- #1
VMP
- 25
- 3
Hello everyone, I have an issue solving the following problem:
You're on a mathematical Olympiad, there are m medals and it lasts for n days.
First day committee gives [itex]U_{1}=1+\frac{1}{7}(m-1)[/itex] medals.
On the second day [itex]U_{2}=2+\frac{1}{7}(m-2-U_{1})[/itex] medals, and so on...
On the last day [itex]U_{n}=n[/itex].
Attempt at the solution:
My assumption is: [tex]U_{i}=i+\frac{1}{7}(m-i-\sum_{j=1}^{i}U_{j}+U_{i})[/tex]
If i take the difference of [itex]U_{i}-U_{(i-1)}[/itex],
I find this way to write the recursion:[tex]U_{i}=\frac{6}{7}(1+U_{(i-1)}),\;(i=2,3,...,n)[/tex]
If I set [itex]i=n[/itex] and do some algebra, then I get the following:[tex]13n=6+6m\;\; (1)[/tex]
...and this is where I'm not sure what to do, number of medals must be discrete...
Every [itex]U_{i}[/itex] must be discrete.
I tried from equation (1) to derive what m and n must be:[tex]n=6k \\ m=-1+13k,\;k\epsilon \mathbb{N}[/tex] However I am not sure how to capture this notion that every [itex]U_{i}[/itex] must be discrete.
Thank you for reading. :)
You're on a mathematical Olympiad, there are m medals and it lasts for n days.
First day committee gives [itex]U_{1}=1+\frac{1}{7}(m-1)[/itex] medals.
On the second day [itex]U_{2}=2+\frac{1}{7}(m-2-U_{1})[/itex] medals, and so on...
On the last day [itex]U_{n}=n[/itex].
Attempt at the solution:
My assumption is: [tex]U_{i}=i+\frac{1}{7}(m-i-\sum_{j=1}^{i}U_{j}+U_{i})[/tex]
If i take the difference of [itex]U_{i}-U_{(i-1)}[/itex],
I find this way to write the recursion:[tex]U_{i}=\frac{6}{7}(1+U_{(i-1)}),\;(i=2,3,...,n)[/tex]
If I set [itex]i=n[/itex] and do some algebra, then I get the following:[tex]13n=6+6m\;\; (1)[/tex]
...and this is where I'm not sure what to do, number of medals must be discrete...
Every [itex]U_{i}[/itex] must be discrete.
I tried from equation (1) to derive what m and n must be:[tex]n=6k \\ m=-1+13k,\;k\epsilon \mathbb{N}[/tex] However I am not sure how to capture this notion that every [itex]U_{i}[/itex] must be discrete.
Thank you for reading. :)