <math>4</math> cases where her running total can equal <math>3</math>:
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https://youtu.be/ShFMyFBxMcY
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1. She rolled <math>1</math> for three times consecutively from the beginning. Probability: <math>\frac{1}{6^3} = \frac{1}{216}</math>
 
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2.  She rolled a <math>1</math>, then <math>2</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
 
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3.  She rolled a <math>2</math>, then <math>1</math>. Probability: <math>\frac{1}{6^2} = \frac{1}{36}</math>
 
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4.  She rolled a <math>3</math> at the beginning. Probability: <math>\frac{1}{6}</math>
 
  
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Add them together to get <math>\boxed{\textbf{(B)} \frac{49}{216}}.</math>
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
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~d_code
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==See Also==
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{{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}}
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{{AMC12 box|year=2023|ab=A|num-b=6|num-a=8}}
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{{MAA Notice}}

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.

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 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)

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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