Question 30
Let $$a_0 = 1$$ and for $$n \ge 1$$, define $$a_n = 3a_{n-1} + 1$$. Find the remainder when $$a_{11}$$ is divided by 97.
Correct Answer: 37
Solution
Adding $$\frac{1}{2}$$ to both sides shows $$a_n + \frac{1}{2} = 3\left(a_{n-1} + \frac{1}{2}\right)$$, so $$a_n = \frac{3^{n+1}-1}{2}$$. Hence $$a_{11} = \frac{3^{12}-1}{2} = \frac{531440}{2} = 265720$$. Since $$265720 = 97 \times 2739 + 37$$, the remainder is 37.
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