Difference between revisions of "2010 AMC 12B Problems/Problem 16"
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== Solution == | == Solution == | ||
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− | + | We group this into groups of <math>3</math>, because <math>3|2010</math>. | |
− | <math> | + | If <math>3|a</math>, we are done. There is a probability of <math>\frac{1}{3}</math> that that happens. |
− | <math>b(c+1)\ | + | 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>. |
− | + | The grand total is <cmath>\frac{1}{3} + \frac{4}{27} = \boxed{\text{(E) }\frac{13}{27}.}</cmath> | |
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== See also == | == See also == |
Revision as of 18:44, 9 October 2017
Problem 16
Positive integers ,
, and
are randomly and independently selected with replacement from the set
. What is the probability that
is divisible by
?
Solution
We group this into groups of , because
.
If , we are done. There is a probability of
that that happens.
Otherwise, we have , which means that
. So either
or
which will lead to the property being true. There are a
chance for each bundle of cases to be true. Thus, the total for the cases is
. But we have to multiply by
because this only happens with a
chance. So the total is actually
.
The grand total is
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 |
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