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1972 USAMO Problems/Problem 3

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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( \fra....

See also

1972 USAMO (Problems)
Preceded by
Problem 2
1 2 3 4 5 Followed by
Problem 4
All USAMO Problems and Solutions
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