Difference between revisions of "1972 USAMO Problems/Problem 3"
m (→Solution) |
m (→Solution) |
||
| Line 12: | Line 12: | ||
The only possibility left is getting a 2 and a 5, making the product divisible by 10. | The only possibility left is getting a 2 and a 5, making the product divisible by 10. | ||
| − | By complementarity and principle of onclusion-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=\boxed{1-(8/9)^n-(5/9)^n+(4/9)^n}</math>. | + | By complementarity and principle of onclusion-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>. |
Revision as of 22:48, 25 March 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 onclusion-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. 
