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Difference between revisions of "2010 AMC 12B Problems/Problem 16"

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{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}
 
{{duplicate|[[2010 AMC 12B Problems|2010 AMC 12B #16]] and [[2010 AMC 10B Problems|2010 AMC 10B #18]]}}
  
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== Problem 16 ==
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== Problem ==
 
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?
 
Positive integers <math>a</math>, <math>b</math>, and <math>c</math> are randomly and independently selected with replacement from the set <math>\{1, 2, 3,\dots, 2010\}</math>. What is the probability that <math>abc + ab + a</math> is divisible by <math>3</math>?
  
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If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens.
 
If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens.
  
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Otherwise, we have <math>3|bc+b+1</math>, which means that <math>b(c+1) \equiv 2\pmod{3}</math>. So either <cmath>b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}</cmath> or <cmath>b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3</cmath> which will lead to the property being true. There are a <math>\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}</math> chance for each bundle of cases to be true. Thus, the total for the cases is <math>\frac{2}{9}</math>. But we have to multiply by <math>\frac{2}{3}</math> because this only happens with a <math>\frac{2}{3}</math> chance. So the total is actually <math>\frac{4}{27}</math>.
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Otherwise, we have <math>3|bc+b+1</math>, which means that <math>b(c+1) \equiv 2\pmod{3}</math>. So either <cmath>b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}</cmath> or <cmath>b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3</cmath> which will lead to the property being true. There is a <math>\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}</math> chance for each bundle of cases to be true. Thus, the total for the cases is <math>\frac{2}{9}</math>. But we have to multiply by <math>\frac{2}{3}</math> because this only happens with a <math>\frac{2}{3}</math> chance. So the total is actually <math>\frac{4}{27}</math>.
  
 
The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath>
 
The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath>
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Just like solution 1, we see that there is a <math>\frac{1}{3}</math> chance of <math>3|a</math> and <math>\frac{2}{9}</math> chance of <math>3|1+b+bc</math>  
 
Just like solution 1, we see that there is a <math>\frac{1}{3}</math> chance of <math>3|a</math> and <math>\frac{2}{9}</math> chance of <math>3|1+b+bc</math>  
  
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Now, we can just use PIE (Principles of Inclusion and Exclusion) to get our answer to be <math>\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{E) \frac{13}{27}}</math>
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Now, we can just use PIE (Principles of Inclusion and Exclusion) to get our answer to be <math>\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{\text{(E) } \frac{13}{27}}</math>
  
 
-Conantwiz2023
 
-Conantwiz2023
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== See also ==
 
== See also ==
 
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}
 
{{AMC12 box|year=2010|num-b=15|num-a=17|ab=B}}
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{{AMC10 box|year=2010|num-b=17|num-a=19|ab=B}}
  
 
[[Category:Introductory Combinatorics Problems]]
 
[[Category:Introductory Combinatorics Problems]]
 
{{MAA Notice}}
 
{{MAA Notice}}

Latest revision as of 17:56, 22 June 2024

The following problem is from both the 2010 AMC 12B #16 and 2010 AMC 10B #18, so both problems redirect to this page.

Problem

Positive integers $a$, $b$, and $c$ are randomly and independently selected with replacement from the set $\{1, 2, 3,\dots, 2010\}$. What is the probability that $abc + ab + a$ is divisible by $3$?

$\textbf{(A)}\ \dfrac{1}{3} \qquad \textbf{(B)}\ \dfrac{29}{81} \qquad \textbf{(C)}\ \dfrac{31}{81} \qquad \textbf{(D)}\ \dfrac{11}{27} \qquad \textbf{(E)}\ \dfrac{13}{27}$

Solution 1

We group this into groups of $3$, because $3|2010$. This means that every residue class mod 3 has an equal probability.

If $3|a$, we are done. There is a probability of $\frac{1}{3}$ that that happens.

Otherwise, we have $3|bc+b+1$, which means that $b(c+1) \equiv 2\pmod{3}$. So either \[b \equiv 1 \pmod{3}, c \equiv 1 \pmod{3}\] or \[b \equiv 2 \pmod {3}, c \equiv 0 \pmod 3\] which will lead to the property being true. There is a $\frac{1}{3}\cdot\frac{1}{3}=\frac{1}{9}$ chance for each bundle of cases to be true. Thus, the total for the cases is $\frac{2}{9}$. But we have to multiply by $\frac{2}{3}$ because this only happens with a $\frac{2}{3}$ chance. So the total is actually $\frac{4}{27}$.

The grand total is \[\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}\]

Solution 2 (Minor change from Solution 1)

Just like solution 1, we see that there is a $\frac{1}{3}$ chance of $3|a$ and $\frac{2}{9}$ chance of $3|1+b+bc$

Now, we can just use PIE (Principles of Inclusion and Exclusion) to get our answer to be $\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{\text{(E) } \frac{13}{27}}$

-Conantwiz2023

Solution 3 (Fancier version of Solution 1)

As with solution one, we conclude that if $a\equiv0\mod 3$ then the requirements are satisfied. We then have: \[a(bc+c)+a\equiv0 \mod 3\] \[a(bc+c)\equiv-a \mod 3\] \[c(b+1)\equiv-1 \mod 3\] \[b+1\equiv \frac{-1}{c} \mod 3\] Which is true for one $b$ when $c\not\equiv 0 \mod 3$ because the integers$\mod 3$ form a field under multiplication and addition with absorbing element $0$.

This gives us $P=\frac{1}{3}+\frac{4}{27}=\boxed{\text{(E) }\frac{13}{27}}$.

~Snacc

Video Solution

https://youtu.be/FQO-0E2zUVI?t=437

~IceMatrix

See also

2010 AMC 12B (Problems • Answer Key • Resources)
Preceded by
Problem 15 		Followed by
Problem 17
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
2010 AMC 10B (Problems • Answer Key • Resources)
Preceded by
Problem 17 		Followed by
Problem 19
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

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