Difference between revisions of "1990 AHSME Problems/Problem 30"
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== Problem == | == Problem == | ||
− | If <math>R_n=\ | + | 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 | ||
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== 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 where and , and then is an integer. Its units digit is
Solution
but and , so this means that . Since and , all terms are integers and we can continue the recurrence to get the repeating sequence . The number is divisible by three but not by six, so the required element in the sequence is which is answer
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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