Difference between revisions of "1972 USAMO Problems/Problem 3"
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For our formula for <math>c_n</math>, we can substitute to find that <math>c_n=\frac{5}{9}c_{n-1}+\frac{1}{9}\cdot\left(\frac{4}{9}\right)^{n-1}</math>. Solving this recursion yields <math>c_n=\left(\frac{5}{9}\right)^n-\left(\frac{4}{9}\right)^n</math>. | For our formula for <math>c_n</math>, we can substitute to find that <math>c_n=\frac{5}{9}c_{n-1}+\frac{1}{9}\cdot\left(\frac{4}{9}\right)^{n-1}</math>. Solving this recursion yields <math>c_n=\left(\frac{5}{9}\right)^n-\left(\frac{4}{9}\right)^n</math>. | ||
| − | Finally, substituting the values into the <math>a_n</math> formula yields <math>a_n=a_{n-1}+\frac{1}{9}\left(\left(\frac{8}{9}\right)^{n-1}+4\cdot\left(\frac{5}{9}\right)^{n-1}-5\cdot\left(\frac{4}{9}\right)^{n-1}\right)</math>. Solving this recursion yields the final answer, <math>\boxed{1-\left(\frac{8}{9}\right)^n | + | Finally, substituting the values into the <math>a_n</math> formula yields <math>a_n=a_{n-1}+\frac{1}{9}\left(\left(\frac{8}{9}\right)^{n-1}+4\cdot\left(\frac{5}{9}\right)^{n-1}-5\cdot\left(\frac{4}{9}\right)^{n-1}\right)</math>. Solving this recursion yields the final answer, <math>\boxed{1-\left(\frac{8}{9}\right)^n-\left(\frac{5}{9}\right)^n+\left(\frac{4}{9}\right)^n}</math>. |
~eevee9406 | ~eevee9406 | ||
Latest revision as of 11:26, 29 July 2024
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 1
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 inclusion-exclusion, the probability of that is 
.
Solution 2 (Recursion)
Define 
as the probability that the product is divisible by 
after selection . Likewise, define 
and 
with divisibility by 
and 
, respectively. Define 
to be the chance of dividing neither nor (and thus not either) similarly.
It is clear that 
. Now we can define other recursive formulas:
We are able to reach a 
state via selecting a non- from a state and selecting an even number from a 
state. Thus its formula is 
.
We are able to reach a 
state via selecting a non-even number from a state and selecting a from a state. Thus its formula is 
.
Finally, to reach an 
state, we can select a from a state and select an even number from a state. We can also reach from 
because of the fact that once the product is divisible by , it will always be divisible by regardless of the following selections. Thus its formula is 
.
For our formula for , we can substitute to find that 
. Solving this recursion yields 
.
For our formula for , we can substitute to find that 
. Solving this recursion yields 
.
Finally, substituting the values into the formula yields 
. Solving this recursion yields the final answer, 
.
~eevee9406
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. 
