Brilliant Math Contest

Hi,

Let's consider the sequence $\left\{ { u }_{ m } \right\}$ defined by the recursive formula:

$\begin{cases} { u }_{ 1 }=1+\frac { 1 }{ 7 } (n-1) \\ { u }_{ m }=m+\frac { 1 }{ 7 } (n-\sum _{ j=1 }^{ m-1 }{ { u }_{ j } } -m)\quad for\quad m=2,3,4,... \end{cases}$

(For $m\le k$ , ${ u }_{ m }$ represents the number of medals that was awarded in the ${ m }^{ th }$ round.)

$\Rightarrow { u }_{ m+1 }=\frac { 6 }{ 7 } ({ u }_{ m }+1)\quad for\quad m=1,2,3,...$

Now, we construct a new sequence $\left\{ { v }_{ m } \right\}$ defined by ${ v }_{ m }={ u }_{ m }-6\quad for\quad m=1,2,3,...$

$\Rightarrow { v }_{ m+1 }=\frac { 6 }{ 7 } { v }_{ m }$

$\Rightarrow { v }_{ m }={ v }_{ 1 }{ \left( \frac { 6 }{ 7 } \right) }^{ m-1 }$

We know that ${ u }_{ k }=k$ , so ${ v }_{ k }=k-6={ v }_{ 1 }{ \left( \frac { 6 }{ 7 } \right) }^{ k-1 }$ . Let's call it $(1)$ .

The problem here is to solve $(1)$ for $(k,{ v }_{ 1 })\in { { Z } }_{ >0 }\times { Z }$

Though $k=1\Rightarrow { v }_{ 1 }=-5$ satisfies the conditions, it leads to $n=1$ which we don't expect here.

So we look for a value of $k>1$ , $(1)\Rightarrow { v }_{ 1 }=\frac { { 7 }^{ k-1 }(k-6) }{ { 6 }^{ k-1 } }$ , together with $gcd(6,7)=1$ we can deduce that $\frac { k-6 }{ { 6 }^{ k-1 } }$ must be an integer.

However, it can be proved that $0\le \left| \frac { k-6 }{ { 6 }^{ k-1 } } \right| <1\quad for\quad k\ge 2$ . Thus, the only way that ${ v }_{ 1 }$ is an integer is $k=6\Rightarrow { v }_{ 1 }=0\Rightarrow { v }_{ m }=0\Rightarrow { u }_{ m }=6\Rightarrow n=36$