Difference between revisions of "1972 USAMO Problems/Problem 3"
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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 <math>n</math> selections (<math>n>1</math>), the product of the <math>n</math> numbers selected will be divisible by 10. | 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 <math>n</math> selections (<math>n>1</math>), the product of the <math>n</math> numbers selected will be divisible by 10. | ||
− | ==Solution== | + | ==Solution 1== |
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. | ||
− | The probability that there is | + | The probability that there is no 5 is <math>\left( \frac{8}{9}\right)^n</math>. |
− | ==See | + | The probability that there is no 2 is <math>\left( \frac{5}{9}\right)^n</math>. |
+ | |||
+ | The probability that there is neither a 2 nor 5 is <math>\left( \frac{4}{9}\right)^n</math>, 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 <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>. | ||
+ | |||
+ | ==Solution 2 (Recursion)== | ||
+ | Define <math>a_n</math> as the probability that the product is divisible by <math>10</math> after selection <math>n</math>. Likewise, define <math>b_n</math> and <math>c_n</math> with divisibility by <math>2</math> and <math>5</math>, respectively. Define <math>d_n</math> to be the chance of dividing neither <math>2</math> nor <math>5</math> (and thus not <math>10</math> either) similarly. | ||
+ | |||
+ | It is clear that <math>d_n=\left(\frac{4}{9}\right)^n</math>. Now we can define other recursive formulas: | ||
+ | |||
+ | We are able to reach a <math>b</math> state via selecting a non-<math>5</math> from a <math>b</math> state and selecting an even number from a <math>d</math> state. Thus its formula is <math>b_n=\frac{8}{9}b_{n-1}+\frac{4}{9}d_{n-1}</math>. | ||
+ | |||
+ | We are able to reach a <math>c</math> state via selecting a non-even number from a <math>c</math> state and selecting a <math>5</math> from a <math>d</math> state. Thus its formula is <math>c_n=\frac{5}{9}c_{n-1}+\frac{1}{9}d_{n-1}</math>. | ||
+ | |||
+ | Finally, to reach an <math>a</math> state, we can select a <math>5</math> from a <math>b</math> state and select an even number from a <math>c</math> state. We can also reach <math>a_n</math> from <math>a_{n-1}</math> because of the fact that once the product is divisible by <math>10</math>, it will always be divisible by <math>10</math> regardless of the following selections. Thus its formula is <math>a_n=a_{n-1}+\frac{1}{9}b_{n-1}+\frac{4}{9}c_{n-1}</math>. | ||
+ | |||
+ | For our formula for <math>b_n</math>, we can substitute to find that <math>b_n=\frac{8}{9}b_{n-1}+\left(\frac{4}{9}\right)^n</math>. Solving this recursion yields <math>b_n=\left(\frac{8}{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-\left(\frac{5}{9}\right)^n+\left(\frac{4}{9}\right)^n}</math>. | ||
+ | |||
+ | ~eevee9406 | ||
+ | |||
+ | {{alternate solutions}} | ||
+ | |||
+ | ==See Also== | ||
{{USAMO box|year=1972|num-b=2|num-a=4}} | {{USAMO box|year=1972|num-b=2|num-a=4}} | ||
+ | {{MAA Notice}} | ||
[[Category:Olympiad Combinatorics Problems]] | [[Category:Olympiad Combinatorics Problems]] |
Latest revision as of 12:26, 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.