Difference between revisions of "1990 AHSME Problems/Problem 30"

 
Line 1: Line 1:
 
== Problem ==
 
== Problem ==
  
If <math>R_n=\frac{1}{2}(a^n+b^n)</math>  where <math>a=3+2\sqrt{2}</math> and <math>b=3-2\sqrt{2}</math>, and <math>n=0,1,2,\cdots,</math> then <math>R_{12345}</math> is an integer.  Its units digit is
+
If <math>R_n=\tfrac{1}{2}(a^n+b^n)</math>  where <math>a=3+2\sqrt{2}</math> and <math>b=3-2\sqrt{2}</math>, and <math>n=0,1,2,\cdots,</math> then <math>R_{12345}</math> is an integer.  Its units digit is
  
 
<math>\text{(A) } 1\quad
 
<math>\text{(A) } 1\quad
Line 10: Line 10:
  
 
== Solution ==
 
== Solution ==
<math>\fbox{E}</math>
+
<math>(a+b)R_n=\tfrac12(a^{n+1}+b^{n+1})+ab\cdot\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}</math>
 +
but <math>a+b=6</math> and <math>ab=1</math>, so this means that <math>R_{n+1}=6R_n-R_{n-1}</math>. Since <math>R_0=1</math> and <math>R_1=3</math>, all terms are integers and we can continue the recurrence <math>\rm{mod}\ 10</math> to get the repeating sequence <math>1,3,7,9,7,3</math>. The number <math>12345</math> is divisible by three but not by six, so the required element in the sequence is <math>9</math> which is answer <math>\fbox{E}</math>
  
 
== See also ==
 
== See also ==

Latest revision as of 00:15, 5 February 2016

Problem

If $R_n=\tfrac{1}{2}(a^n+b^n)$ where $a=3+2\sqrt{2}$ and $b=3-2\sqrt{2}$, and $n=0,1,2,\cdots,$ then $R_{12345}$ is an integer. Its units digit is

$\text{(A) } 1\quad \text{(B) } 3\quad \text{(C) } 5\quad \text{(D) } 7\quad \text{(E) } 9$

Solution

$(a+b)R_n=\tfrac12(a^{n+1}+b^{n+1})+ab\cdot\tfrac12(a^{n-1}+b^{n-1})=R_{n+1}+abR_{n-1}$ but $a+b=6$ and $ab=1$, so this means that $R_{n+1}=6R_n-R_{n-1}$. Since $R_0=1$ and $R_1=3$, all terms are integers and we can continue the recurrence $\rm{mod}\ 10$ to get the repeating sequence $1,3,7,9,7,3$. The number $12345$ is divisible by three but not by six, so the required element in the sequence is $9$ which is answer $\fbox{E}$

See also

1990 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 29
Followed by
Problem 30
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Invalid username
Login to AoPS