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

(Created page with "Let <math>R_n=\frac{1}{2}(a^n+b^n)</math> for each non-negative integer <math>n</math> where <math>a=3+2\sqrt{2}</math> and <math>b=3-2\sqrt{2}</math>. The value of <math>R_{1234...")
 
 
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Let <math>R_n=\frac{1}{2}(a^n+b^n)</math> for each non-negative integer <math>n</math> where <math>a=3+2\sqrt{2}</math> and <math>b=3-2\sqrt{2}</math>. The value of <math>R_{12345}</math> is an integer. What is its units digit?
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== Problem ==
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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
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<math>\text{(A) } 1\quad
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\text{(B) } 3\quad
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\text{(C) } 5\quad
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\text{(D) } 7\quad
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\text{(E) } 9</math>
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== Solution ==
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<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>
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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>
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== See also ==
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{{AHSME box|year=1990|num-b=29|num-a=30}} 
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[[Category: Intermediate Algebra Problems]]
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{{MAA Notice}}

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 (Problems • Answer Key • Resources)
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

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