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Solution. Let $ m_{k}$ be the number of medals awarded on the $ k^{\text{th}}$ day. Notice that after the $ \ell^{\text{th}}$ day, $ 6(m_{\ell}-\ell)$ medals are left for day $ \ell+1$. By the given information in the problem, \[ \begin{eqnarray*}m_{x+1}&=&x+1+\frac{6(m_{x}-x)-x-1}{7}\\ &=&\frac{6}{7}(m_{x}+1)\\ &=&\left(\frac{6}{7}\right)^{x}(m_{1}-6)+6\\ &=&\frac{6^{x}}{7^{x+1}}(m-36)+6,\] using $ m_{1}=1+\frac{m-1}{7}$. Since $ m_{n}=n$, $ 7^{n}(n-6)=6^{n-1}(m-36)\implies 6^{n-1}|n-6\implies n=6\wedge m=36$. As a result, $ (m,n)=(36,6)$. $ \Box$

EDIT: not sure what I was thinking, the above is utter nonsense. Apologies.

johnmichaelwu wrote: Assume that $(m, n) = (36, 6)$. Then there will be $\frac{36-1}{7}=5$ medals after the first day, and then $\frac{5-2}{7} = \frac{3}{7}$ medals after the second day, impossible as there must be an integer number of medals. Thus, we conclude the only answer to be $(m, n) = (1, 1)$ You read it wrong. With $(m,n)=(36,6)$, there will be $1 + \frac{36-1}{7}=6$ medals awarded the first day, thus $36-6 = 30$ left. Then there will be $2 + \frac{30-2}{7}=6$ medals awarded the second day, thus $30-6 = 24$ left. And so on, everything working fine.

Let K(i) denote the number of medals awarded on ith day. So, medals awarded on tth day, K(t) = t + 1/7[m - {t + \sum\limits_{i=1}^t-1 K(i)}] K(t+1) = (t+1) + 1/7 [m - {t+1 + \sum\limits_{i=1}^t K(i)}] K(t+1) - K(t) = 1 - 1/7 - 1/7K(t) K(t+1) = 6/7 (K(t) +1) Therefore, K(n-1) = 6/7 (K(n-2) +1) \sum\limits_{i=1}^n-1 K(i) = m - n K(1) * (\sum\limits_{i=1}^n-2 (6/7)^i-1) + n * (\sum\limits_{i=1}^n-2 (6/7)^i) - (\sum\limits_{i=1}^n-2 (i+1)*(6/7)^i) = m-n m*(6/7)^n-1 - 7n = 36*(6/7)^n-1 - 7(6) Therefore, m=36; n=6

We can let $f(a)$ be the number of medals awarded on day $a$. We know that after day $x$ there will be $6(f(x)-x)$ medals left. We know that if there are $6(f(x)-x)$ after day $x$ then \begin{align*} f(x+1)&=x+1+\frac{6(f(x)-x)-x-1}7 \\ &=\frac{6f(x)-6x-x-1+7x+7}7 \\ &=\frac{6f(x)+6}7 \\ &=\frac67f(x)+\frac67 \end{align*}so we can prove with induction that $f(x+1)=(\frac{6}{7})^x(f(1)-6)+6$. We know that $f(1)=1+\frac{m-1}{7}$ so \begin{align*} f(x+1)&=(\frac{6}{7})^x(1+\frac{m-1}{7}-6)+6\\ &=(\frac{6}{7})^x(\frac{m-36}7)+6\\ &=\frac{6^x}{7^{x+1}}(m-36)+6. \end{align*} We know that $f(n)=n$ and $f(n)=\frac{6^{n-1}}{7^n}(m-36)+6$ so \begin{align*} n&=\frac{6^{n-1}}{7^n}(m-36)+6\\ 7^nn&=6^{n-1}(m-36)+6\cdot7^n\\ 7^n(n-6)&=6^{n-1}(m-36)\\ \end{align*}and $n=6$ and $m=36$.