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

Difference between revisions of "1982 AHSME Problems/Problem 30"

m (→‎Solution)
(→‎Solution)
Line 8: Line 8:
 
\textbf{(E)}\ \text{none of these}</math>   
 
\textbf{(E)}\ \text{none of these}</math>   
  
−
== Solution ==
+
== Solution 1 (Binomial Expansion) ==
 
Let <math>A=15+\sqrt{220}</math> and <math>B=15-\sqrt{220}.</math> Note that <math>A^{19}+B^{19}</math> and <math>A^{82}+B^{82}</math> are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
 
Let <math>A=15+\sqrt{220}</math> and <math>B=15-\sqrt{220}.</math> Note that <math>A^{19}+B^{19}</math> and <math>A^{82}+B^{82}</math> are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.
  
Line 28: Line 28:
  
 
~MRENTHUSIASM
 
~MRENTHUSIASM
 +
 +
== Solution 2 (Characteristic Polynomials) ==
 +
 +
Let <math>a_n = \left(15 + \sqrt{220}\right)^n + \left(15 - \sqrt{220}\right)^n</math>.  Since <math>\left(15 + \sqrt{220}\right) + \left(15 - \sqrt{220}\right) = 30</math> and <math>\left(15 + \sqrt{220}\right)\left(15 - \sqrt{220}\right) = 225 - 220 = 5</math>, by Vieta's rules, <math>15 + \sqrt{220}</math> and <math>15 - \sqrt{220}</math> are roots of the polynomial <math>x^2 - 30x + 5</math>.  This must be the characteristic polynomial of <math>a_n</math>, and therefore <math>a_{n} = 30a_{n-1} - 5a_{n-2}</math> for all <math>n \geq 2</math>.
 +
 +
So <math>a_0 = 2</math>, <math>a_1 = 30</math>, and <math>a_2 = 30 \cdot 30 - 5 \cdot 2 = 890</math>.  Since <math>a_1</math> and <math>a_2</math> are divisible by <math>10</math>, <math>a_3</math> must also be divisible by <math>10</math>.  By similar logic, <math>a_n</math> is divisible by <math>10</math> for every integer <math>n \geq 1</math>.  So <math>a_{19} + a_{82}</math> must be divisible by <math>10</math>.
 +
 +
Now <math>15 - \sqrt{220} < 15 - \sqrt{210.25} = 15 - 14.5 = 0.5</math>, so <math>\left(15 - \sqrt{220}\right)^n < (0.5)^n</math>.  Let <math>x = \left(15 - \sqrt{220}\right)^{19} + \left(15 - \sqrt{220}\right)^{82}</math>.  Then <math>x < (0.5)^{19} + (0.5)^{82} < 0.5 + 0.5 = 1</math>.
 +
 +
Therefore,
  
 
== See Also ==
 
== See Also ==
 
{{AHSME box|year=1982|num-b=29|after=Last Problem}}
 
{{AHSME box|year=1982|num-b=29|after=Last Problem}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 12:07, 2 July 2025

Problem

Find the units digit of the decimal expansion of \[\left(15 + \sqrt{220}\right)^{19} + \left(15 + \sqrt{220}\right)^{82}.\]

$\textbf{(A)}\ 0\qquad \textbf{(B)}\ 2\qquad \textbf{(C)}\ 5\qquad \textbf{(D)}\ 9\qquad \textbf{(E)}\ \text{none of these}$

Solution 1 (Binomial Expansion)

Let $A=15+\sqrt{220}$ and $B=15-\sqrt{220}.$ Note that $A^{19}+B^{19}$ and $A^{82}+B^{82}$ are both integers: When we expand (Binomial Theorem) and combine like terms for each expression, the rational terms are added and the irrational terms are canceled.

We have \begin{align*} A^{19}+B^{19} &= \left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{1}15^{18}\sqrt{220}^1+\cdots+\binom{19}{19}15^0\sqrt{220}^{19}\right] + \left[\binom{19}{0}15^{19}\sqrt{220}^0-\binom{19}{1}15^{18}\sqrt{220}^1+\cdots-\binom{19}{19}15^0\sqrt{220}^{19}\right] \\ &= 2\left[\binom{19}{0}15^{19}\sqrt{220}^0+\binom{19}{2}15^{17}\sqrt{220}^2+\cdots+\binom{19}{18}15^1\sqrt{220}^{18}\right] \\ &= 2\left[\binom{19}{0}15^{19}+\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9\right]. \end{align*} Similarly, we have \[A^{82}+B^{82}=2\left[\binom{82}{0}15^{82}+\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}\right].\] We add the two equations and take the sum modulo $10:$ \begin{align*} \left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right) &= 2\Biggl[\binom{19}{0}15^{19}+\phantom{ }\underbrace{\binom{19}{2}15^{17}220+\cdots+\binom{19}{18}15^1 220^9}_{0\pmod{10}}\phantom{ }\Biggr]+2\Biggl[\binom{82}{0}15^{82}+\phantom{ }\underbrace{\binom{82}{2}15^{80}220+\cdots+\binom{82}{82}220^{41}}_{0\pmod{10}}\phantom{ }\Biggr] \\ &\equiv 2\left[\binom{19}{0}15^{19}\right]+2\left[\binom{82}{0}15^{82}\right] \\ &\equiv 2\left[5\right]+2\left[5\right] \\ &\equiv 0\pmod{10}. \end{align*} It is clear that $0<B^{82}<B^{19}<B<0.5,$ from which $0<B^{19}+B^{82}<1.$ We conclude that the units digit of the decimal expansion of $B^{19}+B^{82}$ is $0.$ Since the units digit of the decimal expansion of $\left(A^{19}+A^{82}\right)+\left(B^{19}+B^{82}\right)$ is $0,$ the units digit of the decimal expansion of $A^{19}+A^{82}$ is $\boxed{\textbf{(D)}\ 9}.$

~MRENTHUSIASM

Solution 2 (Characteristic Polynomials)

Let $a_n = \left(15 + \sqrt{220}\right)^n + \left(15 - \sqrt{220}\right)^n$. Since $\left(15 + \sqrt{220}\right) + \left(15 - \sqrt{220}\right) = 30$ and $\left(15 + \sqrt{220}\right)\left(15 - \sqrt{220}\right) = 225 - 220 = 5$, by Vieta's rules, $15 + \sqrt{220}$ and $15 - \sqrt{220}$ are roots of the polynomial $x^2 - 30x + 5$. This must be the characteristic polynomial of $a_n$, and therefore $a_{n} = 30a_{n-1} - 5a_{n-2}$ for all $n \geq 2$.

So $a_0 = 2$, $a_1 = 30$, and $a_2 = 30 \cdot 30 - 5 \cdot 2 = 890$. Since $a_1$ and $a_2$ are divisible by $10$, $a_3$ must also be divisible by $10$. By similar logic, $a_n$ is divisible by $10$ for every integer $n \geq 1$. So $a_{19} + a_{82}$ must be divisible by $10$.

Now $15 - \sqrt{220} < 15 - \sqrt{210.25} = 15 - 14.5 = 0.5$, so $\left(15 - \sqrt{220}\right)^n < (0.5)^n$. Let $x = \left(15 - \sqrt{220}\right)^{19} + \left(15 - \sqrt{220}\right)^{82}$. Then $x < (0.5)^{19} + (0.5)^{82} < 0.5 + 0.5 = 1$.

Therefore,

See Also

1982 AHSME (Problems • Answer Key • Resources)
Preceded by
Problem 29 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30
All AHSME 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=1982_AHSME_Problems/Problem_30&oldid=253128"