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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.
  
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The probability that there is not a factor of 2 or 5 in there is <math>\left( \frac{4}{9}\right)^n</math>. The probability that there is no 5 is <math>\left( \frac{8}{9}\right)^n</math>, so the probability that there is a 2 but no 5 is <math>\left( \frac{8}{9}\right)^n-\left( \frac{4}{9}\right)^n</math>. The probability that there is no 2 is <math>\left( \frac{5}{9}\right)^n</math>, so the probability that there is a 5 but no 2 is <math>\left( \frac{5}{9}\right)^n-\left( \frac{4}{9}\right)^n</math>. 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 <math>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}</math>.
+
The probability that there is neither a factor of 2 nor 5 in there is <math>\left( \frac{4}{9}\right)^n</math>. The probability that there is no 5 is <math>\left( \frac{8}{9}\right)^n</math>, so the probability that there is a 2 but no 5 is <math>\left( \frac{8}{9}\right)^n-\left( \frac{4}{9}\right)^n</math>. The probability that there is no 2 is <math>\left( \frac{5}{9}\right)^n</math>, so the probability that there is a 5 but no 2 is <math>\left( \frac{5}{9}\right)^n-\left( \frac{4}{9}\right)^n</math>. 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 <math>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}</math>.
  
  

Revision as of 18:24, 1 March 2018

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 neither a factor of 2 nor 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.

See Also

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

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