1972 USAMO Problems/Problem 3


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.


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 $\left( \frac{8}{9}\right)^n$.

The probability that there is no 2 is $\left( \frac{5}{9}\right)^n$.

The probability that there is neither a 2 nor 5 is $\left( \frac{4}{9}\right)^n$, 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 $1- \left( \left( \frac{8}{9}\right)^n + \left( \frac{5}{9}\right)^n - \left( \frac{4}{9}\right)^n\right)=\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}$.

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 (ProblemsResources)
Preceded by
Problem 2
Followed by
Problem 4
1 2 3 4 5
All USAMO Problems and Solutions

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