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Difference between revisions of "2023 AMC 12A Problems/Problem 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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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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<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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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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==Solution==
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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 have to be <math>1</math> and <math>2</math> after consideration). After the case work, we get <math>9</math>, meaning the answer <math>\boxed{\textbf{(E)}~9}</math>.
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For those who are wondering, the numbers are:
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<math>20230113</math>, <math>20230131</math>, <math>20230223</math>, <math>20230311</math>, <math>20230322</math>, <math>20231013</math>, <math>20231031</math>, <math>20231103</math>, <math>20231130</math>.
  
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There are <math>4</math> cases where her running total can equal <math>3</math>:
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== Video Solution 1 by OmegaLearn ==
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https://youtu.be/xguAy0PV7EA
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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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~d_code
 
  
 
==See Also==
 
==See Also==
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{{AMC12 box|year=2023|ab=A|num-b=6|num-a=4}}
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{{AMC10 box|year=2023|ab=A|num-b=8|num-a=10}}
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{{AMC10 box|year=2023|ab=A|num-b=8|num-a=4}}
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{{AMC12 box|year=2023|ab=A|num-b=6|num-a=8}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 23:24, 9 November 2023

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: $20230113$, $20230131$, $20230223$, $20230311$, $20230322$, $20231013$, $20231031$, $20231103$, $20231130$.

Video Solution 1 by OmegaLearn

https://youtu.be/xguAy0PV7EA

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

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