Difference between revisions of "2010 AMC 12B Problems/Problem 16"
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<cmath>c(b+1)\equiv-1 \mod 3</cmath> | <cmath>c(b+1)\equiv-1 \mod 3</cmath> | ||
<cmath>b+1\equiv \frac{-1}{c} \mod 3</cmath> | <cmath>b+1\equiv \frac{-1}{c} \mod 3</cmath> | ||
− | Which is true for | + | Which is true for one <math>b</math> when <math>c\not\equiv 0 \mod 3</math> because the integers <math>\mod 3</math> form a field under multiplication and addition with absorbing element <math>0</math>. |
+ | |||
+ | This gives us <math>P=\frac{1}{3}+\frac{4}{27}=(b)\frac{13}{27}</math>. | ||
==Video Solution== | ==Video Solution== |
Revision as of 13:23, 30 October 2020
- The following problem is from both the 2010 AMC 12B #16 and 2010 AMC 10B #18, so both problems redirect to this page.
Contents
Problem 16
Positive integers , , and are randomly and independently selected with replacement from the set . What is the probability that is divisible by ?
Solution 1
We group this into groups of , because . This means that every residue class mod 3 has an equal probability.
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
Solution 2 (Minor change from Solution 1)
Just like solution 1, we see that there is a chance of and chance of
Now, we can just use PIE (Principals of Inclusion and Exclusion) to get our answer to be
-Conantwiz2023
Solution 3 (Fancier version of Solution 1)
As with solution one, we conclude that if then the requirements are satisfied. We then have: Which is true for one when because the integers form a field under multiplication and addition with absorbing element .
This gives us .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.