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

Revision as of 17:54, 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 $\frac{1}{3}+\frac{2}{9}-\frac{1}{3}\cdot\frac{2}{9} = \boxed{(E) \frac{13}{27}}$

-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

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 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>
-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>
+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>
 -Conantwiz2023

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