Difference between revisions of "2010 AMC 10B Problems/Problem 18"

(Redirected page to 2010 AMC 12B Problems/ Problem 16)
(Tag: New redirect)
Line 1: Line 1:
#redirect [[2010 AMC 12B Problems/ Problem 16]]
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>?
<math>\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}</math>
==Solution 1==
First we factor <math>abc + ab + a</math> as <math>a(bc + b + 1)</math>, so in order for the number to be divisible by 3, either <math>a</math> is divisible by <math>3</math>, or <math>bc + b + 1</math> is divisible by <math>3</math>.
We see that <math>a</math> is divisible by <math>3</math> with probability <math>\frac{1}{3}</math>. We only need to calculate the probability that <math>bc + b + 1</math> is divisible by <math>3</math>.
We need <math>bc + b + 1 \equiv 0\pmod 3</math> or <math>b(c + 1) \equiv 2\pmod 3</math>. Using some modular arithmetic, <math>b \equiv 2\pmod 3</math> and <math>c \equiv 0\pmod 3</math> or <math>b \equiv 1\pmod 3</math> and <math>c \equiv 1\pmod 3</math>. The both cases happen with probability <math>\frac{1}{3} * \frac{1}{3} = \frac{1}{9}</math> so the total probability is <math>\frac{2}{9}</math>.
Then the answer is <math>\frac{1}{3} + \frac{2}{3}\cdot\frac{2}{9} = \frac{13}{27}</math> or <math>\boxed{E}</math>.
==Solution 2==
We see that since <math>2010</math> is divisible by <math>3</math>, the probability that any one of <math>a</math>, <math>b</math>, or <math>c</math> being divisible by <math>3</math> is <math>\frac{1}{3}</math>. Because of this, we can shrink the set of possibilities for <math>a</math>, <math>b</math>, and <math>c</math> to the set <math>\{1,2,3\}</math> without affecting the probability in question.
Listing out all possible combinations for <math>a</math>, <math>b</math>, and <math>c</math>, we see that the answer is <math>\bold{\boxed{\left( E\right) \frac{13}{27}}}</math>.
==See Also==
{{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}}
{{MAA Notice}}

Revision as of 20:45, 26 May 2020

Invalid username
Login to AoPS