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| − | ==Problem==
| + | #REDIRECT [[2000 AMC 12 Problems/Problem 4]] |
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| − | The Fibonacci sequence <math>1, 1, 2, 3, 5, 8, 13, 21, \ldots</math> starts with two <math>1</math>s, and each term afterwards is the sum of its two predecessors. Which one of the ten digits is the last to appear in the units position of a number in the Fibonacci sequence?
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| − | <math>\mathrm{(A)}\ 0 \qquad\mathrm{(B)}\ 4 \qquad\mathrm{(C)}\ 6 \qquad\mathrm{(D)}\ 7\qquad\mathrm{(E)}\ 9</math>
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| − | ==Solution==
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| − | The pattern of the units digits are
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| − | <math>1,1,2,3,5,8,3,1,4,5,9,4,3,7,0,7,7,4,1,5,6</math>
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| − | In order of appearance:
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| − | <math>1,2,3,5,8,4,9,7,0,6</math>.
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| − | <math>6</math> is the last.
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| − | <math>\boxed{\text{C}}</math>
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| − | ==See Also==
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| − | {{AMC10 box|year=2000|num-b=5|num-a=7}}
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