Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2023 AMC 12A Problems/Problem 5"
(→‎Video Solution by Math-X (First understand the problem!!!))
 
(39 intermediate revisions by 14 users not shown)
Line 1: Line 1:
 +
{{duplicate|[[2023 AMC 10A Problems/Problem 7|2023 AMC 10A #7]] and [[2023 AMC 12A Problems/Problem 5|2023 AMC 12A #5]]}}
 +
 
==Problem==
 
==Problem==
−
How many digits are in the base-ten representation of <math>8^5 \cdot 5^{10} \cdot 15^5</math>?
+
Janet rolls a standard <math>6</math>-sided die <math>4</math> times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal <math>3</math>?
 +
 
 +
<math>\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{49}{216}\qquad\textbf{(C) }\frac{25}{108}\qquad\textbf{(D) }\frac{17}{72}\qquad\textbf{(E) }\frac{13}{54}</math>
 +
 
 +
==Solution 1 (Casework)==
 +
 
 +
There are <math>3</math> cases where the running total will equal <math>3</math>: one roll; two rolls; or three rolls:
 +
 
 +
'''Case 1:'''
 +
The chance of rolling a running total of <math>3</math>, namely <math>(3)</math> in exactly one roll is <math>\frac{1}{6}</math>.
 +
 
 +
'''Case 2:'''
 +
The chance of rolling a running total of <math>3</math> in exactly two rolls, namely <math>(1, 2)</math> and <math>(2, 1)</math> is <math>\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}</math>.
 +
 
 +
'''Case 3:'''
 +
The chance of rolling a running total of 3 in exactly three rolls, namely <math>(1, 1, 1)</math> is <math>\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}</math>.
 +
 
 +
Using the rule of sum we have <math>\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}</math>.
 +
 
 +
~walmartbrian ~andyluo ~DRBStudent ~MC_ADe
 +
 
 +
==Solution 2 (Brute Force)==
 +
 
 +
Because there is only a maximum of 3 rolls we must count (running total = 3 means there can't be a fourth roll counted), we can simply list out all of the probabilities.
 +
 
 +
If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of <math>\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}</math>.
 +
 
 +
If we roll a 2 on the first, the roll that follows must be 1, resulting in a probability of <math>\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}</math>.
 +
 
 +
If we roll a 3 on the first, the following rolls do not matter, resulting in a probability of <math>\frac{1}{6}</math>.
 +
Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored.
 +
Summing the answers, we have <math>\frac{7}{216} + \frac{1}{36} + \frac{1}{6} = \frac{7+6+36}{216} = \boxed{\textbf{(B) }\frac{49}{216}}</math>.
 +
 
 +
~Failure.net
 +
 
 +
==Solution 3==
 +
 
 +
Consider sequences of 4 integers with each integer between 1 and 6, the number of permutations of 6 numbers is <math>6^4=1296</math>.
 +
 
 +
The following 4 types of sequences that might generate a running total of the numbers to be equal to 3 (x, y, or z denotes any integer between 1 and 6).
 +
 
 +
Sequence #1, (1, 1, 1, x): there are <math>6</math> possible sequences.
 +
 
 +
Sequence #2, (1, 2, x, y): there are <math>6^2 = 36</math> possible sequences.
 +
 
 +
Sequence #3, (2, 1, x, y): there are <math>6^2 = 36</math> possible sequences.
 +
 
 +
Sequence #4, (3, x, y, z): there are <math>6^3 = 216</math> possible sequences.
 +
 
 +
Out of 1296 possible sequences, there are a total of <math>6 + 36 + 36 + 216 = 294</math> sequences that qualify. Hence, the probability is <math>294 / 1296 = \boxed{\textbf{(B) }\frac{49}{216}}</math>
 +
 
 +
~sqroot
 +
 
 +
 
 +
==Video Solution by Little Fermat==
 +
https://youtu.be/h2Pf2hvF1wE?si=HU7yfXpTg9yBLwSA&t=1203
 +
~little-fermat
 +
==Video Solution by Math-X (First understand the problem!!!)==
 +
https://youtu.be/GP-DYudh5qU?si=xtdXIlJmV_1ivP3T&t=1507
 +
~Math-X
 +
 
 +
==Video Solution (easy to understand) by Power Solve==
 +
https://youtu.be/YXIH3UbLqK8?si=3p9Ap7UQ376Tfi1W&t=312
 +
 
 +
== Video Solution by CosineMethod [🔥Fast and Easy🔥]==
 +
 
 +
https://www.youtube.com/watch?v=FyVQW-1z60w
 +
 
 +
==Video Solution==
 +
 
 +
https://youtu.be/bqLA67R_5Bk
  
−
<cmath>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</cmath>
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
−
==Solution 1==
+
==Video Solution (⚡ Just 3 min ⚡)==
−
Prime factorizing this gives us <math>2^{15}\cdot3^{5}\cdot5^{15}</math> Pairing <math>2^{15}</math> and <math>5^{15}</math> gives us a number with <math>15</math> zeros giving us 15 digits. <math>3^5=243</math> and this adds an extra 3 digits. <math>15+3=\text{\boxed{(E)18}}</math>
+
https://youtu.be/8G5FfjUkQkQ
  
−
~zhenghua
+
~Education, the Study of Everything
  
 
==See Also==
 
==See Also==
−
{{AMC12 box|year=2023|ab=A|num-b=3|num-a=5}}
+
{{AMC10 box|year=2023|ab=A|num-b=6|num-a=8}}
−
{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
+
{{AMC12 box|year=2023|ab=A|num-b=4|num-a=6}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 06:23, 5 November 2024

The following problem is from both the 2023 AMC 10A #7 and 2023 AMC 12A #5, so both problems redirect to this page.

Problem

Janet rolls a standard $6$-sided die $4$ times and keeps a running total of the numbers she rolls. What is the probability that at some point, her running total will equal $3$?

$\textbf{(A) }\frac{2}{9}\qquad\textbf{(B) }\frac{49}{216}\qquad\textbf{(C) }\frac{25}{108}\qquad\textbf{(D) }\frac{17}{72}\qquad\textbf{(E) }\frac{13}{54}$

Solution 1 (Casework)

There are $3$ cases where the running total will equal $3$: one roll; two rolls; or three rolls:

Case 1: The chance of rolling a running total of $3$, namely $(3)$ in exactly one roll is $\frac{1}{6}$.

Case 2: The chance of rolling a running total of $3$ in exactly two rolls, namely $(1, 2)$ and $(2, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot2=\frac{1}{18}$.

Case 3: The chance of rolling a running total of 3 in exactly three rolls, namely $(1, 1, 1)$ is $\frac{1}{6}\cdot\frac{1}{6}\cdot\frac{1}{6}=\frac{1}{216}$.

Using the rule of sum we have $\frac{1}{6}+\frac{1}{18}+\frac{1}{216}=\boxed{\textbf{(B) }\frac{49}{216}}$.

~walmartbrian ~andyluo ~DRBStudent ~MC_ADe

Solution 2 (Brute Force)

Because there is only a maximum of 3 rolls we must count (running total = 3 means there can't be a fourth roll counted), we can simply list out all of the probabilities.

If we roll a 1 on the first, the rolls that follow must be 2 or {1,1}, with the following results not mattering. This leaves a probability of $\frac{1}{6}\times\frac{1}{6} + \frac{1}{6}\times\frac{1}{6}\times\frac{1}{6} = \frac{7}{216}$.

If we roll a 2 on the first, the roll that follows must be 1, resulting in a probability of $\frac{1}{6}\times\frac{1}{6} = \frac{1}{36}$.

If we roll a 3 on the first, the following rolls do not matter, resulting in a probability of $\frac{1}{6}$. Any roll greater than three will result in a running total greater than 3 no matter what, so those cases can be ignored. Summing the answers, we have $\frac{7}{216} + \frac{1}{36} + \frac{1}{6} = \frac{7+6+36}{216} = \boxed{\textbf{(B) }\frac{49}{216}}$.

~Failure.net

Solution 3

Consider sequences of 4 integers with each integer between 1 and 6, the number of permutations of 6 numbers is $6^4=1296$.

The following 4 types of sequences that might generate a running total of the numbers to be equal to 3 (x, y, or z denotes any integer between 1 and 6).

Sequence #1, (1, 1, 1, x): there are $6$ possible sequences.

Sequence #2, (1, 2, x, y): there are $6^2 = 36$ possible sequences.

Sequence #3, (2, 1, x, y): there are $6^2 = 36$ possible sequences.

Sequence #4, (3, x, y, z): there are $6^3 = 216$ possible sequences.

Out of 1296 possible sequences, there are a total of $6 + 36 + 36 + 216 = 294$ sequences that qualify. Hence, the probability is $294 / 1296 = \boxed{\textbf{(B) }\frac{49}{216}}$

~sqroot


Video Solution by Little Fermat

https://youtu.be/h2Pf2hvF1wE?si=HU7yfXpTg9yBLwSA&t=1203 ~little-fermat

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

https://youtu.be/GP-DYudh5qU?si=xtdXIlJmV_1ivP3T&t=1507 ~Math-X

Video Solution (easy to understand) by Power Solve

https://youtu.be/YXIH3UbLqK8?si=3p9Ap7UQ376Tfi1W&t=312

Video Solution by CosineMethod [🔥Fast and Easy🔥]

https://www.youtube.com/watch?v=FyVQW-1z60w

Video Solution

https://youtu.be/bqLA67R_5Bk

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

Video Solution (⚡ Just 3 min ⚡)

https://youtu.be/8G5FfjUkQkQ

~Education, the Study of Everything

See Also

2023 AMC 10A (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 10 Problems and Solutions
2023 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
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

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AMC_12A_Problems/Problem_5&oldid=231683"