Difference between revisions of "1972 USAMO Problems/Problem 3"

Revision as of 17:52, 3 July 2013

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 $n$ selections ( $n>1$ ), the product of the $n$ 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 not a factor of 2 or 5 in there is $\left( \frac{4}{9}\right)^n$ . The probability that there is no 5 is $\left( \frac{8}{9}\right)^n$ , so the probability that there is a 2 but no 5 is $\left( \frac{8}{9}\right)^n-\left( \frac{4}{9}\right)^n$ . The probability that there is no 2 is $\left( \frac{5}{9}\right)^n$ , so the probability that there is a 5 but no 2 is $\left( \frac{5}{9}\right)^n-\left( \frac{4}{9}\right)^n$ . Thus the only possibility left is getting a 2 and a 5, and thus making the product divisible by 10. The probability of that is $1-\left( \frac{4}{9}\right)^n-\left( \frac{8}{9}\right)^n+\left( \frac{4}{9}\right)^n-\left( \frac{5}{9}\right)^n+\left( \frac{4}{9}\right)^n=\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}$ .

Alternate solutions are always welcome. If you have a different, elegant solution to this problem, please add it to this page.

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 ==See Also==
 {{USAMO box|year=1972|num-b=2|num-a=4}}
+{{MAA Notice}}
 [[Category:Olympiad Combinatorics Problems]]

1972 USAMO (Problems • Resources)
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Difference between revisions of "1972 USAMO Problems/Problem 3"

Revision as of 17:52, 3 July 2013

Problem

Solution

See Also