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))
m (→‎Solution 2 (Simple): Fixed lack of divide by 2 in exponent)
 
(13 intermediate revisions by 5 users not shown)
Line 3: Line 3:
 
Let <math>A</math> be an acute angle such that <math>\tan A = 2 \cos A.</math> Find the number of positive integers <math>n</math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer whose units digit is <math>9.</math>
 
Let <math>A</math> be an acute angle such that <math>\tan A = 2 \cos A.</math> Find the number of positive integers <math>n</math> less than or equal to <math>1000</math> such that <math>\sec^n A + \tan^n A</math> is a positive integer whose units digit is <math>9.</math>
  
−
==Solution==
+
==Solution 1==
  
 
Denote <math>a_n = \sec^n A + \tan^n A</math>.
 
Denote <math>a_n = \sec^n A + \tan^n A</math>.
Line 12: Line 12:
 
& = \left( \sec^{n-k} A + \tan^{n-k} A \right) \left( \sec^k A + \tan^k A \right)
 
& = \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 \\
 
- \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 \sec^{n-k} A \cos^k A - 2^k \tan^{n-k} A \cot^k A \\
 
& = a_{n-k} a_k - 2^k a_{n-2k} .
 
& = a_{n-k} a_k - 2^k a_{n-2k} .
 
\end{align*}
 
\end{align*}
Line 84: Line 84:
 
</cmath>
 
</cmath>
  
−
~Steven Chen (Professor Chen Education Palace, www.professorchenedub.com)
+
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
  
 
==Solution 2 (Simple)==
 
==Solution 2 (Simple)==
−
<math>\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}.</math>
+
<cmath>\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</cmath>
−
<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>
+
<cmath>\implies  \cos^2 A = \frac {\sqrt {17} - 1}{8}.</cmath>
 +
<cmath>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}} =</cmath>
 +
<cmath>= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.</cmath>
 +
 
 +
It is clear, that <math>c_n</math> is not integer if <math>n \ne 4k, k > 0.</math>
  
−
It is clear, that <math>c_n</math> is not integer if <math>n \ne 4k.</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>
+
<cmath>x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.</cmath>
 +
 
 +
<cmath>c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.</cmath>
 +
<cmath>c_{4k+4} = x^{2k+2} + y^{2k+2} = (x^{2k} + y^{2k})(x^2+y^2)- (x^2 y^2)(x^{2k-2}+y^{2k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies</cmath>
 +
<cmath>c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.</cmath>
 +
<cmath>c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.</cmath>
 +
<cmath>c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4}  \pmod{10} =</cmath>
 +
<cmath>= (9 \cdot 7 - 6 \cdot 9)  \pmod{10} = (3 - 4)  \pmod{10} = 9.</cmath>
 +
<cmath>c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =</cmath>
 +
<cmath>= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.</cmath>
 +
<cmath>c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =</cmath>
 +
<cmath>= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies</cmath>
 +
 
 +
The condition is satisfied iff <math>n = 12 k + 4</math> or <math>n = 12k + 8.</math>
 +
 
 +
If <math>n \le N</math> then the number of possible n is <math>\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.</math>
  
−
<math>c_8 = x^4 + y^4 = (x^2 + y^2)^2 – 2x^2 y^2 = 9^2 – 2 \cdot 16 = 49.</math>
+
For <math>N = 1000</math> we get <math>\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.</math>
  
−
<math>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</math>
+
'''vladimir.shelomovskii@gmail.com, vvsss'''
  
−
<math>c_{12} = 9 c_8 – 16 c_4 = 9 \cdot 49 – 16 \cdot 9 = 9 \cdot 33 = 297.</math>
+
==Note==
 +
A key idea in this solution is realizing that to calculate values of <math>a^n+b^n</math> is difficult directly, so we try to think about it recursively.
  
−
<math>c_{16} = 9 c_{12} – 16 c_8 = 9 \cdot 297 – 16 \cdot 49 = 1889.</math>
+
We then see how we can go from <math>a^n+b^n</math> to <math>a^{n+1}+b^{n+1}</math>, and learn that <math>(a^n+b^n)(a+b)-a^nb-ab^n=a^{n+1}b^{n+1}</math>.
  
−
<cmath>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.</cmath>
+
So, letting <math>a^i+b^i=s_i</math>, <math>s_i=(a+b)s_{i-1}-(ab)s_{i-2}</math> where we set <math>n+1</math> as <math>i</math>, and <math>(a^n+b^n)=s_{i-1}</math>, as well as <math>(a^nb+ab^n)=(ab)(a^{n-1}+b^{n-1})=(ab)s_{i-2}</math>.
−
<cmath>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.</cmath>
 
−
<cmath>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</cmath>
 
  
−
The condition is satisfied iff <math>n = 12 k + 4</math> or <math>n = 12k + 8.</math>
+
~Bigbrain_2009
  
−
If <math>n \le N</math> then the number of possible n is <math>[\frac {N}{4}] - [\frac {N}{12}].</math>
+
==Video Solution==
 +
https://youtu.be/5Dpdi8IiUiw
  
−
For <math>N = 1000</math> we get <math>[\frac {1000}{4}] - [\frac {1000}{12}] = 250 – 83 = \boxed{167}.</math>
+
~MathProblemSolvingSkills.com
  
−
'''vladimir.shelomovskii@gmail.com, vvsss'''
 
  
 
== See also ==
 
== See also ==

Latest revision as of 12:22, 1 February 2025

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 1

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 \cot^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)

Solution 2 (Simple)

\[\tan A = 2 \cos A \implies \sin A = 2 \cos^2 A \implies \sin^2 A + \cos^2 A = 4 \cos^4 A + \cos^2 A = 1\] \[\implies \cos^2 A = \frac {\sqrt {17} - 1}{8}.\] \[c_n = \sec^n A + \tan^n A = \frac {1}{\cos^n A} + 2^n \cos^n A = (4\cos^2 A +1)^{\frac {n}{2}}+(4 \cos^2 A)^{\frac {n}{2}} =\] \[= \left(\frac {\sqrt {17} + 1}{2}\right)^{\frac {n}{2}}+ \left(\frac {\sqrt {17} - 1}{2}\right)^{\frac {n}{2}}.\]

It is clear, that $c_n$ is not integer if $n \ne 4k, k > 0.$

Denote $x = \frac {\sqrt {17} + 1}{2}, y = \frac {\sqrt {17} - 1}{2} \implies$ \[x \cdot y = 4, x + y = \sqrt{17}, x - y = 1 \implies x^2 + y^2 = (x - y)^2 + 2xy = 9 = c_4.\]

\[c_8 = x^4 + y^4 = (x^2 + y^2)^2 - 2x^2 y^2 = 9^2 - 2 \cdot 16 = 49.\] \[c_{4k+4} = x^{2k+2} + y^{2k+2} = (x^{2k} + y^{2k})(x^2+y^2)- (x^2 y^2)(x^{2k-2}+y^{2k-2}) = 9 c_{4k}- 16 c_{4k – 4} \implies\] \[c_{12} = 9 c_8 - 16 c_4 = 9 \cdot 49 - 16 \cdot 9 = 9 \cdot 33 = 297.\] \[c_{16} = 9 c_{12} - 16 c_8 = 9 \cdot 297 - 16 \cdot 49 = 1889.\] \[c_{12m + 4} \pmod{10} = 9 \cdot c_{12m} \pmod{10} - 16 \pmod{10} \cdot c_{12m - 4} \pmod{10} =\] \[= (9 \cdot 7 - 6 \cdot 9) \pmod{10} = (3 - 4) \pmod{10} = 9.\] \[c_{12m + 8}\pmod{10} = 9 \cdot c_{12m+4} \pmod{10} - 16 \pmod{10} \cdot c_{12m } \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 7) \pmod{10} = (1 - 2)\pmod{10} = 9.\] \[c_{12m + 12} \pmod{10} = 9 \cdot c_{12m + 8} \pmod{10} - 16 \pmod{10} \cdot c_{12m + 4} \pmod{10} =\] \[= (9 \cdot 9 - 6 \cdot 9) \pmod{10} = (1 - 4) \pmod{10} = 7 \implies\]

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

If $n \le N$ then the number of possible n is $\left\lfloor \frac{N}{4} \right\rfloor - \left\lfloor \frac{N}{12} \right\rfloor.$

For $N = 1000$ we get $\left\lfloor \frac{1000}{4} \right\rfloor - \left\lfloor \frac{1000}{12} \right\rfloor = 250 - 83 = \boxed{167}.$

vladimir.shelomovskii@gmail.com, vvsss

Note

A key idea in this solution is realizing that to calculate values of $a^n+b^n$ is difficult directly, so we try to think about it recursively.

We then see how we can go from $a^n+b^n$ to $a^{n+1}+b^{n+1}$, and learn that $(a^n+b^n)(a+b)-a^nb-ab^n=a^{n+1}b^{n+1}$.

So, letting $a^i+b^i=s_i$, $s_i=(a+b)s_{i-1}-(ab)s_{i-2}$ where we set $n+1$ as $i$, and $(a^n+b^n)=s_{i-1}$, as well as $(a^nb+ab^n)=(ab)(a^{n-1}+b^{n-1})=(ab)s_{i-2}$.

~Bigbrain_2009

Video Solution

https://youtu.be/5Dpdi8IiUiw

~MathProblemSolvingSkills.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

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=241611"