Difference between revisions of "1972 USAMO Problems/Problem 3"
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For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there. | For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there. | ||
− | The probability that there is | + | The probability that there is no 5 is <math>\left( \frac{8}{9}\right)^n</math>. |
− | ==See | + | The probability that there is no 2 is <math>\left( \frac{5}{9}\right)^n</math>. |
+ | |||
+ | The probability that there is neither a 2 nor 5 is <math>\left( \frac{4}{9}\right)^n</math>, which is included in both previous cases. | ||
+ | |||
+ | The only possibility left is getting a 2 and a 5, making the product divisible by 10. | ||
+ | By complementarity and principle of inclusion-exclusion, the probability of that is <math>1- \left( \left( \frac{8}{9}\right)^n + \left( \frac{5}{9}\right)^n - \left( \frac{4}{9}\right)^n\right)=\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}</math>. | ||
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+ | {{alternate solutions}} | ||
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+ | ==See Also== | ||
{{USAMO box|year=1972|num-b=2|num-a=4}} | {{USAMO box|year=1972|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Latest revision as of 19:11, 3 June 2023
Problem
A random number selector can only select one of the nine integers 1, 2, ..., 9, and it makes these selections with equal probability. Determine the probability that after selections (
), the product of the
numbers selected will be divisible by 10.
Solution
For the product to be divisible by 10, there must be a factor of 2 and a factor of 5 in there.
The probability that there is no 5 is .
The probability that there is no 2 is .
The probability that there is neither a 2 nor 5 is , which is included in both previous cases.
The only possibility left is getting a 2 and a 5, making the product divisible by 10.
By complementarity and principle of inclusion-exclusion, the probability of that is .
Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.
See Also
1972 USAMO (Problems • Resources) | ||
Preceded by Problem 2 |
Followed by Problem 4 | |
1 • 2 • 3 • 4 • 5 | ||
All USAMO Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.