Difference between revisions of "2010 AMC 10B Problems/Problem 18"
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| − | First we factor <math>abc+bc+c</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be | + | ==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>? | ||
| + | |||
| + | <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== | ||
| + | First we factor <math>abc+bc+c</math> into <math>a(b(c+1)+1)</math>. For <math>a(b(c+1)+1)</math> to be divisible by three we can either have <math>a</math> be a multiple of 3 or <math>b(c+1)+1</math> be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is <math>\boxed{\textbf{(E)}\ \frac{13}{27}}</math> | ||
| + | |||
| + | ==See Also== | ||
| + | {{AMC10 box|year=2010|ab=B|num-b=17|num-a=19}} | ||
Revision as of 23:33, 25 November 2011
Problem
Positive integers 
, 
, and 
are randomly and independently selected with replacement from the set 
. What is the probability that 
is divisible by 
?
Solution
First we factor 
into 
. For to be divisible by three we can either have be a multiple of 3 or 
be a multiple of three. Adding the probability of these two being divisible by 3 we get that the probability is 
See Also
| 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 | ||
