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2023 AMC 12A Problems/Problem 5: Difference between revisions

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==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 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?
 
<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>
 


<cmath>\textbf{(A)}~14\qquad\textbf{(B)}~15\qquad\textbf{(C)}~16\qquad\textbf{(D)}~17\qquad\textbf{(E)}~18\qquad</cmath>


==Solution 1==
==Solution 1==
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>


~zhenghua
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 <math>3</math> in one roll is <math>1/6</math>.
 
Case 2:
The chance of rolling a running total of <math>3</math> in two rolls is <math>1/6\cdot 1/6\cdot 2</math> since the dice rolls are a 2 and a 1 and vice versa.
 
Case 3:
The chance of rolling a running total of 3 in three rolls is <math>1/6\cdot 1/6\cdot 1/6</math> since the dice values would have to be three ones.
 
Using the rule of sum, <math>1/6 + 1/18 + 1/216 = 49/216</math> <math>\boxed{\textbf{(B) }\frac{49}{216}}</math>.
 
~walmartbrian ~andyluo
 
==Solution 2 (Slightly different to Solution 1)==
There are 3 cases where the running total will equal 3.
 
Case 1: Rolling a one three times
 
Case 2: Rolling a one then a two
 
Case 3: Rolling a three immediately
 
The probability of Case 1 is <math>1/216</math>, the probability of Case 2 is (<math>1/36 * 2) = 1/18</math>, and the probability of Case 3 is <math>1/6</math>
 
Using the rule of sums, adding every case gives the answer <math>\boxed{\textbf{(B) }\frac{49}{216}}</math>
 
 
~DRBStudent


==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}}
{{AMC10 box|year=2023|ab=A|num-b=4|num-a=6}}
{{MAA Notice}}
{{MAA Notice}}

Revision as of 22:16, 9 November 2023

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

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 $3$ in one roll is $1/6$.

Case 2: The chance of rolling a running total of $3$ in two rolls is $1/6\cdot 1/6\cdot 2$ since the dice rolls are a 2 and a 1 and vice versa.

Case 3: The chance of rolling a running total of 3 in three rolls is $1/6\cdot 1/6\cdot 1/6$ since the dice values would have to be three ones.

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

~walmartbrian ~andyluo

Solution 2 (Slightly different to Solution 1)

There are 3 cases where the running total will equal 3.

Case 1: Rolling a one three times

Case 2: Rolling a one then a two

Case 3: Rolling a three immediately

The probability of Case 1 is $1/216$, the probability of Case 2 is ($1/36 * 2) = 1/18$, and the probability of Case 3 is $1/6$

Using the rule of sums, adding every case gives the answer $\boxed{\textbf{(B) }\frac{49}{216}}$


~DRBStudent

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 10A (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 10 Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing

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