2025 SSMO Team Round Problems/Problem 12

Problem

Let $a_n=(4+3\sqrt2)^n$ for all nonnegative integers $n$ . Let $\sum_{k=0}^\infty\frac{\lfloor a_k\rfloor}{10^k}=\frac{m}{n},$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$ .

Solution

We utilize radical conjugates to solve this problem. Recall that $(4+3\sqrt{2})^k + (4-3\sqrt{2})^k$ is always an integer for any positive integer $k$ . Furthermore, notice that $-1 < 4-3\sqrt{2} < 0$ . This motivates us to consider the parity of $k$ .

If $k$ is even, we have $0 < (4-3\sqrt{2})^k < 1$ . This means that $(4+3\sqrt{2})^k$ is less than the integer $(4+3\sqrt{2})^k + (4-3\sqrt{2})^k$ while being within less than $1$ of it. If $k$ is odd, $-1 < (4-3\sqrt{2})^k < 0$ . Then, $(4+3\sqrt{2})^k$ is greater than the integer $(4+3\sqrt{2})^k + (4-3\sqrt{2})^k$ while being within less than $1$ of it. Therefore, $\lfloor a_k \rfloor = \begin{cases} 1 & \text{if }k=0 \\ (4+3\sqrt{2})^k + (4-3\sqrt{2})^k - 1 & \text{if }k>0\text{ is even} \\ (4+3\sqrt{2})^k + (4-3\sqrt{2})^k & \text{if }k>0\text{ is odd}. \end{cases}$ Using this, we can evaluate the sum in the problem statement. We have: \begin{align*} \sum_{k=0}^{\infty} \frac{\lfloor a_k \rfloor}{10^k} &= 1 + \sum_{k=1}^\infty \frac{(4+3\sqrt{2})^k}{10^k} + \sum_{k=1}^\infty \frac{(4-3\sqrt{2})^k}{10^k} - \sum_{k=1}^\infty \frac{1}{10^{2k}} \\ &= 1 + \frac{\tfrac{4+3\sqrt{2}}{10}}{1 - \tfrac{4+3\sqrt{2}}{10}} + \frac{\tfrac{4-3\sqrt{2}}{10}}{1 - \tfrac{4-3\sqrt{2}}{10}} - \frac{\tfrac{1}{100}}{1 - \tfrac{1}{100}} \\ &= 1 + \frac{4+3\sqrt{2}}{6-3\sqrt{2}} + \frac{4-3\sqrt{2}}{6+3\sqrt{2}} - \frac{1}{99} \\ &= 1 + \frac{14}{3} - \frac{1}{99} \\ &= \frac{560}{99}. \end{align*} The answer is $560+99 = \boxed{659}$ .

~Sedro

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2025 SSMO Team Round Problems/Problem 12

Problem

Solution