Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus
During AMC testing, the AoPS Wiki is in read-only mode and no edits can be made.

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

(→‎Solution 2 (Simple))
(→‎Solution 2 (Simple))
Line 90: Line 90:
 
<math>c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =  (\frac {\sqrt {17} + 1}{2})^{\frac {n}{2}}+ (\frac {\sqrt {17} – 1}{2})^{\frac {n}{2}}.</math>
 
<math>c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =  (\frac {\sqrt {17} + 1}{2})^{\frac {n}{2}}+ (\frac {\sqrt {17} – 1}{2})^{\frac {n}{2}}.</math>
  
−
It is clear, that <math>c_n</math> is not integer if <math>n \ne 4k.</math>
+
It is clear, that <math>c_n</math> is not integer if <math>n \ne 4k, k > 0.</math>
 
Denote <math>x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies</math>
 
Denote <math>x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies</math>
 
<math>x \cdot y = 4, x + y = \sqrt{17}, x – y = 1 \implies x^2 + y^2 = (x – y)^2 + 2xy = 9 = c_4.</math>
 
<math>x \cdot y = 4, x + y = \sqrt{17}, x – y = 1 \implies x^2 + y^2 = (x – y)^2 + 2xy = 9 = c_4.</math>

Revision as of 21:50, 4 March 2023

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)

Solution 2 (Simple)

$\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1 \implies \cos^2 A = \frac {\sqrt {17} – 1}{8}.$ $c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} = (\frac {\sqrt {17} + 1}{2})^{\frac {n}{2}}+ (\frac {\sqrt {17} – 1}{2})^{\frac {n}{2}}.$

It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$ Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$ $x \cdot y = 4, x + y = \sqrt{17}, x – y = 1 \implies x^2 + y^2 = (x – y)^2 + 2xy = 9 = c_4.$

$c_8 = x^4 + y^4 = (x^2 + y^2)^2 – 2x^2 y^2 = 9^2 – 2 \cdot 16 = 49.$

$c_{4k+4} = x^{4k+4} + y^{4k+4} = (x^{4k} + y^{4k})(x^2+y^2)- (x^2 y^2)(x^{4k-2}+y^{4k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies$

$c_{12} = 9 c_8 – 16 c_4 = 9 \cdot 49 – 16 \cdot 9 = 9 \cdot 33 = 297.$

$c_{16} = 9 c_{12} – 16 c_8 = 9 \cdot 297 – 16 \cdot 49 = 1889.$

\[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} – 16 \pmod{10} \cdot c_{12m – 4} \pmod{10} = (9 \cdot 7 – 6 \cdot 9) \pmod{10} = (3 – 4) \pmod{10} = 9.\] \[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} – 16 \pmod{10} \cdot c_{12m } \pmod{10} = (9 \cdot 9 – 6 \cdot 7) \pmod{10} = (1 – 2)\pmod{10} = 9.\] \[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m+8} \pmod{10} – 16 \pmod{10} \cdot c_{12m +4} \pmod{10} = (9 \cdot 9 – 6 \cdot 9) \pmod{10} = (1 – 4) \pmod{10} = 7 \implies\]

The condition is satisfied iff $n = 12 k + 4$ or $n = 12k + 8.$

If $n \le N$ then the number of possible n is $[\frac {N}{4}] - [\frac {N}{12}].$

For $N = 1000$ we get $[\frac {1000}{4}] - [\frac {1000}{12}] = 250 – 83 = \boxed{167}.$

vladimir.shelomovskii@gmail.com, vvsss

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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2023_AIME_II_Problems/Problem_13&oldid=190628"