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Difference between revisions of "2023 AIME II Problems/Problem 13"

(Created page with "==Solution== Denote <math>a_n = \sec^n A + \tan^n A</math>. For any <math>k</math>, we have <cmath> \begin{align*} a_n & = \sec^n A + \tan^n A \\ & = \left( \sec^{n-k} A + \t...")
 
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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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== See also ==
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{{AIME box|year=2023|num-b=12|num-a=14|n=II}}
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{{MAA Notice}}

Revision as of 19:11, 16 February 2023

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.professorchenedu.com)

See also

2023 AIME II (Problems • Answer Key • Resources)
Preceded by
Problem 12 		Followed by
Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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