2023 AIME II Problems/Problem 13

Problem

Let $A$ be an acute angle such that $\tan A = 2 \cos A.$ Find the number of positive integers $n$ less than or equal to $1000$ such that $\sec^n A + \tan^n A$ is a positive integer whose units digit is $9.$

Solution

Denote $a_n = \sec^n A + \tan^n A$ . For any $k$ , we have $\begin{align*} a_n & = \sec^n A + \tan^n A \\ & = \left( \sec^{n-k} A + \tan^{n-k} A \right) \left( \sec^k A + \tan^k A \right) - \sec^{n-k} A \tan^k A - \tan^{n-k} A \sec^k A \\ & = a_{n-k} a_k - 2^k \sec^{n-k} A \cos^k A - 2^k \tan^{n-k} A \tan^k A \\ & = a_{n-k} a_k - 2^k a_{n-2k} . \end{align*}$

Next, we compute the first several terms of $a_n$ .

By solving equation $\tan A = 2 \cos A$ , we get $\cos A = \frac{\sqrt{2 \sqrt{17} - 2}}{4}$ . Thus, $a_0 = 2$ , $a_1 = \sqrt{\sqrt{17} + 4}$ , $a_2 = \sqrt{17}$ , $a_3 = \sqrt{\sqrt{17} + 4} \left( \sqrt{17} - 2 \right)$ , $a_4 = 9$ .

In the rest of analysis, we set $k = 4$ . Thus, $\begin{align*} a_n & = a_{n-4} a_4 - 2^4 a_{n-8} \\ & = 9 a_{n-4} - 16 a_{n-8} . \end{align*}$

Thus, to get $a_n$ an integer, we have $4 | n$ . In the rest of analysis, we only consider such $n$ . Denote $n = 4 m$ and $b_m = a_{4n}$ . Thus, $\begin{align*} b_m & = 9 b_{m-1} - 16 b_{m-2} \end{align*}$ with initial conditions $b_0 = 2$ , $b_1 = 9$ .

To get the units digit of $b_m$ to be 9, we have $\begin{align*} b_m \equiv -1 & \pmod{2} \\ b_m \equiv -1 & \pmod{5} \end{align*}$

Modulo 2, for $m \geq 2$ , we have $\begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv b_{m-1} . \end{align*}$

Because $b_1 \equiv -1 \pmod{2}$ , we always have $b_m \equiv -1 \pmod{2}$ for all $m \geq 2$ .

Modulo 5, for $m \geq 5$ , we have $\begin{align*} b_m & \equiv 9 b_{m-1} - 16 b_{m-2} \\ & \equiv - b_{m-1} - b_{m-2} . \end{align*}$

We have $b_0 \equiv 2 \pmod{5}$ , $b_1 \equiv -1 \pmod{5}$ , $b_2 \equiv -1 \pmod{5}$ , $b_3 \equiv 2 \pmod{5}$ , $b_4 \equiv -1 \pmod{5}$ , $b_5 \equiv -1 \pmod{5}$ , $b_6 \equiv 2 \pmod{5}$ . Therefore, the congruent values modulo 5 is cyclic with period 3. To get $b_m \equiv -1 \pmod{5}$ , we have $3 \nmid m \pmod{3}$ .

From the above analysis with modulus 2 and modulus 5, we require $3 \nmid m \pmod{3}$ .

For $n \leq 1000$ , because $n = 4m$ , we only need to count feasible $m$ with $m \leq 250$ . The number of feasible $m$ is $\begin{align*} 250 - \left\lfloor \frac{250}{3} \right\rfloor & = 250 - 83 \\ & = \boxed{\textbf{(167) }} . \end{align*}$

~Steven Chen (Professor Chen Education Palace, www.professorchenedub.com)

2023 AIME II (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2023 AIME II Problems/Problem 13

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